ICSE Class 9 Maths Sample Question Paper 6 with Answers

ICSE Class 9 Maths Sample Question Paper 6 with Answers

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) Prove that log (1 + 2 + 3) = log 1 + log 2 + log 3.
Answer:
log (1 + 2 + 3) = log 6 = log (1 x 2 x 3)
= log 1 + log 2 + log 3.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) In the given figure, CD is a diameter which meets the chord AB in E such that
AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 1
Answer:
Given : AE = BE = 4 cm, CE = 3 cm
Let r be the radius (OB = OC)
OE = OC – CE = r – 3.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 6
⇒ OB2 = OE2 + BE2 (Pythagoras theorem)
⇒ r2 = (r – 3)2 + 42 r2
⇒ r2 – 6r + 9 + 16
⇒ 6r = 25
\(r=\frac{25}{6}=4 \frac{1}{6} \mathrm{~cm}\)

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) If ₹ 6,400 is invested at 6 \(\frac{1}{4}\) % p.a. compound interest, find (i) the amount after 2 years (ii) the interest earned in 2 years.
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 7

Question 2.
(a) Evaluate tan x and cos y from the given figure.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 2
Answer:
In ΔACD, AC2 = AD2 + CD2
132 – 52 + CD2
⇒ CD2 = 169 – 25 = 144
⇒ CD = 12.
In A BCD, BC2 = CD2 + BD2
= 144 + 162 = 144 + 256 = 400
BC =20
ICSE Class 9 Maths Sample Question Paper 6 with Answers 8

(b) ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that the altitudes are equal.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 3
Answer:
Given: AC=AB
∴ In ΔBEC and ΔCFB,
∠C=∠B (∵ AB = AC)
∠BEC = ∠CFB (Each being a right angle)
BC = BC (Common side)
∴ ΔBFC  ≅ ΔCFB (AAS axiom)
∴ BE = CF (c.p.ct.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) The mean of 5 observations is 15. If the mean of first three observations is 14 and that of the last three is 17, find the third observation.
Answer:
Mean of 5 observations = 15
∴ Sum of 5 observations = 15 x 5 = 75
Mean of first 3 observations = 14
∴ Sum of first 3 observations = 14 x 3 = 42
Mean of last 3 observations = 17
∴ Sum of last 3 observations = 17 x 3 = 51
∴ The third observation = (42 + 51) – 75 = 18.

Question 3.
(a) Factorize : x4 + 4
Answer:
x4 + 4 = (x4 + 4x2 + 4) – 4x2 = {(x2)2 + 2 .x. 2 + (2)2} – (2x)2
= (x2 + 2)2 – (2x)2 = (x2 + 2 + 2x) (x2 + 2 – 2x)
= (x2 + 2x+ 2) (x2 – 2x+ 2)

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) Evaluate : \(\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \cdot \tan 60^{\circ}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 9

(c) Simplify:\((81)^{3 / 4}-3 \times(7)^{0}-\left(\frac{1}{27}\right)^{-2 / 3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 10

Question 4.
(a) If x \(\frac{2}{x}\) = 5, find the value of \(x^{3}-\frac{8}{x^{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 11

(b) If the hypotenuse of a right angled triangle is 6 m more than twice the shortest side and third side is 2 m less than hypotenuse, find the sides of the triangle.
Answer:
Let the shortest side be x m.
Then, Hypotenuse =(2x+6)cm,thirdside=2x+6-2=(2x+4)m.
∴ (2x + 6)2 = (2x + 4)2 + x2 (Using Pythagoras theorem)
= (2x)2 + 2.2x.6 + 62 = (2x)2 + 2.2xA +42 + x2
= 4x2 + 24x + 36 = 4x2 + 16x + 16 + x2
= 24x – 16x = x2 + 16 – 36
= x2 – 8x – 20 = 0
= x2 – (10 – 2) x – 20 =0
= x2 – 10x + 2x – 20 =0
= x (x – 10) + 2 (x – 10) = 0
= (x – 10) (x + 2) = 0
= x – 10 =0 or x + 2 = 0
= x = 10 or x = – 2
∴ x = 10 (∵ x cannot be negative)
∴ 2x + 6 = 2 x 10 + 6 = 26
and 2x + 4 = 2 x 10 + 4 = 24
Therefore, the sides are 10 m, 26 m and 24 m.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) Simplify: \(\frac{3}{\sqrt{6}+\sqrt{3}}-\frac{4}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 12
ICSE Class 9 Maths Sample Question Paper 6 with Answers 13

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Solve : log10 6 + log10 (4x + 5) = log10 (2x + 7) +1
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 14

(b) 3 men and 4 women can do a piece of work in 14 days while 4 men and 6 women can do it in 10 days. How long would it take 1 woman to finish the work ?
Answer:
Let 1 man take x days and 1 woman take y days to finish the work.
∴ In 1 day, 1 man does = \(\frac{1}{x} \) work and 1 woman does = \(\frac{1}{y}\) work.
So, 3 men and 4 women do=\(3 \times \frac{1}{x}+4 \times \frac{1}{y}=\frac{3}{x}+\frac{4}{y}\)
It is given that 3 men and 4 women finish the work in 14 days.
\(\frac{3}{x}+\frac{4}{y}=\frac{1}{14}\) ………….(i)
Also, 4 men and 6 women do the work in 10 days.
= \(\frac{4}{x}+\frac{6}{y}=\frac{1}{10}\) …………(ii)
Multiplying equation (i) by 4 and equation (ii) by 3, we get
ICSE Class 9 Maths Sample Question Paper 6 with Answers 16

∴ One woman finish the work in 140 days.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) There are two regular polygons with number of sides equal to (n – 1) and (n + 2). Their exterior angles differ by 6°. Find the value of n
Answer:
For first polygon,
ICSE Class 9 Maths Sample Question Paper 6 with Answers 17
\(\frac{3}{n^{2}+2 n-n-2}=\frac{1}{60}\)
n2  + n – 2 = 180
n2 + n- 182 = 0
n2  + (14 – 13) n – 182 = 0
n2  + 14n – 13n – 182 = 0
n(n + 14) -13 (n + 14) = 0
(n + 14) (n – 13) = 0
n + 14 = 0 or n – 13 =0
n = -14 = 0  or n = 13 (∵n cannot be negative)
∴ n = 13.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

Question 6.
(a) If \(a^{2}+\frac{1}{a^{2}}=7\), find the value of \(a^{2}-\frac{1}{a^{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 18

(b) Construct a trapezium ABCD in which AD || BC, Z B = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.
Answer:
Given : AD||BC, ZB = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 19
Steps of construction :
(1) Draw BC = 6.2 cm.
(2) At B, draw ∠CBX = 60° and cut off BA = 5 cm.
(3) At A, draw exterior ∠XAY = 60° such that AY||BC.
(4) From C, cut-off AY at D such that CD = 4.8 cm and join CD.
Hence, ABCD is the required trapezium.

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) The inner dimensions of a closed wooden box are 2 m, 1.2 m and 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1m3 of wood costs ₹ 5400.
Answer:
The inner dimensions of the closed box are 2 m, 1.2 m, 0.75 m.
Inner volume = (2 x 1.2 x 0.75) m3 = 1.8 m3
Thickness of the box = 2.5 cm = 2.5/100m= 0.025 m
∴ Outer dimensions are (2 + 2 x 0.025) m, (1.2 + 2 x 0.025) m, (0.75 + 2 x 0.025) m
i.e. 2.05 m, 1.25, 0.8 m.
∴ Outer volume = (2.05 x 1.25 x 0.8) m3 = 2.05 m3
Volume of wood = (2.05 – 1.8) m3 0.25 m3
Cost of 1 m3 of wood = ₹ 5400
Cost of 0.25 m3 of wood = ₹5400 x 0.25
= ₹ 1350

Question 7.
(a) Solve : 4x2 + 15 =16x
Answer:
4x2+15 =16x
4x2 – 16x+15 =0
4x2 – (10-t-6)x+15 =0
4x2 10x – 6x+15 =0
= 2x(2x – 5)- 3(2x – 5) =0
(2x – 5)(2x – 3) =0
2x – 5 =0 or 2x – 3=0
2x =5 or 2x=3
ICSE Class 9 Maths Sample Question Paper 6 with Answers 20

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) Find graphically the vertices of the triangle whose sides have equations
2y – x = 8, 5y – x = 14 and y – 2x = 1.
Answer:
Given equations are,
2y – x =8 ……….(i)
5y – x =14 …(ii)
and y – 2x =1 …(iii)
From(i), x =2y – 8
ICSE Class 9 Maths Sample Question Paper 6 with Answers 21
∴ (- 6, 1), (- 4, 2), (- 2, 3)
From (ii), x=2 y-8
ICSE Class 9 Maths Sample Question Paper 6 with Answers 22
(- 4, 2), (1, 3), (6, 4)
From (iii) y=2 x+1
ICSE Class 9 Maths Sample Question Paper 6 with Answers 23
∴ (1, 3), (2, 5), (- 1, – 1)
These points are plotted on the graph.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 24
The three lines intersect at point (- 4, 2), (1, 3) and (2, 5) which are the required vertices of triangle formed by them

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) If 3tan2 θ-1=0′, find cos 2θ, given that θ is acute.
Answer:
Given:  3tan2 θ-1=0
tan2θ = 1/3
tan θ   =\(\frac{1}{\sqrt{3}}\)]
⇒ tanθ = tan 30°
θ = 30°
cos2θ = cos (2 x 30°) = cos 60° =\(\frac{1}{2}\)

Question 8.
(a) Solve for x : 3(2x + 1) – 2x+2 + 5 = 0.
Answer:
⇒ 3(2x + 1) – 2x + 2 + 5 =0
⇒ 3.2x + 3 – 2x. 22 + 5 =0
⇒ 3.2x – 4.2x + 8=0
⇒ -2x = – 8
⇒ 2x = 23
⇒ x =3

(b) Find the area of a triangle whose perimeter is 22 cm, one side is 9 cm and the difference of the other two sides is 3 cm.
Answer:
One side = 9 cm, perimeter = 22 cm.
Let other two sides be a cm and b cm and a > b.
According to the question,
a + b + 9 = 22
⇒ a + b = 13 ………..(i)
and a – b =3 (Given) ………(ii)
Adding equations (i) and (ii), we have
2 a = 16 ⇒ a = 8
Subtracting equation (ii) from equation (i), we have
2b = 10 ⇒ b = 5
The sides are a = 8 cm, b = 5 cm, c = 9 cm
ICSE Class 9 Maths Sample Question Paper 6 with Answers 25

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(c) Insert four irrational numbers between 2√3 and 3√2
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 26

Question 9.
(a) Form a cumulative frequency distribution table from the following data by exclusive method taking 4 as the magnitude of class intervals.
31, 23, 19, 29, 20, 16, 10, 13, 34, 38, 33, 28, 21, 15, 18, 36, 24, 18, 15, 12, 30, 27, 23, 20, 17, 14, 32, 26, 25, 18, 29, 24, 19, 16, 11, 22, 15, 17, 10, 25.
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 27

(b) Solve simultaneously : \(\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{4}{y}=-\frac{1}{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 28
ICSE Class 9 Maths Sample Question Paper 6 with Answers 33

(c) The diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that area of ΔOAD = area of ΔOBC. Prove that ΔBCD is a trapezium.
ICSE Class 9 Maths Sample Question Paper 6 with Answers 4
Answer:
Given: Area of ΔOAD = area of ΔOBC.
Draw DM ⊥ AB,CN ⊥AB.
∴ DM || CN (∵Both DM, CN are perpendicular to AB)
Now,
Area of ΔOAD = Area of ΔOBC
Area of ΔOAD + Area of ΔOAB = Area of ΔOBC + Area of ΔOAB
ICSE Class 9 Maths Sample Question Paper 6 with Answers 34

ICSE Class 9 Maths Sample Question Paper 6 with Answers

Question 10.
(a) If the interest is compounded half yearly, calculate the amount when the principal is ₹ 7400, the rate of interest is 5% p. a. and the duration is one year.
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 35

(b) Find the coefficient of x2 and x in the product of (x – 2) (x – 3) (x – 4).
Answer:
Given :  (x -2) (x – 3) {x – 4)
Here,  a = – 2, b = – 3, c = -4
Coefficient of x2 = a + b + c = (- 2) + (- 3) + (- 4) = – 9
Coefficient of x = ab + be + ca = (- 2) (- 3) + (- 3) (- 4) + (- 4) (- 2)
= 6 + 12 + 8 = 26

(c) If the figure given, ABCD is a trapezium in which AB || DC. P is the mid-point of AD and PR || AB. Prove that PR = \(\frac{1}{2} (AB + CD)\).
ICSE Class 9 Maths Sample Question Paper 6 with Answers 5
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 36

Question 11.
(a) Factorize : a3 + 3a2b + 3ab2 + 2b3.
Answer:
a3 + 3a2b + 3 ab2 + 2b3 = (a3 + 3 a2b + 3ab2 + b3) + b3
= (a + b)3 + (b)3 = (a + b + b) {(a + b)2 – (a + b)b + b2}
= (a + 2b) (a2 + 2ab + b2 – ab – b2 + b2)
= (a + 2b) {a2 + ab + b2)

ICSE Class 9 Maths Sample Question Paper 6 with Answers

(b) In the point A (2, – 4) is equidistant from the points P (3, 8) and Q (- 10, y), find the values of y.
Answer:
Given points are A (2, – 4), P (3, 8), Q (- 10, y)
AQ = AP
⇒ AQ2 = AP2
(- 10 – 2)2 + (y + 4)2
= (3 – 2)2 + (8 + 4)2 144 + (y + 4)2
= 1 + 144 (y + 4)2 = 1 y + 4 = ±1
y + 4= 1 or y + 4 = -1 y = – 3 or y = – 5
y = – 3 or – 5

(c) Simplify: \(\sqrt[a b]{\frac{x^{a}}{x^{b}}} \cdot b \sqrt[x]{\frac{x^{b}}{x^{c}}} \cdot \sqrt[c a]{\frac{x^{c}}{x^{a}}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 6 with Answers 37

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 5 with Answers

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) At what rate % p.a. will sum of ₹ 4000 yield ₹ 1324 as compound interest in 3 years ?
Answer:
(a) Given : P = ₹ 4,000, C.I. = ₹  1,324, n = 3 years.
Let r be the rate % p.a.
Now A = P + C.I. = ₹ (4,000 + 1,324) = ₹ 5,324.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 5
ICSE Class 9 Maths Sample Question Paper 5 with Answers 6

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) If x = 2 + √3 , prove that x2 – 4x + 1 = 0.
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 7

(c) How many times will the wheel of a car having radius 28 cm, rotate in a journey of 88 km
Answer:
Given : r = 28 cm and
Distance = 88 km = 88 x 1,000 x 100 cm = 88,00,000 cm
Now, Distance covered in 1 rotation = Circumference of wheelICSE Class 9 Maths Sample Question Paper 5 with Answers 8

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Question 2.
(a) Factorize : \(x^{2}+\frac{1}{x^{2}}-11\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 9

(b) From the adjoining figure, find the value of x.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 1
Answer:
In ΔACD,
AC = CD  (Given)
⇒ ∠ADC = ∠CAD (Angles opposite to equal sides)
Now, ∠ADC + ∠CAD + ∠ACD = 180° (Sum of angles in a triangle is 180°)
2 ∠ADC + 56° = 180°    (∠ADC – ∠CAD)
⇒ 2∠ADC = 180° – 56°
⇒ ∠ADC \(\frac{124^{\circ}}{2}\)
∠ADC = \(\frac{124^{\circ}}{2}\) = 62°

In ΔABD, AD = BD (Given)
∠ABD = ∠ BAD (Angles opposite to equal sides)
Now, ∠ABD + ∠BAD = ∠ADC
(Exterior angle is equal to sum of interior opposite angles)
2∠ABD =62°
∠ABD\(\frac{62^{\circ}}{2}\)=31°
In ΔABC, ∠A + ∠B + ∠ C = 180° (Sum of angles in a triangle is 180°)
x°+31°+56°=180°
x° = 180°- 87°= 93°

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Simplify : \(\frac{5 .(25)^{n+1}-25 .(5)^{2 n}}{5 \cdot(5)^{2 n+3}-(25)^{n+1}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 10

Question 3.
(a) If θ = 30°, verify that cos 3 θ = 4 cos3 θ -3 cos θ .
Answer:
L.H.S. = cos 3θ = cos (3 x 30°) = cos 90° = 0 ………..(1)
L.H.S. = 4cos3 θ -3cosθ= 4 cos3 30° – 3 cos 30°
ICSE Class 9 Maths Sample Question Paper 5 with Answers 11
From (i) and (ii)
L.H.S. = R.H.S.
Hence Proved.

(b) Solve by cross multiplication method :
x – 3y – 7 = 0; 3x – 3y = 15.
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 12
ICSE Class 9 Maths Sample Question Paper 5 with Answers 13

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Prove that: \(\log \frac{11}{5}+\log \frac{14}{3}-\log \frac{22}{15}=\log 7\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 14
= log 11 – log 5 + log 14 – log 3 – log 22 + log 15
= log 11 – log 5 + log (2 x 7)- log 3 – log (2 x 11) + log (3 x 5)
= log 11 – log 5 + log 2 + log 7 – log 3 – log 2 – log 11 + log 3 + log 5
= log 7 = R.H.S
Hence Proved.

Question 4.
(a) If a2 – 3a – 1 = 0, find the value of a + \(a^{2}+\frac{1}{a^{2}}\)
Answer:
(a)
ICSE Class 9 Maths Sample Question Paper 5 with Answers 15

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) Construct a combined histogram and frequency polygon for the following data :
ICSE Class 9 Maths Sample Question Paper 5 with Answers 2
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 16
ICSE Class 9 Maths Sample Question Paper 5 with Answers 17

(c) Of two unequal chords of a circle, prove that longer chord is nearer to the centre of the circle.
Answer:
Given : AB > CD, OM⊥AB, ON ⊥ CD.
Join OA and OC.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 18

We know perpendicular drawn from the centre to the chord bisects the chord.
∴ AM= \(\frac{1}{2}\)AB
and CN = \(\frac{1}{2}\)CD
Now, AM > CN(∵ AB > CD)
In ΔOAM OA2 = AM2 + OM2 (Pythagoras theorem)
In ΔOCN OC2 = CN2 + ON2 (Pythagoras theorem)
AM2 + OM2 =CN2 +ON2(∵ OA = OC, Radii)
⇒ OM2 – ON2 = – (AM2 – CN2)
⇒ OM2 – ON2 <O (∵AM > CN)
⇒ OM2 < ON2
⇒ OM < ON
i.e., longer chord is nearer to the centre.
Hence Proved

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Factorize : \(\frac{y^{6}}{343}+\frac{343}{y^{6}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 19

(b) The diagonals AC and DB of a parallelogram intersect at O. If P is the mid-point of AD,
prove that (i) PO || AB (ii) PO = \(\frac{1}{2}\)CD.
Answer:
Given : In parallelogram ABCD, diagonals AC and BD intersect at O. P is the mid-point of AD.
(i) Since diagonals of a parallelogram bisect each other
∴ O is the mid-point of DB
Now, in ΔADC
P and O are mid-points of sides AD and BD respectively
By mid-point theorem,
ICSE Class 9 Maths Sample Question Paper 5 with Answers 20

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(ii) Also, by mid-point theorem
ICSE Class 9 Maths Sample Question Paper 5 with Answers 21

(c) If θ is acute and 3sin θ = 4cos θ, find the value of 4sin2 θ – 3cos2 θ + 2.
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 22
ICSE Class 9 Maths Sample Question Paper 5 with Answers 23

Question 6.
(a) If the points A (4,3) and B (x, 5) are on the circle with centre C (2, 3), find the value of x
Answer:
Given : A (4, 3), B (x, 5), C (2, 3).
∵ C (2, 3) is the centre,
∴ AC = BC (Redii)
AC2 = BC2
(4 – 2)2 + (3 – 3)2 =(x- 2) + (5 – 3)2
= 4 + 0 =(x – 2)2 + 4
(x-2)2 =0
x – 2 =0
x =2.

(b) ABCD is a trapezium with AB | | CD, and diagonals AC and BD meet at O.
Prove that area of ΔDAO = area of ΔOBC.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 3
Answer:
Given : AB||CD, diagonals AC and BD meet at O.
AB||DC
∴  Area of ΔABD = Area of ΔABC (Triangles on same base and between same parallels are equal in area)
∴ Area of ΔDAO + Area of ΔOAB = Area of ΔOBC + Area of Δ OAB (Addition area axiom)
⇒ Area of ΔDAO = Area of Δ OBC.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Simplify: \(\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 24

Question 7.
(a) If x + y = 10 and x2 + y1 = 58, find the value of x3 + y3.
Answer:
Given :
x + y =10, x2 + y2 = 58.
(x + y)2 = x2 + y2 + 2xy
⇒ 102 = 58 + 2xy
2xy = 100 – 58
⇒ xy = \(\frac{42}{2} \) = 21
x3 + y3 = (x + y)3 – 3xy (x + y)
= 103 – 3 x 21 x 10
= 1000 – 630 = 370.

(b) The larger of two supplementary angles exceeds the smaller by 18°. Find the angles.
Answer:
Let the larger supplementary angle be
Then, smaller supplementary angle = 180° – x According to the question,
x – (180° – x) = 18°
⇒ x – 180° + x =18°
⇒ 2x = 18° + 180°
⇒ x= \(\frac{198^{\circ}}{2}\) = 99°
∴ 180° –  x = 1800 99° = 81°
The supplementary angles are 990 and 81°.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.
Answer:
Mean of 5 numbers = 20
∴Sum of 5 numbers = 20 x 5 = loo.
1f one number is excluded,
Then, Mean of 4 numbers = 23
Sum of 4 numbers = 23 x 4 = 92
The excluded number = 100 – 92 = 8.

Question 8.
(a) Solve for x : 9 x 3X = (27)2x-5
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 25

(b) In a right-angled triangle, if hypotenuse is 20 cm and the ratio of the other two sides is
4 : 3, find the sides.
Answer:
Given: Hypotenuse = 20 cm
and ratio of the other two sides = 4:3
Let the other two sides be 4x and 3x.
∴ By Pythagoras theorem,
ICSE Class 9 Maths Sample Question Paper 5 with Answers 26
4x – 4 x 4 = 16
3x = 3 x 4 = 12
∴ The required sides are 16 cm and 12 cm.

(c) Without using tables, find the value of :
ICSE Class 9 Maths Sample Question Paper 5 with Answers 35
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 27

ICSE Class 9 Maths Sample Question Paper 5 with Answers

Question 9.
(a) In what time will a sum of ₹ 8000 becomes ₹ 9261 at the rate of 10% p. a., if the interest is compounded semi-annually?
Answer:
Given : P = X 8,000, A = ? 9,261, r = 10% p.a.
Let n be the number of years.
∵ C.I. is compounded semi-annually,
ICSE Class 9 Maths Sample Question Paper 5 with Answers 28

(b) Construct a regular hexagon of side 2.2 cm.
Answer:
Each side = 2.2 cm.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 29
Steps of construction :
(1) Draw AB 2.2 cm
(2) At A and B, draw angle of 120°.
(3) From A and B, cut-off arcs of 2.2 cm each.
(4) At C, draw 120° and cut it off at D so that CD = 2.2 cm.
(5) At D, draw 120° and cut-off DE = 2.2 cm.
(6) Join EF.
Then, ABCDEF is the required hexagon.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(c) Solve graphically : 2x – 3y + 2 = 4x + 1=3x-y + 2
Answer:
Given: 2x – 3y + 2 = 4x + 1=3x-y + 2
∴ 2x – 3y + 2 = 4x + 1 and 4x + 1= 3x-y + 2
⇒ 4x – 2x = – 3y + 2 – 1 and 4x-3x = -y + 2- 1
⇒  2x = 1 – 3y
ICSE Class 9 Maths Sample Question Paper 5 with Answers 30
ICSE Class 9 Maths Sample Question Paper 5 with Answers 31
The two lines intersect at the point (2, – 1).
x =2, y = -1

Question 10.
(a) Express (x2 -5x + 7) (x2 + 5x – 7) as a difference of two squares.
Answer:
(x2 – 5x + 7) (x2 + 5x – 7) = {x2 – (5x – 7)} {x2 + (5x – 7)}
= (x2)2 – (5x – 7)2

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) Simplify: \((64)^{2 / 3}-\left(\frac{1}{81}\right)^{-1 / 4}+8^{2 / 3} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 3^{0}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 32

(c) In a pentagon ABCDE, BC | | ED and ∠B: ∠A: ∠E = 5:3:4. Find ∠B.
ICSE Class 9 Maths Sample Question Paper 5 with Answers 4
Answer:
Given: ∠B: ∠A: ∠E =5:3:4
Let Now, ∠B = 5x, ∠A = 3x, ∠E = 4x.
∠C + ∠D = 180°(Co-interior angles; BC||ED)
Sum of angles in a figure with number of sides ‘n’ = (n – 2) x 180°
In pentagon, sum of angles = (5 – 2) x 180°
= 3 x 180° = 540°
∴ ∠ A + ∠B + ∠C + ∠D + ∠E = 540°
⇒ 3x + 5x + 180° + 4x = 540°
⇒ 12x = 540° – 180°
⇒ \(x=\frac{360^{\circ}}{12}\)
⇒ x = 30°
∴ ∠B = 5x = 5 x 30° = 150°

Question 11.
(a) If p + q = 10 and pq = 21, find 3 (p2 + q2).
Answer:
p + q = 10, pq = 21.
∴ p2 + q2 = (p + q)2 – 2pq = 102 – 2 x 21 = 100 – 42 = 58
3 (p2 + q2) = 3 x 58
= 174.

ICSE Class 9 Maths Sample Question Paper 5 with Answers

(b) Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm
Answer:
Let length of each of equal sides be a and that of base be b.
b  = 6 cm(Given)
and Perimeter   = 16 cm
⇒ a + b  = 16
⇒ 2a   + 6  = 16
⇒ 2a = 16 – 6
ICSE Class 9 Maths Sample Question Paper 5 with Answers 33

(c) Prove that : \(\frac{1}{1+\tan ^{2} \theta}+\frac{1}{1+\cot ^{2} \theta}=1\)
Answer:
ICSE Class 9 Maths Sample Question Paper 5 with Answers 34
= cos2 θ + sin2 θ=1
Hence Proved.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 4 with Answers

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Expand : \(\left(\frac{2}{3} x-\frac{3}{2 x}-1\right)^{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 5

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(b) A person invests ₹ 10,000 for two years at a certain rate of interest, compounded annu­ally. At the end of one year, this sum amounts to ₹ 11,200. Calculate :
(i) The rate of interest p. a.
(ii) The amount at the end of second year.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 3

(c) Factorize : 64x6 – 729y
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 4

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 2.
(a) If a2 + b2 = 7ab, prove that 2 log (a + b) = log 9 + log a + log b.
Answer:
Given : a2 + b2 .= 7ab
⇒ a2 + b2 – 9ab – 2ab
⇒ a2 + b2 + 2ab = 9 ab
⇒ (a + b)2 = 9 ab
Taking log of both sides, we get
log (a + b)2 = log (9ab)
⇒ 2 log (a + b) = log 9 + log a + log b.

(b) Prove that: \(\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0\)
Answer:
To prove:
\(\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0\)
Consider L.H.S. = \(\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1\)
= tan2 θ- sec2 θ +1
= (1 + tan2 θ) – sec2 θ (∵ 1 + tan2 0 = sec2 0)
= sec2 θ – sec2 θ = θ= R.H.S.
Hence Proved

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) In the given figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.
Answer:
AB = CD
Minor arc AB = Minor arc CD
Minor arc AB – minor arc BD = Minor arc CD – Minor arc BD
⇒ arc AD = arc CB.
Hence Proved.

Question 3.
(a) In the given figure, ∠ BCD = ∠ADC and ∠BCA = ∠ADB.
Show that: (i)ΔACD ≅ ΔBDC (ii) BC = AD (iii) ∠A = ∠B.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 1
Answer:
Given: ∠BCD = ∠ADC
and ∠BCA = ∠ADB
=> ∠BCA + ∠BCD = ∠ADB + ∠BCD
=> ∠BCA + ∠BCD = ∠ADB + ∠ADC (v ∠BCD = ∠ADC)
=> ∠ACD = ∠BDC
In ΔACD and ΔBDC
(i) ∠ADC = ∠BCD (Given)
CD = CD (Common side)
⇒ ∠ACD = ∠BDC (Proved above)
∴ ΔACD ≅ ΔBDC (ASA axiom)
(ii) ∴BC = AD (c.p.c.t.)
(iii) ∴ ∠A = ∠B. (c.p.c.t.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(b) \(a=\frac{2-\sqrt{5}}{2+\sqrt{5}} \text { and } b=\frac{2+\sqrt{5}}{2-\sqrt{5}}, \text { find } a^{2}-b^{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 6
ICSE Class 9 Maths Sample Question Paper 4 with Answers 7

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) Prove that the points A (2, 3), B (-2, 2), C (-1, -2) and D (3, – 1) are the vertices of a square ABCD.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 8

Question 4.
(a) The mean height of 10 girls in a class is 1.38 m and the mean height of 40 boys is 1.44 m. Find the mean height of 50 students of the class.
Answer:
Given : Mean height of 10 girls = 1.38 m
∴ Sum of heights of 10 girls = 1.38 x 10 = 13.8 m
and Mean height of 40 boys = 1.44 m
∴ Sum of heights of 40 boys = 1.44 x 40 = 57.6 m.
∴ Sum of heights of 50 students = 13.8 +57.6 = 71.4 m.
∴ Mean heights of 50 students \(\frac{71.4}{50}\)
=1.428m

(b) In the given figure, AABC is a right triangle with ∠C = 90° and D is mid-point of side BC. Prove that AB2 = 4AD2 – 3AC2.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 2
Answer:
Given ∠C = 90°, D is the mid-point of BC.
∴ CD = BD
∴  In ΔABC,
AB2 = AC2 + BC(Pythagoras theorem)
= AC2 + (2CD)2 (∵ CD = BD = \(\frac{1}{2}\) BC)
AB2 = AC2 + 4 CD2  …(i)
In ΔACD,
AD2 = AC2 + CD2  (Pythagoras theorem)
⇒ CD2 = AD2 – AC2 …(ii)
From equations (i) and (ii), we get
AB2 = AC2 + 4 (AD2 – AC2)
= AC2 + 4AD2 – 4AC2
= 4AD2 – 3AC2
 Hence Proved.

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) The following observation have been arranged in ascending order.
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28 . If the median of the data is 13, find the value of x.
Answer:
Given : Numbers in ascending order : 3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.
Median = 13
Here, n = 20.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 9

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Factorize : 5x2 + 17xy – 12y2.
Answer:
5x2 + 17xy – 12y2 = 5x2 + (20 – 3) xy – 12y2
= 5x2 + 20xy – 3xy – 12y2
= 5x(x + 4y) – 3y (x + 4y)
= (x + 4y) (5x – 3y).

(b) If twice the son’s age in years is added to the father’s age, the sum is 70. But if twice the  father’s age is added to the son’s age, the sum is 95. Find the ages of father and son.
Answer:
Let father’s age be x years and that of son’s be y years.
By 1st condition, x + 2y = 70 …(i)
By 2nd condition, 2x + y = 95 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y = 190
Subtracting equation (iii) from equation (i), we get
ICSE Class 9 Maths Sample Question Paper 4 with Answers 10
⇒ 40 +2y = 70
⇒ 2y = 70 – 40
⇒ \(y=\frac{30}{2}=15\)
∴ Father’s age is 40 years and son’s age is 15 years.

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.
Answer:
Given: E, F are mid-points of sides AC and AB, respectively.
ICSE Class 9 Maths Sample Question Paper 4 with Answers 11
We know, the line joining the mid-points of any two sides of a triangle is parallel to the third side
FE || BC
⇒ FQ || BP
Now, since F is mid-point of AB and FQ || BP
∴ By converse of mid-point theorem,
⇒ Q is the mid-point of AP.
AQ = QP. Hence Proved.

Question 6.
(a) If a + \(\frac{1}{a} = p,\) prove that \(a^{3}+\frac{1}{a^{3}}=p\left(p^{2}-3\right).\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 12

(b) The side of a square exceeds the side of another square by 3 cm and the sum of the areas of the two squares is 549 cm2. Find the perimeters of the squares.
Answer:
Let the side of one square be x cm
Then, side of other square = (x + 3) cm.
Area of two squares are x2 cm2 and (x + 3)2 cm2, respectively.
According to the question,
x2 + (x + 3)2 = 549 ⇒ x2 + x2 + 6x + 9 = 549
⇒ 2x2 + 6x – 540 = 0
⇒ x2 + 3x – 270 = 0
⇒ x2 + (18 – 15)x – 270 =0
⇒ x2 + 15x – 15x – 270 =0
⇒ x (x + 18) – 15 (x + 18) =0
⇒ (x + 18) (x – 15) =0
⇒ x = – 18 or 15
∴ x – 15      (∵ x cannot be negative)
∴ x+ 3 =15+ 3 = 18
∴ Perimeter of one square = 4 x 15 = 60 cm
and Perimeter of other square = 4 x 18 = 72 cm

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) Simplify :
\(\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \left(90^{\circ}-\theta\right)}{\sec \left(90^{\circ}-\theta\right)}-3 \tan ^{2} 30^{\circ}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 13

Question 7.
(a) If log10 a = b, express 102b3 in terms of a.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 14

(b) ΔABC and ΔDBC are on the same base BC with A, D on opposite sides of BC. If area of ΔABC = area of ΔDBC, prove that BC bisects AD.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 15

(c) If cos θ + sec θ = 2, show that cos8 θ + sec8 θ = 2.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 16

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 8.
(a) If each interior angle is double the exterior angle, find the number of sides.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 17

(b) Solve by the substitution method :
5x + 4y – 4 = 0; x – 20 = 12y.
Answer:
Given: 5x+4y – 4 =0 ………..(i)
and x – 20 =12y ………… (2)
From (ii), x = 12y + 20 ……. (3)
Putting x = 12 + 20 in equation (i), we have
5(12y+20)+4y – 4 =0
⇒ 60y+100+4y – 4=0
⇒ 64y = – 96
⇒ \(y=-\frac{96}{64}=-\frac{3}{2}\)
From (iii), x = 12 x \(\left(\frac{-3}{2}\right)\) +20 = 2
x = 2,y = \(\frac{-3}{2}\)

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) If x = 2 + √3 , find the value of x – \(\frac{1}{x}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 18

Question 9.
(a) Evaluate : \(x^{2 / 3} \cdot y^{-1} \cdot z^{1 / 2}\) when x – 8, y = 4 and z = 25
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 19
(b) Construct a ΔABC is which base AB 5 cm, ∠A = 30° and AC – BC – 2.5 cm.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 20
Steps of construction :
(1) Draw base AB = 5 cm.
(2) Draw ∠BAX = 30°
(3) From AX, cut-off AD = 2.5 cm
(4) Join BD.
(5) Draw the perpendicular bisector of BD to cut AX at C.
(6) Join BC.
Thus, ABC in the required triangle.

(c) Simplify: \(\left(2 x-\frac{1}{2 x}\right)^{2}-\left(2 x+\frac{1}{2 x}\right)\left(2 x-\frac{1}{2 x}\right)\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 21

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 10.
(a) Simplify: \(\left(a^{m-n}\right)^{m+n} \cdot\left(a^{n-l}\right)^{n+l} \cdot\left(a^{l-m}\right)^{l+m}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 22

(b) If area of a semi-circular region is 1232 cm2, find its perimeter.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 23

(c) If x = 15°, evaluate : 8 sin cos 4x. sin 6x.
Answer:
Given: x=15°
∴ 8 sin 2xcos 4x.sin 6x = 8 sin (2 x 15°). cos (4 x 15°) . sin (6 x 15°)
= 8 sin 30° . cos 60° . sin 90°
\(=8 \times \frac{1}{2} \times \frac{1}{2} \times 1=2\)

ICSE Class 9 Maths Sample Question Paper 4 with Answers

Question 11.
(a) A farmer increases his output of wheat in his farm every year by 8%. This year he pro­duced 2187 quintals of wheat. What was his yearly produce of wheat 2 years ago?
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 24

(b) Draw the graph of the equation 3x – y = 4.
Answer:
ICSE Class 9 Maths Sample Question Paper 4 with Answers 25
ICSE Class 9 Maths Sample Question Paper 4 with Answers 26

ICSE Class 9 Maths Sample Question Paper 4 with Answers

(c) Evaluate : (99.9)2 – (0.1)1
Answer:
(99.9)2 – (0.1)2 = (99.9 + 0.1)
(99.9 – 0.1) = 100 x 99.8
= 9980.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 3 with Answers

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Factorize : 8(a – 2b)2 – 2a + 4b – 1
Answer:
8 (a – 2b)2 -2a + 4b – 1 = 8 (a- 2b)2 -2(a – 2b) – 1
a – 2b = x
Then, the expression becomes
= 8x2 – 2x – 1
= 8x2 – 4x + 2x – 1
= 4x (2x – 1) + 1 (2x – 1)
= (2x – 1) (4x + 1)
= {2 (a- 2b) – 1} {4 (a – 2b) + 1} (Putting x-a-2b)
= (2a – 4b – 1) (4a- 8b + 1)

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(b) The mean of 6 observations is 17.5. If five of them are 14, 9, 23, 25 and 10, find the sixth observation.
Answer:
Mean of 6 observations is 17.5
5 observations are 14, 9, 23, 25, 10
Let 6th observation be x
ICSE Class 9 Maths Sample Question Paper 3 with Answers 5

(c) If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ-1.
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 6

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 2.
(a) A man borrowed ₹15,000 for 2 years. The rate of interest for the two successive years are 8% and 10% respectively. If he repays ₹6,200 at the end of first year, find the outstanding amount at the end of the second year.
Answer:
(a) For 1st year : P = ₹ 15,000, R = 8% p.a.
\(\mathrm{I}=\frac{15000 \times 8 \times 1}{100}\) = ₹ 16,200
A = P + I = 15,000 + 1,200 =₹16,200
Amount of money repaid = ₹ 6,200
For 2nd year : P = 16,200 – 6,200 = ₹10,000, R = 10% p.a.
\(\mathrm{I}=\frac{10,000 \times 10 \times 1}{100}\) = = ₹1,000
∴ A =P+I=10,000+1,000=11,000
∴ The amount outstanding at the end of 2nd year = 11, 000.

(b) Solve for x if log2 (x2 – 4) = 5.
Answer:
Given: log2(x2 – 4) = 5
x2 – 4 =25
x2 =32 + 4
X = ±√36= ±6.

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) In the following figure, AB is a diameter of a circle with centre O. If chord AC = chord AD,
prove that (i) arc BC = arc DB (ii) AB is bisector of ∠CAD.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 1

Answer:
(i) Given: chord AC = chord AD
⇒ arc AC = arc AD …(i)
Also, arc ACB = arc ADB (AB is a diameter) …(ii)
Subtracting (i) from (ii),
arc ACB – arc AC = arc ADB – arc AD
⇒ arc BC = arc DB. Hence Proved.

(ii) ∵ arc BC = arc BD (Proved above)
∴ ∠BAC = ∠DAB
⇒ AB is bisector of ∠CAD.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 3.
(a) Prove that: \(\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}=\frac{3}{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 7

(b) Given that 16 cot A = 12, find the value of \(\frac{\sin A+\cos A}{\sin A-\cos A}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 8

(c) If the altitudes from two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 9

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 4.
(a) If a + b + 2c = 0, prove that a3 + b3 + 8c3 – 6abc = 0.
Answer:
Given : a + b + 2c = 0
⇒ a + b = – 2c
Cubing both sides, we get
(a + b)3 = (- 2c)3
⇒ a3 + b3 + 3ab (a + b) = – 8c3
⇒ a3 + b3 + 3ab (- 2c) = – 8c3
⇒ a3 + b3 + 8c3 – 6abc = 0.
Hence Solved.

(b) Express \(0.1 \overline{34}\) in the form \(\frac{p}{q}\) ,p, q ∈ Z and q ≠ 0.
Answer:
Let x = \(0.1 \overline{34}\) = 0.1343434…
Multiplying both sides of (i) by 10, we get
10x = 1.343434 ………(i)
Multiplying both sides of (ii) by 100, we get
1000x = 134.3434 ………….(ii)
Subtracting (ii) from (iii), we get
1000x – 10x= 134.3434 … – 1.3434 ……………
ICSE Class 9 Maths Sample Question Paper 3 with Answers 10

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) If the sides are in the ratio 5 : 3 : 4, prove that it is a right angled triangle.
Answer:
Ratio of sides = 5:3:4
Let the length of sides be 5x, 3x, 4x.
Here,(3x)2 + (4x)2 = 9x2 + 16x2 = 25x2 = (5x)2
∴ By Pythagoras theorem, the triangle is right angled
Hence Proved.

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) On what sum of money will the difference between compound interest and simple inter­est for 2 years be equal to ₹25 if the rate of interest charged for both is 5% p.a. ?
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 12

(b) Show by distance formula that the points A (-1, -1), B (2, 3) and C (8,11) are collinear.
Answer:
Given points are A (-1, -1), B (2, 3), C (8, 11).
Now
ICSE Class 9 Maths Sample Question Paper 3 with Answers 11
AB + BC = 5 + 10 = 15
⇒ AB + BC = AC
The points are collinear.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) Factorize : a6 – 26a3 – 27.
Answer:
a6 – 26a3 – 27 = a6 – (27 – 1) a3-27 = a6 – 27a3 + a3 – 27
= a3 (a3 – 27) + 1 (a3 – 27) = {a3 – 27) (a3 + 1)
= (a3 – 33) (a3 + 13)
= (a – 3) (a2 + 3a + 9) (a + 1) (a2 – a + 1)

Question 6.
(a) Simplify: \(\frac{\left(x^{a+b}\right)^{2} \cdot\left(x^{b+c}\right)^{2} \cdot\left(x^{c+a}\right)^{2}}{\left(x^{a} \cdot x^{b} \cdot x^{c}\right)^{4}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 13

(b) In the given figure AABC, D is the mid-point of AB, E is the mid-point of AC. Calculate :
(i) DE, if BC = 8 cm.
(ii) ∠ADE, if ∠DBC = 125°.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 2
Answer:
Given : D is mid-point of AB, E is the mid-point of AC, BC = 8 cm, ∠DBC = 125°.
The line joining the mid-points of any two sides of a triangle is parallel to the third and is equal the half of it.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 14

(c) If a and b are rational numbers, find the values of a and b :
\(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 16
ICSE Class 9 Maths Sample Question Paper 3 with Answers 17

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 7.
(a) Draw a histogram from the following data :

Weight (in kg) 40-44 45-49 50-54 55-59 60-64 65-69
No. of students 2 8 12 10 6 4

Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 18
ICSE Class 9 Maths Sample Question Paper 3 with Answers 19

(b) Solve:
83x – 67y = 383
67x – 83y = 367.
Answer:
83x – 67y = 383 ………….(i)
67a – 83y = 367 ………….(ii)
Adding (i) and (ii), we get
150x – 150y = 750
x – y = 5 ………. (iii)
Subtracting (ii) from (i), we get
16x + 16y = 16
x + y = 1 ……… (iv)
Adding (iii) and (iv), we get
2x = 6
X = 3
Putting x = 3 in (iv), we get
3 + y = 1
⇒ y = 1 – 3 = -2.
x = 3, y = – 2

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) In the following figure, area of parallelogram AFEC is 140 cm2. State, giving reason, the area of (i) parallelogram BFED (ii) ABFD.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 3
Answer:
Given : Area of parallelogram AFEC = 140 cm2.
(i)  Area of parallelogram BFED = Area of parallelogram AFEC
( ∵ They are on same base and between same parallels)
= 140 cm2

(ii) Area of Δ BFD = \(\frac{1}{2}\) x Area of parallelogram BFED
(∵ They are on same base and between same parallels)
\(\frac{1}{2}\) x 140 cm2 = 70 cm2.

Question 8.
(a) If log10 a = m and log10 b = n, express \(\frac{a^{3}}{b^{2}}\) in terms of m and n
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 20

(b) Draw the graph of 2x + y = 6 and 2x – y + 2 = 0. Hence, find the area of the region bounded by these lines and X-axis.
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 21
ICSE Class 9 Maths Sample Question Paper 3 with Answers 22
ICSE Class 9 Maths Sample Question Paper 3 with Answers 23

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) Factorize : \(8 x^{3}-\frac{1}{27 y^{3}}\)
Answre:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 24

Question 9.
(a) If \(\frac{x^{2}+1}{x}=2 \frac{1}{2}\) find the values of \((i) x-\frac{1}{x}
(ii) x^{3}-\frac{1}{x^{3}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 25

(b) In the following figure, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 4
Answer:
Given : OA = 3.5 cm, OD = 2 cm
Area of shaded region = Area of quadrant AOB – Area of ΔAOD.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 26
= 9.625 – 3.5 = 6.125 cm2.

(c) If a + b + c = 9 and ab + be + ca = 40, find the value of a2 + b2 + c2.
Answer:
Given: a + b + c = 9 ab + bc + ca = 40
We know, (a+b+c)2 =a2+ b2 + c2 +2(ab+bc+ca)
(9)2 =a2+ b2 + c2+ 2 x 40
a2+ b2 + c2 = 81 – 80 = 1

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 10.
(a) Solve: \((\sqrt{2})^{2 x+4}=8^{x-6}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 27

(b) Construct a rhombus whose diagonals are 5 cm and 6.8 cm.
Answer:
Given, diagonals are 5 cm and 6.8 cm.
Steps of construction :
(1) Draw AC = 5 cm.
(2) Draw perpendicular bisector PQ of AC which intersect AC at O.
(3) From POQ, cut-off OB = OD = \(\frac{6.8}{2}\) = 3.4 cm.
(4) Join the points A, B, C, D.
Then, ABCD is the required rhombus.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 28

(c) In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B respectively.
Prove that ∠AOB = \(\frac{1}{2}\) (∠C + ∠D).
Answer:
Given, AO and BO are bisectors of ∠A and ∠B respectively.
ICSE Class 9 Maths Sample Question Paper 3 with Answers 29
ICSE Class 9 Maths Sample Question Paper 3 with Answers 30

ICSE Class 9 Maths Sample Question Paper 3 with Answers

Question 11.
(a) Find the value of log5√5 (125).
Answer:
ICSE Class 9 Maths Sample Question Paper 3 with Answers 31

(b) The sum of a two-digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Answer:
Let the digits in tens and units place be x and y respectively.
The number = 10x + y
The number obtained by reversing digits = 10y + x By 1st condition,
(10 + y) + (10y + x) = 165
⇒ 11x + 11y = 165
⇒ x + y =15 …(ii)
By 2nd condition, x-y =3 …(iii)
or y-x =3 …(iv)
Adding (ii) and (iii), we get 2x = 18
⇒ x =9
Putting x = 9 in (ii), we get y = 15 – 9 = 6
Again, adding (ii) and (iv), we get
2 y =18
⇒ y =9
Putting y = 9 in (ii), we get x = 15 – 9 = 6.
Substituting these values in (i), we get
The number = 10 x 9 + 6 or 10 x 6 + 9 = 96 or 69

ICSE Class 9 Maths Sample Question Paper 3 with Answers

(c) If the area of an equilateral triangle is 81√3 cm2, find its perimeter.
Answer:
Given : Area of equilateral triangle = 81√3 cm2
Let the length side of each of equilateral triangle be a cm.ICSE Class 9 Maths Sample Question Paper 3 with Answers 32

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 2 with Answers

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) If x – 3 – 2√2, find the value of x2 + -y.
Answer:
Given = ICSE Class 9 Maths Sample Question Paper 2 with Answers 7
ICSE Class 9 Maths Sample Question Paper 2 with Answers 8

(b) Factorize : 9x2 – 4 (y + 2x)2
Answer:
9x2 -4(y + 2x)2 = (3x)2 – {2 (y + 2x)}2
= (3x)2 – (2y + 4x)2
= (3x + 2y + 4x) (3x – 2y – 4x)
= (7x + 2y) (-x -2y)
= – (x + 2y) (7x + 2y).

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(c) The area enclosed between the concentric circles is 770 cm2. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.
Answer:
Given : The area enclosed between the circles = 770 cm2
Radius of outer circle (R) = 21 cm.
Let radius of inner circle be r.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 9

Question 2.
(a) A man invests ₹46,875 at 4% p.a. compound interest for 3 years. Calculate :
(i) The interest for the first year.
(ii) The amount at the end of the second year.
(iii) The interest for the third year.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 10

(b) If ax = by = cz and b2 = ac, Prove that \( y=\frac{2 x z}{x+z}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 11

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(c) In the following figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 1
Answer:
Given: AD = BC,
∠OAD = ∠OBC = 90°
In Δ OAD and ΔOBC,
AD =BC (Given)
ΔOAD = ΔOBC (Given)
Δ AOD =Δ BOC (Vertically opposite angles)
∴ ΔOAD ≅ ΔOBC (SAS axiom)
∴ OA = OB (c.p.c.t.)
Hence, CD bisects AB. Hence Proved.

Question 3.
(a) Solve the following equations by cross multiplication method :
3x – 7y = – 10, – 2x + y = 3.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 13

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(b) Find the value of :
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
Answer:
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
\(=2 \sqrt{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}+2 \sqrt{3} \times \frac{1}{2} \times \sqrt{3}-1\)

(c) Construct a frequency polygon for the following frequency distribution using a graph sheet.

Marks 40 – 50 50-60 60 – 70 70-80 80 – 90 90 – 100
No. of Students 5 8 13 9 7 5

Use 1 cm – 10 marks and 1 cm = 5 students.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 14
ICSE Class 9 Maths Sample Question Paper 2 with Answers 15

Question 4.
(a) Express as a single logarithm :
2 log 3 – \(\frac{1}{2}\) log 64 + log 16.
(b) If \(x+\frac{1}{x}=3\),evaluate \(x^{3}+\frac{1}{x^{3}}\)
(c) Prove that the line joining mid-points of two parallel chords of a circle passes through the centre of the circle.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 16

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(c) Given: AB || CD, M and N are mid-points of sides AB and CD respectively.
Construction: Join OM, ON and draw a straight line parallel to AB and CD.
Since, line segment joining the mid-point of the chord with centre of the circle is perpendicular to the chord
ICSE Class 9 Maths Sample Question Paper 2 with Answers 17
∴ OM ⊥ AB and ON ⊥ CD
⇒ ∠AMO = 90° and ∠NOE = 90°
Now, ∠MOE = 90° (Co-interior angles, OE || AB)
∠NOE = 90° (Co-interior angles, OE || CD)
∠MOE + ∠NOF =90° + 900 = 1800
So, MON is a straight line passing through the centre of the circle.
Hence Proved.

Section – B
(Attempt any four questions from this Section)

Question 5.
(a) Find a point on the Y-axis which is equidistant from the points A (6, 5) and B (- 4, 3).
Answer:
Given : A (6, 5), B (- 4, 3).
Let the point on the Y-axis be P (0, b).
According to the question,
AP = BP
⇒ AP2 = BP2
⇒ (6 – 0)2 + (5 – b)2 = (- 4 – 0)2 + (3 – b)2
⇒ 36 + 25 – 10b + V- = 16 + 9 – 6fo + b2
⇒ – 10b + 6b = 25 – 61
⇒  -4b =-36
⇒ b = 9
Required Point = (0,9)

(b) In the following figure, area of parallelogram ABCD is 29 cm2. Calculate the height of parallelogram ABEF, if AB = 5.8 cm.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 2
Answer:
Area of parallelogram ABCD = 29 cm2, AB = 5.8 cm.
Area of parallelogram ABEF = 29 cm2 (area of parallelograms on same base are equal)
⇒ AB x Height = 29 cm2
⇒ \(\text { Height }=\frac{29 \mathrm{~cm}^{2}}{\mathrm{AB}}=\frac{29 \mathrm{~cm}^{2}}{5.8 \mathrm{~cm}}=5 \mathrm{~cm}\)

(c) A sum of money doubles itself at compound interest in 15 years. In how many years will it become eight times ?
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 18
ICSE Class 9 Maths Sample Question Paper 2 with Answers 19

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Question 6.
(a) Construct the quadrilateral ABCD, given that AB = 5 cm, BC = 2.5 cm, CD = 6 cm,
∠BAD = 90° and the diagonal AC = 5.5 cm.
Answer:
Given : AB = 5 cm, BC = 2.5 cm, CD = 6 cm, Z BAD 90°,
AC = 5.5 cm.
Steps of construction :
(1) Draw AB 5 cm.
(2) At A, draw ∠BAP = 90°.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 20

(3) From B and A, draw arcs of lengths 2.5 cm and 5.5 cm, respectively which intersect at C.
(4) From C, cut-off AP at D such that CD = 6 cm.
Thus, ABCD is the required quadrilateral.

(b) Factorize : (a + 1) (a + 2) (a + 3) (a + 4) – 3.
Answer:
(a + 1) (a + 2) (a + 3) (a + 4) – 3 = (a + 1) (a + 4) (a + 2) (a + 3) – 3
= (a2 + 5a + 4) (a2 + 5a + 6) – 3
= (p + 4) (p + 6) – 3  (Putting a2 + 5a = p)
= p2 + 6p + 4p + 24 – 3
= p2 + 10p + 21
= p2 + (7 + 3) p + 21
= p2 + 7p + 3p + 21
= p2 (p + 7) + 3 (p + 7) = (p + 7) (p + 3)
= (a2 + 5a + 7) (a2 + 5a + 3) (∵ p = a2 + 5a)

(c) In the following figure, D and E are mid-points of the sides AB and AC respectively. If BC = 6 cm and ∠B = 72°, compute (i) DE (ii) ∠ADE.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 3
Answer:
Given : BC= 5.6 cm, ∠B = 72° and D, E are mid-points of sides AB, AC, respectively, (i)
The line joining the mid-points of any two sides of a triangle is parallel to the third side and is equal to half of it.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 21

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Question 7.
(a) Evaluate without using tables :
\(\left(\frac{\cos 47^{\circ}}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin 72^{\circ}}{\cos 18^{\circ}}\right)^{2}-2 \cos ^{2} 45^{\circ}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 22

(b) Solve:
\(\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 23
ICSE Class 9 Maths Sample Question Paper 2 with Answers 24

(c) In ΔABC, ∠ACB = 90°, AB = c unit, BC = a unit,
AC = b unit, C perpendicular to AB and CD = p unit.
Prove that \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 4
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 25

Question 8.
(a) If \(x^{2}+\frac{1}{x^{2}}=27\),find the values of :
(i) \(x+\frac{1}{x}\)
(ii) \(x-\frac{1}{x}\)
(iii) \(x^{2}-\frac{1}{x^{2}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 27

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(b) If 1 is added to the numerator of a fraction, it becomes \(\frac{1}{5}\). If 1 is subtracted from the denominator, it becomes \(\frac{1}{7}\). Find the fraction.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 28

(c) Find the mean and median of the numbers :
41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.
Answer:
41, 39, 52, 48, 54, 62, 46, 52, 40, %, 42, 40, 98, 60, 52.
∴ ∑ x =822, n=15
∴ \(\text { Mean }=\frac{\sum x}{n}=\frac{822}{15}=54.8\)
Rearranging in ascending order, we get
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
ICSE Class 9 Maths Sample Question Paper 2 with Answers 29

Question 9.
(a) The volume of a cuboidal block of silver is 10368 cm3. If its dimensions are in the ratio 3:2:1, find :
(i) Dimensions of the block.
(ii) Cost of gold polishing its entire surface at ₹0.50 per cm2.
Answer:
(a) (i) Ratio of dimensions = 3:2:1
Let its length, breadth and height be 3x cm, 2x cm and x cm respectively.
Volume of block = 3× x 2x × x = 10368
⇒ 6x3 = 10368
\(x^{3}=\frac{10368}{6}=1728 \Rightarrow x=12\)
Length (l) =3x = 3 × 12=36cm
Breadth (b) = 2x = 2 × 12=24cm
Height (h) =x = 12 cm

ICSE Class 9 Maths Sample Question Paper 2 with Answers

(ii) Total surface area =2(lb+lh+bh)=2(36 x 24+36 x 12+24x 12)
= 2 (864 + 432 + 288)
= 2 x 1584 = 3168 cm2
∵ Rate of gold polishing = ₹ 0.50 = ₹\(\frac{1}{2}\)
Total cost of gold polishing of entire surface
ICSE Class 9 Maths Sample Question Paper 2 with Answers 30

(b) Factorize : 2 – y (7 – 5y).
Answer:
2 – y (7 – 5y) = 2-7y + 5y2
= 2 – (5 + 2) y + 5y2 = 2 – 5y – 2y + 5y2
= 1 (2 – 5y) – y (2 – 5y) = (2 – 5y) (1 – y)

(c) Solve graphically : x – 2y = 1; x + y – 4.
Answer:
x – 2y =1 ……….(i)
x + y = 4 ……………(ii)
from (i)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 31
∴ The points are (1, 0), (3, 1), (5, 2)

From (ii)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 32
The points are (4, 0), (3, 1), (2, 2)
These points are plotted on the graph.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 33
The two straight lines intersect at (3, 1)
∴ x=3,y=1

ICSE Class 9 Maths Sample Question Paper 2 with Answers

Question 10.
(a) From the adjoining figure, find the values of :
(i) cot2 x – cosec2 x
(ii) \(\tan ^{2} y-\frac{1}{\cos ^{2} y}\)
Answer:
(a) In AABD,
AB2 = AD2 + BD2 (By Pythagoras theorem)
= 42 + 32 = 16 + 9 = 25
AB = √25 = 5.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 38

(b) \(\text { If } \frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}, \text { prove that }: a^{a} . b^{b} \cdot c^{c}=1 \text { . }\)
Answer:
\(\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k(\text { say })\)
log a =k(b – c); log b = k(c-a); log c = k(a-b)
Now, a log a + b log c + c log c = ak (b – c) + bk (c – a) + ck (a – b)
⇒ log ab + log bb + log cc = kab – kac + kbc – kab + kac – kbc
⇒ log (aa. bb . cc) = 0
⇒ log (aa. bb . cc)= log 1
⇒ aa. bb. cc = 1.
Hence Proved.

(c) Prove that √2 is not a rational number.
ICSE Class 9 Maths Sample Question Paper 2 with Answers 5
Answer:
Let us assume that√2 is a rational number.
If \(\sqrt{2}=\frac{p}{q}, p, q \in \mathrm{I}\) have no comman factor and C”q≠ 0.
\(2=\frac{p^{2}}{q^{2}} \Rightarrow p^{2}=2 q^{2} \Rightarrow p^{2}\) C”is an even integer
⇒ p is an even integer
⇒ p = 2m, where m∈I
⇒ p2 = 4m2 ⇒ 2y2 = 4m2 ⇒ q2 = 2m2
⇒ q2 is an even integer
⇒ y is an even integer.
Thus, p and q are both even integers and therefore, have a common factor 2 which contradicts that p and q have no common factor.
√2 is not a rational number.
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 2 with Answers
Question 11.
(a) Simplify: \(\left(a+\frac{1}{a}\right)^{2}-\left(a-\frac{1}{a}\right)^{2}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 35

(b) In the given figure, ABCD is a trapezium. Find the values of x and y.
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 36

(c) Simplify: \(\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}\)
ICSE Class 9 Maths Sample Question Paper 2 with Answers 6
Answer:
ICSE Class 9 Maths Sample Question Paper 2 with Answers 37

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 9 Maths Sample Question Paper 1 with Answers

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Max Marks :80
[2 Hours]

General Instructions

  • Answers to this Paper must be written on the paper provided separately.
  • You will not be allowed to write during the first 15 minutes.
  • This time is to be spent in reading the question paper.
  • The time given at the head of this Paper is the time allowed for writing the answers.
  • Section A is compulsory. Attempt any four questions from Section B.
  • The intended marks for questions or parts of questions are given in brackets [ ].

Section – A [40-Marks]
(Attempt all questions from this Section)

Question 1.
(a) Rationalize the denominator : \(\frac{14}{5 \sqrt{3}-\sqrt{5}}\) [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 11

(b) Factorize the given expression completely : 6×2 + 7x – 5 [3]
Answer:
6 x2+ 7x-5 = 6x2 + (10 – 3)* – 5
– 6x2 + 10x- 3x – 5
= 2x(3x + 5) – 1(3x + 5)
= (3x + 5) (2x – 1).

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) In the given figure, AB = \(\frac{1}{2}\) BC, where BC = 14 cm. Find : [4]
(i) Area of quadrilateral AEFD
(ii) Area of ΔABC
(iii) Area of semicircle
Hence find the area of shaded region. Use 7π = \(\left(\text { Use } \pi=\frac{22}{7}\right)\)
ICSE Class 9 Maths Sample Question Paper 1 with Answers 1
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 12

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 2.
(a) Mr. Ravi borrows ₹ 16,000 for 2 years. The rate of interest for the two successive years are 10% and 12% respectively. If he repays ₹ 5,600 at the end of first year, find the amount outstanding at the end of the second year. [3]

(b) Simplify: \(\left(\frac{8}{27}\right)^{-\frac{1}{3}} \times\left(\frac{25}{4}\right)^{\frac{1}{2}} \times\left(\frac{4}{9}\right)^{0}+\left(\frac{125}{64}\right)^{\frac{1}{3}}\) [3]

(c) In the given figure, ABCD is a parallelogram. AB is produced to P, such that AB = BP and PQ is drawn parallel to BC to meet AC produced at Q.
Given AB = 8 cm, AD = 5 cm, AC = 10 cm,
(i) Prove that point C is mid-point of AQ.
(ii) Find the perimeter of quadrilateral BCQP. [4]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 2
Answer:
(a) Here, P = ₹ 16000
For first year: R = 10% , T = 1 year
∴ \(\text { Interest }=\frac{16000 \times 10 \times 1}{100}= 1600\)
Amount = ₹ (16000 + 1600) = ₹ 17600
∴ Amount repaid = ₹ 5600.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

For Second Year :
P = (17600 – 5600) = 12000, R = 12% , T = 1 year
∴ Intrest = \(\frac{12000 \times 12 \times 1}{100}\) = ₹1440
∴ Amount =(12000+1440) = ₹13440

(b)
ICSE Class 9 Maths Sample Question Paper 1 with Answers 13
(c) Given : ABCD is parallelogram, AB = BP, AB = 8 cm, AD = 5 cm, AC = 10 cm.
(i) ∵ AB = BP
∴ B is mid-point of AP
Also, BC || PQ (Given)
AC = CQ (By mid-point theorem)
∴ C is mid-point of AQ. Hence Proved.

(ii) BP = AB = 8 cm (Given)
BC = AD = 5 cm (∵ ABCD is a parallelogram)
CQ = AC = 10 cm [From part, (i)]
PQ = 2 BC = 2×5 = 10 cm(By mid-point theorem)
∴ Perimeter of quadralateral BCQP = BP + PQ + CQ + BC

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 3.
(a) Solve following pairs of linear equations using cross-multiplication method : [3]
5x – 3y = 2
4x + 7y = – 3
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 14

(b) Without using tables, evaluate : [3]
\(4 \tan 60^{\circ} \sec 30^{\circ}+\frac{\sin 31^{\circ} \sec 59^{\circ}+\cot 59^{\circ} \cot 31^{\circ}}{8 \sin ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}\)
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 15

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) Construct a frequency polygon for the following frequency distribution, using a graph sheet. [4]

Marks 40-50 50-60 60-70 70-80 80-90 90-100
No. of students 7 18 26 37 20 6

Use : 1 cm = 10 marks, 1 cm = 5 students
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 16
ICSE Class 9 Maths Sample Question Paper 1 with Answers 17

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 4.
(a) Evaluate : 3 log 2 – \(\frac{1}{3}log 27 + log 12 – log 4 + 3 log 5\). [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 18

(b) If x –\(\frac{1}{x}\) =3, evaluate x3 – \(\frac{1}{x^{3}} \)[3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 19

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) In the given diagram, O is the centre of the circle and AB is parallel to CD. AB = 24 cm
and distance between the chords AB and CD is 17 cm. If the radius of the circle is 13 cm, find the length of the chord CD.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 3
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 20
ICSE Class 9 Maths Sample Question Paper 1 with Answers 21

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Section – B [40 Marks]
(Attempt any four questions from this Section)

Question 5.
(a) Find the coordinates of the points on Y-axis which are at a distance of 5√2 units from
the point (5, 8). [3]
Answer:
(a) Let the coordinates of the point on Y-axis be (0, y).
Distance = 5 √2
⇒\( \sqrt{(0-5)^{2}+(y-8)^{2}}=5 \sqrt{2} \)
Squaring both sides, we get
(0-5)2 + (y-8)2 = (5-√2)2
⇒ 25 + y2 – 2 y-8 + 64 = 50
⇒ y2 – 16y + 89 – 50 = 0
⇒ y2 – 16y + 39 = 0
⇒ y2 – (13 + 3)y + 39 = 0
⇒ y2-13y-3y+ 39 =0
⇒ y(y – 13) – 3(y – 13) = 0
⇒ (y – 13) (y – 3) = 0
⇒ y-13=0 or y-3 = 0
⇒ y = 13 or y = 3.
.’. The required point is (0,13) or (0, 3).

(b) In the given figure, BC is parallel to DE. Prove that area of ΔABE = Area of ΔACD. [3]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 4
Answer:
Given: BC || DE
∴ Area of ΔBCE = Area of ΔBCD
(Triangles, on same base and between the same parallels are equal in area)
⇒ Area of ΔBCE + Area of ΔABC = Area of ΔBCD + Area of ΔABC
(Adding area of ΔABC to both sides) .
⇒ Area of ΔABE = Area of ΔACD. Hence Proved.

(c) A stun of ₹ 12,500 is deposited for 1 \(\frac{1}{2}\) years, compounded half-yearly. It amounts to ₹ 13,000 at the end of first half year. Find : [4]
(i) The rate of interest
(ii) The final amount. Give your answer correct to the nearest rupee.
Answer:
P = ₹ 12,500, A = ₹ 13,000, T = – year.
∴ Interest for \(\frac{1}{2}\) year
= ₹ (13000 – 12500) = ₹ 500.
(i) Let R be the rate of interest.
∴ \(\mathrm{R}=\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{500 \times 100}{12500 \times \frac{1}{2}}=8\)
∴ The rate of interest = 8 % p.a.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(ii) Now, n = 1 \(\frac{1}{2}\)years =\( \frac{3}{2}\)years.
C.I. is calculated half-yearly,
\(\mathrm{R}=\frac{\mathrm{I} \times 100}{\mathrm{P} \times \mathrm{T}}=\frac{500 \times 100}{12500 \times \frac{1}{2}}=8\)

Question 6.
(a) Construct a parallelogram ABCD in which AB = 6.4 cm, AD = 5.2 cm and the
perpendicular distance between AB and DC is 4 cm. [3]
Answer:
(a) Given : AB = 6.4 cm, AD = 5.2 cm,
Perpendicular distance between AB and DC is 4 cm.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 23
Steps of construction :
(1) Draw a line segment XY and take any point P on it.
(2) At P, draw a perpendicular PZ and cut-off PD = 4 cm.
(3) From D, cut-off XY at A such that DA = 5.2 cm.
(4) From A, cut-off XY at B such that AB = 6.4 cm.
(5) From B and D, draw arcs of 5,2 cm and 6.4 cm radii respectively which intersect at C.
(6) Join AD, BC and CD to obtain the required parallelogram ABCD.

(b) Factorize : 4a2 – 9b2 – 16c2 + 24be [3]
Answer:
4a2 – 9b2 – 16c2 + 24 be =4a2– (9b2 – 14bc + 16c2)
= (2a)2 – {(3b)2 – 2-3b-4c + (4c)2}
= (2a)2 – (3b – 4c)2
= (2a + 3b – 4c) (2a – 3b + 4c).

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) In the given diagram, ABCD is a parallelogram, ΔAPD and ΔBQC are equilateral triangles.
Prove that: . [4]
(i) ∠PAB = ∠QCD
(ii) PB = QD

ICSE Class 9 Maths Sample Question Paper 1 with Answers 5
Answer:
Given : ABCD is parallelogram, ΔAPD and ΔBQC are equilateral triangles.
(i) ∠DAB = ZBCD (Opp. angles of a || gm are equal)
⇒ ∠DAB + ∠PAD = ∠BCD + ∠BCQ (∠PAD = ∠BCQ = 60°)
⇒ ∠PAB = ∠DCQ. Hence Proved.

(ii) In ΔPAB and ΔQCD,
AB DC (Opp. sides of aIgm are equal)
∠PAB = ∠QCD [From (i)
AP = CQ (∵AP=AD=BC=CQ)
∠PAB ≅ ΔQCD (SAS axiom)
PB = QD (c.p.c.t.)
Hence Proved.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 7.
(a) Solve for x : sin2 x + cos2 30° = \(\frac{5}{4}\); where 0° ≤ x ≤ 90° [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 24

(b) Evaluate for x :\(\left(\sqrt{\frac{5}{3}}\right)^{x-8}=\left(\frac{27}{125}\right)^{2 x-3}\) [3]
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 25

(c) In the given figure, triangle ABC is a right angle triangle with ∠B = 90° and D is mid­point of side BC. Prove that AC2 = AD2 + 3 CD2. [4]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 6
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 26
ICSE Class 9 Maths Sample Question Paper 1 with Answers 27

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 8.
(a) In the given figure, ∠ABC = 66°, ∠DAC = 38°. CE is perpendicular to AB and AD is perpendicular to BC. Prove that CP > AP. [3]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 7
Answer:
Given: ∠ABC = 66°, ∠DAC = 38°, CE ⊥AB, AD ⊥ BC.
In ∠ABD, ∠BAD + ∠ABD = ∠ADC (Exterior angle is equal to sum
of interior opposite angles)
∠BAD+66°=90°
∠BAD=90°- 66°=24°.
In ∠SACE, ∠ACE + ∠AEC + ∠CAE = 180° (Sum of angles in a triangle is 180°)
∠ACE + 90° + (24° + 38°) = 180°
∠ACE + 152° = 180°
∠ACE = 180° – 152° = 28°.
Now, ∠CAP > ∠ACP ( 38°> 28°)
CP > AP (In a triangle, greater angle has greater side opposite to it)
Hence Proved.

(b) Mr. Mohan has ₹ 256 in the form of ₹ 1 and ₹ 2 coins. If the number of ₹ 2 coins are three more than twice the number of ₹ 1 coins, find the total value of ₹ 2 coins. [3]
Answer:
Total amount = ₹ 256
Let the no. of ₹ 1 coins be x and that of ₹ 2 coins be y.
∴ Value of x coins = ₹ 1 × x = ₹  x
Value of y coins = ₹ 2 x y = ₹ 2y.
∴ x + 2y = 256
Also, y = 3 + 2x
Using equation (ii) in (i), we have
Also, y=3+2x
Using equation (ii) in (i), we have
⇒ x+2(3+2x)= 256
⇒ x+6+4x= 256
⇒ 5x =256 – 6
⇒ x=\(\frac{250}{5}\)=50.
Putting the value of x in equation (ii), we get
y =3+2x 50 =3+ 100 = 103.
∴ Total value of ₹ 2 coins = ₹ 2y
=₹ 2x 103
=₹ 206.

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) Find (i) mean and (ii) median for the following observations : [4]
10, 47, 3, 9, 17, 27, 4, 48, 12, 15
Answer:
Given observations are 10, 47, 3, 9, 17, 27, 4, 48, 12, 15.
Here, n 10
(i) Σx = 192
\(\text { Mean }=\frac{\Sigma x}{n}=\frac{192}{10}=19.2\)

(ii) Rearranging the observations in ascending order, we have
3, 4, 9, 10, 12, 15, 17, 27, 47, 48
ICSE Class 9 Maths Sample Question Paper 1 with Answers 28

ICSE Class 9 Maths Sample Question Paper 1 with Answers

Question 9.
(a) Three cubes are kept adjacently, edge to edge. If the edge of each cube is 7 cm, find total surface area of the resulting cuboid. [3]
Answer:
Given : Length of each side of cube = 7 cm
For cuboid, 7cm
l= (7 + 7 + 7) cm = 21 cm
b = 7 cm, h = 7 cm.
We know, Total surface area = 2 (lb + bh + Ih)
= 2 (21 x 7 + 7 x 7 + 21 x 7)
= 2 (147+ 49 + 147) = 2 x 343 = 686 cm2
ICSE Class 9 Maths Sample Question Paper 1 with Answers 29

(b) In the given figure, arc AB = twice (arc BC) and ∠AOB = 80°. Find : [3]
(i) ∠BOC
(ii) ∠OAC
ICSE Class 9 Maths Sample Question Paper 1 with Answers 8
Answer:
(i) Given: Arc AB = 2 (arc BC),∠AOB =80°
∠AOB=2∠BOC
∠BOC = \(\frac{1}{2}\) ∠AOB
\(\frac{1}{2}\) × 80°
=40°

(ii) In ΔAOC
OA = OC (Radii)
⇒ ∠OCA = ∠OAC (Angles opposite to equal
sides are equal)
Now, ∠OAC + ∠AOC + ∠OCA = 180° (Angle sum property)
∠OAC + (∠AOB + ∠BOC) + ∠OAC = 180° (∠OAC= ∠OCA)
= 2∠OAC + (80° + 40°) = 180°
2∠OAC + 120° = 180°
2∠OAC = 180° – 120° = 60°
∴ ∠OAC =\(\frac{60^{\circ}}{2}\) 3o°

(c) Solve graphically the following system of linear equations (use graph sheet): [4]
x – 3y = 3
2x + 3y = 6
Also, find the area of the triangle formed by these two lines and the Y-axis.
Answer:
x – 3y = 3 …………….. (i)
2x + 3y = 6 ………. (ii)
from equation (i)
x = 3y + 3

X 3. 0 -3
y 0 -1 -2

The points are (3, 0), (0, – 1), (- 3, – 2).
From equation (ii),
⇒ 2x = 6 – 3y
⇒ \(x=\frac{6-3 y}{2}\)

ICSE Class 9 Maths Sample Question Paper 1 with Answers

X 3 0 -3
y 0 2 4

The points are (3, 0), (0, 2), (- 3, 4).
These points are plotted on the graph.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 30
The two lines intersect at the point (3, 0).
∴ x=3,y=0
Triangle formed by the lines (i), (ii) and Y-axis is ABC.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 31

Question 10.
(a) Each interior angle of a regular polygon is 135°. Find : [3]
(i) The measure of each exterior angle.
(ii) Number of sides of the polygon.
(iii) Name the polygon.
Answer:
(a) Given: Each interior angle = 135°
(i) Exterior angle = 180° – 135° = 45°
ICSE Class 9 Maths Sample Question Paper 1 with Answers 32
(iii) The polygon is a regular octagon.

(b) If log 4 = 0.6020, find the value of log 80. [3]
Answer:
Given : log 4 = 0.6020
⇒ log 22 = 0.6020
⇒ 2 log 2 = 0.6020
⇒ log 2
\(=\frac{0.6020}{2}\)
Now, log 80 = log (8 x 10) = log 8 + log 10
= log 23 + log 10 = 3 log 2 + log 10
= 3 x 0.3010 + 1 = 1.9030

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(c) Evaluate x and y from the figure diagram. [4]
ICSE Class 9 Maths Sample Question Paper 1 with Answers 9
Answer:
ICSE Class 9 Maths Sample Question Paper 1 with Answers 33

Question 11.
(a) ΔABC is an isosceles triangle such that AB = AC. D is a point on side AB such that
BC = CD. Given ∠BAC = 28°. Find the value of ∠DCA. [3]
(b) Prove that opposite angles of a parallelogram are equal. [3]
(c) The cross-section of a 6 m long piece of metal is shown in the figure. Calculate : [4]
(i) The area of the cross-section
(ii) The volume of the piece of metal in cubic centimetres.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 10
Answer:
(a) Given : AB = AC, BC = CD, ∠BAC = 28°
ICSE Class 9 Maths Sample Question Paper 1 with Answers 34
Since, AB = AC
∠ABC = ∠ACB. (Equal sides have equal angles opposite to them)
∠ABC + ∠ACB + ∠BAC = 1800 (Sum of angleš in a triangle is 1800)
∠ABC + ∠ABC + 28° = 180°
2∠ABC =180°-28°
∠ABC= \(\frac{152^{\circ}}{2}\)=76°
∠BDC = ∠CBD = 76°
Now, ∠ACD + ∠CAD = ∠BDC (Exterior angle is equal to sum of interior opposite angles)
∠ACD + 28° = 76°
∠ACD = 76° – 28° = 480

(b) Given : A parallelogram ABCD.
ICSE Class 9 Maths Sample Question Paper 1 with Answers 35

To prove:∠A = ∠C and ∠B = ∠D.
Proof: AB II DC, AD II BC ( ABCD is a parallelogram)
∠A + ∠D = 1800 (Co-interior angles) …(i)
and ∠D + ∠C = 180° (Co-interior angles) …(ii)
From (i) and (ii),∠A + ∠D =∠D + ∠C
∠A=∠C
Similarly,∠B = ∠D.Hence Proved.

(c) In triangle, length of equal sides (a) = 5 cm, base (b) = 8 cm.
In rectangle, Length (L) = 8 cm, Breadth (B) = 6.5 cm.
(i) The area of cross-section = Area of rectangle + Area of triangle
ICSE Class 9 Maths Sample Question Paper 1 with Answers 36

ICSE Class 9 Maths Sample Question Paper 1 with Answers

(ii)  Length of metal = 6 m = 600 cm.
Volume = Area of cross-section x Length
= 64 cm2 x 600 cm
= 38400 cm3.

ICSE Class 9 Maths Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Section -A (40 Marks)
Attempt all questions

Question 1
(a) What do you mean by reusability feature.
(b) Write the java expression for the roots of the quadratic equation.
(c) Name any two jump statments and their usages.
(d) (i) Name the mathematical function which is used to find the cosine of an angle given in radians.
(ii) Name a string functions which removes the blank spaces provided in the prefix and suffix of a string.
(e) What do you mean by dynamic initialization of an array? Give an example.
Answer:
(a) Reusability feature is related to inheritance in java where derived class can use the methods and common data from the base class.

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(b) double r1 = (-b + Math.sqrt(b * b – 4 * a * c)) / (2 * a);
double r2 = (-b – Math.sqrt(b * b – 4 * a * c)) / (2 * a);

(c) (i) break;
(ii) continue
for(int i =1; i< 10; i++)
{
if(i*3 == 9)
break;
}
for(int i =1; i< 10; i++)
{
if(i%2 == 0)
continue;
System.out.println(i);
}

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(d) Math.cos(b) trim( ).
Answer:
Math.cos(b)
trim( ).

(e) When the array is initialized during the run time of program its called dynamic initialization of array.
E.g.
int a[ ] = new int[10];

Question 2
(a) Consider the following code
class lol
{
public static int m=3,y=4; public int a =10,b = 15;
}
(i) Name the variables for which each of object of the class will have its own distinct copy.
(ii) Name the variables that are common to all the objects of the class.
Answer:
(i) m and y
(ii) a and b.

(b) Distinguish between constructor and method.
Answer:

Constructor Method
Automatically called during the creation of the object. Class Object needs to explicitly call the member functions.
It bears the same name as that of class. It doesn’t have same name.

(c) What are the values of a and b after the following function is executed if values passed are 30 and 50.
void pass(inta, int b)
{
a = a+b; b= a-b; a = a-b;
System.out.println(“a=”+a+””b=”+b);
}
Answer:
a=50 b=30

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(d) Rewrite the following statement using if-else statement amount = (x!=50)?((x<50)?(4.0/100*x):(l 0.0/100*x)):500;
Answer:

if(x!=50)
{
if(x<50)
amount = 4.0/100*x; 
else
amount = 10/100*x;
}
else
amount = 500;

(e) Name any two tokens of java.
Answer:
variables and constants.

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 3
(a) What are the different keywords that checks the visibility of a member of the class? What are they called?
(b) Determine how many times loop will be executed and print the output.
int a=1, b=2; while(++b<6)a*b;
System.out.println(a);
(c) Attempt the following:
(i) A package that is involved to manipulate character as well as String?
(ii) Name a data type that can hold 16 bit Unicode characters.
(iii) What is the output of code below:
charc = ‘A’; int m = 5;
System.out.println(char(c+m));
System.out.println(c+m);
(iv) Write statements to show how finding the length of a character array char[ ] differs from finding the lengths of string object str.
(v) Give the output of the following functions System.out. println(“MALAYALAN”.indexOf(‘A,)+”Sidharth”.lastlndexOf(‘h’));
(vi) double a = Math.pow(“200”.index0f(‘0’),2);
System.out.println(a);
(vii) Differentiate between linear search and binary search.
(viii) Evaluate the following expression:
int p, k= 8,m=11,r=7; p = (r++%7)+(-m%5)+k*(++k-8);
Answer:
(a) private, public, protected
(b) 3 times 60
(c) (i) java.lang
(ii) byte
(iii) F 70
(iv) array.length and str.length( )
(v) 8
(vi) 1.0
(vii)

Linear Search Binary Search
It compares each element of the array with rest of the elements in the array. It’s based on divide and conquer rule. Array is sorted in this search. Element is searched only in the selected halves.

(viii) 8

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 4
Write a program in java to input a number and check whether it is a pronic number or Heteromecic number or not.
Pronic number: A pronic number, oblong number, rectangular number or heteromecic number is a number which is the product of two consecutive integers i.e n(n-t-l)
The first few Pronic numbers are:
0,2,6,12.20,30,42,56,72.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.print("lnput a number:"); 
int n = sc.nextlnt( ); 
int result = 0; for(int i=0; i<n; i++)
{
if(i*(i+1) == n)
{
result = 1; break;
}
}
if(result == 1)
System.out.println(''Pronic Number."+n); 
else
System.out.println("Not a Pronic Number."+n);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 5
Write a program to input a word from the user and remove the duplicate characters present in it.
Example:
input: crcricicket
output: criket
Answer:

import java.util.*; 
class Main
{
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in); 
System.out.printlnC'Enter a sentence:"); 
String str=sc.nextLine();
String word =
char c = str.charAt(O);
word+=c;
for(int i=0;i<str.iength();i++)
{
char ch=str.charAt(i);
boolean flag=false;
for(int j = 0; j<word.length();j++)
{
if(ch==word.charAt(j))
{
flag = true;
}
}
if(flag == false) 
word+=ch;
}
System.out.println(word);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 6
Write a program to accept a sentence and print only the first letter of each word of the sentence in capital letters separated by a full stop.
e.g.
INPUT Sentence: Sid is a cricket output: S.I.A.C
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in);
System.out.println(''Enter a sentence: ");
String str=sc.nextLine( );
str = "" +str;
str = str.toUpperCase( );
String word =
for(int i = 0; i < str.length( ); i++)
{
char ch = str.charAt(i); 
if(ch =='')
{
word+=str.charAt(i+1
}
}
System.out.println(word);
} //end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 7
Write a program to accept name and corresponding age in two different single dimensional array. Display the records in descending order of age using bubble sort.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
int temp;
String temp 1;
String a[ ] = new String[15];
int age[ ] = new int[15];
Scanner sc = new Scanner(System.in); 
System.out.printlnC'Enter a 15 names and their ages:"); 
for(int i =0; i<a.length; i++)
{
System.out.print("Enter name"+ (i+1) 
a[i] =sc.next( );
System.out.print("Enter age"+ (i+1) +":"); 
age[i]=sc.nextlnt( );
}
for(int i = 0;i<age.length; i++)
{
for(int j =0; j<age.length - i - 1 ;j++)
{
if(age[j]<(age[j+1]))
{
temp = age[j]; tempi = a[j]; 
age[j]=age[j+1]; 
a[j] = a[j+1 ]; 
age[j+1] = temp; 
a[j+1]=temp1;
}
}
}
for(int i =0; i <a.length;i++)
{
System.out.println(a[i]+"\t"+age[i]);
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Question 8
Write a menu driven program to find area of an Equilateral triangle, an isosceles triangle and a scalene triangle as per the users choice.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in); int n;
float a,c,s,b; double area;
System.out.priintln("1 .Area of equilateral triangle"); System.out.priintln("2.Area of isosceles triangle");
System.out.priintln("3.Area of scalene triangle");
n=sc.nextlnt( );
switch(n)
{

case 1:
System.out.printlnC'Enter side of an equilateral triangle");
s=sc.nextFloat( );
area=Math.sqrt(3.0*s*s)/4.0;
System.out.println("Area="+area);
break;

case 2:
System.out.printlnC'Enter the side and base of isosceles triangle”);
a=sc.nextFloat( );
b=sc.nextFloat( );
area=b/4.0*(Math.sqrt(4.0*a*a-b*b));
System.out.println("Area="+area);
break;

case 3:
System.out.printlnC'Enter the 3 sides of scalene triangle");
a=sc.nextFloat( );
b=sc.nextFloat( );
c= sc.nextFloat( );
s= (a+b+c)/2;
area=Math.sqrt(s*(s-a)*(s-b)*(s-c));
System.out.println(”Area="+area);
break-
default:
System.out.println("Wrong choice");
}
}//end of main 
}//end of class

Question 9
Define a class to overload a function Sum() as follows:
(i) int Sum(int a, int b) – with integer arguments a and b.
Calculate and return sum of all even numbers in the range of a and b.
Sample input: a = 4, b = 16
Sample output: sum = 4 + 6 + 8 + 10 + 12 + 16

(ii) double Sum(double n) – with one double argument n. Calculate and return the product of the following series:
sum = 1.0*1.2*1.4*…’*n

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

(iii) int Sum(int n) – with one integer argument n. Calculate and return sum of only odd digits of the number n.
Sample input: n = 43961
Sample output: sum = 3 + 9+1=13
Answer:

import java.util.*;
class Main
{
public int sum(int a, int b)
{
int sum= 0; for(int i=a; i<=b;i++)
{
if(i%2==0)
sum+=i;
}
return sum;
Wend of sum
public double sum(double n)
{
double prod=1.0;
for(double i =1.0;i<=n; i+=0.2)
prod*=i; 
return prod;
}
public int sum(int n)
{
int sum = 0; 
int a = n; 
int digit; 
while(a>0)
{
digit = a%10; 
if(digit%2!=0) 
sum+=digit; 
a=a/10;
}
return sum;
}
public static void main(String args[ ])
{
Main s = new Main( );
System.out.println(s.sum(1,5));
System.out.println(s.sum(10.0));
System.out.println(s.sum(10935));
}//end of main
} //end of class

ICSE Class 10 Computer Applications Sample Question Paper 12 with Answers

Variable Data Type Description
n int Users input integer value
result int Flag to check pronic number
c char To store first character
flag boolean To check character is present in word or not
str String User input string.
word String Modified string
a[ ] string To store 15 names
age[ ] int To store integer number
temp int Temporary variable for swapping
tempi String Temporary variable for swapping
n int Users choice
a c s b float S = side of equilateral triangle A b side and base of issocless triangle Abe sides of scalene triangle
area double To store area
prod double Store product of decimal number in series
digit int To store digit of number

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Section- A (40 Marks)
Attempt all questions

Question 1
(a) Give one similarity and one difference between do while and while loop.
(b) Write the java statement to find largest of 3 numbers using conditional operator.
(c) What is polymorphism?
(d) Write a function to round off any fractional number? Give one example.
(e) What is difference between infinite loop and fall through statement?
Answer:
(a) Similarity – Both while and do while loop is used when the no. of iterations are not specified. Difference: while loop is entry controlled loop and while do while loop is exit controlled.

(b) int a = (b>c)?b:c; int d = (a>e)?a:e;

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(c) Polymorphism is the identity to represent an object in more than one form.

(d) Math.round( )
double k = Math.round(2.9);

(e)

Infinite Loop Fall Through
Occurs for iteration statement Occurs for conditional statement.
Occurs when conditions is always satisfied. Occurs when the break statement is missing in every case statement.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 2
(a) Compare local, instance and class variables on the basis of the given criteria.
(i) identification in program
(ii) Usage of access specifier with them.
(iii) provision of initial values.
(iv) Number of copies for each object of class.
Answer:
(a) (i) Local variables are usually used in a block of statement. Instance variables are used by a method while class variable has only one copy throughout out the class.

(ii) Local variable scope remains inside the compound statement. Inside the curly braces.
Instance variable scope remains inside the method. All access specifiers can be used in class variables. Scope is throughout the class.

(iii) Local variables is initialized in its scope.
Instance variables initialized in the method where it is used in.
Class variables are initialized after the class declaration.

(iv) local variables multiple copies can exists.
Instance variables multiple copies can exists.
Class variables single copy exists.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(b) Which OOPs principle is implemented when you:
(i) use the keyword extends
(ii) create an interface
(iii) create various functions
(iv) use private data members?
Answer:
(i) Inheritance
(ii) Abstraction
(iii) Polymorphism
(iv) Encapsulation

(c) What are functions? Give one advantage using functions.
Answer:
Functions are a set of statements used to perform certain tasks. Advantage: It organizes the code into a blocks and helps in performing specific task.

(d) Differentiate between classes and objects. Why is object referred as instance of class?
Answer:
(i) Object is an identifiable entity with some characteristics and behaviour. Class is a collection of object with similar characteristics.

(ii) Object is called instance of class as object helps in accessing the class members and functions to get the required output.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(e) Differentiate between linear search and binary search.
Answer:

Linear Search Binary Search
The search element is compared with each element of the array. The array is sorted and divided into two halves and then search element is found accordingly.
Array need not be sorted. Array has to be sorted.

Question 3
(a) What is other name of java interpreter? Explain it.
Answer:
JVM (Java virtual machine): It is used to convert java byte code to machine code which can be understood by the computer running on any operating system.

(b) Name the java keyword
(i) used to finish the execution of the method.
(ii) used to implement the concept of inheritance.
Answer:
(i) exit or return
(ii) extends

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(c) Attempt the following:
(i) Assign the constant value pi as 3.14 to variable using suitable data type?
(ii) Give difference between unary and binary operators.
(iii) Give examples of each
(i) Composite date type.
(ii) Escape sequence.
(iii) Comment lines
(iv) Wrapper class
(iv) Write any two rules of naming variable
(v) Differentiate between type conversion and coericion.
(vi) State the difference between final and finally.
(vii) a. State the use of toString( ) and valueOf( ).
b. Write a statement to extract the last word of the string str.
(viii) a. What is called byte code?
b. Explicit and implicit type conversion.
Answer:
(i) double const pi =3.14;
(ii) Unary operator is is used for one operand.
Binary operator is used for two operands.
(iii) (i) Array, class
(ii) “\n”‘;\\’
(iii) //or/*java*/
(iv) Character, Integer
(iv) Variables should not be keyword and it should not start with digit or special character except for the $ and & and _. It must start with alphabet.

Type conversion Coercion
Assigning one data type to another Promotion of data type
It’s called implicit type casting Also called type promotion.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

(v) Final makes a variable constant throughout the program. Finally is a block which is always executed even if exception occurred or not.

(vi) a.

toString( ) valueOf( )
Converts integer to string. Converts strings to integer.

b. String word = str.substring(sr.lastlndexOf(“)+1);

(vii) a. Byte code is an specific set of instruction to be executed by the JVM.
b. Implicit type conversion: Java compiler will automatically convert one data type to another when data types are compatible.
E.g. int 1 = 10;
double b = I;
Explicit type conversion: Java compiler is to be told explicitly to do the conversion form one data type to another.
E.g.
double d = 10.0;
float f = (float)d;

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 4
Using switch case write a menu driven program to print the patterns.
(a)
0 0 0 0 0
2 2 2 2
6 6 6
12 12
20

(b)
X
YY
XXX
YYYY
XXXXX
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter the choice:");
int a1 = sc.nextlntO;
switch(al)
{
case 1: int a =0; int b = 2;
for(int i =5;i>= 1;i—)
{
for(int j=1; j<=i;j++)
System.out.print(a+""); -
a+=b;
b+=2;
System.out.println( );
}
break;
case 2: for(int i =1 ; i<=5; i++)
{
for(intj=1;j<=i;j++)
{
if(j%2 != 0)
System.out.printC'X");
else
System.out.printC'Y");
}
System.out.printlnO;
}
break;
default: System.out.println("lnvalid Choice");
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 5
Write a program to print the frequency of the digits occurring in number.
Eg.2566773
Output:
2 1
5 1
6 2
7 2
3 1
Answer:

import java.util.*; 
class Main {
static int frequency(int number, int digit)
{
int count = 0;
while (number > 0)
{
if (number % 10 == digit) count++;
number = number / 10;
}
return count;
}
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in);
System.out.println(''Input the number:");
int n = sc.nextlnt( );
int a = n;
int search = 0;
int countdigit = 0;
int []arr = new int[10];
while(a >0)
{
search = a%10;
countdigit = frequency(n,search);
if(countdigit>0)
{
arr[search]=countdigit;
countdigit=0;
}
a=a/10;
}
for(int i =0; 
i<arr.length;i++)
{
if(arr[i]>0)
System.out.println(i+"\t"+arr[i]);
}
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 6
Write a class with a special member function to input a multi-digit number(max 9 digit) and print the following:
(i) total number of even and odd digits
(ii) the reverse number
(iii) total number of zeros present.
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in);
System.out.priintln("lnput the number:");
long a = sc.nextlntQ;
int sumeven =0;
int sumodd =0, sum3 = 0;
String str - long a1 = a;
while(a1 <=999999999&&a1 !=0)
{
int k= (int)al %10;
str+=lnteger.toString(k);
if(k%2 == 0)
sumeven++;
else if(k%2!=0)
sumodd++;
else if (k==0)
sum3++;
else;
a 1 =a 1/10;
}
System.out.println("No. even: "+sumeven+"\t"+ "odd:"+sumodd+"\ t"+"zero: "+sum3+"\t"+"Reverse no:"+str);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 7
Write a class to input a string (combination of letters and digits) and replace the repecated characters with star(*) sign. Then display both old and newly created string.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in); 
System.out.println("Enter a sentence:"); 
String str=sc.nextLine( );
for(int i=0;i<str.length( );i++)
{
char ch=str.charAt(i); if(ch!='*')
{
for(int j=str.length( )-1;j>i;j—)
{
char ch 1 =str.charAt(j); jf(ch—chi)
{
StringBuffer strl =new StringBuffer(str);
str1.setCharAt(j,'*');
str=""+strl;
}
}
}
}
System.out.println(str);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 8
Write a program to print the following pattern by taking the input n from the user.
Note: n determines number of rows.
Eg. n = 4
Output:
* * * *
* * *
* * * *
* * *
Answer:

import java.util.*;
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in); 
System.out.println("lnput the number:"); 
int n = sc.nextlnt( ); 
for(int i = 1; i <=n;i++)
{
if(i%2==0)
{
for(int j =1; j<=3;j++) 
System.out.printC *");
}
else
{
for(int j =1; j<=4;j++)
System.out.printC*");
}
System.out.println( );
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 11 with Answers

Question 9
Write a java class to calculate and print the electricity bill to be paid by a customer.
Assume that the customer pays a rent of Rs.350.00
No. of units — Charge per unit
Upto 100 units — Rs. 1.50
For the next 100 units — Rs. 2.00
For next 50 units — Rs. 2.50
Beyond 250 units — Rs. 4.00
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc=new Scanner(System.in); 
double fixed_price = 350.0; 
double amt = 0.0;
System.out.println("Enter a unit of electricity consumption: ");
int n=sc.nextlnt( );
if(n<=100)
amt+= amt+1.5*n;
else if(n>100&&n<=200)
amt+= 100*1,5+(n-100)*2.0;
else if(n>200&&n<=250)
amt+= 100*1.5+0 00)*2.0 + (n-200)*2.5;
else
amt+= 100*1.5+0 00)*2.0 + (50)*2.5+(n-250)*4.0;
System.out.println("Electricity bill amount: "+amt);
}//end of main 
}//end of class
Variable Data Type Description
al int Store users choice
a b int For printing pattern
ij int For looping
number int Number in which digits to be searched
digit int Digit which is to be searched.
count int Counting number of digits
n int Users input number
arr[] int Array of 10 integers.
search, countdigit int Digit to be searched and count of that digit
sumeven,sumodd,

sum3

int Sum of even digits, sum of odd digits and count of zero.
str String User input string
strl String Buffer To store modified string
ch char To store character
fixed „price, amt double For fixed price and calculating amount

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Section – A
Attempt all questions

Question 1
(a) What do you mean by abstraction.
(b) Using an example explain the term object.
(c) What is wrapper class.
(d) State the use of new operator.
(e) Write a java expression for square root of sinx.
Answer:
(a) Abstraction is the act of representing the essential features of the program without involving in its complexity.

(b) Eg. While using computer user is not concerned about the different parts of the computer only needs to know about usage. Characteristics (big, small, flat etc)

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

(c) Wrapper class is a class which helps to convert a primitive data type to object type. E.g. Integer, Short etc.

(d) New operator supports instantiations. It is used to allocate memory space to the newly created object.
E.g. Scanner sc = new Scanner(System.in)

(e) double d = Math.sin(x*3.14/180);

Question 2
(a) Differentiate between local variable and global variable.
Answer:

Local Variable Global Variable
Scope remains only inside a block or a function. Scope remains throughout the class.
Multiple copies of the variable are used throughout the class. Only a single copy of the variable is used throughout the class.

(b) What is the purpose of default in switch statement?
Answer:
When none of the cases of the switch statement are satisfied, the default case is used and executed in the program.

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

(c) Define an impure function.
Answer:
An impure function is a function where values of the arguments keep changing during the execution of the program.

(d) What is called default constructor?
Answer:
A default constructor is a constructor without any parameters. It is used to assign a default value to the class variables.

(e) Explain the concept of constructor overloading using an example.
Answer:

class lol
{
int a; int n;
lol( )
{
a=0;
}
lol(int c)
{
a=1;
}
public static void main( )
{
lol obj = new lol( );
 lol obj1=new lol(2);
}
}

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 3
(a) How java make use of formal and actual parameter?
Answer:
Formal parameters are used in the function definition of the program while actual parameters are used in the function call statement.

(b) State the java concept that is implemented into smaller groups
(i) Dividing a long set of instruction into smaller groups and modules.
(ii) Wrapping up of data and its associated function into a class.
Answer:
object oriented programming Encapsulation

(c) Attempt the following:
(i) What is role of access specifier?
(ii) Why is java platform independent?
(iii) Differentiate between public and private visibility label.
(iv) Int a = 7,b = 6; b = a>b?a!=0?3:b<10?9:4:8; System.out.println(b);
(v) Int a = 7,b = 6; b=++a+ ++a/ ++a; System.out.println(b);
(vi) What are functions? Give one advantage of using functions.
(vii) How do objects communicate with each other?
(viii) What are two types of methods? Give two concrete differences between them and include example of each type of function from java.lang.Math class.
Answer:
(i) Access specifiers are used to moderate the accessibility of the data according to user’s choice.
(ii) Java is platform-independent because the JVM can interpret the byte code which can run on any platform.
(iii) Public: It can be accessed anywhere.
Private: It can be accessed only inside the class.
(iv) 3
(v) b=8+9/10=8+0=8
(vi) A function is block of statements in a java program used to perform certain tasks. It organizes the data into segments and helps in performing tasks easily.
(vii) Objects communicate with each other using methods.
(viii)

Pure Function Impure Function
Functions where the value of variables don’t change Function where the values of the variables changes.
Depends on the actual parameters passed in the function call. Depends on where created object calls the functions.
E.g. Math.sqrt( ) E.g. Math.random(i)

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 4
A Smith number is a composite number, the sum of whose digits is the sum of the digits of its prime factors obtained as a result of prime factorization(excluding 1).
The first few such numbers are 4,22,27,58,85,94,121
eg.666
Prime factors are 2,3,3 and 37
Sum of the digits are (6+6+6) = 18
Sum of the digits of the factors (2+3+3+3+7)=18
Write a program to input a number and display whether the number is a Smith number or not.
Answer:

import java.util.*; 
class Main {
int sumDigfint n) 
{
int sum=0; while(n>0)
{
sum+=n%10;
n=n/10;
}
return sum;
} 
int sumPrimeFact(int n)
{
int i=2, sum=0; while(n>1)
{
if(n%i-=0)
{
sum=sum+sumDig(i);
n=n/i;
}
else
i++;
}
return sum;
}
public static void main(String[ ] args)
{
Main ob=new Main( );
Scanner sc = new Scanner(System.in);
System.out.print("Enter a Number:");
int n=sc.nextlnt( );
int a=ob.sumDig(n);
int b=ob.sumPrimeFact(n);
if(a==b)
System.out.print("lt is a Smith Number"); else
System.out.print("lt is Not a Smith Number"); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 5
Write a program to assign a full path and file name as given below. Using library functions, extract and output the file path, file name and file extension separately as shown.
Input: C:\users\sid\pictures\cricket.jpg
output: C:\users\sid\pictures
File name: Cricket
Extension: jpg
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in); 
System.out.priintlnflnput the file path:");
String str = sc.nextLine( );
String strl = str.substring(0,str.lastlndexOf('\V)); 
String str2 = str.substring(str.lastlndexOf('\V)+1); 
String str3 = str2.substring(str2.indexOf(' ’)+1); 
System.out.println("Path:" +str1); 
System.out.println("Filename:"+str2); 
System.out.println("Extension:"+str3);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 6
Write a program to input a word from user and remove the consecutive repeated characters by replacing the sequence of repeated characters by its single occurrence.
input: ssiiddaarrtth
Output: sidarth
Answer:

import java.util.*;
class Main {
public static void main(String args[ ])
{
String word ="";
Scanner sc = new Scanner(System.in);
System.out.println("lnput the sentence:");
String str = sc.nextLine( ); char chprev='';
for(int i =0;i<str.length( );i++)
{
char ch = str.charAt(i); 
if(chprev!=ch)
{
word+=ch; chprev = ch;
}
else
continue;
}
System.out.println("word without successive repeated characters is:"+word);
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 7
Write a program in java to input a number and check whether it is Duck number or not.
Note: A Duck number is a number which has zeroes present in it, but there should be no zero present in the beginning of the number. For example 3210, 7056, 8430709 are all duck numbers whereas 08237,04309 are not.
Answer:

import java.util .*;
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in);
System.out.printlnC'Input the number:");
int a = sc.nextlnt( );
int a 1 = a;
int count =0;
String str = Integer.toString(a);
while(a1!=0)
{
if (str.charAt(0)==0)
break;
else
{
int k = a 1 %10; 
if(k==0) count++; a1/=10;
}
}
if(count>=1)
System.out.println("Duck no."); else
System.out.println("Not Duck no.");
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 8
Write a program in java to input a number in Decimal number system and convert it into its equivalent number in the octal number system.
Note: Octal number system is a number system which can represent a number in any other number system in terms of digits ranging from 0 to 7 only. This number system consists of only eight basic digits i.e. 0,1,2,3,4,5,6 and 7.
eg. 25 in the decimal number system can be represented as 31 in the octal number system.
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
Scanner sc = new Scanner(System.in); 
System.out.println("lnput the decimal number:"); 
int a = sc.nextlnt( ); 
int a1 =a%8; 
int a2 = a/8;
String str=" "; 
int a3=a2%8;
String strl = Integer.toString(al);
String str2 = lnteger.toString(a3); 
str=str2+str1;
int ah = Integer.parselnt(str); 
System.out.print(ah);
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 10 with Answers

Question 9
Using Scanner class, write a program to input a string and display all those words of the strings which begins with capital letter and end with a small letter.
Sample Input: We all love Java for School Students because of its Uniqueness
Sample Output: We Java School Students Uniqueness
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
String word =
Scanner sc = new Scanner(System.in);
System.out.println("lnput the sentence:");
String str = sc.nextLine( ); str+="";
for(int i =0;i<str.length( );i++)
{
char ch = str.charAt(i);
if(ch!='') word+=ch; 
else {
jf(Character.isUpperCase(word.charAt(0))&&Character.isLowerCase(word. charAt(word.lengthO-1)))
{
System.out.print(word+"");
}
word =
}
}
} //end of main 
}//end of class
Variable Data Type Description
sum int To store sum of digits of number
n int Number whose digits are to be added
sum int To store sum
i int To start with 2
b int Store sum of prime factors
a int Store sum of digits
str String Store the string
strl String To store the path
str2 String To store filename
str3 String Store file extension
word String To store modified word
ch char to store character
chprev char To store previous character
a al count int Users input number, modified number and count number of 0
a1 int To store octal number last digit
a2 int Quotient after dividing by 8
a3 int Quotient after dividing by 8
strl String To store a 1 as string
str2 String To store a2 as string
ah int Store converted integer from string.

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Section- A (40 Marks)
Attempt all questions

Question 1
(a) What is the return of following function.
(i) equals( )
(ii) rint( ).
(b) What do you mean by compound statement. When do you need it.
(c) Explain difference between character constant and string constant.
(d) State any two features of constructor.
(e) Define abstraction.
Answer:
(a) (i) boolean
(ii) double

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

(b) A set or a block of statement in a program is called compound statement. It can be used to divide the program into segments and organise the data accordingly.

(c)

Character Constant String Constant
Enclosed in single quotes E.g. ‘A’, ‘e’ Enclosed in double quotes
E.g. “hello”, “1 like JAVA”

(d) It has same name as that of class. It has no return type.
(e) Abstraction is the act of representing the essential features of a program without involving in its complexity.

Question 2
(a) Explain the term object using example?
(b) What is the wrapper class. Justify with example?
(c) State the purpose of new operator?
(d) Write java expression for
\(\sin x+\sqrt[2]{a x^{2}-b x+c}\)
Answer:
(a) Object is an identifiable entity with some particular characteristics and behaviour.
E.g.TV, characteristics: Big, small etc.
Behaviour: Show different channels, off/on etc.
(b) The class which helps in converting primitive data types to object type is called wrapper class. Eg. Character, Integer etc.
(c) The ‘new’ keyword supports instantiation. It allocates memory for the newly created object.
(d) double z = Math.sin(x)+Math.sqrt(a*x*x + b*x + c);.
(e) class lol

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 3
(a) Explain the concept of constructor overloading with example.
(b) Differentiate between equals( ) and compareTo( ).
(c) Attempt the following:
(i) Differentiate between call by value and call by reference.
(ii) What is an identifier. Give example
(iii) Name the package that contains the scanner class. Which unit or class gets called when object is created.
(iv) What is meant by encapsulation.
(v) What is meant by inheritance.
(vi) char a[ ]={‘a’,’b’,,c’}; System.out.println(a); What will be the output.
(vii) char a[ ]={‘a’,’b’,’c’}; System.out.println(a[1]++); System.out. println(++a[1]);
(viii) inta[ ]={1,2,3};System.out.println(a[1]++);System.out.println(++a[1]);
Answer:
(a) Same as above.

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

(b)

Equals( ) CompareTo( )
It returns boolean value It returns integer value
Compares the whole string at once. It compares the ASCII values of each character of the string.

(c) (i)

Call by Value Call by Reference
Actual parameters are copied into the formal parameters. Actual parameters alias are created in the formal parameters.
Changes made by the formal parameters is not reflected back to the actual parameter Changes made by the formal parameter directly affect the actual parameters.

(ii) Identifiers are those quantities which change their values during the execution of the program. E.g. Int a, b;

(iii) java.util and constructor.

(iv) Encapsulation: Wrapping of data and its associated functions into a single unit.

(v) Inheritance is the method of inheriting the properties of one class from another class. The class which derives its own properties is called base class and the class which derives the properties of another class is called sub class.

(vi) Address of array a[ ]

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

(vii) b d

(viii) 2

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

Question 4
Write a program to print the sum of prime digits from given number.
E.g. 134667
Output: 1+3+7=11
Answer:

import java.util.*;
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("enter the number:"); 
int a = sc.nextlnt( ); 
int a 1 = a; int sum = 0; while (a 11=0)
{
int count = 0; int k = a1%10;
for(int i=1; i<=k; i++)
{
if(k%i==0) count++;
}
if(count==2) sum+=k; a 1 /=10;
}
System.out.println("sum = "+sum); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 5
Write a program to print the longest word from the sentence and print the number of characters of the longest word.
Answer:

import java.util.*; 
class Main
{
public static void main(String [ ] largs)
{
int count = 0; String temp;
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a sentence:"); 
String str = sc.nextLine( ), 
word =""; str - str+" ";
for(int i = 0; i<str.length( ); i++)
{
char ch = str.charAt(i);
if(ch=='') count++;
}
String a[ ] = new Stringfcount]; 
for(int i = 0; i<count;i++)
{
for(int j = 0;j< str.length( ); j++)
{
char ch = str.charAt(j);
if(ch!='') word+=ch; 
else {
a[i] =word; word ="
}
}
}
for(int i = 0;i<a.length; i++)
{
for(int j = 0;j<a.length-i-1;j++)
{
if(a[j].length() > a[j+1].length())
{
temp = a[j]; a[j]=a[j+1]; a[j+1]=temp;
}
}
}
System.out.printlnfLONGESTWORD: "+a[a.length-1]); 
System.out.println("No. of characters:"+a[a. length-1 ].length()); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 6
Write a program to find sum of negative numbers, sum of positive odd numbers and sum of positive even numbers entered by user and list terminates when user enters 0.
Answer:

import java.util class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
int sum = 0, sum1 = 0, sum2 = 0;
System.out.println("Enter the number to calculate and 0 for exit:"); 
int n = sc.nextlnt( ); int a = n; while(a!=0)
{
if(a>0)
{
if(a%2 == 0)
sum1+=a;
else
sum2+=a;
}
else
{
sum+=a;
}
System.out.println(''Enter the next number and 0 to exit:"); 
a = sc.nextlnt( );
}
System.out.println("Positive even sum: "+suml+"\t"+"Positive odd sum:"+sum2+"\t"+"Negative sum:"+sum);
} //end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 7
Using switch case write a menu driven program to print the patterns.
(a)
0 0 0 0 0
2 2 2 2
6 6 6
12 12
2 0

(b)
X
Y Y
X X X
Y Y Y Y
X X X X X
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println(''Enter the choice:");
intal = sc.nextlnt( ); switch(al)
{
case 1: inta = 0; int b = 2;
for(int i =5;i>= 1;i—)
{
for(int j=1; j<=i;j++)
System.out.print(a+" ");
a+=b;
b+=2;
System.out.println( );
}
break; case 2:
for(int i =1; i<=5; i++)
{
for(int j=1;j<=i;j++)
{
if(j%2 != 0)
System.out.printC'X");
else
System.out.printC'Y'');
}
System.out.println( );
}
break;
default: System.out.println("lnvalid Choice");
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 8
Write a program to find using binary search method from list of roll numbers entered by user in ascending order. If the search is successful print “you are selected to go” else print “Try next time”.
Answer:

import java.util.*; 
class Main {
public static void main(String args[ ])
{
int first, last, middle, n, search, array! ];
Scanner sc = new Scanner(System.in);
System.out.println(''Enter total number of students");
n = sc.nextlntO; array = new int[n];
System.out.printlnC'Enter" + n + " integers"); 
for (int i = 0; i<n; i++) arrayti] = sc.nextlnt( );
System.out.printlnC'Enter value to find");
search = sc.nextlnt( ); first = 0; last = n - 1;
middle = (first + last)/2; while(first <= last)
{
if (array[middle] < search)
first = middle + 1;
else if (array[middle] == search)
{
System.out.println(search + " you are selected to go " + (middle + 1) + "•"); break;
}
else
last = middle - 1; 
middle = (first + last)/2;
}
if (first > last)
System.out.println(search + " Try next timeAn");
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Question 9
Write a program to input sentence and print in the lowercase letters and replace all the words like “is” and “are” with “were” and “had” and “has” with “had”.
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in); 
System.out.println("Enter the sentence:"); 
String str = sc.nextLine( );
String str1 = str.toLowerCaseQ;
String word = strl +=" ";
String str2 =" ";
for(int i =0;i<str1 Jength(); i++)
{
char ch = strl .charAt(i);
if(ch!='') word+=ch;
else {
if(word ,equals("is")) str2=str2+"were"+"";
else if(word.equals("are")) str2=str2 + "had"+""; 
else if(word.equals("has")) str2=str2+"had" +""; else
str2=str2+word+" "; word =
}
}
System.out.println(str2);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 9 with Answers

Variable Data Type Description
a int Store input integer from user
a1,sum int For modifying digits and to calculate sum of digits
str String Store input string from user
ch char Store the character
count int Count number of blank spaces
word String To store new word
temp String Temporary storage for swapping strings.
a[ ] String Array of strings.
n int Store integer from user
a int For modifying value
sum1,sum2,sum3 int Even number sum Positive odd sum Negative sum
b int For printing even pattern
‘j int For looping
first last middle n search arrayj] int First last and middle are Index of array N is total number of elements Search is number to be searched in array. Array[ ] array of sorted roll numbers
strl String Store lowercase word
str2 String To store the modified sentence.

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Section – A
Attempt all questions

Question 1
(a) What is an array. Write the ground syntax for declaring the array.
(b) What is conditional /selection statement in java? Name the operator used for condition checking in java.
(c) What is composite data type. Explain using example.
(d) Which elements of java program represents characteristics and functions of an object.
(e) Define Polymorphism. Which part of the function differentiates between overloaded functions?
Answer:
(a) Array is set of data of the same type.
Eg. inta[ ] = {1,2,3,4,5};
(b) Conditional statements are used to check a condition and print the output accordingly. Ternary operator.
(c) A set of primitive data type is a composite datatype. Eg. Array, Class etc.
(d) Methods and data
(e) The ability to represent a thing in more than one form. The parameters of the function differentiates between the overloaded functions.

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 2
(a) What do you mean by function signature in java?
(b) Public static void main( ) why is main method declared as static?
(c) int i=3; i+=(i++*2)+ – – i*i++ + ++i/2; System.out.println(i); what is the output of the above code?
(d) Why is the string class called as immutable class.
(e) What’s the use of exception handling in java code:
try{ //code here }catch(IO Exception e)
{//some code here }what is the use of e.
Answer:
(a) Function signature is the part which defines the return type, parameters and visibility of a method.
(b) Main method is declared static because only one copy of the main method is used in the program.
(c) 3 + 6 + 9 + 2 = 20.
(d) String is immutable because once it is stored, it cannot be altered during the execution of the program.
(e) It is used to find the run time errors and catch those errors gracefully, e is object of IO exception class it is used as input to the catch function.

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 3
(a) Why is binary search called as divide and conquer.
Answer:
Binary search is called divide and conquer because it divides the array into the equal halves and finds the search element in the lower half and upper half of array accordingly.

(b) Differentiate between private and protected access modifiers.
Answer:
Private: accessibility of data or methods is only inside class.
Protected: accessibility of data or methods is only in subclass and package.

(c) Attempt the following:
(i) Find the output of the following code
int a=5,b=6,c=9; int g = a>b?a>c?a:c:(b>c)?b:c;
(ii) Differentiate between length and lengthO function.
(iii) Define class variable vs instance variable
(iv) Write java expression for the following;
\(\frac{\cos x+\sin x}{3.14^{*} \sqrt{x}}\)
(v) What is the source code in java and what is its extension.
(vi) Write one difference between
(i) char type and String type
(ii) “true” and true.
(vii) State the escape sequence for vertical tab and new line feed.
(viii) Which unit of the class gets called when object is created? Give an example
Answer:
(i) 9
(ii) length is used to find the length of array.
Length( ) is used to find length of string.

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

(iii)

Class Variable Instance Variable
Only one copy of these variable is used through out the class. Multiple copies of these variables are used throughout the class.
Delcared inside the class and outside the methods Declared inside the methods.

(iv) double x = (Math.cos(x*3.14/180) + Math.sin(x*3.14/180))/ (3.14*Math.sqrt(x));

(v) Source code in java is the set of commands in java to be executed by the JVM. Extension is .java.

(vi) (i)

CharType String Type
Declared using single quotes”. Declared using double quotes” “
Compared by operator == Compared by the help of string functions

(ii) “true” – String value
true – boolean value
(vii) “\t”and”\n‘
(viii) Constructor
E.g.

Class lol
{
Int a;
lol( )
{
a=10;
}
Public static void main( )
{
lol obj = new lol( );
}
}

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

Question 4
Write a program to accept the word from user and print the pattern as shown below:
Sample input: COMP
Output:
C
CO
COM
COMP
COM
CO
C
Answer:

import java.util.*;
class lol {
public static void main()
{
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a word"); 
String str = sc.next( ); 
for(int i =0; i<str.length( ); i++)
{
for(int j =0; j<=i; j++)
{
char ch = str.charAt(i);
System.out.print(ch+" ");
}
System.out.println( );
}
for(int i = str.length( )-2; i<=0; i- -)
{
for(int j = 0; j<=i; j- -)
{
charch = str.charAt(j);
System.out.print(ch+" ");
}
System.out.println( );
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 5
Using switch…case write a menu driven program to print
‘p’:
S
SS
ASS
LASS
CLASS
x’:
\(x^{1}+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\ldots \frac{1}{n !} x^{n}\)
Answer:

import java.util.*; 
class Main {
public static void main(String[ ] args)
{
Scanner sc = new Scanner(System.in);
System.out.println(''Enter p for printing pattern and x for summation series");
charch = sc.next( ).charAt(0); switch(ch)
{
case 'p': String str ="CLASS"; for(int i = 0; i<str.length(); i++)
{
for(int j =str.length()-1 -i; j<str.length(); j++)
{
System.out.print(str.charAt(j));
}
System.out.println( );
}
break;
case 'x': System.out.println("Enter the value of x and n"); 
double x = sc.nextDouble( ); 
int n = sc.nextlnt( ); 
double sum = 0.0; 
for(int i =1; i <=n; i++)
{
double fact = 1; for(int j = 1; j <=i; j++) fact*=j;
sum=sum+Math.pow(x/i)/fact;
}
System.out.println("sum:" +sum); break;
default: System.out.println("Error"); break;
}
}
}

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 6
Write a program to accept sentence from user and print number of words which starts with vowel and ends with vowel.
Input: india is greatest country
Output: 1
Answer:

import java.util.*; 
class lol {
public static void main( )
{
int count =0;
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a sentence:");
String str = sc.nextLine( ), word =""; 
for(int i =0; i<str.length(); i++)
{
char ch = str.charAt(i);
if(ch!='')
word = word + ch; 
else {
char chi =word.charAt(0);
char ch2 = word.charAt(word.length()-1);
if((ch1=='a' || ch1=='A' || chi == 'e'|| ch1 == 'E* || chi =='i,||ch1==T|| 
ch1 -='u'||ch1 =='U'||ch1 =='o'||chl =='0')&&(ch2=='a' || ch2==A' || ch2 == 'e'|| ch2 == 'E' || ch2=='i,||ch2==T||
ch2=-,u,||ch2=='U'||ch2==,o,|| ch2=='0'))
{
count++;
}
word
}
}
System.out.println(count);
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 7
Write a program to accept a sentence from the user and convert it into uppercase and display the isogram words available in the sentence and also display the count of such words.
(Isogram are words without repetition of the characters)
Input: An important part of life
Output: An part of life
Answer:

import java.util.*; 
class Main {
public static void main(String [jargs)
{
int count =0;
Scanner sc = new Scanner(System.in);
System.out.println("Enter a sentence:");
String str_ = sc.nextLine( ), word =" ";
String str = str_.toUpperCase( );
for(int i =0; i<str.length( ); i++)
{
char ch = str.charAt(i); if(ch!='')
word = word + ch; 
else 
{
char [ ]str1 = word.toCharArray( ); 
for(int k = 0; k<str1 .length-1; k++)
{
for(int j = 0; j<str1 .length—k—1; j++)
{
if(str1[j]>str1[j+1])
{
char temp = strltj]; 
str 1 [j] = str 1 [j+1 ];
str1 [j+1] = temp;
}
}
}
int flag = 0;
for(int m =0; m < str1 .length-1; i++)
{
if(str1[m] == strl [m+1 ])
{
flag = 1; break;
}
}
if(flag==0)
{
System.out.print(str1+""); count++;
} 
word
}
}
System.out.priintln(count);
}//end of main }//end of class

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 8
Write a program to accept names of 20 students and the total marks scored by them. Arrange the names as per the rank list with maximum marks at the top. Display the final names of the students along with their total marks scored.
Answer:

import java.util.*; 
class lol {
public static void main( )
{
String temp; double tempi;
Scanner sc = new Scanner(System.in);
String at ] = new String[20];
double b[ ] = new double[20]; 
for(int i =0; ka.length; i++)
{
System.out.println("Enter name & marks:"); 
a[i] = sc.nextLineO; 
b[i] = sc.nextDoubleO;
}
for(int i =0; ka.length; i++)
{
for(int j=0;j<a. length-1 -i; j++)
{
if(b[j] <b[j+1])
{
temp = a[j]; a[j] = a[j+1 ]; 
a[j+1] = temp; tempi = b[j];
b[j] = b[j+l]; b[j+1] = temp1;
}
}
}
System.out.println("Name" + "\t" + "Marks"); 
for(int i =0; i<a.length; i++) 
System.out.println(a[i]+"\t"+b[i]);
} //end of function main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 8 with Answers

Question 9
Write a program to accept a word in upper case and display the position of each alphabet i.e A as 1 B as 2 …Y as 25 and Z as 26.
Sample Input: CAT
Sample Output:
C3
A 1
T 20
Answer:

import java.util.*; 
class lol 
{
public static void main( )
{
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a word:");
String str = sc.next ( );
String strl = str.toUpperCase( );
forfint i =0; i<str.length( ); i++)
{
char ch = str.charAt(i); inta = ch-64;
System.out.println(ch+"\t"+a);
}//end of function main 
}//end of class
Variable Data Type Description
ch char For option porx
str String To store “CLASS”
X double Variable in the series
sum double Stores the sum of series
fact double Factorial of number it stores
Str String Stores the word given by user
Str1[ ] char String converted into the character array.
temp char Temporary storage for character for swapping
flag int To check for next consecutive same characters.
m int For looping
a[] String String array to store 20 strings.
b[] double Array to store 20 marks
str1 String Store all the capital letter word.
a int Store the ascii value of character

ICSE Class 10 Computer Applications Question Papers with Answers

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Section – A (40 Marks)
Attempt all questions

Question 1
(a) (i) Name the keyword that makes the variable constant in a program.
(ii) Name the keyword that makes the system to report an error.
a. Differentiate between equals!) and equalslgnoreCase( )
b. What is the use of access specifier. Name any two.
c. String si = “computer” , s2 =”computers”; System.out.
i println(s1.compareTo(s2);
d. Differentiate between linear search and binary search.
Answer:
(a) (i) final
(ii) throws

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

(b)

equals( ) equalslgnoreCase( )
Compare two strings character by character.

Eq: String si = “lol”, s2 = “loll”; System.out.print(s1.equals(s2));

Compares two strings ignoring the case of the letters.

Eg: String s1 = “lol”,s2=”lol”;
System.out.print(s1.
equalslgnoreCase(s2);

(c) Access specifiers are used to limit the visibility of data according to the users choice.
Eg. private, public.
(d) -1
(e)

Linear Search Binary Search
One by one search element is compared with all elements of the array. The array is sorted and then divided into two halves and then search element is compared to the middle element of the array, if its greater than middle element then search is to be done in upper halves and its its lesser than middle element search to be done in lower halves.
The array need not to be sorted. The array must be sorted.

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 2
(a) Write the memory occupied in computer in bytes for array a[5] of character data type.
Answer:
10 bytes 24 bytes

(b) Differentiate between length!) and length functions.
Answer:

Length( ) Length
The method is applicable for finding the length of string. This is used for finding the array length.

(c) What is error? How logical error differ from syntax error?.
Answer:
When there is an improper use of statement or statement is logically incorrect then an error is displayed.
Syntax error is displayed when there is improper use of basic java statements.
Logical error is displayed when statement is not giving correct output after execution.

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

(d) Name the methods of scanner class that:
(i) Is used to input an integer data from standard input stream.
(ii) Is used to input a string data from standard input stream.
Answer:
(i) nextlnt( )
(ii) nextLine( ) or next( )

(e) Write the statement to perform the following task:
(i) To find and display the position of the first space in the string str
(ii) To find and display the position of the last space in string str.
Answer:
(i) System.out.printlnC’COMP IS FUN”.indexOf(”));
(ii) System.out.printlnC’COMP IS FUN”.lastindexOf(”));

Question 3
(a) The arguments of the function given in the function definition are called …………
Answer:
formal parameters

(b) Why JVM is required for running java program. Justify your answer.
Answer:
JVM is used to convert the byte code into machine language which can be run on any machine. In java, the compiler produces machine code for a particular system. Java compiler produces code for a Virtual Machine known as Java Virtual Machine.

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

(c) Attempt the following:
(i) What is called type casting in java. Explain with example.
(ii) Write all the possible correct syntaxes of declaring array.
(iii) What is the purpose of new operator. Why is it useful?.
(iv) double as =Math.pow(“214”.indexOf(‘4’,3));
(v) char a[ ] = {‘j’/a’/v’/a’};
int c = 1;
System.out.pritnln(a[++c]);
System.out.pritnln(a[c]++);
(vi) Explain the use of return statement in the java.
(vii) State the purpose and return data type of the following String functions:
(i) indexOf( )
(ii) compareTo(  )
(viii) State the output of the code:
int m = 25;
int n = 100;
for(int 1 = 1; i<5; i++);
m++; -n;
System.out.println(“m=”+m);
System.out.println(“n=”+n);
Answer:
(i) The conversion of data from one data type to another datatype is called typecasting.
Eg. char ch =’A’;
Int chi = ch;
(ii) int a[ ] = new int [10]; int a[ ];
int a[ ] = {1,2,3};
(iii) The purpose of new operator is to allocate memory to the object. It is helpful in the creation of new object.
(iv) 8.0
(v) (i) v (ii) a
(vi) It is the keyword in java to return from the method with value or without value.
(vii) (i) Find the index of a character return type is integer.
(ii) Compares two strings return type is integer.
(viii) m=20,n=9

Section – B (60 Marks)
Attempt any four questions from this Section

The answers in this Section should consist of the Programs in either Blue J environment or any program environment with Java as the base. Each program should be written using Variable descriptions/Mnemonic Codes such that the logic of the program is clearly depicted. Flow-Charts and Algorithms are not required.

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 4
Write a program to accept the word from user and print the pattern as shown below:
Sample Input: COMP
Output:
C
CO
COM
COMP
COM
CO
C
Answer:

import java.util.*; 
class lol {
public static void main( )
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter a word");
String str = sc.next( );
for(int i =0; i<str.length( ); i++)
{
for(int j =0; j<=i;j++)
{
charch = str.charAt(i);
System.out.print(ch+" ");
}
System.out.println( );
}
for(int i = str.length( ) - 2; i<=0; i- -)
{
for(int j = 0; j<=i; j- -)
{
char ch = str.charAt(i); 
System.out.print(ch+" ");
}
System.out.println( );
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 5
Write a program to accept the 15 names in the array and arrange them in alphabetical order by bubble sort technique only. And print the output list.
Answer:

import java.util.*; 
class lol
{
public static void main()
{
String temp;
String a[ ] = new Stringfl 5];
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a 15 names");
for(int i =0; ka.length; i++)
{
a[i] =sc.nextLine();
}
forfint i = 0; i<a.length; i++)
{
for(int j =0; j<a.length-i-1; j++)
{
if(a[j].compareTo(a[j++])>0)
{
temp = a[j]; a[j]=a[j+1]; 
a[j+1] = temp;
}
}
}
for(int i =0; ka.length; i++)
{
System.out.println(a[i]);
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 6
Write a program to accept sentence from user and print number of words which starts with vowel and ends with vowel.
Input: india is greatest country
Output: 1
Answer:

import java.util.*; 
class lol
 {
public static void main( )
{
int count =0;
Scanner sc = new Scanner(System.in);
System.out.priintln("Enter a sentence:");
String str = sc.nextLine( ), word ="";
 for(int i =0; i<str.length( ); i++)
{
char ch = str.charAt(i); if(ch!='')
word = word + ch; else {
char ch1 =word.charAt(0);
char ch2 = word.charAt(word.length()-1);
if((ch1=='a' || ch1=='A' || chi == 'ej| chi == 'E' || chi ==,i,||chl ==T|| ch1=='u'||ch1=='Uj|ch1=='oj|ch1==,0')&&(ch2=='a' || ch2=='A' || ch2
== 'e'|| ch2 == 'E' || ch2=='i'||ch2==THch2=='uj|ch2=='Uj|ch2=-oj| ch2=='0')) 
{ . count++;
}
word
}
}
System.out.println(count);
}//end of main
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 7
Write a program to arrange the words in sentence given by user in alphabetical order. Assume input is in lower case letters only.
Input: india is my country
Output: country india is my
Answer:

import java.util.*; 
class lot
{
public static void main( )
{
int count =0;
Scanner sc = new Scanner(System.in); 
System.out.println("Enter a sentence:"); 
String str = sc.nextLine( ), 
word = str = str+" ";
forfint i =0; i<str.length(); i++)
{
char ch= str.charAt(i); if(ch =='') count++;
}
String a[ ] = new Stringfcount];
String temp;
forfint i =0; i<str.length(); i++)
{
char ch = str.charAt(i); if(ch!='') word+= ch;
else
{
int j = 0; a[j]=word; word ="
j++;
}
}
for(int i =0; i<a.length; i++)
{
for(int j =0;j<a.length-i-1; j++)
{
if(atj].compareTo(a[j+1 ])>0)
{
temp = a[i]; a[j] = a[j+1]; 
a[j+1] = temp;
}
}
}
for(int j =0; j<a. length; j++)
{
System.out.print(a[j]+" ");
}
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 8
Write a program to accept integer values in the array and find the second lowest value from the array without sorting the array.
inta[ ] = {3.10.4,8,12,6,23}
Output: 4
Answer:

import java.util.*; 
class lol {
public static void main( )
{
int count = 0; 
int sum = 0;
Scanner sc = new Scanner(System.in);
int a[ ] = new int[10]; 
System.out.println("Enter 10 values:"); 
for(int i =0; ka.length; i++)
{
a[i] = sc.nextlnt( );
}
int low=( );
int secondlow=low;
for(int i =0; i<a.length; i++)
{
if(low >a[i])
{
secondlow=low; low = a[i];
}
}
System.out.println(secondlow); 
}//end of main 
}//end of class

ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers

Question 9
Write a function to display the given pattern: void pattern (int n)
Note: n is the number of rows:
ICSE Class 10 Computer Applications Sample Question Paper 7 with Answers 1
Answer

import java.util.*; 
class lol
{
public void pattern(int n)
{
int n = 5; 
int 1 = 0;
for(int i =1; i<=n; i++)
{
int k = 1; 
for(int j = 1; j<=i; j++)
{
System.out.print(k+" "); 
k+=2;
}
int 11 = I;
for(int j = 2;j<=i;j++)
{
System.out.print(l1+" "); 11 ++;
}
1+=2;
System.out.println( );
}
}
//end of function pattern
//end of class
Variable Data type Description
sc Scanner Object for reading data from input stream.
str String Store word from user
ij int For looping
ch char To reach one character from word.
a[ ] String String array for storng 15 names.
temp String Storing temporary strings.
count int For counting the vowels
word String Temporary for storing string
low secondlow int Store lowest value and store second lowest value.

ICSE Class 10 Computer Applications Question Papers with Answers