## ICSE Class 9 Maths Sample Question Paper 10 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Evaluate without using tables :
sin 38° sin 52° – cos 38° cos 52°.
sin 38° sin 52° – cos 38° cos 52° = sin 38° sin (90° – 38°) – cos 38° cos (90° – 38°)
= sin 38° cos 38° – cos 38° sin 38°
= 0.

(b) In the figure, CD is a diameter which meets the chord AB at E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle

Given : CD is a diameter, AE = BE = 4 cm, CE = 3 cm.
∠OEB = 90° (∵ E is the mid-point of BA).
Let the radius OB = OC = r cm.
Then, OE = OC – CE = (r – 3) cm.

In ΔOBE, by Pythagoras theorem,
OB2 = OE2 + BE2
⇒ r2 = (r – 3)2 + 42
⇒ r2 = r2 – 2.r.3 + 9 + 16
⇒ 6r = 25
⇒ $$r=\frac{25}{6}=4 \frac{1}{6}$$
The required radius = $$4 \frac{1}{6}$$ cm.

(c) If a = 1 – √3, find the value of $$\left(a-\frac{1}{a}\right)^{3}$$

Question 2.
If $$a=\frac{1}{a-5}$$,Find
(i) $$a-\frac{1}{a}$$
(ii) $$a+\frac{1}{a}$$
(iii) $$a^{2}-\frac{1}{a^{2}}$$

(b) The water bills (in ₹) of 32 houses in a locality are given below.
80, 48, 52, 78, 103, 85, 37, 94, 72, 73, 66, 52, 92, 85, 78, 81, 64, 60, 75, 78, 108, 63, 71, 54, 59, 75, 100, 103, 35, 89, 95, 73.
Construct a frequency distribution table with a class size of 10.
Here, minimum value = 35
Maximum value = 108
Class size =10.

(c) Factorize : (a2 – b2) (c2 – d2) – 4 abcd.

Question 3.
(a) Show that:

(b) In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that:
(i) ΔACE ≅ ΔDBE
(ii) AE = DF.

Given: AB = CD,CE = BF and ∠ACE= ∠DBF.
AB = CD
AB + BC = BC + CD (Adding BC on both sides)
⇒ AC = BD.
(i) Now, in ΔACE and ΔDBF
AC = BD (Proved above)
∠ACE = ∠DBF (Given)
CE = BF (Given)
ΔACE ≅ ΔDBF (SAS axiom)
Hence Proved.

(ii) AE = DF (c.p.c.t.)
Hence Proved.

(c) If log10y = x, find the value of 102x in terms of y.
Given:
log10 y = x
⇒ y = 10x
Or 10x = y
∴ 102x = (10x)2 = y2

Question 4.
(a) Given, 5 cos A -12 sin A = 0, find the value of $$\frac{\sin A+\cos A}{2 \cos A-\sin A}$$

(b) Solve by the substitution method :
2x – $$\frac{3}{4}$$y – 3 = 0; 5x – 2y – 7 = 0

(c) What sum of money will amount to ₹ 3630 in 2 years at 10% p. a. compound interest ?

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

Question 5.
(a) If x= $$7-4 \sqrt{3}$$,find the value of $$\sqrt{x}+\frac{1}{\sqrt{x}}$$

(b) In the given figure, X and Y are mid-points of sides AB and AC respectively of ΔABC. If
= 6 cm, AB = 7.4 cm and AC = 6.4 cm, then find the perimeter of the trapezium XBCY.

Given : BC = 6 cm, AB = 7.4 cm and AC = 6.4 cm X,
∴ Y are mid-points of AB and AC, respectively
∵ By mid-points theory.

(c) The mean weight of 8 students is 46.5 kg. Three more students having weight 41.7 kg, 52.8 kg and 51.8 kg joined the group. What is the new mean weight of the students ?
Given : No. of students = 8, mean weight = 46.5 kg.
∴ Total weight of 8 students = 8 x 46.5 kg = 372 kg
∵ 3 more students joined the group,
∴ Total weight of 11 students = 372 + 41.7 + 52.8 + 51.8 = 518.3 kg.
∴ New mean of 11 students = $$\frac{518.3}{11}$$ = 47.12 kg.

Question 6.
(a) Factorize : (x2 – 4x) {x2 – 4x – 1} – 20.
(x2 – 4x) (x2 – 4x – 1) – 20
Let x2 – 4x = a
⇒ a (a – 1) – 20 = a2 – a – 20
⇒ a2 – (5 – 4) a – 20
⇒ a2 -5a + 4a -20
⇒ a (a – 5) + 4 (a – 5)
⇒ (a – 5) (a + 4)
Putting the value of a = x2 – 4x, we get
(x2 – 4x – 5) (x2 – 4x + 4)
= (x2 – 5x + x -5) (x2 – 1. x .2 +22)
= {x {x – 5) + 1 (x – 5)} (x – 2)2
= (x -5) (x + 1) (x – 2)2.

(b) The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them saves ₹ 200 per month, find their monthly incomes.
The ratio of incomes = 9 : 7 and the ratio of expenditures = 4 : 3
Let their incomes be 9x and 7x and their expenditures be 4y and 3y.
:. Their savings are 9x – 4y and 7x – 3y respectively.
According to the question,
9x – = 200 ———- (i)
7x – = 200 ———- (ii)
Multiplying equation (i) by 3 and equation (ii) by 4, we get

Their incomes are ₹ 1800 and ₹ 1400.

(c) The centre of a circle is C (2α -1,3α +1) and it passes through the point A (- 3, – 1). If the diameter of the circle is of length 20 units, find the value (s) of a.
Given points are C(2α – 1, 3α+1) and A(-3,-1)
Diameter 20 units.

Question 7.
(a) Insert 3 rational numbers between $$\frac{1}{4}$$ and $$\frac{1}{2}$$.

(b) In the figure, ABCD is a quadrilateral in which AD = 13 cm, DC = 12 cm, BC = 3 cm and
∠ABD = ∠BCD = 90°. Calculate the length of AB.

Given : AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90°. (Pythagoras theorem)
In ΔBCD, BD2 = BC2 + CD2
= 32 + 122 = 9 + 144 = 153
In ΔBCD, AD2 = BD2 + AB2
132= 153 + AB2
169 = 153 + AB2
AB2 = 169 – 153 = 16
AB = √16 = 4 cm.

(c) Prove that $$: \frac{1}{1+x^{b-a}+x^{c-a}}+\frac{1}{1+x^{a-b}+x^{c-b}}+\frac{1}{1+x^{b-c}+x^{a-c}}=1$$

Question 8.
(a) A sum amounts to ₹ 9680 in 2 years and to ₹ 10648 in 3 years compounded annually. Find the sum and the rate of interest per annum.

(b) A triangle is formed by the lines x + 2y – 3 = 0, 3x – 2y + 7 = 0 and y + 1=0. Find graphically :
(i) The coordinates of vertices of the triangle.

Let the lines intersect at the points A, B and C.
(i) The coordinates of vertices of triangle are A (- 1, 2), B (- 3, – 1) and C (5, – 1).
(ii) From A, draw AM ⊥ BC.
Area of Δ ABC
= $$\frac{1}{2}$$ × BC × AM
= $$\frac{1}{2}$$ × 8 × 3
= 12 sq. units.

(c) Solve : $$\frac{8}{x+3}-\frac{3}{2-x}=2$$

Question 9.
(a) In the figure, ABCD is a parallelogram and P is any point on BC. Prove that area of ΔABP + area of ΔDPC = area of ΔAPD.

Given : ABCD is a parallelogram
We know, triangles on same base and between same parallels are equal.
∴ Area of AAPD = Area of AABD ……. (i)
and Area of AABP = Area of ABDP ……. (ii)
Also, since diagonal of a parallelogram divides it into two triangles of equal area
∴ Area of ΔBCD = Area of ΔABD …… (iii)
⇒ Area of ΔBDP + Area of ΔDPC = Area of ΔABD
⇒ Area of ΔABP + Area of ΔDPC = Area of ΔAPD [Using (i) and (ii)]
Hence Proved.

(b) Factorize : 1+a+b+c+ab+bc+ca+abc.
1+a+b+c+ab+bc+ca+abc=(1+a)+(b+ab)+(c+ca)+(bc+abc)
=(1+a)+b(1+a)+c(1+a)+bc(1+a)
=(1+a)(1+b+c+bc)
=(1+a)((1+b)+c(1+b)
= (1+ a) (1+ b) (1+ c).

(c) The area of cross-section of a pipe is 5.4 cm2 and water is pumped out of it at the rate of 27 km/h. Find in litres the volume of water which flows out of the pipe in one minute.

Question 10.
(a) If 4 cos2 x° – 1 = 0 and 0 ≤ x°≤ 90°,find the value of cos2 x° sin2 x°.

(b) Draw the parallelogram ABCD in which AB = BC = 4.8 cm and AC = 7.5 cm. Find the angle between the two diagonals. What special name can you give it to this parallelogram?
Given : AB = BC = 4.8 cm, AC = 7.5cm

Steps of construction:
(1) Draw BC = 4.8 cm.
(2) From B and C, draw arcs of length 4.8 cm and 7.5 cm, respectively which intersect at A.
(3) From A and C, draw arcs each of length 4.8 cm which intersect at D.
(4) Join BA, AD and CD to complete the required parallelogram.
(5) Join AC and BD which intersect at O.
∴ ∠AOD=90°
The special name of this parallelogram is rhombus as all sides are equal.

(c) Elavuate $$\frac{5^{10+n} \times 25^{3 n-4}}{5^{7 n}}$$

Question 11.
(a) Prove that: $$x^{\log y-\log z} \times y^{\log z-\log x} \times z^{\log x-\log y}=1$$

(b) In the figure, ABCD is a rhombus in which the diagonal DB is produced to E. If ∠ABE = 160°, then find x, y and z.

(c) If a + b – c = 4 and a2 + b2 + c2 = 38, find the value of ab – bc – ca.

## ICSE Class 9 Maths Sample Question Paper 9 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Find xy, if x + y = 6 and x – y = 4.
Given: x + y =6,x – y=4.
Now, 4xy =(x+y)2-(x-y)2
= 62 – 42
=20
xy=$$\frac{20}{4}$$=5.

(b) Find the mean of first 10 prime numbers.
First lo prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
∴ ∑ =129,n=10
Mean = $$\frac{\sum x}{n}=\frac{129}{10}=12.9$$

(c) If x = acosθ + bsinθ and y = asinθ – bcosθ, prove that x2 + y2 = a2 + b2.

Question 2.
(a) Simplify : $$\left(\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{5}+2}\right)\left(\frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}}\right)$$

(b) The area of a trapezium is 540 cm2. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.
Given : Area of trapezium = 510 cm2
Perpendicular distance between parallel lines = 18 cm
Ratio of parallel sides = 7 : 5
Let the length of parallel sides be 7x and 5x.
Area of trapezium = $$\frac{1}{2}$$ (Sum of parallel sides) x Perpendicular distance

7x = 7 x 5 = 35
and 5x = 5 x 5 = 25
Hence, the length of parallel sides are 35 cm and 25 cm.

(c) Solve : (3x + 1) (2x + 3) = 3.
(3x + 1) (2x + 3) = 3
6 x 2 + 9x + 2x + 3 – 3 = O
6 x 2 +11x = 0
x (6x + 11) = 0
x=0 or 6x+11=0
x = 0 or x=$$\frac{-11}{6}$$
x = 0 or $$\frac{-11}{6}$$

Question 3.
(a) Express $$\log _{10}\left(\frac{a^{3} c^{2}}{\sqrt{b}}\right)$$
A ladder 13 cm long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the in terms of log10 a, log10 b and log10 c.

(b) A ladder 13 cm long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.
Let AC be the ladder and BC be the height of the wall.
Then, AC = 13m, AB= 15m, ∠B=90°
∴ By Pythagoras theorem,

(c) Factorize : a4 + b4 – 11a2 b2.

Question 4.
(a) Simplify: $$\sqrt{\frac{1}{4}}+(0.04)^{-1 / 4}-(8)^{2 / 3}$$

(b) In the figure, ABCD is a trapezium in which DA || CB. AB has been produced to E. Find the angles of the trapezium.

Given: DA || CB
x + + 100 = 180° (Co-interior angles) … (i)
⇒ x+2y = 170°
Also, x + 25° = y (Corresponding angles) … (ii)
x – y= – 25°
Subtracting equation (ii) from equation (i), we get
3y = 195°
= $$y=\frac{195^{\circ}}{3}=65^{\circ}$$
Putting y = 65° in equation (ii), we get
X – 65° = – 25°
x = 65 – 25°=40°
∠A = x + 25° = 40°+ 25° = 65°
∠B = 180° – y= 180 – 650= 115°
∠C = 2y + 10° =2 x 65° + 10° = 1400
∠D = x = 40°

(c) Calculate the compound interest for the second year on ₹ 8000 when invested for 3 years at 10% p.a.

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

Question 5.
(a) Solve for x : 25x – 1 = 52x – 1 – 100.

(b) In a ΔABC, ∠A = 80°, ∠B = 40° and bisectors of ∠B and ∠C meet at O. Find ∠BOC.

(c) Without using tables, evaluate :

Question 6.
(a) If x = 3 + 2√2 , find the value of $$x^{3}-\frac{1}{x^{3}}$$

(b) Construct a rectangle each of whose diagonals measures 6 cm and the diagonals intersect at an angle of 45°.

Given: Each diagonal = 6 cm, diagonals intersect at 45°.
Steps of construction:
(1) Draw diagonal AC = 6 cm.
(2) Bisect AC at O
(3) At O, draw ∠COX = 45° and extend XO to Y.
(4) From O, cut-off XY at D and B such that OD = OB = 3 cm,
i.e., BD = 6 cm.
(5) Join A, B, C, D to get the required rectangle ABCD.

(c) KM is a straight line of 13 units. If K has the coordinates (2, 5) and M has the coordinates (x, – 7), find the possible values of x.

Question 7.
(a) If $$x=\frac{\sqrt{7}+1}{\sqrt{7}-1} \text { and } y=\frac{\sqrt{7}-1}{\sqrt{7}+1}$$find the value of $$\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}$$

(b) In ΔABC, AC = 3 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC at N, what is the length of MN ?

Given : AC = 3 cm, M is the mid-point of AB, MN || AC.
Since, M the mid-point of AB and MN || AC
Therefore, by mid-point theorem, we have
MN = $$\frac{1}{2}$$AC= $$\frac{1}{2}$$ x 3cm = 1.5cm

(c) If θ is an acute angle and tan θ = $$\frac{5}{12}$$, find the value of cosθ + cot θ.

Question 8.
(a) Factorize : 1 – 2ab – (a2 + b2)

(b) In the given figure, AOC is a diameter of a circle with centre O and arc A×B = $$\frac{1}{2}$$ arc BYC. Find ∠BOC.

(c) In a class of 90 students, the marks obtained in a weekly test were as under.

Construct a combined histogram and frequency polygon.

Question 9.
(a) Divide ₹ 1,95,150 between A and B so that the amount that A receives in 2 years is the same as that of B receives in 4 years. The interest is compounded annually at the rate of 4% p.a.

A’s share = 1,01,400.
and B’s share = (1,95,150 – 1,01,400) = ₹ 93,750.

(b) Solve simultaneously : $$\frac{3}{x+y}+\frac{2}{x-y}=3 ; \frac{2}{x+y}+\frac{3}{x-y}=\frac{11}{3}$$

(c) If a = cz, b = ax and c = by, prove that xyz = 1.

Question 10.
(a) If sin (A + B) = 1 and cos (A – B) = $$\frac{\sqrt{3}}{2}$$, 0° < A + B ≤ 90°, A > B, then find A and B.
sin (A + B) = 1 = sin 90°
A + B = 90°
cos (A – B) =$$\frac{\sqrt{3}}{2}$$ = cos 30°
A – B = 30°
Adding (i) and (ii), we get
A + B + A – B = 90° + 30°
⇒ 2A = 120°
A = $$\frac{120^{\circ}}{2}$$ = 60°
Putting A= 60° in (i), we get
60° + B = 90°
⇒ B = 90° – 60° = 30°
∴ A = 60°, B = 30°

(b) The sum of the digits of a two-digit number is 5. The digit obtained by increasing the digit in ten’s place by unity is one-eighth of the number. Find the number.

Let the digit at ten’s place be  and that at unit’s place be y.
∴ The number = 10x + y
By 1st condition,
x + y = 5
By 2nd condition,
x + 1 = $$\frac{1}{8}$$ (10x + y)
⇒ 8 (x + 1) = 10x + y
⇒ 8x + 8 = 10x + y
⇒ 2x + y = 8
Subtracting (i) from (ii), we get
x = 3.
Putting x = 3 in (i), we get
3 + y = 5
⇒ y = 5 – 3 = 2
The required number = 10x + y = 10 x 3 + 2 = 32

(c) Given, log10 x = a and log10 y =
(i) Write down 10a-1 in terms of x
(ii) Write down 102b in terms of y.
(iii) If log10 P = 2a-b, express P in terms of x and y.

Question 11.
(a) Draw the graph of the equations 2x – 3y = 7 and x + 6y = 11 and find their solutions.

The two lines intersect each other at the point (5, 1).
∴ x = 5, y = 1

(b) In the figure, area of ΔABD = 24 sq. units. If AB = 8 units, find the height of ΔABC.
Given : Area of A ABD = 24 sq. units, AB = 8 units, DC || AB.
Area of ΔABC = Area of ΔABD
(∵ They are on same base and between same parallels)
= 24 sq. units.
⇒ $$\frac{1}{2}$$ x AB x Height of ΔABC = 24
⇒ $$\frac{1}{2}$$ x 8 x Height of ΔABC = 24
⇒ Height of Δ ABC = $$\frac{24}{4}$$ = 6 units.

(c) If x – y = 8 and xy = 20, evaluate : (i) x + y (ii) x2 – y2.
Given x – y – 8, xy = 20.
(i) (x + y)2 = (x – y)2 + 4 xy = 82 + 4 x 20 = 64 + 80 = 144
x + y = ± √144 = ± 12
x2 – y2 = (x + y) (x – y)
= (± 12) x 8
= ± 96.

## ICSE Class 9 Maths Sample Question Paper 8 with Answers

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

Question 1.
(a) The mean of 100 observations was found to be 30. If two observations were wrongly taken as 32 and 12 instead of 23 and 11, find the correct mean.
Here, n = 100, $$\bar{x}$$= 30
∴ Incorrect = Σx= $$\bar{x}$$ n = 30 x 100 = 3000.
∴ Correct Σ x = 3000 – (32 + 12) + (23 + 11)
= 3000 – 44 + 34 = 2990
∴ Correct mean = $$\frac{2990}{100}=29.9$$

(b) Determine the rate of interest for a sum that becomes $$\frac{216}{125}$$ times of itself in 3 years, compounded annually.
Let principal be ? P and rate of interest be r% p. a. So,

(c) Without using tables, find the value of :

Question 2.
(a) If $$x=\frac{3+\sqrt{7}}{2}$$ find the value of $$4 x^{2}+\frac{1}{x^{2}}$$

(b) In the given figure, ABCD is a rectangle with sides AB = 8 cm and AD = 5 cm. Compute : (i) area of parallelogram ABEF, (ii) area of ΔEFG.

Given : AB 8 cm, AD = 5 cm.
(i) Area of parallelogram ABEF = Area of rectangle ABCD
(∵ they are on same base and between same parallels)
= (8 x 5) cm2 = 40 cm2
Area of ΔEFG = $$\frac{1}{2}$$ x Area of parallelogram ABEF
(∵ both are on same base and between same parallels)
= $$\frac{1}{2}$$ x 40 cm2 20 cm2

(c) Without using tables, find the value of :
$$\frac{(b+c)^{2}}{b c}+\frac{(c+a)^{2}}{c a}+\frac{(a+b)^{2}}{a b}$$

Question 3.
(a) Solve for x : 2x +3 + 2x+1 = 320.

(b) In the given figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm respectively.

Given: radius = OA = OC = 15 cm, AB || CD.
Let AB = 24 cm, CD = 18 cm.
We know perpendicular drawn from centre to the chord, bisects the chord
∴ M and N are mid-point of sides AB and CD respectively.
AM= $$\frac{1}{2}$$ AB= $$\frac{1}{2}$$ x24=12cm.

(c) Factorize : 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2.
Given expression is, 4 (2a – 3)2 – 3 (2a – 3) (a – 1) – 7 (a – 1)2
Let 2a – 3 = x and a – 1 = y
The expression becomes
= 4a2 – 3xy – 7y2 = 4x2 – (7 – 4) xy – 7y2
= 4x2 – 7xy + 4xy – 7y2
= x (4x – 7y) + y (4x – 7y)
= (4x – 7y) (x + y)
Substituting values of x and y, we have
= {4 (2a – 3) – 7 (a – 1)} {2a – 3 + a – 1)
= (8a -12 -7a + 7) (3a – 4) = (a – 5) (3a – 4).

Question 4.
(a) Solve for x : log (x + 5) + log (x – 5) = 4 log 2 + 2 log 3.
log {x + 5) + log {x – 5)
=4 log 2 + 2 log 3 log (x + 5) + log (x – 5)
= log 24 + log 32 log (x + 5) + log (x – 5)
= log 16 + log 9 log [(x + 5) (x – 5)]
= log (16 x 9) log(x2 – 25) – log 144
⇒ x2 – 25 = 144
⇒ x2 = 144 + 25 = 169
⇒ x = √169 = 13

(b) Solve simultaneously : $$2 x+\frac{x-y}{6}=2 ; x-\frac{(2 x+y)}{3}=1$$

(c) If 8 cot 915, find the value of: $$\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}$$

Question 5.
(a) The cost of enclosing a rectangular garden with a fence all around at the rate of ₹ 15 per metre is ₹ 5400. If the length of the garden is 100 m, find the area of the garden.
Total cost of fendng = ₹ 5400
Rate = ₹ 15 per metre
Perimeter = $$\frac{5400}{15}$$ = 360 m
Length, l= 100 m
2 (l + b) = 360
b = 180 $$\frac{360}{2}$$ 100 = 80 m
Area = i x b = 100 m x 80 m = 8000 m2

(b) If 4 sin2 x° – 3 = 0 and x° is an acute angle, find (i) sin x° (ii) x°.
Given: 4 sin2 x° – 3 = O
(i) 4 sin2 x° =3
sin2 x° = $$\frac{3}{4}$$
sin x°= $$\frac{\sqrt{3}}{2}$$

(ii) Now sin x°= $$\frac{\sqrt{3}}{2}$$
⇒ sin x°= sin 60°
⇒ x° = 60°

(c) Draw a frequency polygon from the following data :

 Age (in years) 25-30 30-35 35-40 40-45 45-50 No. of doctors 40 60 50 35 20

Question 6.
(a) If $$\frac{(x-\sqrt{24})(\sqrt{75}+\sqrt{50})}{\sqrt{75}-\sqrt{50}}=1$$ find the value of x.

(b) From the given figure, find the values of a and b.

∠DBC = ∠ADB = a° (Alternate angles)
Now, a + 28° = 75° (Exterior angle is equal to sum of interior opposite angles)
⇒ a = 75° – 28° = 47°.
Also, ∠ABC + ∠BAD = 180° (Co-interior angles)
⇒ a + b + 90° = 180°
⇒ 47° + b + 90° = 180°
⇒ b = 180° – 137° = 43°
a = 47°, b = 43°

(c) Show that the points A (2, – 2), B (8, 4), C (5, 7) and D (- 1, 1) are the vertices of a
rectangle. Also, find the area of the rectangle.

i.e., opposite sides are equal and diagonals are equal
ABCD is a rectangle.
Hence Proved.
Area of rectangle = AB x BC = 6√2 x 3√2 = 36 sq. units.

Question 7.
(a) By using suitable identity, evaluate : (9.8).
(9.8)1 = (10 – 0.2)3 = 103 – 3 x 102 x 0.2 + 3 x 10 x (0.2)2 – (0.2)3
= 1000 – 60 + 1.2 – 0.008 = 941.192.

(b) ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB
and parallel to BC intersects AC at D. Show that (i) D is mid-point of AC (ii) MD L AC
(iii) CM = MA = $$\frac{1}{2}$$ AB.
Given : M is mid-point of AB, ∠C = 90°, MD||BC.
Join MC.
(i) ∵ M is mid-point of AB and MD|| BC

∴ By the converse of mid-point theorem,
D is mid-point of AC.
Hence Proved.

(ii) ∠BCD + ∠CDM = 1800 (Co-interior angles, MDIIBC)
⇒ 90° + ∠CDM = 180° (∠BCD = 90°)
∠CDM= 180°-90°=90°
MD ⊥ AC. Hence Proved.

(iii) In ΔAMD and ΔCMD,
AD = CD (D is mid-point of AC)
∠ADM = ∠CDM (Each being 90°)
MD = MD (Common side)
∴ ΔAMD ≅ ΔCMD. (SAS axiom)
∴ AM = CM (c.p.c.t.)
Also, AM = $$\frac{1}{2}$$ AB ( M is mid-point of AB)
∴ CM = AM = $$\frac{1}{2}$$ AB.  Hence Proved.

(c) Solve $$x+\frac{1}{x}=2 \frac{1}{2}$$

Question 8.
(a) If : a = b2x, b – c2y and c = a2z, show that 8xyz = 1.

(b) In the given figure, ABC is a right triangle at C. If D is the mid-point of BC, prove that AB2 = 4AD2 – 3AC2.

Given :∠C = 90°, D is mid-point of BC.
In ΔABC, In ΔACD, ⇒ AB2 = AC2 + BC2 AD2
⇒ AC2 + CD2 CD2
⇒ AD2 – AC2 (Pythagoras theorem) … (i) (Pythagoras theorem)
⇒ $$\left(\frac{1}{2} \mathrm{BC}\right)^{2}$$ = AD2 – AC2 (∵ D is mid-point of BC)
⇒ $$\frac{1}{4}$$ BC2 = AD2 – AC2 4
⇒ BC2 = 4AD2 – 4AC2 …(h)
Using (i) and (ii), we have
AB2 = AC2 + 4AD2 – 4AC2
⇒ AB2 = 4AD2 – 3AC2
Hence Proved.

(c) Find the value of x, if tan 3x = sin 45° cos 45° + sin 30°.
Given: tan 3x = sin 45° cos 45° + sin 30°
⇒  tan 3x $$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{2}$$
⇒ tan 3x = $$=\frac{1}{2}+\frac{1}{2}$$
⇒ tan 3x = 1
⇒ tan 3x = tan 45°
⇒ 3x = 45° ,
⇒ 45°
x = $$\frac{45^{\circ}}{3}$$ = 15°

Question 9.
(a) Factorize : 12 – (x + x1) (8 – x – x2).
12 – (x + x2) (8 – x – x2) = 12 – (x + x2) {8 – (x + x2)}
Let  x + x2 = a
⇒ 12 – a (8 – a) = 12 – 8a + a2
⇒ a2 – 8a + 12
⇒ a2 – (6 + 2) a + 12
= a2 – 6a – 2m + 12
= a (a – 6) – 2 (a – 6)
⇒ (a-6) (a- 2)
Substituting a = x + x2, we get
⇒ (x + x2 – 6) (x + x2 – 2)
⇒ (x2 + x – 6) (x2 + x – 2)
⇒ (x2 + 3x – 2x – 6) (x2 + 2x – x – 2)
⇒ {x (x + 3) – 2 (x + 3)} {x (x + 2) – 1 (x + 2)}
⇒ (x + 3) (x – 2) (x + 2) (x – 1)
⇒ (x – 1) (x + 2) (x – 2) (x + 3)

(b) A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Let the speed of the train be x km/h and that of the car be y km/h.
Total distance travelled = 370 km.
Case I : Distance travelled by train = 250 km.
Distance travelled by car = (370 – 250) km = 120 km
∴ Time taken by train $$=\frac{250}{x} \mathrm{~h}$$

(c) Express as a single logarithm :
2 log10 5 – log10 2 + 3log10 4+1
2 log10 5 – log10 2 + 3 log10 4 + 1 = log10 52 – log10 2 + log10 43 + log10 10
$$=\log _{10}\left(\frac{5^{2} \times 4^{3} \times 10}{2}\right)$$
= log10 8000 = log10 (20)3 = 3 log10 20

Question 10.
(a) The value of a car purchased 2 years ago depreciates by 10% every year. Its present value is ₹ 1,21,500. Find the cost price of the car. What will be its value after 2 years ?

(b) Construct a quadrilateral ABCD given that AB = 4.5 cm, ∠BAD = 60°, ∠ABC = 105°, AC = 6.5 cm and AD = 5 cm.
Given : AB 4.5 cm, Z BAD = 60°, Z ABC 105°, AC = 6.5 cm and AD – 5 cm. Steps of
construction :
(1) Draw AB = 4.5 cm.
(2) At A, draw ∠BAX = 60°.
(3) At B, draw ∠ABY = 105°.

(4) From A, cut BY at C such that AC = 6.5 cm.
(5) From A, cut AX at D such that AD = 5 cm.
(6) Join CD.
Hence, ABCD is the required quadrilateral.

(c) Factorize : x9 + y9.
= x9 + y9 = (x3)3 + (y3)3 = (x3 + y3) {(x3)2 – x3y3 + (y3)2}
= (x + y) (x2 – xy + y2) (x6 – x3y3 + y6).

Question 11.
(a) In the given figure, ABCD is a parallelogram. Find the values of x, y and z.

Given : ABCD is a parallelogram.
AB = CD
⇒ 3x – 1 = 2x + 2
⇒ 3x – 2x =2 + 1
⇒ x = 3
Also, ∠D = ∠B = 102° ( ∵ Opposite angles are equal) Exterior
In ΔACD, y = 50° + 102° (∵ angle is equal to sum of interior opposite angles)
= 152°
and ∠A + ∠D = 180°
⇒ z + 50° + 102° = 180°
⇒ z = 180° – 152° = 28°
⇒ x = 3, y = 152° z = 28°

(b) Draw the graph of 3x + 2 = 0 and 2y – 1 = 0 on the same graph sheet. Do these lines intersect ? If yes, find the point of intersection.
Given: 3x+2=0 ……..(i)
and 2y – 1 = 0 ……. (ii)
From (i), 3x = – 2
= $$x=\frac{-2}{3}$$
It is a straight line parallel to Y-axis at $$x=\frac{-2}{3}$$
From (ii), 2y = 1
⇒ y = $$\frac{1}{2}$$
It is a straight line parallel to Y-axis at y = $$\frac{1}{2}$$

(c) Prove that (sin A + cos A)2 + (sin A – cos A)2 = 2.
L. H.S. = (sin A + cos A)2 + (sin A – cos A)2
= sin2A + cos2 A + 2 sin A cos A + sin2A + cos2A – 2 sin A cos A = 2 (sin2A + cos2A)
= 2 x 1
= 2 = R.H.S.

## ICSE Class 9 Maths Sample Question Paper 7 with Answers

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

Question 1.
(a) If, in a ∆ABC, AB = 3 cm, BC = 4 cm and ∠ABC = 90°, find the values of cos C, sin C and
tan C.
Given : AB = 3 cm, BC = 4 cm, ∠ABC = 90°
By Pythagoras theorem,
AC2 = AB2 + BC2 = 32 + 42 = 25

(b) A man purchased an old scooter for ₹ 16,000. If the cost of the scooter after 2 years depreciates to ₹ 14,440, find the rate of depreciation.
Present value (V0) = ₹ 16,000
Value after 2 year (V1) = ₹ 14,440
∴ n =2
Let r be the rate of depreciation.

(c) Prove that √2 + √5 is irrational.
Let us assume that √2 + √5 is a rational number.
Then $$\sqrt{2}+\sqrt{5}=\frac{a}{b}$$
Where a and b co-prime positive integers.
$$\frac{a}{b}-\sqrt{2}=\sqrt{5}$$

Question 2.
(a) If $$x=\frac{1}{x-2 \sqrt{3}}$$ , find the values of (i) x – $$\frac{1}{x}$$ (ii) x + $$\frac{1}{x}$$.

(b) In the given figure, ABC is an equilateral triangle. Find the measures of angles marked by x, y and z.

Given : ABC is an equilateral triangle.
∠ABC = ∠ACB = ∠B AC = 60°.
Now, ∠BAD + ∠ADB = ∠ABC (Ext. angle is equal to sum of int. opp. angles)
⇒ x + 40° = 60°
⇒ x = 60° – 40°
⇒ x = 20°.
Also, ∠CAE + ∠AEC = ∠ACB (Ext. angle is equal to sum of int. opp. angles)
⇒ y + 30° = 60°
⇒ y = 60° – 30°
⇒ y = 30°
and ∠ACE +∠ACB = 180° (Linear Pair)
⇒ z + 60° = 180°
⇒ z = 180° – 60°
⇒ z = 120°

(c) Solve $$\frac{2}{3} x^{2}-\frac{1}{3} x-1=0$$
$$\frac{2}{3} x^{2}-\frac{1}{3} x-1=0$$
$$3 \times \frac{2}{3} x^{2}-3 \times \frac{1}{3} x-3 \times 1=3 \times 0$$
⇒ 2x2 – x – 3 = 0
⇒ 2 x 2 – (3 – 2)x -3=0
⇒ 2x2 – 3x + 2x – 3 = 0
⇒ x (2x – 3) + 1 (2x – 3) = 0
⇒ (2x – 3) (x + 1) = 0
⇒ 2x-3=0 or x + 1= 0
⇒ x= $$\frac{3}{2}$$ or x =-1
⇒ x= $$\frac{3}{2}$$ or -1

Question 3.
(a) Factorize : a3 – b3 – a + b.
a3 -b3 – a + b = (a-b) (a2 + ab + b2) – (a-b) = (a-b) (a2 + ab + b2 – 1).

(b) Draw a histogram to represent the following :

 Class Interval 40 – 48 48-56 56-64 64-72 72 – 80 Frequency 15 25 35 30 10

(c) Prove that $$\sqrt{\frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}}}=\tan 30^{\circ}$$

Question 4.
(a) Simplify: $$\frac{5^{2(x+6)} \times(25)^{-7+2 x}}{(125)^{2 x}}$$

(b) In the figure, DE||BC. Prove that (i) Area of ΔACD = Area of ΔABE (ii) Area of ΔOBD = Area of ΔOCE.

Given DE || BC
Area of ΔBCD = Area of ΔBCE
(Triangles on same base and between same parallels have equal area) Now, Area of ΔACD + Area of ΔBCD = Area of ΔABE + Area of ΔBCE
⇒ Area of ΔACD = Area of ΔABE (∵ Area of ABCD = Area of ABCE).
Hence Proved.

(ii) Area of ABCD = Area of ABCE [From (i)]
⇒ Area of ABCD – Area of ΔOBC = Area of ΔBCE – Area of ΔOBC
(Subtracting area of ΔOBC from both side)
⇒ Area of ΔOBD = Area of ΔOCE.
Hence Proved

(c) If log10 x + $$\frac{1}{3}$$ log10 y = 1, express y in terms of x.
Given log10 x + $$\frac{1}{3}$$ log10y = 1
log10 x + log10 y1/3 = log10 10
log10 (xy1/3) = log10 10

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

Question 5.
(a) The mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.
We know,
Σx =$$\bar{x}$$ x n
Incorrect ∑ x = 35 x 9 = 315
Correct ∑ x =315 – 18 + 81 = 378
Correct mean = $$\frac{378}{9}=42$$

(b) In the given figure, O is the centre of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find (i) radius of the circle (ii) Length of chord CD.

⇒ 169 =144 ÷ CN2
⇒ CN2 = (169 – 144) = 25
⇒ CN= √25 =5
⇒ CD =2 CN (∵ N is mid-point of CD)
⇒ 2 x 5 = 10cm.

(c) If $$x^{2}+\frac{1}{x^{2}}=83$$ find the value of $$x^{3}-\frac{1}{x^{3}}$$

Question 6.
(a) A cumulative frequency distribution is given below. Convert this into a frequency distribution table.

 Marks Below 45 Below 60 Below 75 Below 90 Below 105 Below 120 No. of Students 0 8 23 48 85 116

 Marks No. of Students Class Interval Frequency Below 45 0 0-45 0 Below 60 8 45 – 60 8 (8-0) Below 75 23 60 – 75 15 (23 – 8) Below 90 48 75 – 90 25 (48 – 23) Below 105 85 90 -105 37 (85 – 48) Below 120 116 105 – 120 31 (116 – 85)

(b) Half the perimeter of a garden, whose length is 4 more than its width, is 36 m. Find the dimensions of the garden.
Let length and breadth of the garden be x m and y m respectively.
According to the question,
x = 4 + y …(i)
and x + y = 36 …(ii)
Substituting x = 4 + y in equation (ii), we get
4 + y + y = 36
2y = 36 – 4
y = $$\frac{32}{2}$$ = 16
Substituting y= 16 in equation (i), we get
x = 4 + 16 = 20
∴ Length = 20 m and breadth = 16 m.

(c) If x and y are rational numbers and $$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=x-y \sqrt{3}$$

Question 7.
(a) Factorize : (x2 + y2 – z2)2 – 4x2y2.
(x2 + y2 – z2)[1] – 4x2y2 = (x2 + y1 – z2)2 – (2xy)2
= (x2 + y2 – z2 + 2xy) (x2 + y2 – z2 – 2xy)
= {(x2 + y2 + 2xy) – z2} {(x2 + y2 – 2xy) – z2}
= {(x + y)2 – (z)2}  – y)2 – (z)2}
= {x + y + z) {x + y – z) {x – y + z) {x – y – z).

(b) Prove that in a right angled triangle, the median drawn to the hypotenuse is half the hypotenuse in length.

(c) Find the value of x if 3 cot2 (x – 5°) = 1.
3 cot2 (x – 5°) =1
1 cot2 (x – 5°) = $$\frac{1}{3}$$
cot (x – 5°) = $$\frac{1}{\sqrt{3}}$$
cot (x – 5°) = cot 60°
x – 5°= 60°
x = 60° + 5°
x = 65°

Question 8.
(a) Solve: $$\frac{x+y}{x y}=2 ; \frac{x-y}{x y}=1$$

(b) Construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.
Given: AB =5.1cm, BC = 7cm and ∠ABC = 75°
Steps of construction:
(1) Draw BC=7cm.
(2) At B, draw ∠ XBC = 75°
(3) From B, cut-off BA = 5.1 cm on BX.
(4) From C, draw an arc of radius 5.1 cm.
(5) From A, draw an arc of 7 cm to cut the arc from C at D.
Hence, ABCD is the required parallelogram.

(c) Calculate the distance between A (7, 3) and B on the X-axis whose abscissa is 11.
Given : A (7, 3)
∵ B lies on the X-axis whose abscissa is 11, the coordinates of B are (11, 0)
$$\mathrm{AB}=\sqrt{(11-7)^{2}+(0-3)^{2}}=\sqrt{4^{2}+(-3)^{2}}=\sqrt{16+9}=\sqrt{25}$$
= 5 Units.

Question 9.
(a) A sum of money ₹ 15,000 amounts to ₹ 16,537.50 in x years at the rate of 5% p.a. compounded annually. Find x.

(b) In the given figure, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.

Given: ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm.
∴ In Δ PQ2, PQ2 = PS2 + QS2 (Pythagoras theorem)
102 =PS2+62
PS2 = 100 – 36
PS = √64 = 8cm
In ΔPRS, PR2 = PS2 + RS2 (Pythagoras theorem)
PR2 = 8 + (9 + 6)2 = 64 + 225 = 289
PR=√289=17cm.

(c) In the given figure, ACB is a semicircle whose radius is 10.5 cm and C is a point on the semicircle at a distance of 7 cm from B. Find the area of the shaded region.

For semi-circle,
r = 10.5 cm
∴ Area =$$\text { Area }=\frac{1}{2} \pi r^{2}=\frac{1}{2} \times \frac{22}{7} \times(10.5)^{2}=173.25 \mathrm{~cm}^{2}$$
For triangle ABC,
AB2 = BC2 + AC2 (Pythagoras theorem, ∠C = 90°)
(2 x 10.5)2 = 72 ÷ AC2
AC2 =441 – 49 =392
AC = 19.8 cm.
Area = x BC x AC = x 7 x 19.8 = 69.3 cm2
The area of shaded region = (173.25 – 69.3) cm2
= 103.95 cm2

Question 10.
(a) If a2 + b2 + c2 – ab – be – ca = 0, prove that a = b = c.
Given a2 + b2 + c2 – ab – be – ca =0
⇒ 2 (a2 + b2 + c2 – ab – be – ca) = 0
⇒  2a2 + 1b2 + 2c2 -2ab – 2bc – 2ca = 0
⇒ (a2 – 2ab + b2) + (b2 – 2be + c2) + (c2 – 2ca + a2) = 0
⇒ (a – b)2 + (b – c)2 + (c – a)2 =0
The above expression is possible only if
⇒ (a- b)2 = 0 Ab- c)2 = 0, (c – a)2 = 0
a-b =0, b – c = 0, c-a = 0
a = b,b = c, c = a
a = b = c.
Hence Prove.

(b) Solve graphically x + 3y = 6; 2x – 3y = 12 and hence find the value of a, if Ax + 3y = a
x+3y=6 ………. (i)
2x – 3y = 12 ….. (ii)
from (i)  x = 6 – 3y

 X 6 3 0 y 0 1 2

∴ (6, 0), (3, 1), (0, 2)
From (ii),
2x = 3y + 12
x = $$\frac{3 y+12}{2}$$

 X 6 3 0 y 0 1 2

(6, 0), (3, – 2), (0, – 4)
These points are piotted in the graph.

The two lines intersect at the point (6, 0).
∴ x = 6, y = 0
Now 4x + 3 y = a
⇒ 4 x 6 + 3 x 0 = a
24 + 0 =a
⇒ a = 24

(c) Given, 1008 = 2p.3q.7r, find the values of p, q, r and hence evaluate 2p.3q.7-r÷192.

Question 11.
(a) If log $$\frac{x-y}{2}=\frac{1}{2}$$(log x + log y), prove that x2 + y2 = 6xy.

x2 + y2 – 2xy = 4xy
x2 + y2 – 4xy + 2 xy
x2 + y2 = 6 xy.
Hence Proved

(b) In a pentagon ABCDE, AB||ED and ∠B = 140°. Find ∠C and ∠D if ∠C: ∠D = 5:6.

Given : AB||ED, ZB = 140°, ∠C : ∠D = 5:6.
Let  ∠C =5x, ∠D = 6x.
Now,∠A+∠E= 180° (Co-interior angles, AB||ED)
Also, ∠A+ ∠B+ ∠C+ ∠D+ ∠E= (5-2) x 180°
(∠A + ∠E) + ∠B + ∠C + ∠D
= 3 x 180° 180° + 140° + 5x + 6x = 540°
11 x = 540° – 320°
$$x=\frac{220^{\circ}}{11}$$
∠C = 5x = 5 x 20° = 100°
∠D = 6x = 6 x 20° = 120°

(c) Factorize : 4 a3b – 44 a2b + 112
4 a3b  – 44 a2b + 112 ab = 4 ab (a2 – 11a + 28)
= 4 ab {(a2  – (7 + 4) a + 28)}
= 4ab(a2 – 7a – 4a+28)
= 4ab {a (a – 7) – 4(a – 7))
= 4ab (a – 7) (a – 4).

## 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.
log (1 + 2 + 3) = log 6 = log (1 x 2 x 3)
= log 1 + log 2 + log 3.

(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.

Given : AE = BE = 4 cm, CE = 3 cm
Let r be the radius (OB = OC)
OE = OC – CE = r – 3.

⇒ OB2 = OE2 + BE2 (Pythagoras theorem)
⇒ r2 = (r – 3)2 + 42 r2
⇒ r2 – 6r + 9 + 16
⇒ 6r = 25
$$r=\frac{25}{6}=4 \frac{1}{6} \mathrm{~cm}$$

(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.

Question 2.
(a) Evaluate tan x and cos y from the given figure.

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

(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.

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.

(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.
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
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)

(b) Evaluate : $$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \cdot \tan 60^{\circ}}$$

(c) Simplify:$$(81)^{3 / 4}-3 \times(7)^{0}-\left(\frac{1}{27}\right)^{-2 / 3}$$

Question 4.
(a) If x $$\frac{2}{x}$$ = 5, find the value of $$x^{3}-\frac{8}{x^{3}}$$

(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.
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.

(c) Simplify: $$\frac{3}{\sqrt{6}+\sqrt{3}}-\frac{4}{\sqrt{6}+\sqrt{2}}+\frac{1}{\sqrt{3}+\sqrt{2}}$$

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

Question 5.
(a) Solve : log10 6 + log10 (4x + 5) = log10 (2x + 7) +1

(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 ?
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

∴ One woman finish the work in 140 days.

(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
For first polygon,

$$\frac{3}{n^{2}+2 n-n-2}=\frac{1}{60}$$
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.

Question 6.
(a) If $$a^{2}+\frac{1}{a^{2}}=7$$, find the value of $$a^{2}-\frac{1}{a^{2}}$$

(b) Construct a trapezium ABCD in which AD || BC, Z B = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.
Given : AD||BC, ZB = 60°, AB = 5 cm, BC = 6.2 cm, and CD = 4.8 cm.

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.

(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.
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
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

(b) Find graphically the vertices of the triangle whose sides have equations
2y – x = 8, 5y – x = 14 and y – 2x = 1.
Given equations are,
2y – x =8 ……….(i)
5y – x =14 …(ii)
and y – 2x =1 …(iii)
From(i), x =2y – 8

∴ (- 6, 1), (- 4, 2), (- 2, 3)
From (ii), x=2 y-8

(- 4, 2), (1, 3), (6, 4)
From (iii) y=2 x+1

∴ (1, 3), (2, 5), (- 1, – 1)
These points are plotted on the graph.

The three lines intersect at point (- 4, 2), (1, 3) and (2, 5) which are the required vertices of triangle formed by them

(c) If 3tan2 θ-1=0′, find cos 2θ, given that θ is acute.
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.
⇒ 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.
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

(c) Insert four irrational numbers between 2√3 and 3√2

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.

(b) Solve simultaneously : $$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{4}{y}=-\frac{1}{2}$$

(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.

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

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.

(b) Find the coefficient of x2 and x in the product of (x – 2) (x – 3) (x – 4).
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 = $$\frac{1}{2} (AB + CD)$$.

Question 11.
(a) Factorize : a3 + 3a2b + 3ab2 + 2b3.
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)

(b) In the point A (2, – 4) is equidistant from the points P (3, 8) and Q (- 10, y), find the values of y.
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: $$\sqrt[a b]{\frac{x^{a}}{x^{b}}} \cdot b \sqrt[x]{\frac{x^{b}}{x^{c}}} \cdot \sqrt[c a]{\frac{x^{c}}{x^{a}}}$$

## 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 ?
(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.

(b) If x = 2 + √3 , prove that x2 – 4x + 1 = 0.

(c) How many times will the wheel of a car having radius 28 cm, rotate in a journey of 88 km
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 wheel

Question 2.
(a) Factorize : $$x^{2}+\frac{1}{x^{2}}-11$$

(b) From the adjoining figure, find the value of x.

In ΔACD,
AC = CD  (Given)
Now, ∠ADC + ∠CAD + ∠ACD = 180° (Sum of angles in a triangle is 180°)
⇒ 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)
(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°

(c) Simplify : $$\frac{5 .(25)^{n+1}-25 .(5)^{2 n}}{5 \cdot(5)^{2 n+3}-(25)^{n+1}}$$

Question 3.
(a) If θ = 30°, verify that cos 3 θ = 4 cos3 θ -3 cos θ .
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°

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.

(c) Prove that: $$\log \frac{11}{5}+\log \frac{14}{3}-\log \frac{22}{15}=\log 7$$

= 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 + $$a^{2}+\frac{1}{a^{2}}$$
(a)

(b) Construct a combined histogram and frequency polygon for the following data :

(c) Of two unequal chords of a circle, prove that longer chord is nearer to the centre of the circle.
Given : AB > CD, OM⊥AB, ON ⊥ CD.
Join OA and OC.

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

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

Question 5.
(a) Factorize : $$\frac{y^{6}}{343}+\frac{343}{y^{6}}$$

(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.
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
P and O are mid-points of sides AD and BD respectively
By mid-point theorem,

(ii) Also, by mid-point theorem

(c) If θ is acute and 3sin θ = 4cos θ, find the value of 4sin2 θ – 3cos2 θ + 2.

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
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.

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.

(c) Simplify: $$\frac{2 \sqrt{3}-\sqrt{5}}{2 \sqrt{2}+3 \sqrt{3}}$$

Question 7.
(a) If x + y = 10 and x2 + y1 = 58, find the value of x3 + y3.
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.
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°.

(c) The mean of 5 numbers is 20. If one number is excluded, mean of the remaining numbers becomes 23. Find the excluded number.
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

(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.
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,

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 :

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?
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,

(b) Construct a regular hexagon of side 2.2 cm.
Each side = 2.2 cm.

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.

(c) Solve graphically : 2x – 3y + 2 = 4x + 1=3x-y + 2
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

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.
(x2 – 5x + 7) (x2 + 5x – 7) = {x2 – (5x – 7)} {x2 + (5x – 7)}
= (x2)2 – (5x – 7)2

(b) Simplify: $$(64)^{2 / 3}-\left(\frac{1}{81}\right)^{-1 / 4}+8^{2 / 3} \cdot\left(\frac{1}{2}\right)^{-1} \cdot 3^{0}$$

(c) In a pentagon ABCDE, BC | | ED and ∠B: ∠A: ∠E = 5:3:4. Find ∠B.

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).
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.

(b) Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm
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

(c) Prove that : $$\frac{1}{1+\tan ^{2} \theta}+\frac{1}{1+\cot ^{2} \theta}=1$$

= cos2 θ + sin2 θ=1
Hence Proved.

## ICSE Class 9 Maths Sample Question Paper 4 with Answers

Section – A
(Attempt all questions from this Section)

Question 1.
(a) Expand : $$\left(\frac{2}{3} x-\frac{3}{2 x}-1\right)^{2}$$

(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.

(c) Factorize : 64x6 – 729y

Question 2.
(a) If a2 + b2 = 7ab, prove that 2 log (a + b) = log 9 + log a + log b.
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: $$\tan ^{2} \theta-\frac{1}{\cos ^{2} \theta}+1=0$$
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

(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.
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.

=> ∠BCA + ∠BCD = ∠ADB + ∠BCD
=> ∠ACD = ∠BDC
In ΔACD and ΔBDC
CD = CD (Common side)
⇒ ∠ACD = ∠BDC (Proved above)
∴ ΔACD ≅ ΔBDC (ASA axiom)
(iii) ∴ ∠A = ∠B. (c.p.c.t.)
Hence Proved.

(b) $$a=\frac{2-\sqrt{5}}{2+\sqrt{5}} \text { and } b=\frac{2+\sqrt{5}}{2-\sqrt{5}}, \text { find } a^{2}-b^{2}$$

(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.

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.
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.

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
Hence Proved.

(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.
Given : Numbers in ascending order : 3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.
Median = 13
Here, n = 20.

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

Question 5.
(a) Factorize : 5x2 + 17xy – 12y2.
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.
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

⇒ 40 +2y = 70
⇒ 2y = 70 – 40
⇒ $$y=\frac{30}{2}=15$$
∴ Father’s age is 40 years and son’s age is 15 years.

(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.
Given: E, F are mid-points of sides AC and AB, respectively.

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 $$a^{3}+\frac{1}{a^{3}}=p\left(p^{2}-3\right).$$

(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.
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

(c) Simplify :
$$\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \left(90^{\circ}-\theta\right)}{\sec \left(90^{\circ}-\theta\right)}-3 \tan ^{2} 30^{\circ}$$

Question 7.
(a) If log10 a = b, express 102b3 in terms of a.

(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.

(c) If cos θ + sec θ = 2, show that cos8 θ + sec8 θ = 2.

Question 8.
(a) If each interior angle is double the exterior angle, find the number of sides.

(b) Solve by the substitution method :
5x + 4y – 4 = 0; x – 20 = 12y.
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 = $$\frac{-3}{2}$$

(c) If x = 2 + √3 , find the value of x – $$\frac{1}{x}$$

Question 9.
(a) Evaluate : $$x^{2 / 3} \cdot y^{-1} \cdot z^{1 / 2}$$ when x – 8, y = 4 and z = 25

(b) Construct a ΔABC is which base AB 5 cm, ∠A = 30° and AC – BC – 2.5 cm.

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: $$\left(2 x-\frac{1}{2 x}\right)^{2}-\left(2 x+\frac{1}{2 x}\right)\left(2 x-\frac{1}{2 x}\right)$$

Question 10.
(a) Simplify: $$\left(a^{m-n}\right)^{m+n} \cdot\left(a^{n-l}\right)^{n+l} \cdot\left(a^{l-m}\right)^{l+m}$$

(b) If area of a semi-circular region is 1232 cm2, find its perimeter.

(c) If x = 15°, evaluate : 8 sin cos 4x. sin 6x.
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°
$$=8 \times \frac{1}{2} \times \frac{1}{2} \times 1=2$$

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?

(b) Draw the graph of the equation 3x – y = 4.

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

## 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
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)

(b) The mean of 6 observations is 17.5. If five of them are 14, 9, 23, 25 and 10, find the sixth observation.
Mean of 6 observations is 17.5
5 observations are 14, 9, 23, 25, 10
Let 6th observation be x

(c) If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ-1.

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.
(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.
Given: log2(x2 – 4) = 5
x2 – 4 =25
x2 =32 + 4
X = ±√36= ±6.

(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.

(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 BC = arc DB. Hence Proved.

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

Question 3.
(a) Prove that: $$\frac{2^{n}+2^{n-1}}{2^{n+1}-2^{n}}=\frac{3}{2}$$

(b) Given that 16 cot A = 12, find the value of $$\frac{\sin A+\cos A}{\sin A-\cos A}$$

(c) If the altitudes from two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.

Question 4.
(a) If a + b + 2c = 0, prove that a3 + b3 + 8c3 – 6abc = 0.
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.
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 ……………

(c) If the sides are in the ratio 5 : 3 : 4, prove that it is a right angled triangle.
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. ?

(b) Show by distance formula that the points A (-1, -1), B (2, 3) and C (8,11) are collinear.
Given points are A (-1, -1), B (2, 3), C (8, 11).
Now

AB + BC = 5 + 10 = 15
⇒ AB + BC = AC
The points are collinear.
Hence Proved.

(c) Factorize : a6 – 26a3 – 27.
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: $$\frac{\left(x^{a+b}\right)^{2} \cdot\left(x^{b+c}\right)^{2} \cdot\left(x^{c+a}\right)^{2}}{\left(x^{a} \cdot x^{b} \cdot x^{c}\right)^{4}}$$

(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°.

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.

(c) If a and b are rational numbers, find the values of a and b :
$$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3}$$

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

(b) Solve:
83x – 67y = 383
67x – 83y = 367.
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

(c) In the following figure, area of parallelogram AFEC is 140 cm2. State, giving reason, the area of (i) parallelogram BFED (ii) ABFD.

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

(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.

(c) Factorize : $$8 x^{3}-\frac{1}{27 y^{3}}$$
Answre:

Question 9.
(a) If $$\frac{x^{2}+1}{x}=2 \frac{1}{2}$$ find the values of $$(i) x-\frac{1}{x} (ii) x^{3}-\frac{1}{x^{3}}$$

(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.

Given : OA = 3.5 cm, OD = 2 cm
Area of shaded region = Area of quadrant AOB – Area of ΔAOD.

= 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.
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

Question 10.
(a) Solve: $$(\sqrt{2})^{2 x+4}=8^{x-6}$$

(b) Construct a rhombus whose diagonals are 5 cm and 6.8 cm.
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.

(c) In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B respectively.
Prove that ∠AOB = $$\frac{1}{2}$$ (∠C + ∠D).
Given, AO and BO are bisectors of ∠A and ∠B respectively.

Question 11.
(a) Find the value of log5√5 (125).

(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.
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

(c) If the area of an equilateral triangle is 81√3 cm2, find its perimeter.
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 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.
Given =

(b) Factorize : 9x2 – 4 (y + 2x)2
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).

(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.
Given : The area enclosed between the circles = 770 cm2
Radius of outer circle (R) = 21 cm.
Let radius of inner circle be r.

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.

(b) If ax = by = cz and b2 = ac, Prove that $$y=\frac{2 x z}{x+z}$$

(c) In the following figure, AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

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.

(b) Find the value of :
2√2 cos 45° cos 60° + 2√3 sin 30° tan 60° – cos 0°
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.

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.

(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

∴ 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).
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.

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 ?

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.
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°.

(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.
(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.

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.

Question 7.
(a) Evaluate without using tables :
$$\left(\frac{\cos 47^{\circ}}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin 72^{\circ}}{\cos 18^{\circ}}\right)^{2}-2 \cos ^{2} 45^{\circ}$$

(b) Solve:
$$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}$$

(c) In ΔABC, ∠ACB = 90°, AB = c unit, BC = a unit,
AC = b unit, C perpendicular to AB and CD = p unit.
Prove that $$\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$$

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) $$x^{2}-\frac{1}{x^{2}}$$

(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.

(c) Find the mean and median of the numbers :
41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.
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

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.
(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

(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

(b) Factorize : 2 – y (7 – 5y).
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.
x – 2y =1 ……….(i)
x + y = 4 ……………(ii)
from (i)

∴ The points are (1, 0), (3, 1), (5, 2)

From (ii)

The points are (4, 0), (3, 1), (2, 2)
These points are plotted on the graph.

The two straight lines intersect at (3, 1)
∴ x=3,y=1

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

(b) $$\text { If } \frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}, \text { prove that }: a^{a} . b^{b} \cdot c^{c}=1 \text { . }$$
$$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=k(\text { say })$$
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.

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.

Question 11.
(a) Simplify: $$\left(a+\frac{1}{a}\right)^{2}-\left(a-\frac{1}{a}\right)^{2}$$

(b) In the given figure, ABCD is a trapezium. Find the values of x and y.

(c) Simplify: $$\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}$$

## 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]

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

(c) In the given figure, AB = $$\frac{1}{2}$$ BC, where BC = 14 cm. Find : [4]
(ii) Area of ΔABC
(iii) Area of semicircle
Hence find the area of shaded region. Use 7π = $$\left(\text { Use } \pi=\frac{22}{7}\right)$$

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]

(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.

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)

(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

Question 3.
(a) Solve following pairs of linear equations using cross-multiplication method : [3]
5x – 3y = 2
4x + 7y = – 3

(b) Without using tables, evaluate : [3]
$$4 \tan 60^{\circ} \sec 30^{\circ}+\frac{\sin 31^{\circ} \sec 59^{\circ}+\cot 59^{\circ} \cot 31^{\circ}}{8 \sin ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}$$

(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

Question 4.
(a) Evaluate : 3 log 2 – $$\frac{1}{3}log 27 + log 12 – log 4 + 3 log 5$$. [3]

(b) If x –$$\frac{1}{x}$$ =3, evaluate x3 – $$\frac{1}{x^{3}}$$[3]

(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.

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]
(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]

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.
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.

(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]
(a) Given : AB = 6.4 cm, AD = 5.2 cm,
Perpendicular distance between AB and DC is 4 cm.

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]
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).

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

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)
∠PAB ≅ ΔQCD (SAS axiom)
PB = QD (c.p.c.t.)
Hence Proved.

Question 7.
(a) Solve for x : sin2 x + cos2 30° = $$\frac{5}{4}$$; where 0° ≤ x ≤ 90° [3]

(b) Evaluate for x :$$\left(\sqrt{\frac{5}{3}}\right)^{x-8}=\left(\frac{27}{125}\right)^{2 x-3}$$ [3]

(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]

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]

Given: ∠ABC = 66°, ∠DAC = 38°, CE ⊥AB, AD ⊥ BC.
In ∠ABD, ∠BAD + ∠ABD = ∠ADC (Exterior angle is equal to sum
of interior opposite angles)
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]
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.

(c) Find (i) mean and (ii) median for the following observations : [4]
10, 47, 3, 9, 17, 27, 4, 48, 12, 15
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

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]
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

(b) In the given figure, arc AB = twice (arc BC) and ∠AOB = 80°. Find : [3]
(i) ∠BOC
(ii) ∠OAC

(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
⇒ ∠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.
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
⇒ $$x=\frac{6-3 y}{2}$$

 X 3 0 -3 y 0 2 4

The points are (3, 0), (0, 2), (- 3, 4).
These points are plotted on the graph.

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.

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.
(a) Given: Each interior angle = 135°
(i) Exterior angle = 180° – 135° = 45°

(iii) The polygon is a regular octagon.

(b) If log 4 = 0.6020, find the value of log 80. [3]
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

(c) Evaluate x and y from the figure diagram. [4]

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.

(a) Given : AB = AC, BC = CD, ∠BAC = 28°

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.

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

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