## 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.

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

Answer:

(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 wheel

Question 2.

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

Answer:

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

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°

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

Answer:

Question 3.

(a) If θ = 30°, verify that cos 3 θ = 4 cos^{3} θ -3 cos θ .

Answer:

L.H.S. = cos 3θ = cos (3 x 30°) = cos 90° = 0 ………..(1)

L.H.S. = 4cos^{3 }θ -3cosθ= 4 cos^{3} 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.

Answer:

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

Answer:

= 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 a^{2} – 3a – 1 = 0, find the value of a + \(a^{2}+\frac{1}{a^{2}}\)

Answer:

(a)

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

Answer:

(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.

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 OA^{2} = AM^{2} + OM^{2 }(Pythagoras theorem)

In ΔOCN OC^{2} = CN^{2} + ON^{2 }(Pythagoras theorem)

AM^{2} + OM^{2} =CN^{2} +ON^{2}(∵ OA = OC, Radii)

⇒ OM^{2} – ON^{2} = – (AM^{2} – CN^{2})

⇒ OM^{2} – ON^{2} <O (∵AM > CN)

⇒ OM^{2} < ON^{2}

⇒ 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 : \(\frac{y^{6}}{343}+\frac{343}{y^{6}}\)

Answer:

(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,

(ii) Also, by mid-point theorem

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

Answer:

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)

AC^{2} = BC^{2}

(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.

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.

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

Answer:

Question 7.

(a) If x + y = 10 and x^{2} + y^{1} = 58, find the value of x^{3} + y^{3}.

Answer:

Given :

x + y =10, x^{2} + y^{2} = 58.

(x + y)^{2} = x^{2} + y^{2} + 2xy

⇒ 10^{2} = 58 + 2xy

2xy = 100 – 58

⇒ xy = \(\frac{42}{2} \) = 21

x^{3} + y^{3} = (x + y)^{3} – 3xy (x + y)

= 10^{3} – 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°.

(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 3^{X} = (27)^{2x-}^{5}

Answer:

(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,

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 :

Answer:

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,

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

Answer:

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

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

The two lines intersect at the point (2, – 1).

x =2, y = -1

Question 10.

(a) Express (x^{2} -5x + 7) (x^{2} + 5x – 7) as a difference of two squares.

Answer:

(x^{2} – 5x + 7) (x^{2} + 5x – 7) = {x^{2} – (5x – 7)} {x^{2} + (5x – 7)}

= (x^{2})^{2} – (5x – 7)^{2}

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

Answer:

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

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 (p^{2} + q^{2}).

Answer:

p + q = 10, pq = 21.

∴ p^{2} + q^{2} = (p + q)^{2} – 2pq = 10^{2} – 2 x 21 = 100 – 42 = 58

3 (p^{2} + q^{2}) = 3 x 58

= 174.

(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

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

Answer:

= cos^{2} θ + sin^{2} θ=1

Hence Proved.