## 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

= 8x^{2} – 2x – 1

= 8x^{2} – 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.

Answer:

Mean of 6 observations is 17.5

5 observations are 14, 9, 23, 25, 10

Let 6^{th} observation be x

(c) If θ is an acute angle and sin θ = cos θ, find the value of 2 tan^{2} θ + sin^{2} θ-1.

Answer:

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 log_{2} (x^{2} – 4) = 5.

Answer:

Given: log_{2}(x^{2} – 4) = 5

x^{2} – 4 =2^{5}

x^{2} =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.

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.

Question 3.

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

Answer:

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

Answer:

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

Answer:

Question 4.

(a) If a + b + 2c = 0, prove that a^{3} + b^{3} + 8c^{3} – 6abc = 0.

Answer:

Given : a + b + 2c = 0

⇒ a + b = – 2c

Cubing both sides, we get

(a + b)^{3} = (- 2c)^{3 }⇒ a^{3} + b^{3} + 3ab (a + b) = – 8c^{3 }⇒ a^{3} + b^{3} + 3ab (- 2c) = – 8c^{3 }⇒ a^{3} + b^{3} + 8c^{3} – 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 ……………

(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} = 9x^{2} + 16x^{2 }= 25x^{2 }= (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:

(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

AB + BC = 5 + 10 = 15

⇒ AB + BC = AC

The points are collinear.

Hence Proved.

(c) Factorize : a^{6} – 26a^{3} – 27.

Answer:

a^{6} – 26a^{3} – 27 = a^{6} – (27 – 1) a^{3}-27 = a^{6} – 27a^{3} + a^{3} – 27

= a^{3} (a^{3} – 27) + 1 (a^{3} – 27) = {a^{3} – 27) (a^{3} + 1)

= (a^{3} – 3^{3}) (a^{3} + 1^{3})

= (a – 3) (a^{2} + 3a + 9) (a + 1) (a^{2} – 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:

(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°.

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.

(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:

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:

(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

(c) In the following figure, area of parallelogram AFEC is 140 cm^{2}. State, giving reason, the area of (i) parallelogram BFED (ii) ABFD.

Answer:

Given : Area of parallelogram AFEC = 140 cm^{2}.

(i) Area of parallelogram BFED = Area of parallelogram AFEC

( ∵ They are on same base and between same parallels)

= 140 cm^{2}

(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 cm^{2} = 70 cm^{2}.

Question 8.

(a) If log_{10} a = m and log_{10} b = n, express \(\frac{a^{3}}{b^{2}}\) in terms of m and n

Answer:

(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:

(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}}\)

Answer:

(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.

Answer:

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 cm^{2}.

(c) If a + b + c = 9 and ab + be + ca = 40, find the value of a^{2} + b^{2} + c^{2}.

Answer:

Given: a + b + c = 9 ab + bc + ca = 40

We know, (a+b+c)^{2} =a^{2}+ b^{2 }+ c^{2 }+2(ab+bc+ca)

(9)^{2} =a^{2}+ b^{2 }+ c^{2}+ 2 x 40

a^{2}+ b^{2 }+ c^{2 }= 81 – 80 = 1

Question 10.

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

Answer:

(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.

(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.

Question 11.

(a) Find the value of log_{5√5} (125).

Answer:

(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

(c) If the area of an equilateral triangle is 81√3 cm^{2}, find its perimeter.

Answer:

Given : Area of equilateral triangle = 81√3 cm^{2 }Let the length side of each of equilateral triangle be a cm.