## 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.

Answer:

Given: x + y =6,x – y=4.

Now, 4xy =(x+y)^{2}-(x-y)^{2}

= 6^{2} – 4^{2}

=20

xy=\(\frac{20}{4}\)=5.

(b) Find the mean of first 10 prime numbers.

Answer:

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 x^{2} + y^{2} = a^{2} + b^{2}.

Answer:

Question 2.

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

Answer:

(b) The area of a trapezium is 540 cm^{2}. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.

Answer:

Given : Area of trapezium = 510 cm^{2}

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.

Answer:

(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 log_{10} a, log_{10} b and log_{10} c.

Answer:

(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.

Answer:

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 : a^{4} + b^{4} – 11a^{2} b^{2}.

Answer:

Question 4.

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

Answer:

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

Answer:

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.

Answer:

Section – B

(Attempt any four questions from this Section)

Question 5.

(a) Solve for x : 25^{x – 1} = 5^{2x – 1} – 100.

Answer:

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

Answer:

(c) Without using tables, evaluate :

Answer:

Question 6.

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

Answer:

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

Answer:

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.

Answer:

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 \(\frac{x^{2}+x y+y^{2}}{x^{2}-x y+y^{2}}\)

Answer:

(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 ?

Answer:

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.

Answer:

Question 8.

(a) Factorize : 1 – 2ab – (a^{2} + b^{2})

Answer:

(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.

Answer:

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

Construct a combined histogram and frequency polygon.

Answer:

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.

Answer:

A’s share = 1,01,400.

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

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

Answer:

(c) If a = c^{z}, b = a^{x} and c = b^{y}, prove that xyz = 1.

Answer:

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.

Answer:

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.

Answer:

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, log_{10} x = a and log_{10} y =

(i) Write down 10^{a-}^{1} in terms of x

(ii) Write down 10^{2b} in terms of y.

(iii) If log_{10} P = 2a-b, express P in terms of x and y.

Answer:

Question 11.

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

Answer:

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.

Answer:

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) x^{2} – y^{2}.

Answer:

Given x – y – 8, xy = 20.

(i) (x + y)^{2} = (x – y)^{2} + 4 xy = 8^{2} + 4 x 20 = 64 + 80 = 144

x + y = ± √144 = ± 12

x^{2} – y^{2} = (x + y) (x – y)

= (± 12) x 8

= ± 96.