## 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.

Answer:

Given : AB = 3 cm, BC = 4 cm, ∠ABC = 90°

By Pythagoras theorem,

AC^{2} = AB^{2} + BC^{2} = 3^{2} + 4^{2} = 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.

Answer:

Present value (V_{0}) = ₹ 16,000

Value after 2 year (V_{1}) = ₹ 14,440

∴ n =2

Let r be the rate of depreciation.

(c) Prove that √2 + √5 is irrational.

Answer:

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 + \(\frac{1}{x}\).

Answer:

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

Answer:

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 \(\frac{2}{3} x^{2}-\frac{1}{3} x-1=0\)

Answer:

\(\frac{2}{3} x^{2}-\frac{1}{3} x-1=0\)

\(3 \times \frac{2}{3} x^{2}-3 \times \frac{1}{3} x-3 \times 1=3 \times 0\)

⇒ 2x^{2} – x – 3 = 0

⇒ 2 x 2 – (3 – 2)x -3=0

⇒ 2x^{2} – 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 : a^{3} – b^{3} – a + b.

Answer:

a^{3} -b^{3} – a + b = (a-b) (a^{2} + ab + b^{2}) – (a-b) = (a-b) (a^{2} + ab + b^{2} – 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 |

Answer:

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

Answer:

Question 4.

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

Answer:

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

Answer:

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 log_{10} x + \(\frac{1}{3}\) log_{10} y = 1, express y in terms of x.

Answer:

Given log_{10} x + \(\frac{1}{3}\) log_{10}y = 1

log_{10} x + log_{10} y^{1/3} = log_{10 }10

log_{10} (xy^{1/3}) = log_{10 }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.

Answer:

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.

Answer:

⇒ 169 =144 ÷ CN^{2}

⇒ CN^{2} = (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 \(x^{3}-\frac{1}{x^{3}}\)

Answer:

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 |

Answer:

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.

Answer:

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 \(\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=x-y \sqrt{3}\)

Answer:

Question 7.

(a) Factorize : (x^{2} + y^{2} – z^{2})^{2} – 4x^{2}y^{2}.

Answer:

(x^{2} + y^{2} – z^{2})^{[1]} – 4x^{2}y^{2} = (x^{2} + y^{1} – z^{2})^{2} – (2xy)^{2 }= (x^{2} + y^{2} – z^{2} + 2xy) (x^{2} + y^{2} – z^{2} – 2xy)

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

= {(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.

Answer:

(c) Find the value of x if 3 cot^{2} (x – 5°) = 1.

Answer:

3 cot^{2} (x – 5°) =1

1 cot^{2} (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: \(\frac{x+y}{x y}=2 ; \frac{x-y}{x y}=1\)

Answer:

(b) Construct a parallelogram ABCD with AB = 5.1 cm, BC = 7 cm and ∠ABC = 75°.

Answer:

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.

(6) Join CD and AD.

Hence, ABCD is the required parallelogram.

(c) Calculate the distance between A (7, 3) and B on the X-axis whose abscissa is 11.

Answer:

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.

Answer:

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

Answer:

Given: ∠PSR = 90°, PQ = 10 cm, QS = 6 cm, RQ = 9 cm.

∴ In Δ PQ^{2}, PQ^{2} = PS^{2} + QS^{2} (Pythagoras theorem)

10^{2} =PS^{2}+6^{2}

PS^{2} = 100 – 36

PS = √64 = 8cm

In ΔPRS, PR^{2} = PS^{2} + RS^{2} (Pythagoras theorem)

PR^{2} = 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.

Answer:

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,

AB^{2} = BC^{2} + AC^{2} (Pythagoras theorem, ∠C = 90°)

(2 x 10.5)^{2} = 72 ÷ AC^{2}

AC2 =441 – 49 =392

AC = 19.8 cm.

Area = x BC x AC = x 7 x 19.8 = 69.3 cm^{2}

The area of shaded region = (173.25 – 69.3) cm^{2}

= 103.95 cm^{2}

Question 10.

(a) If a^{2} + b^{2} + c^{2} – ab – be – ca = 0, prove that a = b = c.

Answer:

Given a^{2} + b^{2} + c^{2} – ab – be – ca =0

⇒ 2 (a^{2} + b^{2} + c^{2} – ab – be – ca) = 0

⇒ 2a^{2} + 1b^{2} + 2c^{2} -2ab – 2bc – 2ca = 0

⇒ (a^{2} – 2ab + b^{2}) + (b^{2} – 2be + c^{2}) + (c^{2} – 2ca + a^{2}) = 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

Answer:

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 = 2^{p}.3^{q}.7^{r}, find the values of p, q, r and hence evaluate 2^{p}.3^{q}.7^{-r}÷192.

Answer:

Question 11.

(a) If log \(\frac{x-y}{2}=\frac{1}{2} \)(log x + log y), prove that x^{2} + y^{2} = 6xy.

Answer:

x^{2} + y^{2} – 2xy = 4xy

x^{2} + y^{2} – 4xy + 2 xy

x^{2} + y^{2} = 6 xy.

Hence Proved

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

Answer:

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 a^{3}b – 44 a^{2}b + 112

Answer:

4 a^{3}b – 44 a2b + 112 ab = 4 ab (a^{2} – 11a + 28)

= 4 ab {(a^{2} – (7 + 4) a + 28)}

= 4ab(a^{2} – 7a – 4a+28)

= 4ab {a (a – 7) – 4(a – 7))

= 4ab (a – 7) (a – 4).