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ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable MCQs

MCQs

Question 1

Which of the following is not a quadratic equation ?

(a) (x+2)²=2(x+3)

(b) x²+3x=(–1) (1–3x)

(c) (x+2)(x–1)=x²–2x–3

(d) x³–x²+2x+1=(x+1)3

Sol :

(a) (x + 2)² = 2(x + 3)

⇒ x² + 4x + 4 = 2x + 6

⇒ x² + 4x – 2x + 4 – 6 = 0

⇒ x² + 2x – 2

It is a quadratic equation.

(b) $x^{2}+3 x=(-1)(1-3 x) $

$\Rightarrow x^{2}+3 x=-1+3 x$

$\Rightarrow x^{2}+1=0$

It is also quadratic equation.

$x^{2}-x+2 x-2=x^{2}-2 x-3$

$x^{2}-x^{2}+x+2 x-2+3=0 \Rightarrow 3 x+1=0$

It is not a quadratic equation.

(d) $x^{3}-x^{2}+2 x+1=(x+1)^{3}$

$=x^{3}+3 x^{2}+3 x+1$

$x^{3}-x^{2}+2 x+1$

$3 x^{2}+x^{2}-2 x-1+3 x+1=0$

$\Rightarrow 4 x^{2}+x=0$

It is a quadratic equation

Ans : (c)

Question 2

Which of the following is a quadratic equation ?

(a) (x – 2) (x + 1) = (x – 1) (x – 3)

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

(c) $x^{2}+3 x+1=(x-2)^{2}$

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

Sol :

(a) (x – 2) (x + 1) = (x – 1) (x – 3)
⇒ x2 + x – 2x – 2 = x2 – 3x – x + 3
⇒ 3x + x – 2x + x = 3 + 2
⇒ 3x = 5
It is not a quadratic equation.

(b) $(x+2)^{3}=2 x\left(x^{2}-1\right)$

$x^{3}+6 x^{2}+12 x+8=2 x^{3}-2 x$

$x^{3}+6 x^{2}+12 x+8-2 x^{3}+2 x=0$

$-x^{3}+6 x^{2}+14 x+8=0$

It is not a quadratic equation.

$x^{2}+3 x+1=x^{2}-4 x+4$

⇒3x+1+4 x-4=0

⇒7x-3=0

It is not a quadratic equation.

(d) $8(x-2)^{3}=(2 x-1)^{3}+3$

$8\left(x^{3}-6 x^{2}+12 x-8\right)$

$=8 x^{3}-12 x^{2}+6 x-1+3$

$8 x^{3}-48 x^{2}+96 x-64-8 x^{3}+12 x^{2}-6 x+1-3=0$

$-36 x^{2}+90 x-66=0$

It is a quadratic equation

Ans : (d)

Question 3

Which of the following equations has 2 as a root ?

(a) $x^{2}-4 x+5=0$

(b) $x^{2}+3 x-12=0$

(c) $2 x^{2}-7 x+6=0$

(d) $3 x^{2}-6 x-2=0$

Sol :

(a) $x^{2}-4 x+5=0$

$\Rightarrow(2)^{2}-4 x^{2}+5=0$

⇒ 4 – 8 + 5 = 0

⇒ 9 – 8 ≠ 0

2 is not its root.

(b) $x^{2}+3 x-12=0$

$ \Rightarrow(2)^{2}-3 \times 2-12=0$

⇒4-6-12=4-18=-14

∴ 2 is not its roots.

$ \Rightarrow 2(2)^{2}-7 \times 2+6=0$

⇒ 8-14+6=0 $

⇒0=0

∴ 2 is its root

(d) $3 x^{2}-6 x-2=0$

$ \Rightarrow 3(2)^{2}-6 \times 2-2=0$

⇒12-12-2=0

⇒12-14=0

∴ 2 is not its root.

Ans : (c)

Question 4

If $\frac{1}{2}$ is a root of the equation $x^{2}+k x-\frac{5}{4}=0$ then the value of k is

(a) 2

(b) – 2

(c) $\frac{1}{4}$

(d) $\frac{1}{2}$

Sol :

$\frac{1}{2}$ is a root of the equation

$x^{2}+k x-\frac{5}{4}=0$

Substituting the value of $x=\frac{1}{2}$ in the equation

$\left(\frac{1}{2}\right)^{2}+k \times \frac{1}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0$

$\Rightarrow \frac{k}{2}-1=0$

⇒k=1×2=2

∴k=2

Ans (a)

Question 5

If $\frac{1}{2}$ is a root of the quadratic equation $4 x^{2}-4 k x+k+5=0$ then the value of k is

(a) – 6

(b) – 3

(c) 3

(d) 6

Sol :

$\frac{1}{2}$ is a root of the equation

$4 x^{2}-4 k x+k+5=0$

Substituting the value of $x=\frac{1}{2}$ in the equation

$4\left(\frac{1}{2}\right)^{2}-4 \times k \times \frac{1}{2}+k+5=0$

1-2k+k+5=0

-k+6=0

k=6

Ans (d)

Question 6

The roots of the equation $x^{2}-3 x-10=0$ are

(a) 2,- 5

(b) – 2, 5

(c) 2, 5

(d) – 2, – 5

Sol :

$x=\frac{-(-3) \pm \sqrt{(-3)^{2}-4 \times 1 \times(-10)}}{2 \times 1}$

$=\frac{3 \pm \sqrt{9+40}}{2}=\frac{3 \pm \sqrt{49}}{2}=\frac{3+7}{2}$

∴$x=\frac{3+7}{2}=5$ and $x=\frac{3-7}{2}=\frac{-4}{2}=-2$

x = 5, – 2 or – 2, 5

Ans (b)

Question 7

If one root of a quadratic equation with rational coefficients is $\frac{3-\sqrt{5}}{2}$, then the other

(a) $\frac{-3-\sqrt{5}}{2}$

(b) $\frac{-3+\sqrt{5}}{2}$

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

(d) $\frac{\sqrt{3}+5}{2}$

Sol :

One root of a quadratic equation is $\frac{3-\sqrt{5}}{2}$ then other root will be $\frac{3+\sqrt{5}}{2}$

Ans (c)

Question 8

If the equation $2 x^{2}-5 x+(k+3)=0$ has equal roots then the value of k is

(a) $\frac{g}{8}$

(b) $-\frac{g}{8}$

(c) $\frac{1}{8}$

(d) $-\frac{1}{8}$

Sol :

$2 x^{2}-5 x+(k+3)=0$

a=2, b=-5, c=k+3

=25-8(k+3)

∴ Roots are equal.

$\therefore b^{2}-4 a c=0$

∴ 25-8(k+3)=0

⇒25-8k-24=0

⇒1-8k=0

⇒8 k=1

$\therefore k=\frac{1}{8}$

Ans (c)

Question 9

The value(s) of k for which the quadratic equation $2 x^{2}-k x+k=0$ has equal roots is (are)

(a) 0 only

(b) 4

(c) 8 only

(d) 0, 8

Sol :

$2 x^{2}-k x+k=0$

a=2, b=-k, c=k

$\therefore b^{2}-4 a c=(-k)^{2}-4 \times 2 \times k$

$\quad=k^{2}-8 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$k^{2}-8 k=0$

⇒k(k-8)=0$

Either k=0

or k-8=0, then k=8

k=0,8

Ans (d)

Question 10

If the equation $3 x^{2}-k x+2 k=0$ roots, then the the value(s) of k is (are)

(a) 6

(b) 0 Only

(c) 24 only

(d) 0

Sol :

$3 x^{2}-k x+2 k=0$

Here, a=3, b=-k, c=2 k

$b^{2}-4 a c=(-k)^{2}-4 \times 3 \times 2 k$

$=k^{2}-24 k$

∴ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore k^{2}-24 k=0$

⇒k(k-24)=0

Either k=0

or k-24=0, then k=24

∴ k=0, 24

Ans (d)

Question 11

If the equation $\{k+1\} x^{2}-2(k-1) x+1=0$ has equal roots, then the values of k are

(a) 1, 3

(b) 0, 3

(c) 0, 1

(d) 0, 1

Sol :

(k + 1)x² – 2(k – 1)x + 1 = 0

Here, a = k + 1, b = -2(k – 1), c = 1

$\therefore b^{2}-4 a c=[-2(k-1)]^{2}-4(k+1)(1)$

$\quad=4\left(k^{2}-2 k+1\right)-4 k-4$

$\quad=4 k^{2}-8 k+4-4 k-4$

$\quad=4 k^{2}-12 k$

∵ Roots are equal. $\therefore b^{2}-4 a c=0$

$\therefore 4 k^{2}-12 k=0$

⇒4 k(k-3)=0

⇒ k(k-3)=0

Either k=0 or k-3=0, then k=3

k=0,3(b)

Question 12

If the equation 2x² – 6x + p = 0 has real and different roots, then the values ofp are given by

(a) $p<\frac{9}{2}$

(b)p $\leq \frac{9}{2}$

(d) $p \geq \frac{9}{2}$

Sol :

2x² – 6x + p = 0

Here, a = 2, b = -6, c = p

$b^{2}-4 a c=(-6)^{2}-4 \times 2 \times p$

=36-8 p

∵ Roots are real and unequal.

$\therefore b^{2}-4 a c>0$

⇒ 36-8p>0

⇒ 36>8p

⇒ $\frac{36}{8}>p$

⇒ $p<\frac{36}{8}$

⇒ $p<\frac{9}{2}$

Ans (a)

Question 13

The quadratic equation 2x² – √5x + 1 = 0 has

(a) two distinct real roots

(b) two equal real roots

(c) no real roots

(d) more than two real roots

Sol :

2x² – √5x + 1 = 0

Here, a = 2, b = -√5, c = 1

$b^{2}-4 a c=(-\sqrt{5})^{2}-4 \times 2 \times 1$

=5-8=-3

$\because b^{2}-4 a c<0$

∵ It has no real roots.

Question 14

Which of the following equations has two distinct real roots ?

(a) $2 x^{2}-3 \sqrt{2 x}+\frac{9}{4}=0$

(b) $x^{2}+x-5=0$

(c) $x^{2}+3 x+2 \sqrt{2}=0$

(d) $5 x^{2}-3 x+1=0$

Sol :

(a) $2 x^{2}-3 \sqrt{2} x+\frac{9}{4}=0$

$b^{2}-4 a c=(-3 \sqrt{2})^{2}-4 \times 2 \times \frac{9}{4}=18-18=0$

∵ Roots are real and equal.

(b) $x^{2}+x-5=0$

$b^{2}-4 a c=(1)^{2}-4 \times 1 \times(-5)$

$=1+20=\sqrt{21}>0$

Roots are real and distinct.

Ans (b)

Question 15

Which of the following equations has no real roots ?

(a) x² – 4x + 3√2 = 0

(b) x² + 4x – 3√2 = 0

(c) x² – 4x – 3√2 = 0

(d) 3x² + 4√3x + 4 = 0

Sol :

(a) x² – 4x + 3√2 = 0

b² – 4ac = ( -4)² – 4 × 1 × 3√2

= 16 – 12√2

= 16 – 12(1.4)

= 16 – 16.8

= -0.8

b² – 4ac < 0

Roots are not real.

Ans (a)

ML AGGARWAL CLASS 10 Chapter 1 GST MCQs

MCQs

Page-17

A retailer purchases a fan for Rs 1500 from a wholesaler and sells t to a consumer at 10%profit if the sales are intra-state and the rate of GST is 12%, then choose the correct answer from the given four options for questions 1 to 6:

OPEN IN YOUTUBE

Question 1

The selling price of the fan by the retailer (excluding tax) is?

(a) Rs.1500

(b) Rs.1650

(c) Rs. 1848

(d) Rs. 1800

Sol :

Cost price of fan for retailer = ₹ 1500

Given profit% = 10%

∴ Selling price of fan by the retailer

= C.P. + 10% of C.P.

$=\left(1500+\frac{10}{100}\times 1500\right)=1650$

OPEN IN YOUTUBE

Question 2

The selling price of the fan including tax (under GST) by retailer is?

(a) Rs.1650

(b) Rs.1800

(c) Rs.1848

(d) Rs.1830

Sol :

Given, GST (rate) = 12%

∴GST=12% of S.P

$=\left(\frac{12}{100}\times 1650\right)=198$

Thus, the required selling price of fan including tax by the retailer (under GST)

= S.P. + GST = ₹ (1650 + 198) = ₹ 1848

OPEN IN YOUTUBE

Question 3

The tax (under GST) paid by the wholesaler to the Central Government is ?

(a)Rs 90

(b)Rs 9

(c)Rs 99

(d) Rs 180

Sol :

The tax (under GST) paid by wholesaler to Central Government

=6% of 1500

$=\left(\frac{6}{100}\times 1500\right)$

=90

[SGST – CGST = ½ × rate of GST – 6%]

OPEN IN YOUTUBE

Question 4

The tax (under GST) paid by the retailer to the State Government ?

(a) Rs 99

(b) Rs 9

(c)Rs 18

(d)Rs 198
Sol :
Amount of input SGST of the retailer = 6% of ₹ 1500

$=\left(\frac{6}{100}\times 1500\right)$

=90

Since, the retailer sells the article to the consumer at 10% profit,

S.P of article $=\left(1500+\frac{10}{100}\times 1500\right)$

=1650

Amount of output SGST of the retailer=6% of 1650

$=\left(\frac{6}{100}\times 1650\right)$

=99

Amount of tax (under GST) paid by retailer to State Government = Output SGST – Input SGST

= (99 – 9) = ₹9

Page-18

OPEN IN YOUTUBE

Question 5

The tax (under GST) received by the Central Government is ?

(a)Rs 18

(b) Rs 198

(c) Rs 90

(d)Rs 99

Sol :

Amount of CGST paid by the retailer = Output CGST – Input CGST

= ₹ (99-90) = ₹9

Thus, amount of tax (under GST) received by Central Government

= CGST paid by distributor + CGST paid by retailer

= (6% of ₹1500) + 9

= 90 + 9 = ₹99

OPEN IN YOUTUBE

Question 6

The cost of the fan to the consumer inclusive of tax is ?

(a)Rs 1650

(b) Rs 1800

(c)Rs 1830

(4) Rs 1848

Sol :

Here, selling price of fan = ₹1650

GST on fan = 12% of ₹1650

$=\left(\frac{12}{100}\times 1650\right)$

=198

Thus, cost of a fan to the consumer inclusive of tax

= ₹ (1650+198) = ₹1848

A shopkeeper bought a TV from a distributor at a discount of 25% of the listed price of Rs 32000. The shopkeeper sells the TV to a consumer at the listed price. if the sales are intra-state and the rate of GST la 18%, then choose the correct er from the given four options for questions 7 to 11:

OPEN IN YOUTUBE

Question 7

The selling price of the TV including tax (under GST?) by the distributor is?

(a) Rs 32000

(b) Rs 24000

(c) Rs 28320

(4) Rs 26160

Sol :

It is case of intra state

Discount = 32000 x 25/100

= 8000

SP= 32000-8000

=24000

SP with GST by distributor

24000 + 24000 x 18/100

=28320

Hence option (c) is correct

OPEN IN YOUTUBE

Question 8

The tax (under GST) paid by the distributor to the State Government is

(a) Rs 4320

(b) Rs 2160

(c) Rs 2880

(d) Rs 720

Sol :

Tax (under GST) paid by distributor to the State Government

= SGST = 9% of ₹24000

$=\left(\frac{9}{100}\times 24000\right)$

=2160

OPEN IN YOUTUBE

Question 9

The tax (under GST) paid by the shopkeeper to the Central Government is?

(a)Rs 720

(b)Rs 1440

(c) Rs 2880

(4)Rs 2160

Sol :

Amount of input CGST by the shopkeeper,

CGST = ₹2160, SGST = ₹2160

Amount of GST collected by the shopkeeper or paid by the consumer

= 18% of ₹32000

$=\left(\frac{18}{100}\times 32000\right)$=5760

SGST = 5760/2 = ₹2880 = CGST

Amount of CGST paid by shopkeeper to Central Government = Output CGST – Input CGST

= 2880-2160 = ₹720

OPEN IN YOUTUBE

Question 10

The tax (under GST) received by the State Government is?

(a)Rs 5760

(b) Rs 4320

(c) Rs 1440

(d) Rs 2880

Sol :

Amount of SGST paid by Shopkeeper to state government = ₹ 720

∴ Total tax (under GST) received by State Government = ₹2160 +₹720 = ₹2880

OPEN IN YOUTUBE

Question 11

The price including tax (under GST) of the TV paid by the consumer is

(a)Rs 28320

(b) Rs 37760

(c) Rs 34880

(4) Rs 32000

Sol :

Consumer buy on list price 32000

It is a case of intra-state transaction of goods and services.

SGST = CGST = ½ GST

Given:

The price inclusive of tax (under GST) which the consumer pays for the TV.

CP of an article for shopkeeper = ₹32000

SP of article = ₹32000 + 18% of ₹32000

= ₹32000 + (18/100) × 32000

= ₹32000 + 5760

= 37760

Hence option (b) is correct

ML Aggarwal Solution Class 10 Chapter 22 Probability MCQs

MCQs

Question 1

Which of the following cannot be the probability of an event?

(a) 0.7

(b) $\frac{2}{3}$

(c) – 1.5

(d) 15%

Sol :

– 1.5 (negative) can not be a probability as a probability is possible 0 to 1. (c)

Question 2

If the probability of an event is p, then the probability of its complementary event will be

(a) p – 1

(b) p

(c) 1 – p

(d) $1-\frac{1}{p}$

Sol :

Complementary of p is 1 – p

Probability of complementary even of p is 1 – p. (c)

Question 3

Out of one digit prime numbers, one selecting an even number is

(a) $\frac{1}{2}$

(b) $\frac{1}{4}$

(c) $\frac{4}{9}$

(d) $\frac{2}{5}$

Sol :

One digit prime numbers are 2, 3, 5, 7 = 4

Probability of an even prime number (i.e , 2)

$=\frac{1}{4}$

Ans (b)

Question 4

Out of vowels, of the English alphabet, one letter is selected at random. The probability of selecting ‘e’ is

(a) $\frac{1}{26}$

(b) $\frac{5}{26}$

(c) $\frac{1}{4}$

(d) $\frac{1}{5}$

Sol :

Vowels of English alphabet are a, e, i, o, u = 4

One letter is selected at random.

The probability of selecting ’e’ $=\frac{1}{5}$

Ans (d)

Question 5

When a die is thrown, the probability of getting an odd number less than 3 is

(a) $\frac{1}{6}$

(b) $\frac{1}{3}$

(c) $\frac{1}{2}$

(d) 0

Sol :

A die is thrown

Total number of events = 6

Odd number less than 3 is 1 = 1

Probability $=\frac{1}{6}$

Ans (a)

Question 6

A fair die is thrown once. The probability of getting an even prime number is

(a) $\frac{1}{6}$

(b) $\frac{2}{3}$

(c) $\frac{1}{3}$

(d) $\frac{1}{2}$

Sol :

A fair die is thrown once

Total number of outcomes = 6

Prime numbers = 2, 3, 5 and even prime is 2

Probability of getting an even prime number

$=\frac{1}{6}$

Ans (a)

Question 7

A fair die is thrown once. The probability of getting a composite number is

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{\dot{2}}{3}$

(d) 0

Sol :

A fair die is thrown once

Total number of outcomes = 6

Composite numbers are 4, 6 = 2

Probability$=\frac{2}{6}=\frac{1}{3}$

Ans (a)

Question 8

If a fair dice is rolled once, then the probability of getting an even number or a number greater than 4 is

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{5}{6}$

(d) $\frac{2}{3}$

Sol :

A fair dice is thrown once.

Total number of outcomes = 6

Even numbers or a number greater than 4 = 2, 4, 5, 6 = 4

Probability$=\frac{4}{6}=\frac{2}{3}$

Ans (d)

Question 9

Rashmi has a die whose six faces show the letters as given below :

If she throws the die once, then the probability of getting C is

(a) $\frac{1}{3}$

(b) $\frac{1}{4}$

(c) $\frac{1}{5}$

(d) $\frac{1}{6}$

Sol :

A die having 6 faces bearing letters A, B, C, D, A, C

Total number of outcomes = 4

Probability of getting C

$=\frac{2}{6}=\frac{1}{3}$

Ans (a)

Question 10

If a letter is chosen at random from the letters of English alphabet, then the probability that it is a letter of the word ‘DELHI’ is

(a) $\frac{1}{5}$

(b) $\frac{1}{26}$

(c) $\frac{5}{26}$

(d) $\frac{21}{26}$

Sol :

Total number of English alphabets = 26

Letter of Delhi = D, E, L, H, I. = 5

Probability$=\frac{5}{26}$

Ans (c)

Question 11

A card is drawn from a well-shuffled pack of 52 playing cards. The event E is that the card drawn is not a face card. The number of outcomes favourable to the event E is

(a) 51

(b) 40

(c) 36

(d) 12

Sol :

Number of playing cards = 52

Probability of a card which is not a face card = (52 – 12) = 40

Number of possible events = 40 (b)

Question 12

A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is

(a) 4

(b) 13

(c) 48

(d) 51

Sol :

Total number of cards = 52

Balance 52 – 1 = 51

Number of possible events = 51 (d)

Question 13

If one card is drawn from a well-shuffled pack of 52 cards, the probability of getting an ace is

(a) $\frac{1}{52}$

(b) $\frac{4}{13}$

(d) $\frac{1}{13}$

Sol :

Total number of cards = 52

Number of aces = 4

Probability of card being an ace

$=\frac{4}{52}=\frac{1}{13}$

Ans (d)

Question 14

A card is selected at random from a well- shuffled deck of 52 cards. The probability of its being a face card is

(a) $\frac{3}{13}$

(b) $\frac{4}{13}$

(c) $\frac{6}{13}$

(d) $\frac{9}{13}$

Sol :

Total number of cards = 52

No. of face cards = 3 × 4 = 12

∴ Probability of face card $=\frac{12}{52}=\frac{3}{13}$

Ans (a)

Question 15

A card is selected at random from a pack of 52 cards. The probability of its being a red face card is

(a) $\frac{3}{26}$

(b) $\frac{3}{13}$

(c) $\frac{2}{13}$

(d) $\frac{1}{2}$

Sol :

Total number of card = 52

No. of red face card = 3 × 2 = 6

∴ Probability $=\frac{6}{52}=\frac{3}{26}$

Ans (a)

Question 16

If a card is drawn from a well-shuffled pack of 52 playing cards, then the probability of this card being a king or a jack is

(a) $\frac{1}{26}$

(b) $\frac{1}{13}$

(c) $\frac{2}{13}$

(d) $\frac{4}{13}$

Sol :

Total number of cards 52

Number of a king or a jack = 4 + 4 = 8

∴ Probability $=\frac{8}{52}=\frac{2}{13}$

Ans (c)

Question 17

The probability that a non-leap year selected at random has 53 Sundays is.

(a) $\frac{1}{365}$

(b) $\frac{2}{365}$

(c) $\frac{2}{7}$

(d) $\frac{1}{7}$

Sol :

Number of a non-leap year 365

Number of Sundays = 53

In a leap year, there are 52 weeks or 364 days

One days is left

Now we have to find the probability of a Sunday out of remaining 1 day

∴ Probability$=\frac{1}{7}$

Ans (d)

Question 18

A bag contains 3 red balk, 5 white balls and 7 black balls. The probability that a ball drawn from the bag at random will be neither red nor black is

(a) $\frac{1}{5}$

(b) $\frac{1}{3}$

(c) $\frac{7}{15}$

(d) $\frac{8}{1}$

Sol :

In a bag, there are

3 red balls + 5 white balls + 7 black balls

Total number of balls = 15

One ball is drawn at random which is neither

red not black

Number of outcomes = 5

Probability$=\frac{5}{15}=\frac{1}{3}$

Ans (b)

Question 19

A bag contains 4 red balls and 5 green balls. One ball is drawn at random from the bag. The probability of getting either a red ball or a green ball is

(a) $\frac{4}{9}$

(b) $\frac{5}{9}$

(c) 0

(d) 1

Sol :

In a bag, there are

4 red balls + 5 green balls

Total 4 + 5 = 9

One ball is drawn at random

Probability of either a red or a green ball

$=\frac{9}{9}=1$

Ans (d)

Question 20

A bag contains 5 red, 4 white and 3 black balls. If a. ball is drawn from the bag at random, then the probability of the ball being not black is

(a) $\frac{5}{12}$

(b) $\frac{1}{3}$

(c) $\frac{3}{4}$

(d) $\frac{1}{4}$

Sol :

In a bag, there are

5 red + 4 white + 3 black balls = 12

One ball is drawn at random

Probability of a ball not black$=\frac{5+4}{12}=\frac{9}{12}=\frac{3}{4}$

Ans (c)

Question 21

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

(a) $\frac{1}{5}$

(b) $\frac{3}{5}$

(c) $\frac{4}{5}$

(d) $\frac{1}{3}$

Sol :

There are t to 40 = 40 tickets in a bag

No. of tickets which is multiple of 5 = 8

(5, 10, 15, 20, 25, 30, 35, 40)

Probability$=\frac{8}{40}=\frac{1}{5}$

Ans (a)

Question 22

If a number is randomly chosen from the numbers 1,2,3,4, …, 25, then the probability of the number to be prime is

(a) $\frac{7}{25}$

(b) $\frac{9}{25}$

(c) $\frac{11}{25}$

(d) $\frac{13}{25}$

Sol :

There are 25 number bearing numbers 1, 2, 3,…,25

Prime numbers are 2, 3, 5, 7, 11, 13, 17 19, 23 = 9

Probability being a prime number$=\frac{9}{25}$

Ans (b)

Question 23

A box contains 90 cards numbered 1 to 90. If one card is drawn from the box at random, then the probability that the number on the card is a perfect square is

(a) $\frac{1}{10}$

(b) $\frac{9}{100}$

(c) $\frac{1}{9}$

(d) $\frac{1}{100}$

Sol :

In a box, there are

90 cards bearing numbers 1 to 90

Perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81 = 9

Probability of being a perfect square$=\frac{9}{90}=\frac{1}{10}$

Ans (a)

Question 24

If a (fair) coin is tossed twice, then the probability of getting two heads is

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{3}{4}$

(d) 0

Sol :

A coin is tossed twice

Number of outcomes = 2 x 2 = 4

Probability of getting two heads (HH = 1)

$=\frac{1}{4}$

Ans (a)

Question 25

If two coins are tossed simultaneously, then the probability of getting atleast one head is

(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) $\frac{3}{4}$

(d) 1

Sol :

Two coins are tossed

Total outcomes = 2 × 2 = 4

Probability of getting atleast one head (HT,TH,H,H)

$=\frac{3}{4}$

Ans (c)

Question 26

Lakshmi tosses two coins simultaneously. The probability that she gets almost one head

(a) 1

(b) $\frac{3}{4}$

(c) $\frac{1}{2}$

(d) $\frac{1}{7}$

Sol :

Two coins are tossed

Total number of outcomes = 2 × 2 = 4

Probability of getting atleast one head = (HT, TH, RH = 3)

$=\frac{3}{4}$

Ans (b)

Question 27

The probability of getting a bad egg in a lot of 400 eggs is 0.035. The number of bad eggs in the lot is

(a) 7

(b) 14

(c) 21

(d) 28

Sol :

Total number of eggs 400

Probability of getting a bad egg = 0.035

Number of bad eggs = 0.035 of 400

$=400 \times \frac{35}{1000}=14$

Ans (b)

Question 28

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets she has bought?

(a) 40

(b) 240

(c) 480

(d) 750

Sol :

For a girl,

Winning a first prize = 0.08

Number of total tickets = 6000

Number of tickets she bought = 0.08 of 6000

$=6000 \times \frac{8}{100}=480$

Ans (c)

ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency MCQs

MCQs

Question 1

If the classes of a frequency distribution are 1-10, 11-20, 21-30, …, 51-60, then the size of each class is

(a) 9

(b) 10

(c) 11

(d) 5.5

Sol :

In the classes 1-10, 11-20, 21-30, …, 51-60,

the size of each class is 10. (b)

Question 2

If the classes of a frequency distribution are 1-10, 11-20, 21-30,…, 61-70, then the upper limit of the class 11-20 is

(a) 20

(b) 21

(c) 19.5

(d) 20.5

Sol :

In the classes of distribution, 1-10, 11-20, 21-30, …, 61-70,

upper limit of 11-20 is 20-5 as the classes after adjustment are

0.5-10.5, 10.5-20.5, 20.5-30.5, … (d)

Question 3

If the class marks of a continuous frequency distribution are 22, 30, 38, 46, 54, 62, then the class corresponding to the class mark 46 is

(a) 41.5-49.5

(b) 42-50

(c) 41-49

(d) 41-50

Sol :

The class marks of distribution are 22, 30, 38, 46, 54, 62,

then classes corresponding to these class marks 46 is

46.4 – 4 = 42, 46 + 4 = 50

(Class intervals is 8 as 30 – 22 = 8, 38 – 30 = 8

i.e:, 42 – 50 (b)

Question 4

If the mean of the following distribution is 2.6,

x_i	1	2	p	4	5
f_i	3	3	1	1	2

then the value of P is

(a) 2

(b) 3

(c) 2.6

(d) 2.8

Sol :

Mean = 2.6

x_i	1	2	p	4	5	Total
f_i	3	3	1	1	2	10
f_ix_i	3	6	p	4	10	23+p

Mean$=\frac{\sum f_{i} x_{i}}{\sum f}=\frac{23+p}{10}$

=2.6

$23+p=2 \cdot 6 \times 10=26 $

$\Rightarrow p=26-23=3$

Ans (b)

Question 5

The measure of central tendency of statistical data which takes into account all the data is

(a) mean

(b) median

(c) mode

(d) range

Sol :

A measure of central tendency of statistical data is mean. (a)

Question 6

In a grouped frequency distribution, the mid-values of the classes are used to measure which of the following central tendency?

(a) median

(b) mode

(c) mean

(d) all of these

Sol :

In a grouped frequency distribution,

the mid-values of the classes are used to measure Mean (c)

Question 7

In the formula: $\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$ for finding the mean of the grouped data, d’is are deviations from a (assumed mean) of

(a) lower limits of the classes

(b) upper limits of the classes

(c) mid-points of the classes

(d) frequencies of the classes

Sol :

The formula $\bar{x}=a+\frac{\sum f_{i} d_{i}}{\sum f_{i}}$ is the finding of mean of the grouped data, d’is are mid-points of the classes

Question 8

In the formula: $\bar{x}=a+c\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)$, for finding the mean of grouped frequency distribution $\text{u}_{i}$

(a) $\frac{y_{i}+a}{c}$

(b) $c\left(y_{i}-a\right)$

(c) $\frac{y_{i}-a}{c}$

(d) $\frac{a-y_{i}}{c}$

Sol :

In $\bar{x}=a+c\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right)$

for finding the mean of grouped frequency, $\mathrm{u}_{\mathrm{i}}$ is $\frac{y_{i}-a}$

Ans (c)

Question 9

While computing mean of grouped data, we assumed that the frequencies are

(a) evenly distributed over all the classes

(b) centred at the class marks of the classes

(c) centred at the upper limits of the classes

(d) centred at the lower limits of the classes

Sol :

For computing mean of grouped data,

we assumed that frequencies are centred at class marks of the classes.

Ans (b)

Question 10

Construction of a cumulative frequency distribution table is useful in determining the

(a) mean

(b) median

(c) mode

(d) all the three measures

Sol :

Construction of a cumulative frequency distribution table

is used for determining the median,

Ans (b)

Question 11

The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below:

Class	13.8-14	14-14.2	14.2-14-4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	5	71	48	20

The number of athletes who completed the race in less than 14.6 seconds is

(a) 11

(b) 71

(c) 82

(d) 130

Sol :

Time taken in seconds by 150 athletes to run a 110 m hurdle race as given in the sum,

the number of athletes who completed the race in less then 14.6 second is

2 + 4 + 5 + 71 = 82 athletes.

Ans (c)

Question 12

Consider the following frequency distribution:

Class	0-5	6-11	12-17	18-23	24-29
Frequency	13	10	15	8	11

The upper limit of the median class is

(a) 17

(b) 17.5

(c) 18

(d) 18.5

Sol :

From the given frequency upper limit of median class is 17.5

as total frequencies 13 + 10 + 15 + 8 + 11 = 57

$\frac{57+1}{2}=\frac{58}{2}=29$

and 13 + 10 + 15 = 28 where class is 12-17

But actual class will be 11.5-17.5

Upper limit is 17.5

Ans (b)

Question 13

Daily wages of a factory workers are recorded as:

Daily wages(in ₹)	131-136	137-142	143-148	149-154	155-160
No. of workers	5	27	20	18	12

The lower limit of the modal class is

(a) Rs 137

(b) Rs 143

(c) Rs 136.5

(d) Rs 142.5

Sol :

In the daily wages of workers of a factory are 131-136, 137-142, 142-148, …

which are not a proper class

So, proper class will be 130.5-136.5, 136.5-142.5, 142.5-148.5, …

Lower limit of a model class is 136.5 as 136.5-142.5 is the modal class. (c)

Question 14

For the following distribution:

Class	0-5	5-10	10-15	15-20	20-25
Frequency	10	15	12	20	9

The sum of lower limits of the median class and modal class is

(a) 15

(b) 25

(c) 30

(d) 35

Sol :

From the given distribution

Sum of frequencies = 10 + 15 + 12 + 20 + 9 = 66

and median is $\frac{66}{2}=33$

Median class will be 10-15 and modal class is 15-20

Sum of lower limits = 10 + 15 = 25 (b)

Question 15

Consider the following data:

Class	65-85	85-105	105-125	125-145	145-165	14
Frequency	4	5	13	20	14	14

The difference of the upper limit of the median class and the lower limit of the modal class is

(a) 0

(b) 19

(c) 20

(d) 38

Sol :

From the given data

Total frequencies = 4 + 5 + 13 + 20 + 14 + 7 + 4 = 67

Median class $\frac{67+1}{2}=34$

which is (4 + 5 + 13 + 20) 125-145 and modal class is 125-145

Difference of upper limit of median class and the lower limit of the modal class

= 145 – 125 = 20 (c)

Question 16

An ogive curve is used to determine

(a) range

(b) mean

(c) mode

(d) median

Sol :

An ogive curve is used to find median.

Ans (d)

ML Aggarwal Solution Class 10 Chapter 20 Heights and Distances MCQs

MCQs

Question 1

In the given figure, the length of BC is

(a) 2 √3 cm

(b) 3 √3 km

(c) 4 √3 cm

(d) 3 cm

Sol :

In the given figure, $\frac{B C}{A C}=\sin 30^{\circ}$

$\Rightarrow \frac{B C}{6}=\frac{1}{2}$

$\Rightarrow \mathrm{BC}=\frac{6}{2}$

=3 cm

Ans (d)

Question 2

In the given figure, if the angle of elevation is 60° and the distance AB = 10 √3 m, then the height of the tower is

(a) 20 √3 cm

(b) 10 m

(c) 30 m

(d) 30 √3 m

Sol :

In the given figure,

∠A = 60°, AB = 10 √3 m

Let BC = h

$\tan 60^{\circ}=\frac{h}{10 \sqrt{3}}$

$\Rightarrow \sqrt{3}=\frac{h}{10 \sqrt{3}}$

$\Rightarrow h=10 \sqrt{3} \times \sqrt{3}=10 \times 3=30 \mathrm{~m}$

Ans (c)

Question 3

If a kite is flying at a height of 40 √3 metres from the level-ground, attached to a string inclined at 60° to the horizontal, then the length of the string is

(a) 80 m

(b) 60 √3 m

(c) 80 √3 m

(d) 120 m

Sol :

Let K is kite

Height of KT = 40 √3 m

Angle of elevation of string at the ground = 60°

Let length of string AK = x m

Now $\sin 60^{\circ}=\frac{\mathrm{KT}}{\mathrm{AK}}=\frac{40 \sqrt{3}}{x}$

$\frac{\sqrt{3}}{2}=\frac{40 \sqrt{3}}{x}$

$x=\frac{40 \sqrt{3} \times 2}{\sqrt{3}}=80 \mathrm{~m}$

$\therefore$ Length of string $=80 \mathrm{~m}$

Ans (a)

Question 4

The top of a broken tree has its top touching the ground (shown in the given figure) at a distance of 10 m from the bottom. If the angle made by the broken part with ground is 30°, then the length of the broken part is

(a) 10 √3 m

(b) $\frac{20}{\sqrt{3}}$

(c) 20 m

(d) 20 √3 m

Sol :

From the figure, AC is the height of tree and from B, it was broken

AB = A’C

Angle of elevation = 30°

A’C = 10 m

Let AC = hm’

and A’B = x m

BC = h – x m

$\cos \theta=\frac{A^{\prime} C}{A^{\prime} B}$

$\cos 30^{\circ}=\frac{10}{x}$

$\cos \theta=\frac{A^{\prime} C}{A^{\prime} B}$

$\cos 30^{\circ}=\frac{10}{x}$

$\frac{\sqrt{3}}{2}=\frac{10}{x}$

$\Rightarrow x=\frac{2 \times 10}{\sqrt{3}}$

$=\frac{20}{\sqrt{3}} \mathrm{~m}$

Ans (b)

Question 5

If the angle of depression of an object from a 75 m high tower is 30°, then the distance of the object from the tower is

(a) 25 √3 m

(b) 50√ 3 m

(c) 75 √3 m

(d) 150 m

Sol :

Height tower AB = 75 m

C is an object on the ground and angle of depression from A is 30°.

Let BC= x m

Now $\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{75}{x}$

$\frac{1}{\sqrt{3}}=\frac{75}{x}$

$\Rightarrow x=75 \sqrt{3} \mathrm{~m}$

Ans (c)

Question 6

A ladder 14 m long rests against a wall. If the foot of the ladder is 7 m from the wall, then the angle of elevation is

(a) 15°

(b) 30°

(c) 45°

(d) 60°

Sol :

Length of a ladder AB = 14 m

Foot of the ladder is 7m from the wall θ is the angle of elevation

$\therefore \cos \theta=\frac{B C}{A B}$

$=\frac{7}{14}=\frac{1}{2}=\cos 60^{\circ}$

$\therefore \theta=60^{\circ}$

Ans (d)

Question 7

If a pole 6 m high casts shadow 2 √3 m long on the ground, then the sun’s elevation is

(a) 60°

(b) 45°

(c) 30°

(d) 90°

Sol :

Height of pole AB = 6 m

and its shadow BC = 2√3 m

$\tan \theta=\frac{A B}{B C}=\frac{6}{2 \sqrt{3}}=\frac{3}{\sqrt{3}}$

$=\frac{3 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{3 \sqrt{3}}{3}$

$=\frac{3 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$=\frac{3 \sqrt{3}}{3}$

$=\sqrt{3}=\tan 60^{\circ}$

$\therefore \theta=60^{\circ}$

$\therefore$ Angle of elevation $=60^{\circ}$

Ans (a)

Question 8

If the length of the shadow of a tower is √3 times that of its height, then the angle of elevation of the sun is

(a) 15°

(b) 30°

(c) 45°

(d) 60°

Sol :

Let height of a tower AB = h m

Then its shadow BC = √3 hm

$\tan \theta=\frac{A B}{B C}=\frac{h}{\sqrt{3} h}$

$=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$

$\therefore \theta=30^{\circ}$

Angle of elevation $=30^{\circ}$

Ans (b)

Question 9

In ∆ABC, ∠A = 30° and ∠B = 90°. If AC = 8 cm, then its area is

(a) 16 √3 cm²

(b) 16 m²

(c) 8 √3 cm²

(d) 6 √3 cm²

Sol :

In ∆ABC, ∠A = 30°, ∠B = 90°

AC = 8 cm

$\therefore \sin 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}$

$\Rightarrow \frac{1}{2}=\frac{B C}{8}$

$\mathrm{BC}=\frac{8}{2}=4 \mathrm{~cm}$

$\cos 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AB}}{8}$

$\Rightarrow A B=\frac{8 \sqrt{3}}{2}=4 \sqrt{3} \mathrm{~cm}$

Now area $\Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{BC}$

$=\frac{1}{2} \times 4 \sqrt{3} \times 4 \mathrm{~cm}^{2}$

$=8 \sqrt{3} \mathrm{~cm}^{2}$

Ans (c)

ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities MCQs

MCQs

Choose the correct answer from the given four options (1 to 12) :

Question 1

$\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$ is equal to

(a) 1

(b) -1

(c) $\sin ^{2} \theta$

(d) $\sec ^{2} \theta$

Sol :

$\cot ^{2} \theta-\frac{1}{\sin ^{2} \theta}$

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-\frac{1}{\sin ^{2} \theta}$

$\frac{\cos ^{2} \theta-1}{\sin ^{2} \theta}=\frac{-\sin ^{2} \theta}{\sin ^{2} \theta}=-1$

Ans (b)

Question 2

$\left(\sec ^{2} \theta-1\right)\left(1-\operatorname{cosec}^{2} \theta\right)$ is equal to

(a) – 1

(b) 1

(c) 0

(d) 2

Sol :

$\left(\sec ^{2} \theta-1\right)\left(1-\operatorname{cosec}^{2} \theta\right)$

$=\left(\frac{1}{\cos ^{2} \theta}-1\right)\left(1-\frac{1}{\sin ^{2} \theta}\right)$

$=\frac{1-\cos ^{2} \theta}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta-1}{\sin ^{2} \theta}$

$=\frac{-\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta \cos ^{2} \theta}=-1$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Ans (a)

Question 3

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$ is equal to

(a) $2 \sin ^{2} \theta$

(b) $2 \cos ^{2} \theta$

(c) $\sin ^{2} \theta$

(d) $\cos ^{2} \theta$

Sol :

$\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}}$

$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\cos ^{2} \theta}{\sin ^{2} \theta+\cos ^{2} \theta}$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

$=\frac{\sin ^{2} \theta}{1}=\sin ^{2} \theta$

Ans (c)

Question 4

$(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}$ is equal to

(a) – 2

(b) 0

(c) 1

(d) 2

Sol :

$(\cos \theta+\sin \theta)^{2}+(\cos \theta-\sin \theta)^{2}$

$=\cos ^{2} \theta+\sin ^{2} \theta+2 \sin \theta \cos \theta+\cos ^{2} \theta+\sin ^{2} \theta-2 \sin \theta \cos \theta$

$=2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$

$=2 \times 1=2(d)$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Question 5

(sec A + tan A) (1 – sin A) is equal to

(a) sec A

(b) sin A

(c) cosec A

(d) cos A

Sol :

(sec A + tan A) (1 – sin A)

$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$

$=\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}} \times 1-\sin \mathrm{A}$

$=\frac{(1+\sin \mathrm{A})(1-\sin \mathrm{A})}{\cos \mathrm{A}}$

$=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}=\cos A$

Ans (d)

Question 6

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$ is equal to

(a) $\sec ^{2} A$

(b) -1

(c) $\cot ^{2} A$

(d) $\tan ^{2} A$

Sol :

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$

$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}$

$=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}$

$=\frac{1}{\cos ^{2} A} \times \frac{\sin ^{2} A}{1}$

$=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A$

Ans (d)

Question 7

If sec θ – tan θ = k, then the value of sec θ + tan θ is

(a) $1-\frac{1}{k}$

(b) $1-k$

(c) $1+k$

(d) $\frac{1}{k}$

Sol :

sec θ – tan θ = k

$\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}=k$

$\frac{1-\sin \theta}{\cos \theta}=k$

Squaring both sides, we get

$\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}=(k)^{2} \Rightarrow \frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}=k^{2}$

$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta}=k^{2}$

$\Rightarrow \frac{(1-\sin \theta)^{2}}{(1+\sin \theta)(1-\sin \theta)}=k^{2}$

$=\frac{1-\sin \theta}{1+\sin \theta}=k^{2}$

$\Rightarrow \frac{1+\sin \theta}{1-\sin \theta}=\frac{1}{k^{2}}$

$ \Rightarrow \sqrt{\frac{(1+\sin \theta)}{1-\sin \theta}}=\frac{1}{k}$

Multiplying and dividing by $(1+\sin \theta)$

$=\sqrt{\left[\frac{(1+\sin \theta)^{2}}{1-\sin ^{2} \theta}\right]}=\frac{1}{k}$

$\Rightarrow \sqrt{\left[\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}\right]}=\frac{1}{k}$

$\frac{1+\sin \theta}{\cos \theta}=\frac{1}{k}=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\frac{1}{k}$

$\Rightarrow \sec \theta+\tan \theta=\frac{1}{k}$

Ans (d)

Question 8

Which of the following is true for all values of θ (0° < θ < 90°):

(a) $\cos ^{2} \theta-\sin ^{2} \theta=1$

(b) $\operatorname{cosec}^{2} \theta-\sec ^{2} \theta=1$

(c) $\sec ^{2} \theta-\tan ^{2} \theta=1$

(c) $\cot ^{2} \theta-\tan ^{2} \theta=1$

Sol :

$\therefore \sec ^{2} \theta-\tan ^{2} \theta=1$ is true for all values of θ as it is an identity.

(0° < θ < 90°)

Ans (c)

Question 9

If θ is an acute angle of a right triangle, then the value of sin θ cos (90° – θ) + cos θ sin (90° – θ) is

(a) 0

(b) 2 sin θ cos θ

(c) 1

(c) $2 \sin ^{2} \theta$

Sol :

sin θ cos (90° – θ) + cos θ sin (90° – θ)

= sin θ sin θ + cos θ cos θ

{ ∵ sin(90° – θ) = cosθ, cos (90° – θ) = sin θ }

$=\sin ^{2} \theta+\cos ^{2} \theta=1$

Ans (c)

Question 10

The value of cos 65° sin 25° + sin 65° cos 25° is

(a) 0

(b) 1

(b) 2

(d) 4

Sol :

cos 65° sin 25° + sin 65° cos 25°

= cos (90° – 25°) sin 25° + sin (90° – 25°) cos 25°

= sin 25° . sin 25° + cos 25° . cos 25°

$=\sin ^{2} 25^{\circ}+\cos ^{2} 25^{\circ}$

$\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)$

Ans (b)

Question 11

The value of $3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}^{2} 64^{\circ}$ is

(a) 0

(b) 3

(c) – 3

(d) – 1

Sol :

$3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}^{2} 64^{\circ}$

$=3 \tan ^{2} 26^{\circ}-3 \operatorname{cosec}\left(90^{\circ}-26^{\circ}\right)$

$=3 \tan ^{2} 26^{\circ}-3 \sec ^{2} 26^{\circ}$

$=3\left(\tan ^{2} 26^{\circ}-\sec ^{2} 26^{\circ}\right)$

$=3 \times(-1)=-3 $ $\left\{\because \sec ^{2} \theta-\tan ^{2} \theta=1\right\}$

Ans (c)

Question 12

The value of $\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$ is

(a) $-\cot \theta$

(b) $-\sin ^{2} \theta$

(c) $-\cos ^{2} \theta$

(d) $-\operatorname{cosec}^{2} \theta$

Sol :

$\frac{\sin \left(90^{\circ}-\theta\right) \sin \theta}{\tan \theta}-1$

$=\frac{\cos \theta \sin \theta}{\frac{\sin \theta}{\cos \theta}}-1$

$=\frac{\sin \theta \cos \theta \times \cos \theta}{\sin \theta}-1$

$=\cos ^{2} \theta-1=-\left(1-\cos ^{2} \theta\right)$

$=-\sin ^{2} \theta$

Ans (b)

ML Aggarwal Solution Class 10 Chapter 17 Mensuration MCQs

MCQs

Question 1

In a cylinder, if radius is halved and height is doubled then the volume will be

(a) same

(b) doubled

(c) halved

(d) four times

Sol :

Let radius of cylinder = r

and height = h

then volume = πr²h

If the radius is halved and the height is doubled

Then volume $=\pi\left(\frac{r}{2}\right)^{2} \times 2 h$

$=\pi \frac{r^{2}}{4} \times 2 h=\frac{1}{2}\left(\pi r^{2} h\right)$

which is half

Ans (c)

Question 2

In a cylinder, if the radius is doubled and height is halved then its curved surface area will be

(a) halved

(b) doubled

(c) same

(d) four times

Sol :

Let radius of a cylinder = r

and height = h

Then curved surface area = 2πrh

Now if radius is doubled and height is halved,

then curved surface area $=2 \pi \frac{r}{2} \times 2 h=2 \pi r h$ which is same (c)

Question 3

If a well of diameter 8 m has been dug to the depth of 14 m, then the volume of the earth dug out is

(a) 352 $m^{3}$

(b) 704 $m^{3}$

(c) 1408 $m^{3}$

(d) 2816 $m^{3}$

Sol :

Diameter of a well = 8 m

Radius $(r)=\frac{8}{2}=4 m$

Depth (h) = 14 m

Volume of the earth dug put = $\pi r^{2} h$

$=\frac{22}{7} \times 4 \times 4 \times 14 \mathrm{~m}^{3}$

$=704 \mathrm{~m}^{3}$

Ans (b)

Question 4

If two cylinders of the same lateral surface have their radii in the ratio 4 : 9, then the ratio of their heights is

(a) 2 : 3

(b) 3 : 2

(c) 4 : 9

(d) 9 : 4

Sol :

Ratio in two cylinder having same lateral surface area their radii is 4 : 9

Let r1 be the radius of the first and $r_2$ be the second cylinder

and $h_1, h_2$ and their heights

Let $r_{1}=4 x$ and $r_{2}=9 x$

$\therefore 2 \pi r_{1} h_{1}=2 \pi r_{2} h_{2}$

$=2 \pi 4 x \times h_{1}=2 \times \pi \times 9 x h_{2}$

$\frac{h_{1}}{h_{2}}=\frac{9 x}{4 x}=9: 4$

Ratio in their heights = 9 : 4

Ans (d)

Question 5

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is

(a) 10 : 17

(b) 20 : 27

(c) 17 : 27

(d) 20 : 37

Sol :

Radii of two cylinder are in the ratio = 2 : 3

Let radius $(r_1)$ = 2x

and radius $(r_2)$ = 3x

Ratio in their height = 5 : 3

Let height of the first cylinder = 5y

and of second = 3y

Now, volume of the first cylinder

$\pi r_{1}^{2} h=\pi(2 x)^{2} \times 5 h=20 \pi x^{2} y$

and volume of second $=\pi(3 x)^{2} \times 3 y$

$=\pi \times 27 x^{2} y$

$\therefore$ Ratio is $20 \pi x^{2} y: 27 \pi x^{2} y$

$=20: 27$

Ans (b)

Question 6

The total surface area of a cone whose radius is $\frac{r}{2}$ and slant height 2l is

(a) 2πr (l + r)

(b) $\pi r\left(l+\frac{r}{4}\right)$

(c) πr(l + r)

(d) 2πrl

Sol :

Radius of a cone $=\frac{r}{2}$

and slant height = 2l

total surface area of a cone

$=\pi r l+\pi r^{2}$

$=\pi \frac{r}{2} \times 2 l+\pi\left(\frac{r}{2}\right)^{2}$

$=\pi r l+\frac{\pi r^{2}}{4}$

$=\pi r\left(l+\frac{r}{4}\right)$

Ans (b)

Question 7

If the diameter of the base of cone is 10 cm and its height is 12 cm, then its curved surface area is

(a) 60π $cm^2$

(b) 65π $cm^2$

(c) 90π $cm^2$

(d) 120π $cm^2$

Sol :

Diameter of the base of a cone = 10 cm

Radius $(r)=\frac{10}{2}=5 \mathrm{~cm}$

and height (h) = 12 cm

$l=\sqrt{r^{2}+h^{2}}=\sqrt{5^{2}+12^{2}}$

$=\sqrt{25+144}=\sqrt{169}=13 \mathrm{~cm}$

Curved surface area

$=\pi r l=\pi \times 5 \times 13=65 \pi \mathrm{cm}^{2}$

Ans (b)

Question 8

If the diameter of the base of a cone is 12 cm and height is 20 cm, then its volume is ,

(a) 240π $cm^3$

(b) 480π $cm^3$

(c) 720π $cm^3$

(d) 960π $cm^3$

Sol :

Diameter of the base of a cone = 12 cm

Radius $(r)=\frac{12}{2}=6 \mathrm{~cm}$

and height (h) = 20 cm

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 6 \times 6 \times 20 \mathrm{~cm}^{3}$

$=240 \pi \mathrm{cm}^{3}$

Ans (a)

Question 9

If the radius of a sphere is 2r, then its volume will be

(a) $\frac{4}{3} \pi r^{3}$

(b) $4 \pi r^{3}$

(c) $\frac{8 \pi r^{3}}{3}$

(d) $\frac{32 \pi r^{3}}{3}$

Sol :

Radius of a sphere = 2r

Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi \times(2 r)^{3}$

$=\frac{4}{3} \pi \times 8 r^{3}=\frac{32 \pi r^{3}}{3}$

Ans (d)

Question 10

If the diameter of a sphere is 16 cm, then its surface area is

(a) 64π $cm^2$

(b) 256π $cm^2$

(c) 192π $cm^2$

(d) 256 $cm^2$

Sol :

Diameter of a sphere = 16 cm

Radius $(r)=\frac{16}{2}=8 \mathrm{~cm}$

Surface area $=4 \pi^{2}=4 \pi \times 8 \times 8 \mathrm{~cm}^{2}=256 \pi \mathrm{cm}^{2}$

Ans (b)

Question 11

If the radius of a hemisphere is 5 cm, then its volume is

(a) $\frac{250}{3} \pi \quad \mathrm{cm}^{3}$

(b) $\frac{500}{3} \pi \quad c m^{3}$

(c) $75 \pi \mathrm{cm}^{3}$

(d) $\frac{125}{3} \pi \quad c m^{3}$

Sol :

Radius of a hemisphere (r) = 5 cm

Volume $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi(5)^{3} \mathrm{~cm}^{3}$

$=\frac{250}{3} \pi \mathrm{cm}^{3}$

Ans (a)

Question 12

If the ratio of the diameters of the two spheres is 3 : 5, then the ratio of their surface areas is

(a) 3 : 5

(b) 5 : 3

(c) 27 : 125

(d) 9 : 25

Sol :

Ratio in the diameters of two spheres = 3 : 5

Let radius of the first sphere = 3x cm

and radius of the second sphere = 5x cm

Ratio in their surface area

$=4 \pi(3 x)^{2}: 4 \pi(5 x)^{2}$

$ \Rightarrow 9 x^{2}: 25 x^{2}$

=9: 25

Ans (d)

Question 13

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is

(a) 1 : 4

(b) 1 : 3

(c) 2 : 3

(d) 2 : 1

Sol :

Radius of balloon (hemispherical) in the original position = 6 cm

and in increased position = 12 cm

Ratio in their surface areas

$4 \pi(6)^{2}: 4 \pi(12)^{2}$

$=6^{2}: 12^{2}=36: 144$

=1: 4

Ans (a)

Question 14

The shape of a Gilli, in the game of Gilli- danda, is a combination of

(a) two cylinders

(b) a cone and a cylinders

(c) two cones and a cylinder

(d) two cylinders and a cone

Sol :

The shape of a Gilli is the combination of

two cones and a cylinder (as shown in the figure).

Ans (c)

Question 15

If two solid hemisphere of same base radius r are joined together along with their bases, then the curved surface of this new solid is

(a) $4πr^2$

(b) $6πr^2$

(c) $3πr^2$

(d) $8πr^2$

Sol :

Radius of two solid hemispheres = r

These are joined together along with the bases

Curved surface area $= 2π^2 × 2 = 4πr^2$

Ans (a)

Question 16

During conversion of a solid from one shape to another, the volume of the new shape will

(a) increase

(b) decrease

(c) remain unaltered

(d) be doubled

Sol :

During the conversion of a solid into another, the volume of the new shaper will be the same.

i.e. remain unaltered

Ans (c)

Question 17

If a solid of one shape is converted to another, then the surface area of the new solid

(a) remains same

(b) increases

(c) decreases

(d) can’t say

Sol :

If a solid of one shape has conversed into another then

the surface area of the new solid will same or not same

i.e. can’t say.

Ans (d)

Question 18

If a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height 6 cm, then the volume of water that flows out of the cylindrical cup is

(a) 38.8 $cm^3$

(b) 55.4 $cm^3$

(c) 19.4 $cm^3$

(d) 471.4 $cm^3$

Sol :

Radius of a marble = 2.1 cm

Volume of marble $=\frac{4}{3} \pi r^{3} \mathrm{~cm}^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 \mathrm{~cm}^{3}$

$=38.88 \mathrm{~cm}^{3}$

Ans (a)

Question 19

The volume of the largest right circular cone that can be carved out from a cube of edge 4.2 cm is

(a) 9.7 $cm^3$

(b) 77.6 $cm^3$

(c) 58.2 $cm^3$

(d) 19.4 $cm^3$

Sol :

Edge of cube = 4.2 cm

Radius of largest cone cut out $=\frac{42.2}{2}=2.1 \mathrm{~cm}$

and height = 4.2 cm

Volume $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4.2 \mathrm{~cm}^{3}$

= 19.404

= 19.4 $cm^3$

Ans (d)

Question 20

The volume of the greatest sphere cut off from a circular cylindrical wood of base radius 1 cm and height 6 cm is

(a) 288 π $cm^3$

(b) $\frac{4}{3} \pi \mathrm{cm}^{3}$

(c) 6 π $cm^3$

(d) 4 π $cm^3$

Sol :

Radius of cylinder (r) = 1 cm

Height (h) = 6 cm

The largest sphere that can be cut off from the cylinder of radius 1 cm

$\therefore$ Volume $=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(1)^{3}$

$=\frac{4}{3} \pi \mathrm{cm}^{3}$

Ans (b)

Question 21

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

(a) 3 : 4

(b) 4 : 3

(c) 9 : 16

(d) 16 : 9

Sol :

Ratio in volumes of two spheres = 64 : 27

Ratio in their radii $=\frac{r_{1}^{3}}{r_{2}^{3}}=\frac{64}{27}$

$=\left(\frac{r_{1}}{r_{2}}\right)^{3}=\left(\frac{4}{3}\right)^{3}$

$\Rightarrow \frac{r_{1}}{r_{2}}=\frac{4}{3}$

$\therefore$ Ratio in their surface area

$=\frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}}=\left(\frac{4}{3}\right)^{2}=\frac{16}{9}$

$\therefore$ Ratio is 16: 9

Ans (d)

Question 22

If a cone, a hemisphere and a cylinder have equal bases and have same height, then the ratio of their volumes is

(a) 1 : 3 : 2

(b) 2 : 3 : 1

(c) 2 : 1 : 3

(d) 1 : 2 : 3

Sol :

If a cone, a hemisphere and a cylinder have equal bases = r (say)

and height = h in each case and r = h

Ratio in their volumes $=\frac{1}{3} \pi r^{2} h: \frac{2}{3} \pi r^{3}: \pi r^{2} h$

$=\frac{1}{3} \pi r^{2} r: \frac{2}{3} \pi r^{3}: \pi r^{2} r$

$=\frac{1}{3} \pi r^{3}: \frac{2}{3} \pi r^{3}: \pi r^{3}$

$=\frac{1}{3}: \frac{2}{3}: 1=1: 2: 3$

Ans (d)

Question 23

If a sphere and a cube have equal surface areas, then the ratio of the diameter of the sphere to the edge of the cube is

(a) 1 : 2

(b) 2 : 1

(c) √π : √6

(d) √6 : √π

Sol :

A sphere and a cube have equal surface area

Let a be the edge of a cube and r be the radius of the sphere, then

$4 \pi r^{2}=6 a^{2} $

$\Rightarrow \pi(2 r)^{2}: 6 a^{2} \quad(\because d=2 r)$

$\Rightarrow \frac{d^{2}}{a^{2}}=\frac{6}{\pi}$

$ \Rightarrow \frac{d}{a}=\frac{\sqrt{6}}{\sqrt{\pi}}$

Radii $d: a=\sqrt{6}: \sqrt{\pi}$

Ans (d)

Question 24

A solid piece of iron in the form of a cuboid of dimensions 49 cm x 33 cm x 24 cm is moulded to form a sphere. The radius of the sphere is

(a) 21 cm

(b) 23 cm

(c) 25 cm

(d) 19 cm

Sol :

Dimension of a cuboid = 49 cm × 33 cm × 24 cm

Volume of a cuboid = 49 × 33 × 24 $cm^3$

⇒ Volume of sphere = Volume of a cuboid

Volume of a sphere = 49 × 33 × 24 $cm^3$

$\therefore$ Radius $=\left(\frac{\text { Volume }}{\frac{4}{3} \pi}\right)^{\frac{1}{3}}$

$=\left(\frac{49 \times 33 \times 24 \times 3 \times 7}{4 \times 22}\right)^{\frac{1}{3}}$

$=(49 \times 7 \times 3 \times 3 \times 3)^{\frac{1}{3}}$

$=(7 \times 7 \times 7 \times 3 \times 3 \times 3)^{\frac{1}{3}}$

$=7 \times 3=21 \mathrm{~cm}$

Ans (a)

Question 25

If a solid right circular cone of height 24 cm and base radius 6 cm is melted and recast in the shape of a sphere, then the radius of the sphere is

(a) 4 cm

(b) 6 cm

(c) 8 cm

(d) 12 cm

Sol :

Height of a circular cone (h) = 24 cm

and radius (r) = 6 cm

$\therefore$ Volume of a cone $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \pi \times 6 \times 6 \times 24 \mathrm{~cm}^{3}$

Volume of sphere=Volume of a cone

Now volume of sphere $=\frac{1}{3} \pi \times 36 \times 24 \mathrm{~cm}^{3}$

Let r be in radius of sphere

Then $\frac{4}{3} \pi r^{3}=\frac{1}{3} \pi \times 36 \times 24$

$4 r^{3}=36 \times 24 \Rightarrow r^{3}=\frac{36 \times 24}{4}$

$r^{3}=3 \times 3 \times 3 \times 2 \times 2 \times 2=3^{3} \times 2^{3}$

$\therefore r=3 \times 2^{\circ}=6 \mathrm{~cm}$

Ans (b)

Question 26

If a solid circular cylinder of iron whose diameter is 15 cm and height 10 cm is melted and recasted into a sphere, then the radius of the sphere is

(a) 15 cm

(b) 10 cm

(c) 7.5 cm

(d) 5 cm

Sol :

Diameter of a cylinder = 15 cm

Radius $=\frac{15}{2} \mathrm{~cm}$

and height = 10 cm

$\therefore$ Volume $=\pi r^{2} h=\pi \times \frac{15}{2} \times \frac{15}{2} \times 10 \mathrm{~cm}^{3}$

$=\frac{1125 \pi}{2} \mathrm{~cm}^{3}$

$\therefore$ Volume of sphere $=\frac{1125 \pi}{2} \mathrm{~cm}^{3}$

$\therefore$ Radius of sphere $=\left(\frac{\text { Volume }}{\frac{4}{3} \pi}\right)^{\frac{1}{3}}$

$=\left(\frac{1125 \pi \times 3}{2 \times 4 \pi}\right)^{\frac{1}{3}}=\left(\frac{3375}{8}\right)^{\frac{1}{3}}$

$\begin{array}{l|l}3 & 3375 \\\hline 3 & 1125 \\\hline 3 & 375 \\\hline 5 & 125 \\\hline 5 & 25 \\\hline 5 & 5 \\\hline & 1\end{array}$

$=\left(\frac{5^{3} \times 3^{3}}{2^{3}}\right)^{\frac{1}{3}}=\frac{5 \times 3}{2} \mathrm{~cm}$

$=\frac{15}{2}=7.5 \mathrm{~cm}$

Ans (c)

Question 27

The number of balls of radius 1 cm that can be made from a sphere of radius 10 cm is

(a) 100

(b) 1000

(c) 10000

(d) 100000

Sol :

Radius of sphere (R) = 10 cm

Volume of sphere $=\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi(10)^{3} \mathrm{~cm}^{3}$

$=\frac{4}{3} \pi \times 1000 \mathrm{~cm}^{3}$

and radius of one ball $(r)=1 \mathrm{~cm}$

Volume of one ball $=\frac{4}{3} \pi(1)^{3} \mathrm{~cm}^{3}=\frac{4}{3} \pi \mathrm{cm}^{3}$

$\therefore$ Number of

$\frac{4 \pi \times 1000 \times 3}{3 \times 4 \times \pi}=1000$

Ans (b)

Question 28

A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8 cm. The height of the cone is

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 18 cm

Sol :

The internal diameter of the metallic shell = 4 cm

and external diameter = 8 cm

$\therefore$ Internal radius $(r)=\frac{4}{2}=2 \mathrm{~cm}$

Volume of metal used $=\frac{4}{3} \pi\left(\mathrm{R}^{3}-r^{3}\right)$

$=\frac{4}{3} \pi\left(4^{3}-2^{3}\right) \mathrm{cm}^{3}$

$=\frac{4}{3} \times \pi(64-8) \mathrm{cm}^{3}$

$=\frac{4}{3} \pi \times 56 \mathrm{~cm}^{3}$

Diameter of cone $=8 \mathrm{~cm}$

$\therefore$ Radius of cone $=\frac{8}{2}=4 \mathrm{~cm}$

$\Rightarrow$ Height $=\frac{\text { Volume }}{\frac{1}{3} \pi r^{2}}$

$=\frac{4 \pi \times 56 \times 3}{3 \times 1 \times \pi \times 4 \times 4}$

=14 cm

Ans (b)

Question 29

A cubical ice cream brick of edge 22 cm is to be distributed among some children by filling ice cream cones of radius 2 cm and height 7 cm up to its brim. The number of children who will get the ice cream cones is

(a) 163

(b) 263

(c) 363

(d) 463

Sol :

Edge of a cubical icecream brick = 22 cm

Volume = $a^3 = (22)^3$ = 10648 $cm^3$

Radius (r) of ice cream cone (r) = 2 cm

and height (h) = 7 cm

$\therefore$ Volume of one cone $=\frac{1}{3} \pi r^{2} h$

$=\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7 \mathrm{~cm}^{3}=\frac{88}{3} \mathrm{~cm}^{3}$

$\therefore$ Number of cones $=\frac{10648 \times 3}{88}=363$

Ans (c)

Question 30

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is

(a) 4 cm

(b) 3 cm

(b) 2 cm

(d) 6 cm

Sol :

Diameter of cylinder = 2 cm

Radius $=\frac{2}{2}=1 \mathrm{~cm}$

and height = 16 cm

$\therefore$ Volume $=\pi r^{2} h$

$=\frac{22}{7} \times 1 \times 1 \times 16=\frac{352}{7} \mathrm{~cm}^{3}$

$\therefore$ Volume of 12 solid spheres so formed

$=\frac{352}{7} \mathrm{~cm}^{3}$

$\therefore$ Volume of each sphere $=\frac{352}{7 \times 12}=\frac{352}{84} \mathrm{~cm}^{3}$

$\therefore$ Radius of each sphere $=\left(\frac{352 \times 3 \times 7}{84 \times 4 \times 22}\right)^{\frac{1}{3}}$

$=(1)^{\frac{1}{3}}=1 \mathrm{~cm}$

$\therefore$ Diameter $=2 \times 1=2 \mathrm{~cm}$

Ans (c)

Question 31

A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that $\frac{1}{8}$ space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is

(a) 142296

(b) 142396

(c) 142496

(d) 142596

Sol :

Internal edge of a hollow cube = 22 cm

Volume $=(\text { side })^{3}=(22)^{3}=22 \times 22 \times 22 \mathrm{~cm}^{3}=10648 \mathrm{~cm}^{3}$

Diameter of spherical marble $=0.5 \mathrm{~cm}=\frac{1}{2}$

$\therefore$ Radius $=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \mathrm{~cm}$

$\therefore$ Volume $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \mathrm{~cm}^{3}$

$=\frac{11}{168} \mathrm{~cm}^{3}$

Space left unfilled $=10648 \times \frac{1}{8} \mathrm{~cm}^{3}$

$=1331 \mathrm{~cm}^{3}$

$\therefore$ Remaining volume for marbles

$=10648-1331=9317 \mathrm{~cm}^{3}$

$\therefore$ Number of marble to accommodate

$=9317 \div \frac{11}{168}=\frac{9317 \times 168}{11}$

=142296

Ans (a)

Question 32

In the given figure, the bottom of the glass has a hemispherical raised portion. If the glass is filled with orange juice, the quantity of juice which a person will get is

(a) 135 π $cm^3$

(b) 117 π $cm^3$

(c) 99 π $cm^3$

(d) 36 π $cm^3$

Sol :

Radius of base of cylinder (r) $=\frac{6}{2} \mathrm{~cm}=3 \mathrm{~cm}$

and height (h)= 15 cm

$\therefore$ Volume of the glass $=\pi r^{2} h-\frac{2}{3} \pi r^{3}$

$=\pi r\left(r h-\frac{2}{3} r^{2}\right)$

$=\pi \times 3\left(3 \times 15-\frac{2}{3} \times 9\right)$

$=3 \pi(45-6) \mathrm{cm}^{3}$

$=3 \pi \times 39=117 \pi \mathrm{cm}^{3}$

Ans (b)