Ratio and Proportion — Exercise 9(C)

In a ratio which is equal to 5 : 8, the antecedent is 40, then the consequent is:

1. 25

2. 48

3. 64

4. none of these

Answer

Let the consequent be $x$. Then,

$\Rightarrow 5 : 8 = 40 : x \\[1em] \Rightarrow \dfrac{5}{8} = \dfrac{40}{x} \\[1em] \Rightarrow 5x = 8 \times 40 \\[1em] \Rightarrow x = \dfrac{320}{5} = 64$

Hence, option 3 is the correct option.

0.5 of a number is equal to 0.07 of another. The ratio of the numbers is

1. 1 : 14

2. 5 : 7

3. 50 : 7

4. 7 : 50

Answer

Let the two numbers be $x$ and $y$. Then,

$\Rightarrow 0.5 \times x = 0.07 \times y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{0.07}{0.5} = \dfrac{7}{50}$

∴ $x : y$ = 7 : 50.

Hence, option 4 is the correct option.

If $2x = 3y = 4z$, then $x : y : z$ is

1. 2 : 3 : 4

2. 4 : 3 : 2

3. 3 : 4 : 2

4. 6 : 4 : 3

Answer

Let $2x = 3y = 4z = k$. Then,

$x = \dfrac{k}{2}, \quad y = \dfrac{k}{3} \quad \text{and} \quad z = \dfrac{k}{4}$

Calculating L.C.M. of 2, 3 and 4:

$\begin{array}{l|rrr} 2 & 2, & 3, & 4 \\ \hline 2 & 1, & 3, & 2 \\ \hline 3 & 1, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$

∴ L.C.M. of 2, 3 and 4 = 2 × 2 × 3 = 12.

$\Rightarrow x : y : z \\[1em] \Rightarrow \dfrac{k}{2} : \dfrac{k}{3} : \dfrac{k}{4} \\[1em] \Rightarrow \left(\dfrac{k}{2} \times \dfrac{12}{k}\right) : \left(\dfrac{k}{3} \times \dfrac{12}{k}\right) : \left(\dfrac{k}{4} \times \dfrac{12}{k}\right) \\[1em] \Rightarrow 6 : 4 : 3$

Hence, option 4 is the correct option.

If $x : y$ = 7 : 9 and $y : z$ = 5 : 4, then $x : y : z$ is

1. 7 : 45 : 36

2. 35 : 45 : 36

3. 28 : 36 : 35

4. none of these

Answer

Calculating L.C.M. of two values of $y$, namely 9 and 5:

$\begin{array}{l|rr} 3 & 9, & 5 \\ \hline 3 & 3, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \\ \end{array}$

∴ L.C.M. of 9 and 5 = 3 × 3 × 5 = 45.

$\Rightarrow x : y = 7 : 9 = (7 \times 5) : (9 \times 5) = 35 : 45 \\[1em] \Rightarrow y : z = 5 : 4 = (5 \times 9) : (4 \times 9) = 45 : 36$

∴ $x : y : z$ = 35 : 45 : 36.

Hence, option 2 is the correct option.

The fourth proportional to 3, 5 and 21 is

1. 35

2. $\dfrac{7}{5}$

3. $\dfrac{5}{7}$

4. 12.6

Answer

Let the fourth proportional to 3, 5 and 21 be $x$. Then,

$\Rightarrow 3 : 5 :: 21 : x \\[1em] \Rightarrow 3 \times x = 5 \times 21 \\[1em] \Rightarrow 3x = 105 \\[1em] \Rightarrow x = \dfrac{105}{3} = 35$

Hence, option 1 is the correct option.

Mean proportional between 7 and 28 is

1. 17.5

2. 12

3. 14

4. 16

Answer

Mean proportional between 7 and 28 = $\sqrt{7 \times 28} = \sqrt{196} = 14$

Hence, option 3 is the correct option.

Third proportional to 9 and 12 is

1. $6\sqrt{3}$

2. 10.5

3. 16

4. none of these

Answer

Let the third proportional to 9 and 12 be $x$. Then,

$\Rightarrow 9 : 12 :: 12 : x \\[1em] \Rightarrow 9 \times x = 12 \times 12 \\[1em] \Rightarrow 9x = 144 \\[1em] \Rightarrow x = \dfrac{144}{9} = 16$

Hence, option 3 is the correct option.

A fraction bears the same ratio to $\dfrac{1}{27}$ as $\dfrac{3}{7}$ does to $\dfrac{5}{9}$. The fraction is

1. $\dfrac{7}{45}$

2. $\dfrac{1}{35}$

3. $\dfrac{45}{7}$

4. $\dfrac{5}{21}$

Answer

Let the required fraction be $x$. Then,

$x : \dfrac{1}{27} = \dfrac{3}{7} : \dfrac{5}{9}$

Product of extremes = product of means.

$\Rightarrow x \times \dfrac{5}{9} = \dfrac{1}{27} \times \dfrac{3}{7} \\[1em] \Rightarrow \dfrac{5x}{9} = \dfrac{3}{189} = \dfrac{1}{63} \\[1em] \Rightarrow x = \left(\dfrac{1}{63} \times \dfrac{9}{5}\right) \\[1em] \Rightarrow x = \dfrac{9}{315} \\[1em] \Rightarrow x = \dfrac{1}{35}$

Hence, option 2 is the correct option.

What must be added to each term of the ratio 49 : 68 so that it becomes 3 : 4?

1. 8

2. 9

3. 5

4. 3

Answer

Let the required number be $x$. Then,

$\Rightarrow \dfrac{49 + x}{68 + x} = \dfrac{3}{4} \\[1em] \Rightarrow 4(49 + x) = 3(68 + x) \\[1em] \Rightarrow 196 + 4x = 204 + 3x \\[1em] \Rightarrow 4x - 3x = 204 - 196 \\[1em] \Rightarrow x = 8$

Hence, option 1 is the correct option.

What least number must be added to each one of 6, 14, 18, 38 to make them in proportion?

1. 1

2. 2

3. 3

4. 4

Answer

Let the required number be $x$. Then,

$\Rightarrow \dfrac{6 + x}{14 + x} = \dfrac{18 + x}{38 + x} \\[1em] \Rightarrow (6 + x)(38 + x) = (14 + x)(18 + x) \\[1em] \Rightarrow 228 + 6x + 38x + x^2 = 252 + 14x + 18x + x^2 \\[1em] \Rightarrow 228 + 44x = 252 + 32x \\[1em] \Rightarrow 44x - 32x = 252 - 228 \\[1em] \Rightarrow 12x = 24 \\[1em] \Rightarrow x = 2$

Hence, option 2 is the correct option.

What least number must be subtracted from each of the numbers 14, 17, 34, 42 so that the remainders may be proportional?

1. 0

2. 1

3. 2

4. 7

Answer

Let the required number be $x$. Then,

$\Rightarrow \dfrac{14 - x}{17 - x} = \dfrac{34 - x}{42 - x} \\[1em] \Rightarrow (14 - x)(42 - x) = (17 - x)(34 - x) \\[1em] \Rightarrow 588 - 14x - 42x + x^2 = 578 - 17x - 34x + x^2 \\[1em] \Rightarrow 588 - 56x = 578 - 51x \\[1em] \Rightarrow 588 - 578 = 56x - 51x \\[1em] \Rightarrow 5x = 10 \\[1em] \Rightarrow x = 2$

Hence, option 3 is the correct option.

If the ratio of A's money to B's money is 4 : 5 and that of B's money to C's money is 2 : 3 and A has ₹ 800, then C has:

1. ₹ 1,000

2. ₹ 1,200

3. ₹ 1,500

4. ₹ 2,000

Answer

Calculating L.C.M. of two values of B, namely 5 and 2:

$\begin{array}{l|rr} 2 & 5, & 2 \\ \hline 5 & 5, & 1 \\ \hline & 1, & 1 \\ \end{array}$

∴ L.C.M. of 5 and 2 = 2 × 5 = 10.

A : B = 4 : 5 = (4 × 2) : (5 × 2) = 8 : 10

B : C = 2 : 3 = (2 × 5) : (3 × 5) = 10 : 15

∴ A : B : C = 8 : 10 : 15.

Let A's money, B's money and C's money be ₹ $8x, ₹ 10x$ and ₹ $15x$ respectively.

But, A has ₹ 800.

$\Rightarrow 8x = 800 \\[1em] \Rightarrow x = 100$

∴ C's money = ₹ (15 × 100) = ₹ 1,500.

Hence, option 3 is the correct option.

A sum of ₹ 7,000 is divided among A, B and C in such a way that the shares of A and B are in the ratio 2 : 3 and those of B and C are in the ratio 4 : 5. The amount received by C is

1. ₹ 2,600

2. ₹ 2,800

3. ₹ 3,000

4. ₹ 3,900

Answer

Calculating L.C.M. of two values of B, namely 3 and 4:

$\begin{array}{l|rr} 2 & 3, & 4 \\ \hline 2 & 3, & 2 \\ \hline 3 & 3, & 1 \\ \hline & 1, & 1 \\ \end{array}$

∴ L.C.M. of 3 and 4 = 2 × 2 × 3 = 12.

A : B = 2 : 3 = (2 × 4) : (3 × 4) = 8 : 12

B : C = 4 : 5 = (4 × 3) : (5 × 3) = 12 : 15

∴ A : B : C = 8 : 12 : 15.

Sum of ratio terms = (8 + 12 + 15) = 35.

∴ C's share = ₹ $\left(7000 \times \dfrac{15}{35}\right)$ = ₹ 3,000

Hence, option 3 is the correct option.

A sum of ₹ 53 is divided among A, B and C in such a way that A gets ₹ 7 more than what B gets and B gets ₹ 8 more than what C gets. The ratio of their shares is

1. 16 : 9 : 18

2. 25 : 18 : 10

3. 18 : 25 : 10

4. 15 : 8 : 30

Answer

Let C's share be ₹ $x$. Then,

B's share = ₹ ($x$ + 8) and A's share = ₹ ($x$ + 8 + 7) = ₹ ($x$ + 15).

$\Rightarrow (x + 15) + (x + 8) + x = 53 \\[1em] \Rightarrow 3x + 23 = 53 \\[1em] \Rightarrow 3x = 30 \\[1em] \Rightarrow x = 10$

∴ C's share = ₹ 10, B's share = ₹ 18 and A's share = ₹ 25.

∴ Ratio of their shares = 25 : 18 : 10.

Hence, option 2 is the correct option.

A bag contains ₹ 600 in the form of one-rupee, 50-paise and 25-paise coins in the ratio 3 : 4 : 12. The number of 25-paise coins is

1. 600

2. 900

3. 1200

4. 1376

Answer

Let the number of one-rupee, 50-paise and 25-paise coins be $3x, 4x$ and $12x$ respectively. Then,

$\Rightarrow 3x + \dfrac{4x}{2} + \dfrac{12x}{4} = 600 \\[1em] \Rightarrow 3x + 2x + 3x = 600 \\[1em] \Rightarrow 8x = 600 \\[1em] \Rightarrow x = 75$

∴ Number of 25-paise coins = (12 × 75) = 900.

Hence, option 2 is the correct option.

A certain amount was divided between P and Q in the ratio 4 : 3. If Q's share was ₹ 4,800, the total amount was:

1. ₹ 11,200

2. ₹ 6,400

3. ₹ 19,200

4. ₹ 39,200

Answer

Let the total amount be ₹ $x$.

Sum of ratio terms = (4 + 3) = 7.

∴ Q's share = ₹ $\left(x \times \dfrac{3}{7}\right) = ₹\ \dfrac{3x}{7}$

But, Q's share = ₹ 4,800.

$\Rightarrow \dfrac{3x}{7} = 4800 \\[1em] \Rightarrow x = \left(4800 \times \dfrac{7}{3}\right) = 11200$

Hence, option 1 is the correct option.

The ratio of number of boys and girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?

1. 90

2. 120

3. 220

4. 240

Answer

Ratio of boys : girls = 7 : 5.

Sum of ratio terms = (7 + 5) = 12.

No. of boys = $\left(720 \times \dfrac{7}{12}\right) = 420$

No. of girls = $\left(720 \times \dfrac{5}{12}\right) = 300$

Let $x$ more girls be admitted to make the ratio 1 : 1.

$\Rightarrow \dfrac{420}{300 + x} = \dfrac{1}{1} \\[1em] \Rightarrow 300 + x = 420 \\[1em] \Rightarrow x = 120$

Hence, option 2 is the correct option.

The prices of a scooter and a moped are in the ratio 9 : 5. If a scooter costs ₹ 6,800 more than a moped, the price of a scooter is:

1. ₹ 13,600

2. ₹ 15,300

3. ₹ 17,000

4. none of these

Answer

Let the prices of the scooter and the moped be ₹ $9x$ and ₹ $5x$ respectively. Then,

$\Rightarrow 9x - 5x = 6800 \\[1em] \Rightarrow 4x = 6800 \\[1em] \Rightarrow x = 1700$

∴ Price of the scooter = ₹ (9 × 1700) = ₹ 15,300.

Hence, option 2 is the correct option.

The ratio of zinc and copper in a brass piece is 13 : 7. How much zinc will be there in 100 kg of such a piece?

1. 20 kg

2. 35 kg

3. 55 kg

4. 65 kg

Answer

Ratio of zinc : copper = 13 : 7.

Sum of ratio terms = (13 + 7) = 20.

∴ Quantity of zinc = $\left(100 \times \dfrac{13}{20}\right) \text{ kg} = 65 \text{ kg}$

Hence, option 4 is the correct option.

Two numbers are in the ratio 3 : 5. If 9 be subtracted from each, then they are in the ratio 12 : 23. The second number is:

1. 52

2. 53

3. 54

4. 55

Answer

Let the numbers be $3x$ and $5x$. Then,

$\Rightarrow \dfrac{3x - 9}{5x - 9} = \dfrac{12}{23} \\[1em] \Rightarrow 23(3x - 9) = 12(5x - 9) \\[1em] \Rightarrow 69x - 207 = 60x - 108 \\[1em] \Rightarrow 69x - 60x = 207 - 108 \\[1em] \Rightarrow 9x = 99 \\[1em] \Rightarrow x = 11$

∴ Second number = (5 × 11) = 55.

Hence, option 4 is the correct option.

Two whole numbers whose sum is 64 cannot be in the ratio:

1. 5 : 3

2. 7 : 1

3. 3 : 4

4. 9 : 7

Answer

For two whole numbers in the ratio $a : b$ with sum 64, the sum of ratio terms ($a + b$) must divide 64 exactly.

Option 1 : 5 : 3 ⇒ sum of terms = 8, and 64 ÷ 8 = 8. Numbers are 40 and 24.

Option 2 : 7 : 1 ⇒ sum of terms = 8, and 64 ÷ 8 = 8. Numbers are 56 and 8.

Option 3 : 3 : 4 ⇒ sum of terms = 7, and 64 ÷ 7 is not a whole number.

Option 4 : 9 : 7 ⇒ sum of terms = 16, and 64 ÷ 16 = 4. Numbers are 36 and 28.

So, the numbers cannot be in the ratio 3 : 4.

Hence, option 3 is the correct option.

Out of the ratios (7 : 20); (13 : 25); (17 : 30) and (11 : 15), the smallest is:

1. 7 : 20

2. 13 : 25

3. 17 : 30

4. 11 : 15

Answer

Given ratios are $\dfrac{7}{20}, \dfrac{13}{25}, \dfrac{17}{30}$ and $\dfrac{11}{15}$.

Calculating L.C.M. of 20, 25, 30 and 15:

$\begin{array}{l|rrrr} 2 & 20, & 25, & 30, & 15 \\ \hline 2 & 10, & 25, & 15, & 15 \\ \hline 3 & 5, & 25, & 15, & 15 \\ \hline 5 & 5, & 25, & 5, & 5 \\ \hline 5 & 1, & 5, & 1, & 1 \\ \hline & 1, & 1, & 1, & 1 \\ \end{array}$

∴ L.C.M. of 20, 25, 30 and 15 = 2 × 2 × 3 × 5 × 5 = 300.

$\Rightarrow \dfrac{7}{20} = \dfrac{7 \times 15}{20 \times 15} = \dfrac{105}{300};\\[1em] \Rightarrow \dfrac{13}{25} = \dfrac{13 \times 12}{25 \times 12} = \dfrac{156}{300}; \\[1em] \Rightarrow \dfrac{17}{30} = \dfrac{17 \times 10}{30 \times 10} = \dfrac{170}{300}; \\[1em] \Rightarrow \dfrac{11}{15} = \dfrac{11 \times 20}{15 \times 20} = \dfrac{220}{300}$

Clearly, $\dfrac{105}{300}$ is the smallest.

Hence, option 1 is the correct option.

In a ratio which is equal to 5 : 8, the antecedent is 40, then the consequent is

1. 25

2. 48

3. 64

4. none of these

Answer

Let the consequent be $x$. Then,

$\Rightarrow 5 : 8 = 40 : x \\[1em] \Rightarrow \dfrac{5}{8} = \dfrac{40}{x} \\[1em] \Rightarrow 5x = 8 \times 40 \\[1em] \Rightarrow x = \dfrac{320}{5} = 64$

Hence, option 3 is the correct option.

Fill in the blanks:

(i) In $a : b, a$ is called the ......... while $b$ is called the .........

(ii) $a : b \gt c : d$ if ......... > .........

(iii) The simplest ratio of 3 hours : 1 day is .........

(iv) If 10% of $x$ is equal to 20% of $y$, then $x : y$ is equal to .........

(v) The mean proportion between 49 and 81 is .........

(vi) The ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$ is the same as .........

(vii) If a sum of ₹ 455 is divided between A and B in the ratio 4 : 3, then A gets .........

Answer

(i) In $a : b, a$ is called the antecedent while $b$ is called the consequent.

(ii) $a : b \gt c : d$ if $ad$ > $bc$.

(Reason : $\dfrac{a}{b} \gt \dfrac{c}{d} \Leftrightarrow ad \gt bc$.)

(iii) The simplest ratio of 3 hours : 1 day is 1 : 8.

(Reason : 1 day = 24 hours. So, 3 : 24 = $\dfrac{3 \div 3}{24 \div 3} = \dfrac{1}{8}$ = 1 : 8.)

(iv) If 10% of $x$ is equal to 20% of $y$, then $x : y$ is equal to 2 : 1.

(Reason : $\dfrac{10}{100} \times x = \dfrac{20}{100} \times y \Rightarrow \dfrac{x}{y} = \dfrac{20}{10} = \dfrac{2}{1}$.)

(v) The mean proportion between 49 and 81 is 63.

(Reason : Mean proportion = $\sqrt{49 \times 81} = \sqrt{3969} = 63$.)

(vi) The ratio $\dfrac{1}{2} : \dfrac{1}{3} : \dfrac{1}{4}$ is the same as 6 : 4 : 3.

(Reason : Calculating L.C.M. of 2, 3 and 4:

$\begin{array}{l|rrr} 2 & 2, & 3, & 4 \\ \hline 2 & 1, & 3, & 2 \\ \hline 3 & 1, & 3, & 1 \\ \hline & 1, & 1, & 1 \\ \end{array}$

∴ L.C.M. of 2, 3 and 4 = 2 × 2 × 3 = 12.

So, $\left(\dfrac{1}{2} \times 12\right) : \left(\dfrac{1}{3} \times 12\right) : \left(\dfrac{1}{4} \times 12\right) = 6 : 4 : 3$.)

(vii) If a sum of ₹ 455 is divided between A and B in the ratio 4 : 3, then A gets ₹ 260.

(Reason : Sum of ratio terms = (4 + 3) = 7. So, A's share = ₹ $\left(455 \times \dfrac{4}{7}\right)$ = ₹ 260.)

Write true (T) or false (F):

(i) $a : b :: c : d \Leftrightarrow ac = db$

(ii) $\dfrac{1}{x} : \dfrac{1}{y}$ is the same as $y : x$.

(iii) Three numbers $a, b, c$ are in continued proportion if $\dfrac{a}{b} = \dfrac{c}{b}$.

(iv) If $a : b :: b : c$, then $b$ is called the mean proportion between $a$ and $c$.

Answer

(i) False. If $a : b :: c : d$, then product of extremes = product of means, i.e., $ad = bc$ and not $ac = db$.

(ii) True. $\dfrac{1}{x} : \dfrac{1}{y} = \dfrac{1}{x} \times \dfrac{y}{1} = \dfrac{y}{x} = y : x$.

(iii) False. Three numbers $a, b, c$ are in continued proportion if $a : b = b : c$, i.e., if $\dfrac{a}{b} = \dfrac{b}{c}$ and not $\dfrac{a}{b} = \dfrac{c}{b}$.

(iv) True. If $a : b :: b : c$, then $b^2 = ac \Rightarrow b = \sqrt{ac}$, and $b$ is called the mean proportion between $a$ and $c$.

The salaries of three friends—P, Q and R are in the ratio of 5 : 6 : 8 respectively:

1. If P, Q and R save 10%, 15% and 20% of their salaries respectively, what is the ratio of their savings?
   (a) 2 : 5 : 8
   (b) 5 : 6 : 8
   (c) 6 : 8 : 9
   (d) 5 : 9 : 16

2. If P, Q and R save one-half, one-third and one-fourth of their salaries respectively, what is the ratio of their expenditures?
   (a) 2 : 3 : 4
   (b) 5 : 4 : 4
   (c) 5 : 4 : 6
   (d) 5 : 8 : 12

3. If P, Q and R receive increments of 20%, 10% and 15% respectively in their salaries, what will be the ratio of their new salaries?
   (a) 25 : 16 : 23
   (b) 25 : 30 : 42
   (c) 30 : 33 : 46
   (d) cannot be determined

4. If P's salary gets reduced by 10%, Q's salary gets increased by 20% and R's salary gets reduced by 30%, what will be the ratio of their new salaries?
   (a) 45 : 72 : 56
   (b) 55 : 48 : 104
   (c) 12 : 15 : 16
   (d) 8 : 7 : 12

Answer

1. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P's savings = 10% of $5x = \left(\dfrac{10}{100} \times 5x\right) = \dfrac{5x}{10}$

Q's savings = 15% of $6x = \left(\dfrac{15}{100} \times 6x\right) = \dfrac{9x}{10}$

R's savings = 20% of $8x = \left(\dfrac{20}{100} \times 8x\right) = \dfrac{16x}{10}$

∴ Ratio of their savings

$= \dfrac{5x}{10} : \dfrac{9x}{10} : \dfrac{16x}{10} = 5 : 9 : 16$

Hence, option (d) is the correct option.

2. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P saves $\dfrac{1}{2}$ of his salary, so he spends $\left(1 - \dfrac{1}{2}\right) = \dfrac{1}{2}$ of his salary.

Q saves $\dfrac{1}{3}$ of his salary, so he spends $\left(1 - \dfrac{1}{3}\right) = \dfrac{2}{3}$ of his salary.

R saves $\dfrac{1}{4}$ of his salary, so he spends $\left(1 - \dfrac{1}{4}\right) = \dfrac{3}{4}$ of his salary.

P's expenditure = $\left(\dfrac{1}{2} \times 5x\right) = \dfrac{5x}{2}$

Q's expenditure = $\left(\dfrac{2}{3} \times 6x\right) = 4x$

R's expenditure = $\left(\dfrac{3}{4} \times 8x\right) = 6x$

∴ Ratio of their expenditures

$= \dfrac{5x}{2} : 4x : 6x = 5 : 8 : 12 \quad \text{[multiplying each term by 2]}$

Hence, option (d) is the correct option.

3. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P's new salary = 120% of $5x = \left(\dfrac{120}{100} \times 5x\right) = 6x$

Q's new salary = 110% of $6x = \left(\dfrac{110}{100} \times 6x\right) = \dfrac{33x}{5}$

R's new salary = 115% of $8x = \left(\dfrac{115}{100} \times 8x\right) = \dfrac{46x}{5}$

∴ Ratio of their new salaries

$= 6x : \dfrac{33x}{5} : \dfrac{46x}{5} = 30 : 33 : 46 \quad \text{[multiplying each term by 5]}$

Hence, option (c) is the correct option.

4. Let the salaries of P, Q and R be ₹ $5x, ₹ 6x$ and ₹ $8x$ respectively.

P's new salary = 90% of $5x = \left(\dfrac{90}{100} \times 5x\right) = \dfrac{45x}{10}$

Q's new salary = 120% of $6x = \left(\dfrac{120}{100} \times 6x\right) = \dfrac{72x}{10}$

R's new salary = 70% of $8x = \left(\dfrac{70}{100} \times 8x\right) = \dfrac{56x}{10}$

∴ Ratio of their new salaries

$= \dfrac{45x}{10} : \dfrac{72x}{10} : \dfrac{56x}{10} = 45 : 72 : 56$

Hence, option (a) is the correct option.

Assertion (A): The fourth proportional to 7, 12, 28 is 48.

Reason (R): If $a, b, c, d$ are in proportion, then, $ac = bd$.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

Let the fourth proportional to 7, 12 and 28 be $x$. Then,

$\Rightarrow 7 : 12 :: 28 : x \\[1em] \Rightarrow 7 \times x = 12 \times 28 \\[1em] \Rightarrow 7x = 336 \\[1em] \Rightarrow x = \dfrac{336}{7} = 48$

So, Assertion (A) is true.

If $a, b, c, d$ are in proportion, i.e., $a : b :: c : d$, then product of extremes = product of means, i.e., $ad = bc$.

So, Reason (R) is false.

Hence, option 3 is the correct option.

Assertion (A): Mean proportion between 8 and 98 is 28.

Reason (R): Three numbers $a, b, c$ are said to be in continued proportion, if $a : b :: b : c$.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

Mean proportion between 8 and 98 $= \sqrt{8 \times 98} = \sqrt{784} = 28$

So, Assertion (A) is true.

By definition, three numbers $a, b, c$ are said to be in continued proportion if $a : b :: b : c$. So, Reason (R) is true.

Reason (R) only defines continued proportion; it does not by itself explain how the mean proportion $\sqrt{ac}$ is calculated. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

Assertion (A): If we add 3 to each of the numbers 3, 7, 18, 32, then the resulting numbers will be in proportion.

Reason (R): The numbers $a, b, c, d$ are in proportion, if $ab : cd$.

1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

2. Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

3. Assertion (A) is true but Reason (R) is false.

4. Assertion (A) is false but Reason (R) is true.

Answer

On adding 3 to each of the numbers 3, 7, 18 and 32, the resulting numbers are 6, 10, 21 and 35.

Product of extremes = 6 × 35 = 210.

Product of means = 10 × 21 = 210.

Since product of extremes = product of means, the numbers 6, 10, 21, 35 are in proportion. So, Assertion (A) is true.

The numbers $a, b, c, d$ are in proportion if $a : b :: c : d$, i.e., if $ad = bc$, and not if "$ab : cd$". So, Reason (R) is false.

Hence, option 3 is the correct option.

Ratio of number of men and women in an office is 12 : 13.

(i) We can definitely say that there are 13 women in the office.

(ii) We can say that there can be 26 women and 24 men in the office.

Which of the above statements is correct?

1. only (i)

2. only (ii)

3. both are correct

4. none of these

Answer

Let the number of men and women be $12x$ and $13x$ respectively.

Statement (i): The number of women is $13x$, which need not be 13 as $x$ can be any positive integer. So, statement (i) is false.

Statement (ii): Taking $x$ = 2, number of men = (12 × 2) = 24 and number of women = (13 × 2) = 26. So, statement (ii) is true.

Hence, option 2 is the correct option.

Three numbers are in the ratio 2 : 3 : 5 and sum of their squares is 608. The numbers are:

1. 8, 12, 20

2. 8, 15, 40

3. 12, 18, 30

4. 20, 30, 50

Answer

Let the numbers be $2x, 3x$ and $5x$. Then,

$\Rightarrow (2x)^2 + (3x)^2 + (5x)^2 = 608 \\[1em] \Rightarrow 4x^2 + 9x^2 + 25x^2 = 608 \\[1em] \Rightarrow 38x^2 = 608 \\[1em] \Rightarrow x^2 = \dfrac{608}{38} = 16 \\[1em] \Rightarrow x = \sqrt{16} = 4$

∴ The numbers are (2 × 4), (3 × 4) and (5 × 4), i.e., 8, 12 and 20.

Hence, option 1 is the correct option.

0.6 of a number equals 0.09 of another number. The ratio of the numbers is:

1. 2 : 5

2. 1 : 6

3. 3 : 20

4. 20 : 3

Answer

Let the two numbers be $x$ and $y$. Then,

$\Rightarrow 0.6 \times x = 0.09 \times y \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{0.09}{0.6} = \dfrac{9}{60} = \dfrac{3}{20}$

∴ $x : y$ = 3 : 20.

Hence, option 3 is the correct option.

If A : B = 2 : 3 and 2 : A = 1 : 2, then the values of B is:

1. 6

2. 7

3. 8

4. 10

Answer

We have : 2 : A = 1 : 2.

$\Rightarrow \dfrac{2}{\text{A}} = \dfrac{1}{2}$

⇒ A = 4

Also, A : B = 2 : 3.

$\Rightarrow \dfrac{4}{\text{B}} = \dfrac{2}{3} \\[1em] \Rightarrow 2 \times \text{B} = 12 \\[1em] \Rightarrow \text{B} = 6$

Hence, option 1 is the correct option.

A fraction bears the same ratio to $\dfrac{1}{27}$ as $\dfrac{3}{7}$ does to $\dfrac{5}{9}$. The fraction is:

1. $\dfrac{5}{7}$

2. $\dfrac{3}{25}$

3. $\dfrac{6}{35}$

4. $\dfrac{1}{35}$

Answer

Let the required fraction be $x$. Then,

$x : \dfrac{1}{27} = \dfrac{3}{7} : \dfrac{5}{9}$

Product of extremes = product of means.

$\Rightarrow x \times \dfrac{5}{9} = \dfrac{1}{27} \times \dfrac{3}{7} \\[1em] \Rightarrow \dfrac{5x}{9} = \dfrac{3}{189} = \dfrac{1}{63} \\[1em] \Rightarrow x = \left(\dfrac{1}{63} \times \dfrac{9}{5}\right) \\[1em] \Rightarrow x = \dfrac{9}{315} \\[1em] \Rightarrow x = \dfrac{1}{35}$

Hence, option 4 is the correct option.

In a class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girls in the class is:

1. 5 : 11

2. 28 : 25

3. 14 : 11

4. 11 : 14

Answer

Let the total strength of the class be $x$ and let the number of boys and girls be $b$ and $g$ respectively. Then,

$b + g = x$ .........(1)

$b - g$ = 12% of $x$

$b - g = \dfrac{12x}{100}$

$b - g = \dfrac{3x}{25}$ .....(2)

Adding equations (1) and (2), we get :

$\Rightarrow 2b = x + \dfrac{3x}{25} \\[1em] \Rightarrow 2b = \dfrac{28x}{25} \\[1em] \Rightarrow b = \dfrac{14x}{25}$

Subtracting equation (2) from (1), we get :

$\Rightarrow 2g = x - \dfrac{3x}{25} \\[1em] \Rightarrow 2g = \dfrac{22x}{25} \\[1em] \Rightarrow g = \dfrac{11x}{25}$

$\Rightarrow b : g = \dfrac{14x}{25} : \dfrac{11x}{25} = 14 : 11$

Hence, option 3 is the correct option.

The speeds of three cars are in the ratio 3 : 4 : 5. The ratio between the times taken by them to travel the same distance is:

1. 3 : 4 : 5

2. 20 : 15 : 12

3. 5 : 4 : 3

4. 8 : 4 : 3

Answer

We know that, $\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}$.

For the same distance, time taken is inversely proportional to the speed.

∴ Ratio of times taken $= \dfrac{1}{3} : \dfrac{1}{4} : \dfrac{1}{5}$

Calculating L.C.M. of 3, 4 and 5:

$\begin{array}{l|rrr} 2 & 3, & 4, & 5 \\ \hline 2 & 3, & 2, & 5 \\ \hline 3 & 3, & 1, & 5 \\ \hline 5 & 1, & 1, & 5 \\ \hline & 1, & 1, & 1 \\ \end{array}$

∴ L.C.M. of 3, 4 and 5 = 2 × 2 × 3 × 5 = 60.

$\Rightarrow \dfrac{1}{3} : \dfrac{1}{4} : \dfrac{1}{5} \\[1em] \Rightarrow \left(\dfrac{1}{3} \times 60\right) : \left(\dfrac{1}{4} \times 60\right) : \left(\dfrac{1}{5} \times 60\right) \\[1em] \Rightarrow 20 : 15 : 12$

Hence, option 2 is the correct option.

The ratio between two numbers is 3 : 4 and their LCM is 180. The first number is:

1. 15

2. 20

3. 45

4. 60

Answer

Let the two numbers be $3x$ and $4x$, where $x$ is their H.C.F.

Then, L.C.M. of the two numbers = $3 \times 4 \times x = 12x$.

$\Rightarrow 12x = 180 \\[1em] \Rightarrow x = \dfrac{180}{12} = 15$

∴ First number = (3 × 15) = 45.

Hence, option 3 is the correct option.

If $x : y = y : z$, then $x^2 : y^2$ is:

1. $x^2 : z^2$

2. $z : x$

3. $x : z$

4. $x : y$

Answer

Given, $x : y = y : z$.

$\Rightarrow \dfrac{x}{y} = \dfrac{y}{z} \\[1em] \Rightarrow y^2 = xz \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{x^2}{xz} = \dfrac{x}{z}$

∴ $x^2 : y^2 = x : z$.

Hence, option 3 is the correct option.

The sum of three numbers is 174. The ratio of second number to the third number is 9 : 16 and the ratio of first number to the third one is 1 : 4. The second number is:

1. 54

2. 50

3. 36

4. 32

Answer

Let the first, second and third numbers be $a, b$ and $c$ respectively. Then,

$b : c$ = 9 : 16 and $a : c = 1 : 4 = (1 \times 4) : (4 \times 4) = 4 : 16$

∴ $a : b : c$ = 4 : 9 : 16.

Sum of ratio terms = (4 + 9 + 16) = 29.

∴ Second number = $\left(174 \times \dfrac{9}{29}\right) = 54$

Hence, option 1 is the correct option.