Percentage — Exercise 6(C)

What per cent of 7.2 kg is 18 g?

1. 25%

2. 2.5%

3. 0.25%

4. 0.025%

Answer

Both the quantities must be of the same kind (in the same units).

7.2 kg = (7.2 x 1000) g = 7200 g

Required percentage = $\dfrac{18}{7200} \times 100$%

= $\dfrac{1800}{7200}$%

= 0.25%

Hence, option 3 is the correct option.

A number decreased by $27\dfrac{1}{2}$% gives 87. The number is

1. 58

2. 110

3. 120

4. 135

Answer

Let the required number be $x$.

$27\dfrac{1}{2}$% = $\dfrac{55}{2}$%

Since the number is decreased by $\dfrac{55}{2}$%, we get

$\text{New number} = \Big(1 - \dfrac{55}{2 \times 100}\Big) \times x \\[1em] \Rightarrow 87 = \Big(1 - \dfrac{55}{200}\Big) \times x \\[1em] \Rightarrow 87 = \dfrac{145}{200} \times x \\[1em] \Rightarrow x = \dfrac{87 \times 200}{145} \\[1em] \Rightarrow x = \dfrac{17400}{145} \\[1em] \Rightarrow x = 120$

Hence, option 3 is the correct option.

A number increased by $37\dfrac{1}{2}$% gives 33. The number is

1. 22

2. 24

3. 25

4. 27

Answer

Let the required number be $x$.

$37\dfrac{1}{2}$% = $\dfrac{75}{2}$%

Since the number is increased by $\dfrac{75}{2}$%, we get

$\text{New number} = \Big(1 + \dfrac{75}{2 \times 100}\Big) \times x \\[1em] \Rightarrow 33 = \Big(1 + \dfrac{75}{200}\Big) \times x \\[1em] \Rightarrow 33 = \dfrac{275}{200} \times x \\[1em] \Rightarrow x = \dfrac{33 \times 200}{275} \\[1em] \Rightarrow x = \dfrac{6600}{275} \\[1em] \Rightarrow x = 24$

Hence, option 2 is the correct option.

If $37\dfrac{1}{2}$% of a number is 900, then $62\dfrac{1}{2}$% of the number will be

1. 540

2. 1200

3. 1350

4. 1500

Answer

Let the required number be $x$.

$37\dfrac{1}{2}$% = $\dfrac{75}{2}$%

According to the given condition,

$\dfrac{75}{2}$% of $x$ = 900

$\Rightarrow \dfrac{75}{2} \times \dfrac{1}{100} \times x = 900 \\[1em] \Rightarrow \dfrac{75x}{200} = 900 \\[1em] \Rightarrow x = \dfrac{900 \times 200}{75} \\[1em] \Rightarrow x = \dfrac{180000}{75} \\[1em] \Rightarrow x = 2400$

$62\dfrac{1}{2}$% = $\dfrac{125}{2}$%

$62\dfrac{1}{2}$% of 2400 = $\dfrac{125}{2} \times \dfrac{1}{100} \times 2400$

$= \dfrac{125 \times 2400}{200} \\[1em] = \dfrac{300000}{200} \\[1em] = 1500$

Hence, option 4 is the correct option.

By how much per cent is four-fifths of 70 lesser than five-sevenths of 112?

1. 24%

2. 30%

3. 36%

4. 42%

Answer

Four-fifths of 70 = $\dfrac{4}{5} \times 70$

= $\dfrac{280}{5}$

= 56

Five-sevenths of 112 = $\dfrac{5}{7} \times 112$

= $\dfrac{560}{7}$

= 80

As 56 < 80, four-fifths of 70 is lesser than five-sevenths of 112.

Difference = 80 - 56 = 24

Here, 80 is the value with which the comparison is being made, so it is taken as the base.

Required percentage = $\dfrac{24}{80} \times 100$%

= $\dfrac{2400}{80}$%

= 30%

Hence, option 2 is the correct option.

If 11% of a number exceeds 7% of the same by 18, the number is

1. 72

2. 360

3. 450

4. 720

Answer

Let the required number be $x$.

According to the given condition,

11% of $x$ - 7% of $x$ = 18

$\Rightarrow \dfrac{11}{100} \times x - \dfrac{7}{100} \times x = 18 \\[1em] \Rightarrow \dfrac{11x - 7x}{100} = 18 \\[1em] \Rightarrow \dfrac{4x}{100} = 18 \\[1em] \Rightarrow x = \dfrac{18 \times 100}{4} \\[1em] \Rightarrow x = \dfrac{1800}{4} \\[1em] \Rightarrow x = 450$

Hence, option 3 is the correct option.

96% of the population of a town is 23040. The total population of the town is

1. 24000

2. 24936

3. 25640

4. 32256

Answer

Let the total population of the town be $x$.

According to the given condition,

96% of $x$ = 23040

$\Rightarrow \dfrac{96}{100} \times x = 23040 \\[1em] \Rightarrow \dfrac{96x}{100} = 23040 \\[1em] \Rightarrow x = \dfrac{23040 \times 100}{96} \\[1em] \Rightarrow x = \dfrac{2304000}{96} \\[1em] \Rightarrow x = 24000$

Hence, option 1 is the correct option.

In an examination, 65% of the total examinees passed. If the number of failures is 168, the number of those who passed is

1. 218

2. 312

3. 480

4. 648

Answer

Percentage of examinees who passed = 65%

Percentage of examinees who failed = (100 - 65)% = 35%

Let the total number of examinees be $x$.

According to the given condition,

35% of $x$ = 168

$\Rightarrow \dfrac{35}{100} \times x = 168 \\[1em] \Rightarrow \dfrac{35x}{100} = 168 \\[1em] \Rightarrow x = \dfrac{168 \times 100}{35} \\[1em] \Rightarrow x = \dfrac{16800}{35} \\[1em] \Rightarrow x = 480$

Total number of examinees = 480

Number of examinees who passed = 480 - 168 = 312

Hence, option 2 is the correct option.

The price of an item is increased by 20% and then decreased by 20%. The final price as compared to original price is

1. 20% less

2. 20% more

3. 4% less

4. 4% more

Answer

Let the original price of the item be ₹100.

Increased price = $\Big(1 + \dfrac{20}{100}\Big) \times 100$

$= \dfrac{120}{100} \times 100 \\[1em] = ₹120$

Final price after the increased price is decreased by 20% = $\Big(1 - \dfrac{20}{100}\Big) \times 120$

$= \dfrac{80}{100} \times 120 \\[1em] = \dfrac{9600}{100} \\[1em] = ₹96$

As ₹96 < ₹100, the final price is less than the original price.

Decrease in price = ₹100 - ₹96 = ₹4

Percentage decrease = $\dfrac{4}{100} \times 100$% = 4%

Hence, option 3 is the correct option.

A customer asks for the production of $x$ number of goods. The company produces $y$ number of goods daily out of which $z$% are unfit for sale. The order will be completed in

1. $\dfrac{100x}{y(z - 1)}$ days

2. $\dfrac{x}{100y(1 - z)}$ days

3. $\dfrac{100yz}{x}$ days

4. $\dfrac{100x}{y(100 - z)}$ days

Answer

Number of goods produced daily = $y$

Percentage of goods unfit for sale = $z$%

Percentage of goods fit for sale = $(100 - z)$%

Number of goods fit for sale produced daily = $(100 - z)$% of $y$

= $\dfrac{(100 - z)}{100} \times y$

= $\dfrac{y(100 - z)}{100}$

Number of goods required by the customer = $x$

Number of days required to complete the order = $\dfrac{\text{Total goods required}}{\text{Goods fit for sale produced daily}}$

= $\dfrac{x}{\dfrac{y(100 - z)}{100}}$

= $x \times \dfrac{100}{y(100 - z)}$

= $\dfrac{100x}{y(100 - z)}$ days

Hence, option 4 is the correct option.

A man's wages were decreased by 50%. Again the reduced wages were increased by 50%. He has a loss of

1. 0%

2. 2.5%

3. 0.25%

4. 25%

Answer

Let the original wages of the man be ₹100.

Reduced wages = $\Big(1 - \dfrac{50}{100}\Big) \times 100$

$= \dfrac{50}{100} \times 100 \\[1em] = ₹50$

Final wages after the reduced wages are increased by 50% = $\Big(1 + \dfrac{50}{100}\Big) \times 50$

$= \dfrac{150}{100} \times 50 \\[1em] = \dfrac{7500}{100} \\[1em] = ₹75$

As ₹75 < ₹100, the man has a loss.

Loss = ₹100 - ₹75 = ₹25

Loss percentage = $\dfrac{25}{100} \times 100$% = 25%

Hence, option 4 is the correct option.

If A's income is 30% more than B's, then how much per cent is B's income less than A's?

1. $23\dfrac{1}{13}$%

2. 25%

3. 30%

4. $33\dfrac{1}{3}$%

Answer

Let B's income be ₹100.

A's income is 30% more than B's income.

A's income = $\Big(1 + \dfrac{30}{100}\Big) \times 100$

$= \dfrac{130}{100} \times 100 \\[1em] = ₹130$

Difference in the incomes = ₹130 - ₹100 = ₹30

To find how much per cent B's income is less than A's, A's income is taken as the base.

Required percentage = $\dfrac{30}{130} \times 100$%

= $\dfrac{3000}{130}$%

= $\dfrac{300}{13}$%

= $23\dfrac{1}{13}$%

Hence, option 1 is the correct option.

If A's income is 30% less than B's, then how much per cent is B's income more than A's?

1. 30%

2. $32\dfrac{1}{10}$%

3. $42\dfrac{6}{7}$%

4. 51%

Answer

Let B's income be ₹100.

A's income is 30% less than B's income.

A's income = $\Big(1 - \dfrac{30}{100}\Big) \times 100$

$= \dfrac{70}{100} \times 100 \\[1em] = ₹70$

Difference in the incomes = ₹100 - ₹70 = ₹30

To find how much per cent B's income is more than A's, A's income is taken as the base.

Required percentage = $\dfrac{30}{70} \times 100$%

= $\dfrac{3000}{70}$%

= $\dfrac{300}{7}$%

= $42\dfrac{6}{7}$%

Hence, option 3 is the correct option.

If the numerator of a fraction is increased by 20% and its denominator be diminished by 10%, the value of the fraction is $\dfrac{16}{21}$. The original fraction is

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

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

3. $\dfrac{4}{7}$

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

Answer

Let the original fraction be $\dfrac{a}{b}$.

New numerator = $\Big(1 + \dfrac{20}{100}\Big)a$ = $\dfrac{120a}{100}$ = $\dfrac{6a}{5}$

New denominator = $\Big(1 - \dfrac{10}{100}\Big)b$ = $\dfrac{90b}{100}$ = $\dfrac{9b}{10}$

According to the given condition,

$\dfrac{\dfrac{6a}{5}}{\dfrac{9b}{10}} = \dfrac{16}{21} \\[1em] \Rightarrow \dfrac{6a}{5} \times \dfrac{10}{9b} = \dfrac{16}{21} \\[1em] \Rightarrow \dfrac{60a}{45b} = \dfrac{16}{21} \\[1em] \Rightarrow \dfrac{4a}{3b} = \dfrac{16}{21} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{16}{21} \times \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{48}{84} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{4}{7}$

Hence, option 3 is the correct option.

In an examination, it is required to get 36% of maximum marks to pass. A student got 113 marks and was declared failed by 85 marks. The maximum marks are

1. 500

2. 550

3. 640

4. 720

Answer

Marks secured by the student = 113

Marks by which he failed = 85

Passing marks = 113 + 85 = 198

Let the maximum marks be $x$.

According to the given condition,

36% of $x$ = 198

$\Rightarrow \dfrac{36}{100} \times x = 198 \\[1em] \Rightarrow \dfrac{36x}{100} = 198 \\[1em] \Rightarrow x = \dfrac{198 \times 100}{36} \\[1em] \Rightarrow x = \dfrac{19800}{36} \\[1em] \Rightarrow x = 550$

Hence, option 2 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 is

1. 11 : 14

2. 14 : 11

3. 25 : 28

4. 28 : 25

Answer

Let the total strength of the class be 100.

Let the number of boys be $b$ and the number of girls be $g$.

$b + g$ = 100 ...........(1)

The number of boys is more than the number of girls by 12% of the total strength.

$b - g$ = 12% of 100

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

$b - g$ = 12 ...........(2)

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

$2b = 112 \\[1em] \Rightarrow b = \dfrac{112}{2} \\[1em] \Rightarrow b = 56$

Substituting $b$ = 56 in equation (1), we get

$56 + g = 100 \\[1em] \Rightarrow g = 100 - 56 \\[1em] \Rightarrow g = 44$

Ratio of boys to girls = 56 : 44

= $\dfrac{56}{44}$

= $\dfrac{14}{11}$

= 14 : 11

Hence, option 2 is the correct option.

Fill in the blanks:

(i) 240 ml is ________ % of 3 l.

(ii) If $x$% of 35 is 42, then $x$ = ________.

(iii) $x$% of $y$ is $y$% of ________.

(iv) A period of 4 hours 30 min is ________ % of a day.

(v) The number which exceeds 20% of itself by 40, is ________.

(vi) If 24-carat gold is 100% pure, then the percentage of pure gold in 22-carat gold is ________.

Answer

(i) Both the quantities must be in the same units.

3 l = (3 x 1000) ml = 3000 ml

Required percentage = $\dfrac{240}{3000} \times 100$%

= $\dfrac{24000}{3000}$%

= 8%

Hence, 240 ml is 8% of 3 l.

(ii) $x$% of 35 = 42

$\Rightarrow \dfrac{x}{100} \times 35 = 42 \\[1em] \Rightarrow \dfrac{35x}{100} = 42 \\[1em] \Rightarrow x = \dfrac{42 \times 100}{35} \\[1em] \Rightarrow x = \dfrac{4200}{35} \\[1em] \Rightarrow x = 120$

Hence, if $x$% of 35 is 42, then $x$ = 120.

(iii) $x$% of $y = \dfrac{x}{100} \times y = \dfrac{xy}{100}$

$y$% of $x = \dfrac{y}{100} \times x = \dfrac{xy}{100}$

As both are equal, $x$% of $y$ is $y$% of $x$.

Hence, $x$% of $y$ is $y$% of $\bm{x}$.

(iv) Both the quantities must be in the same units.

4 hours 30 min = (4 x 60 + 30) min = 270 min

1 day = 24 hours = (24 x 60) min = 1440 min

Required percentage = $\dfrac{270}{1440} \times 100$%

= $\dfrac{27000}{1440}$%

= $\dfrac{75}{4}$%

= $18\dfrac{3}{4}$%

Hence, a period of 4 hours 30 min is $\mathbf{18\dfrac{3}{4}}$% of a day.

(v) Let the required number be $x$.

According to the given condition,

$x$ - 20% of $x$ = 40

$\Rightarrow x - \dfrac{20}{100} \times x = 40 \\[1em] \Rightarrow \dfrac{100x - 20x}{100} = 40 \\[1em] \Rightarrow \dfrac{80x}{100} = 40 \\[1em] \Rightarrow x = \dfrac{40 \times 100}{80} \\[1em] \Rightarrow x = \dfrac{4000}{80} \\[1em] \Rightarrow x = 50$

Hence, the number which exceeds 20% of itself by 40, is 50.

(vi) 24-carat gold is 100% pure.

Percentage of pure gold in 22-carat gold = $\dfrac{22}{24} \times 100$%

= $\dfrac{2200}{24}$%

= $\dfrac{275}{3}$%

= $91\dfrac{2}{3}$%

Hence, the percentage of pure gold in 22-carat gold is $\mathbf{91\dfrac{2}{3}}$%.

Write true (T) or false (F):

(i) $16\dfrac{2}{3}$% = $0.1\dot{6}$. ________

(ii) 5 g is 5% of 1 kg. ________

(iii) If we double a number, we increase it by 100%. ________

(iv) 36% when expressed as a ratio is equivalent to 9 : 50. ________

(v) If 26% of $x$ = 65, then $x$ = 250. ________

Answer

(i) $16\dfrac{2}{3}$% = $\dfrac{50}{3}$%

$\Rightarrow \dfrac{50}{3} \times \dfrac{1}{100} \\[1em] \Rightarrow \dfrac{50}{300} \\[1em] \Rightarrow \dfrac{1}{6} \\[1em] \Rightarrow 0.1666.....$

In 0.1666..... the digit 6 repeats endlessly, so it is written as $0.1\dot{6}$.

Hence, the given statement is true.

(ii) Both the quantities must be in the same units.

1 kg = (1 x 1000) g = 1000 g

Required percentage = $\dfrac{5}{1000} \times 100$%

= $\dfrac{500}{1000}$%

= 0.5%

As 5 g is 0.5% of 1 kg and not 5% of 1 kg,

Hence, the given statement is false.

(iii) Let the number be $x$.

On doubling it, the new number = $2x$.

Increase in the number = $2x - x = x$

Percentage increase = $\dfrac{x}{x} \times 100$% = 100%

Hence, the given statement is true.

(iv) 36% = $\dfrac{36}{100}$

$\Rightarrow \dfrac{9}{25} \\[1em] \Rightarrow 9 : 25$

As 36% is equivalent to 9 : 25 and not 9 : 50,

Hence, the given statement is false.

(v) 26% of $x$ = 65

$\Rightarrow \dfrac{26}{100} \times x = 65 \\[1em] \Rightarrow \dfrac{26x}{100} = 65 \\[1em] \Rightarrow x = \dfrac{65 \times 100}{26} \\[1em] \Rightarrow x = \dfrac{6500}{26} \\[1em] \Rightarrow x = 250$

Hence, the given statement is true.

Four friends studying in different schools sat down to discuss their scores in their mid-term examinations. Their scores are as under:

(i) A secured 56 marks out of 80.

(ii) B secured 108 marks out of 150.

(iii) C secured 51 marks out of 75.

(iv) D secured 87 marks out of 120.

1. Which of the four friends had the best performance?
   (a) A
   (b) B
   (c) C
   (d) D

2. C got his test paper re-evaluated and his score increased by 12%. His new score is:
   (a) 56
   (b) 57.5
   (c) 58
   (d) 60

3. If the pass percentage in A's test was 40%, how many percent more marks did he obtain than the passing marks?
   (a) 32%
   (b) 40%
   (c) 60%
   (d) 75%

4. If D obtained 45 marks more than the passing marks, then what was the pass percentage in his test?
   (a) 34%
   (b) 35%
   (c) 38%
   (d) 40%

Answer

1. As the four friends were examined out of different maximum marks, their performances can be compared only by expressing their scores as percentages.

Percentage of marks secured by A = $\dfrac{56}{80} \times 100$%

= $\dfrac{5600}{80}$%

= 70%

Percentage of marks secured by B = $\dfrac{108}{150} \times 100$%

= $\dfrac{10800}{150}$%

= 72%

Percentage of marks secured by C = $\dfrac{51}{75} \times 100$%

= $\dfrac{5100}{75}$%

= 68%

Percentage of marks secured by D = $\dfrac{87}{120} \times 100$%

= $\dfrac{8700}{120}$%

= 72.5%

Arranging the percentages in ascending order, we get

68% < 70% < 72% < 72.5%

The highest percentage is 72.5%, which is secured by D.

Hence, option (d) is the correct option.

2. C secured 51 marks out of 75.

Percentage of marks secured by C = $\dfrac{51}{75} \times 100$% = 68%

On re-evaluation his score increased by 12%, that is, by 12% of the maximum marks.

Increase in his marks = 12% of 75

$= \dfrac{12}{100} \times 75 \\[1em] = \dfrac{900}{100} \\[1em] = 9$

New score of C = 51 + 9 = 60

New percentage of marks secured by C = (68 + 12)% = 80%

Hence, option (d) is the correct option.

3. A secured 56 marks out of 80.

Pass percentage in A's test = 40%

Passing marks = 40% of 80

$= \dfrac{40}{100} \times 80 \\[1em] = \dfrac{3200}{100} \\[1em] = 32$

As 56 > 32, A obtained more marks than the passing marks.

Difference in the marks = 56 - 32 = 24

To find how many per cent more marks he obtained than the passing marks, the passing marks are taken as the base.

Required percentage = $\dfrac{24}{32} \times 100$%

= $\dfrac{2400}{32}$%

= 75%

Hence, option (d) is the correct option.

4. D secured 87 marks out of 120.

Marks obtained by D more than the passing marks = 45

Passing marks = 87 - 45 = 42

Pass percentage = $\dfrac{42}{120} \times 100$%

= $\dfrac{4200}{120}$%

= 35%

Hence, option (b) is the correct option.

Assertion (A): The population of a village is 8000. Out of these, 80% are literate and of these literate people, 40% are women. The ratio of the number of literate women to the number of illiterate people is 8 : 5.

Reason (R): Percentage can be converted into ratios and vice-versa.

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

Total population of the village = 8000

Number of literate people = 80% of 8000

$= \dfrac{80}{100} \times 8000 \\[1em] = \dfrac{640000}{100} \\[1em] = 6400$

Number of literate women = 40% of 6400

$= \dfrac{40}{100} \times 6400 \\[1em] = \dfrac{256000}{100} \\[1em] = 2560$

Number of illiterate people = 8000 - 6400 = 1600

Ratio of the number of literate women to the number of illiterate people = 2560 : 1600

= $\dfrac{2560}{1600}$

= $\dfrac{8}{5}$

= 8 : 5

So, Assertion (A) is true.

A percentage is a fraction with denominator 100, and every fraction in its simplest form can be written as a ratio. Conversely, a ratio can be written as a fraction and then multiplied by 100 to get a percentage.

So, Reason (R) is true.

Reason (R) only states that percentages and ratios are interconvertible. The Assertion (A), however, is established by first finding the actual number of literate women and the actual number of illiterate people, and then forming their ratio. So, Reason (R) is not the correct explanation of Assertion (A).

Hence, option 2 is the correct option.

Assertion (A): If 240 is increased by 5%, it becomes 252.

Reason (R): If a value $p$ is increased by $x$%, then the new values becomes $\left(1 + \dfrac{x}{100}\right)p$.

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 increasing 240 by 5%, we get

$\text{New number} = \Big(1 + \dfrac{5}{100}\Big) \times 240 \\[1em] = \dfrac{105}{100} \times 240 \\[1em] = \dfrac{25200}{100} \\[1em] = 252$

So, Assertion (A) is true.

If a value $p$ is increased by $x$%, then the increase = $\dfrac{x}{100} \times p$ and the new value = $p + \dfrac{x}{100} \times p = \left(1 + \dfrac{x}{100}\right)p$.

So, Reason (R) is true.

The new value 252 in Assertion (A) is obtained by applying exactly the formula given in Reason (R) with $p$ = 240 and $x$ = 5. So, Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

If $x$% of $y$ is 1000 and $y$% of $z$ is 2000, then:

1. $x = 3z$

2. $x = 4z$

3. $x = \dfrac{z}{2}$

4. $x = \dfrac{z}{3}$

Answer

$x$% of $y$ = 1000

$\Rightarrow \dfrac{x}{100} \times y = 1000 \\[1em] \Rightarrow xy = 1000 \times 100 \\[1em] \Rightarrow xy = 100000$

$xy$ = 100000 ...........(1)

$y$% of $z$ = 2000

$\Rightarrow \dfrac{y}{100} \times z = 2000 \\[1em] \Rightarrow yz = 2000 \times 100 \\[1em] \Rightarrow yz = 200000$

$yz$ = 200000 ...........(2)

Dividing equation (1) by equation (2), we get

$\dfrac{xy}{yz} = \dfrac{100000}{200000} \\[1em] \Rightarrow \dfrac{x}{z} = \dfrac{1}{2} \\[1em] \Rightarrow x = \dfrac{z}{2}$

Hence, option 3 is the correct option.

In an examination, 27% students failed in Science, and 24% students failed in Mathematics. If 20% students failed in both the subjects, then the percentage of students who failed in one or both the subjects is:

1. 25%

2. 31%

3. 51%

4. 71%

Answer

Percentage of students who failed in Science = 27%

Percentage of students who failed in Mathematics = 24%

Percentage of students who failed in both the subjects = 20%

Percentage of students who failed in Science only = (27 - 20)% = 7%

Percentage of students who failed in Mathematics only = (24 - 20)% = 4%

Percentage of students who failed in one or both the subjects = (7 + 20 + 4)% = 31%

Hence, option 2 is the correct option.

What is 25% of 25% equal to?

1. 6.25

2. 0.625

3. 0.0625

4. 0.00625

Answer

25% = $\dfrac{25}{100}$

25% of 25% = $\dfrac{25}{100} \times \dfrac{25}{100}$

$= \dfrac{625}{10000} \\[1em] = 0.0625$

Hence, option 3 is the correct option.

($x$% of $y + y$% of $x$) is equal to:

1. $x$% of $(x + y)$

2. $y$% of $(x + y)$

3. 50% of $(x + y)$

4. 2% of $xy$

Answer

$x$% of $y = \dfrac{x}{100} \times y = \dfrac{xy}{100}$

$y$% of $x = \dfrac{y}{100} \times x = \dfrac{xy}{100}$

$x$% of $y + y$% of $x = \dfrac{xy}{100} + \dfrac{xy}{100}$

= $\dfrac{2xy}{100}$

= $\dfrac{2}{100} \times xy$

= 2% of $xy$

Hence, option 4 is the correct option.

Subtracting 5% of $x$ from $x$ is equivalent to multiplying $x$ by:

1. 95

2. 0.95

3. 0.5

4. 9.5

Answer

5% of $x$ = $\dfrac{5}{100} \times x = \dfrac{5x}{100}$

$x$ - 5% of $x$ = $x - \dfrac{5x}{100}$

$= \dfrac{100x - 5x}{100} \\[1em] = \dfrac{95x}{100} \\[1em] = 0.95x$

Hence, option 2 is the correct option.

75% of a number when added to 75 is equal to the number. The number is:

1. 200

2. 250

3. 300

4. 400

Answer

Let the required number be $x$.

According to the given condition,

75% of $x$ + 75 = $x$

$\Rightarrow \dfrac{75}{100} \times x + 75 = x \\[1em] \Rightarrow \dfrac{75x}{100} + 75 = x \\[1em] \Rightarrow 75 = x - \dfrac{75x}{100} \\[1em] \Rightarrow 75 = \dfrac{100x - 75x}{100} \\[1em] \Rightarrow 75 = \dfrac{25x}{100} \\[1em] \Rightarrow x = \dfrac{75 \times 100}{25} \\[1em] \Rightarrow x = \dfrac{7500}{25} \\[1em] \Rightarrow x = 300$

Hence, option 3 is the correct option.

3 litres of water is added to 15 litres of a mixture of a 20% solution of milk and water. The strength of milk now is:

1. 50%

2. $\dfrac{50}{3}$%

3. $\dfrac{100}{3}$%

4. $\dfrac{200}{3}$%

Answer

Quantity of the mixture = 15 litres

Strength of milk in the mixture = 20%

Quantity of milk in the mixture = 20% of 15 litres

$= \dfrac{20}{100} \times 15 \\[1em] = \dfrac{300}{100} \\[1em] = 3 \text{ litres}$

On adding 3 litres of water, only the quantity of water changes and the quantity of milk remains 3 litres.

Quantity of the new mixture = (15 + 3) litres = 18 litres

Strength of milk now = $\dfrac{3}{18} \times 100$%

= $\dfrac{300}{18}$%

= $\dfrac{50}{3}$%

Hence, option 2 is the correct option.

If $x = \dfrac{y^2}{z}$ and $y$ and $z$ both are increased in value by 10%, then the value of $x$ is:

1. increased by 10%

2. increased by 11%

3. unchanged

4. none of these

Answer

$x = \dfrac{y^2}{z}$

On increasing $y$ by 10%,

New value of $y = \Big(1 + \dfrac{10}{100}\Big)y = \dfrac{110y}{100} = \dfrac{11y}{10}$

On increasing $z$ by 10%,

New value of $z = \Big(1 + \dfrac{10}{100}\Big)z = \dfrac{110z}{100} = \dfrac{11z}{10}$

New value of $x = \dfrac{\Big(\dfrac{11y}{10}\Big)^2}{\dfrac{11z}{10}}$

$= \dfrac{\dfrac{121y^2}{100}}{\dfrac{11z}{10}} \\[1em] = \dfrac{121y^2}{100} \times \dfrac{10}{11z} \\[1em] = \dfrac{1210y^2}{1100z} \\[1em] = \dfrac{11}{10} \times \dfrac{y^2}{z} \\[1em] = \dfrac{11x}{10}$

Increase in the value of $x = \dfrac{11x}{10} - x$

$= \dfrac{11x - 10x}{10} \\[1em] = \dfrac{x}{10}$

Percentage increase = $\dfrac{\dfrac{x}{10}}{x} \times 100$%

= $\dfrac{x}{10} \times \dfrac{1}{x} \times 100$%

= $\dfrac{100}{10}$%

= 10%

Hence, option 1 is the correct option.

If 35% of A's income is equal to 25% of B's income, then the ratio of their incomes is:

1. 4 : 3

2. 4 : 7

3. 5 : 7

4. 7 : 5

Answer

Let A's income be ₹$a$ and B's income be ₹$b$.

According to the given condition,

35% of $a$ = 25% of $b$

$\Rightarrow \dfrac{35}{100} \times a = \dfrac{25}{100} \times b \\[1em] \Rightarrow 35a = 25b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{25}{35} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{5}{7}$

Ratio of their incomes = 5 : 7

Hence, option 3 is the correct option.

Ratio of number of boys to the number of girls in a class is 2 : $x$. If percentage of girls is 30%, then the value of $x$ is:

1. 3

2. $\dfrac{7}{6}$

3. $\dfrac{6}{7}$

4. $\dfrac{14}{3}$

Answer

Percentage of girls in the class = 30%

Percentage of boys in the class = (100 - 30)% = 70%

Ratio of the number of boys to the number of girls = 70 : 30

Also, the ratio of the number of boys to the number of girls = 2 : $x$

$\dfrac{2}{x} = \dfrac{70}{30} \\[1em] \Rightarrow \dfrac{2}{x} = \dfrac{7}{3} \\[1em] \Rightarrow 7x = 2 \times 3 \\[1em] \Rightarrow 7x = 6 \\[1em] \Rightarrow x = \dfrac{6}{7}$

Hence, option 3 is the correct option.