Percentage

To convert a percentage into a fraction, replace the % sign with $\frac{1}{100}$ and reduce the fraction to simplest form.

$$\begin{aligned}\Rightarrow 68 \times \frac{1}{100}\\[1em] \Rightarrow \frac{68}{100}\\[1em] \Rightarrow \frac{17}{25}\end{aligned}$$

Hence, 68% = $\mathbf{\frac{17}{25}}$.

To convert a percentage into a fraction, replace the % sign with $\frac{1}{100}$ and reduce the fraction to simplest form.

$3\frac{1}{3}$% = $\frac{10}{3}$%

$$\begin{aligned}\Rightarrow \frac{10}{3} \times \frac{1}{100}\\[1em] \Rightarrow \frac{10}{300}\\[1em] \Rightarrow \frac{1}{30}\end{aligned}$$

Hence, $\mathbf{3\frac{1}{3}}$% = $\mathbf{\frac{1}{30}}$.

To convert a percentage into a fraction, replace the % sign with $\frac{1}{100}$ and reduce the fraction to simplest form.

$$\begin{aligned}\Rightarrow 224 \times \frac{1}{100}\\[1em] \Rightarrow \frac{224}{100}\\[1em] \Rightarrow \frac{56}{25}\\[1em] \Rightarrow 2\frac{6}{25}\end{aligned}$$

Hence, 224% = $\mathbf{2\frac{6}{25}}$.

To convert a percentage into a fraction, replace the % sign with $\frac{1}{100}$ and reduce the fraction to simplest form.

$$\begin{aligned}\Rightarrow 0.05 \times \frac{1}{100}\\[1em] \Rightarrow \frac{0.05}{100}\\[1em] \Rightarrow \frac{5}{10000}\\[1em] \Rightarrow \frac{1}{2000}\end{aligned}$$

Hence, 0.05% = $\mathbf{\frac{1}{2000}}$.

To convert a fraction into a percentage, multiply the fraction by 100 and put the % sign.

$\Rightarrow \frac{2}{15} \times 100$%

$\Rightarrow \frac{200}{15}$%

$\Rightarrow \frac{40}{3}$%

$\Rightarrow 13\frac{1}{3}$%

Hence, $\mathbf{\frac{2}{15}} = \mathbf{13\frac{1}{3}}$%.

To convert a fraction into a percentage, multiply the fraction by 100 and put the % sign.

$\Rightarrow \frac{9}{40} \times 100$%

$\Rightarrow \frac{900}{40}$%

$\Rightarrow \frac{45}{2}$%

$\Rightarrow 22\frac{1}{2}$%

Hence, $\mathbf{\frac{9}{40}} = \mathbf{22\frac{1}{2}}$%.

To convert a fraction into a percentage, multiply the fraction by 100 and put the % sign.

$1\frac{2}{3}$ = $\frac{5}{3}$

$\Rightarrow \frac{5}{3} \times 100$%

$\Rightarrow \frac{500}{3}$%

$\Rightarrow 166\frac{2}{3}$%

Hence, $\mathbf{1\frac{2}{3}} = \mathbf{166\frac{2}{3}}$%.

To convert a fraction into a percentage, multiply the fraction by 100 and put the % sign.

$2\frac{2}{5} = \frac{12}{5}$

$\Rightarrow \frac{12}{5} \times 100$%

$\Rightarrow \frac{1200}{5}$%

⇒ 240%

Hence, $\mathbf{2\frac{2}{5}}$ = 240%.

To convert a ratio into a percentage, first convert the given ratio into a fraction and then convert the fraction into a percentage.

⇒ 13 : 20

$\Rightarrow \frac{13}{20}$

$\Rightarrow \frac{13}{20} \times 100$%

$\Rightarrow \frac{1300}{20}$%

⇒ 65%

Hence, 13 : 20 = 65%.

To convert a ratio into a percentage, first convert the given ratio into a fraction and then convert the fraction into a percentage.

⇒ 11 : 18

$\Rightarrow \frac{11}{18}$

$\Rightarrow \frac{11}{18} \times 100$%

$\Rightarrow \frac{1100}{18}$%

$\Rightarrow \frac{550}{9}$%

$\Rightarrow 61\frac{1}{9}$%

Hence, 11 : 18 = $\mathbf{61\frac{1}{9}}$%.

To convert a ratio into a percentage, first convert the given ratio into a fraction and then convert the fraction into a percentage.

⇒ 87 : 25

$\Rightarrow \frac{87}{25}$

$\Rightarrow \frac{87}{25} \times 100$%

$\Rightarrow \frac{8700}{25}$%

⇒ 348%

Hence, 87 : 25 = 348%.

To convert a ratio into a percentage, first convert the given ratio into a fraction and then convert the fraction into a percentage.

$6\frac{1}{4} : 4\frac{3}{8} = \frac{25}{4} : \frac{35}{8}$

$$\begin{aligned}\Rightarrow \frac{\frac{25}{4}}{\frac{35}{8}}\\[1em] \Rightarrow \frac{25}{4} \times \frac{8}{35}\\[1em] \Rightarrow \frac{200}{140}\\[1em] \Rightarrow \frac{10}{7}\end{aligned}$$

Converting the ratio into percentage:

$\Rightarrow \frac{10}{7} \times 100$%

$\Rightarrow \frac{1000}{7}$%

$\Rightarrow 142\frac{6}{7}$%

Hence, $\mathbf{6\frac{1}{4} : 4\frac{3}{8} = 142\frac{6}{7}}$%.

To convert a decimal into a percentage, first convert the given decimal into a fraction and then multiply the fraction by 100 and put the % sign.

0.2 = $\frac{2}{10}$

$\Rightarrow \frac{2}{10} \times 100$%

$\Rightarrow \frac{200}{10}$%

⇒ 20%

Hence, 0.2 = 20%.

To convert a decimal into a percentage, first convert the given decimal into a fraction and then multiply the fraction by 100 and put the % sign.

0.06 = $\frac{6}{100}$

$\Rightarrow \frac{6}{100} \times 100$%

$\Rightarrow \frac{600}{100}$%

⇒ 6%

Hence, 0.06 = 6%.

To convert a decimal into a percentage, first convert the given decimal into a fraction and then multiply the fraction by 100 and put the % sign.

0.008 = $\frac{8}{1000}$

$\Rightarrow \frac{8}{1000} \times 100$%

$\Rightarrow \frac{800}{1000}$%

⇒ 0.8%

Hence, 0.008 = 0.8%.

To convert a decimal into a percentage, first convert the given decimal into a fraction and then multiply the fraction by 100 and put the % sign.

2.4 = $\frac{24}{10}$

$\Rightarrow \frac{24}{10} \times 100$%

$\Rightarrow \frac{2400}{10}$%

⇒ 240%

Hence, 2.4 = 240%.

To convert a percentage into a decimal, first convert the percentage into a fraction by replacing the % sign with $\frac{1}{100}$, then convert the fraction into a decimal.

$\Rightarrow \frac{25}{100}$

⇒ 0.25

Hence, 25% = 0.25.

To convert a percentage into a decimal, first convert the percentage into a fraction by replacing the % sign with $\frac{1}{100}$, then convert the fraction into a decimal.

$\Rightarrow \frac{4}{100}$

⇒ 0.04

Hence, 4% = 0.04.

To convert a percentage into a decimal, first convert the percentage into a fraction by replacing the % sign with $\frac{1}{100}$, then convert the fraction into a decimal.

$3\frac{1}{5}$% = $\frac{16}{5}$%

$$\begin{aligned}\Rightarrow \frac{16}{5} \times \frac{1}{100}\\[1em] \Rightarrow \frac{16}{500}\\[1em] \Rightarrow 0.032\end{aligned}$$

Hence, $\mathbf{3\frac{1}{5}}$% = 0.032.

To convert a percentage into a decimal, first convert the percentage into a fraction by replacing the % sign with $\frac{1}{100}$, then convert the fraction into a decimal.

$$\begin{aligned}\Rightarrow \frac{0.3}{100}\\[1em] \Rightarrow \frac{3}{1000}\\[1em] \Rightarrow 0.003\end{aligned}$$

Hence, 0.3% = 0.003.

To convert a percentage into a ratio, first convert the given percentage into a fraction in the simplest form and then express it as a ratio.

$$\begin{aligned}\Rightarrow \frac{48}{100}\\[1em] \Rightarrow \frac{12}{25}\\[1em] \Rightarrow 12 : 25\end{aligned}$$

Hence, 48% = 12 : 25.

To convert a percentage into a ratio, first convert the given percentage into a fraction in the simplest form and then express it as a ratio.

$26\frac{2}{3}$% = $\frac{80}{3}$%

$$\begin{aligned}\Rightarrow \frac{80}{3} \times \frac{1}{100}\\[1em] \Rightarrow \frac{80}{300}\\[1em] \Rightarrow \frac{4}{15}\\[1em] \Rightarrow 4 : 15\end{aligned}$$

Hence, $\mathbf{26\frac{2}{3}}$% = 4 : 15.

To convert a percentage into a ratio, first convert the given percentage into a fraction in the simplest form and then express it as a ratio.

$$\begin{aligned}\Rightarrow \frac{0.06}{100}\\[1em] \Rightarrow \frac{6}{10000}\\[1em] \Rightarrow \frac{3}{5000}\\[1em] \Rightarrow 3 : 5000\end{aligned}$$

Hence, 0.06% = 3 : 5000.

To convert a percentage into a ratio, first convert the given percentage into a fraction in the simplest form and then express it as a ratio.

$$\begin{aligned}\Rightarrow \frac{120}{100}\\[1em] \Rightarrow \frac{6}{5}\\[1em] \Rightarrow 6 : 5\end{aligned}$$

Hence, 120% = 6 : 5.

To find a percentage of a given quantity, change the percentage into a fraction and multiply it by the given quantity.

$$\begin{aligned}\Rightarrow \frac{33}{100} \times 50\\[1em] \Rightarrow \frac{1650}{100}\\[1em] \Rightarrow ₹16.50\end{aligned}$$

Hence, 33% of ₹50 = ₹16.50.

To find a percentage of a given quantity, change the percentage into a fraction and multiply it by the given quantity.

3 m = (3 x 100) cm = 300 cm

$6\frac{2}{3}$% of 3 m = $\frac{20}{3}$% of 300 cm

$$\begin{aligned}\Rightarrow \frac{20}{3} \times \frac{1}{100} \times 300\\[1em] \Rightarrow \frac{6000}{300}\\[1em] \Rightarrow 20 \text{ cm}\end{aligned}$$

Hence, $\mathbf{6\frac{2}{3}}$% of 3 m = 20 cm.

To find a percentage of a given quantity, change the percentage into a fraction and multiply it by the given quantity.

35 kg = (35 x 1000) g = 35000 g

$$\begin{aligned}\Rightarrow \frac{0.6}{100} \times 35000\\[1em] \Rightarrow \frac{21000}{100}\\[1em] \Rightarrow 210 \text{ g}\end{aligned}$$

Hence, 0.6% of 35 kg = 210 g.

To find a percentage of a given quantity, change the percentage into a fraction and multiply it by the given quantity.

5 l = (5 x 1000) ml = 5000 ml

$3\frac{1}{4}$% of 5 l = $\frac{13}{4}$% of 5000 ml

$$\begin{aligned}\Rightarrow \frac{13}{4} \times \frac{1}{100} \times 5000\\[1em] \Rightarrow \frac{65000}{400}\\[1em] \Rightarrow 162.5 \text{ ml}\end{aligned}$$

Hence, $\mathbf{3\frac{1}{4}}$% of 5 l = 162.5 ml.

To express $x$ as a percentage of $y$, percentage = $\Big(\frac{x}{y} \times 100\Big)$%. Both the quantities must be of the same kind (in the same units).

Required percentage = $\frac{5}{9} \times 100$%

= $\frac{500}{9}$%

= $55\frac{5}{9}$%

Hence, ₹5 is $\mathbf{55\frac{5}{9}}$% of ₹9.

To express $x$ as a percentage of $y$, percentage = $\Big(\frac{x}{y} \times 100\Big)$%. Both the quantities must be of the same kind (in the same units).

Required percentage = $\frac{80}{32} \times 100$%

= $\frac{8000}{32}$%

= 250%

Hence, 80 m is 250% of 32 m.

To express $x$ as a percentage of $y$, percentage = $\Big(\frac{x}{y} \times 100\Big)$%. Both the quantities must be of the same kind (in the same units).

Required percentage = $\frac{65}{50} \times 100$%

= $\frac{6500}{50}$%

= 130%

Hence, 65 kg is 130% of 50 kg.

To express $x$ as a percentage of $y$, percentage = $\Big(\frac{x}{y} \times 100\Big)$%. Both the quantities must be of the same kind (in the same units).

5 litres = (5 x 1000) ml = 5000 ml

Required percentage = $\frac{400}{5000} \times 100$%

= $\frac{40000}{5000}$%

= 8%

Hence, 400 ml is 8% of 5 litres.

Let the required number be $x$.

According to the given condition,

8% of $x$ = 24

$$\begin{aligned}\Rightarrow \frac{8}{100} \times x = 24 \\[1em] \Rightarrow \frac{8x}{100} = 24 \\[1em] \Rightarrow x = \frac{24 \times 100}{8} \\[1em] \Rightarrow x = \frac{2400}{8} \\[1em] \Rightarrow x = 300\end{aligned}$$

Hence, the required number is 300.

Let the required number be $x$.

According to the given condition,

7.25% of $x$ = 2.9

$$\begin{aligned}\Rightarrow \frac{7.25}{100} \times x = 2.9 \\[1em] \Rightarrow \frac{7.25x}{100} = 2.9 \\[1em] \Rightarrow x = \frac{2.9 \times 100}{7.25} \\[1em] \Rightarrow x = \frac{290}{7.25} \\[1em] \Rightarrow x = 40\end{aligned}$$

Hence, the required number is 40.

Let the required number be $x$.

$6\frac{2}{3}$% = $\frac{20}{3}$%

According to the given condition,

$\frac{20}{3}$% of $x$ = 1

$$\begin{aligned}\Rightarrow \frac{20}{3} \times \frac{1}{100} \times x = 1 \\[1em] \Rightarrow \frac{20x}{300} = 1 \\[1em] \Rightarrow \frac{x}{15} = 1 \\[1em] \Rightarrow x = 15\end{aligned}$$

Hence, the required number is 15.

If a value $a$ is increased by $x$%, then new value = $\Big(1 + \frac{x}{100}\Big)a$.

On increasing 75 by 24%, we get

$$\begin{aligned}\text{New number} = \Big(1 + \frac{24}{100}\Big) \times 75 \\[1em] = \frac{124}{100} \times 75 \\[1em] = \frac{31}{25} \times 75 \\[1em] = \frac{2325}{25} \\[1em] = 93\end{aligned}$$

Hence, on increasing 75 by 24% we get 93.

If a value $a$ is decreased by $x$%, then new value = $\Big(1 - \frac{x}{100}\Big)a$.

On decreasing 375 by 8%, we get

$$\begin{aligned}\text{New number} = \Big(1 - \frac{8}{100}\Big) \times 375 \\[1em] = \frac{92}{100} \times 375 \\[1em] = \frac{23}{25} \times 375 \\[1em] = \frac{8625}{25} \\[1em] = 345\end{aligned}$$

Hence, on decreasing 375 by 8% we get 345.

Let the required number be $x$.

Since the number is increased by 15%, we get

$$\begin{aligned}x + \frac{x \times 15}{100} = 276 \\[1em] \Rightarrow \frac{100x + 15x}{100} = 276 \\[1em] \Rightarrow \frac{115x}{100} = 276 \\[1em] \Rightarrow x = \frac{276 \times 100}{115} \\[1em] \Rightarrow x = \frac{27600}{115} \\[1em] \Rightarrow x = 240\end{aligned}$$

Hence, the required number is 240.

Let the required number be $x$.

Since the number is decreased by 8%, we get

$$\begin{aligned}x - \frac{x \times 8}{100} = 345 \\[1em] \Rightarrow \frac{100x - 8x}{100} = 345 \\[1em] \Rightarrow \frac{92x}{100} = 345 \\[1em] \Rightarrow x = \frac{345 \times 100}{92} \\[1em] \Rightarrow x = \frac{34500}{92} \\[1em] \Rightarrow x = 375\end{aligned}$$

Hence, the required number is 375.

Given:

Percentage of girls in the school = 36%

Number of boys in the school = 1440

Percentage of boys in the school = (100 - 36)% = 64%

Let the total strength of the school be $x$.

According to the given condition,

64% of $x$ = 1440

$$\begin{aligned}\Rightarrow \frac{64}{100} \times x = 1440 \\[1em] \Rightarrow \frac{64x}{100} = 1440 \\[1em] \Rightarrow x = \frac{1440 \times 100}{64} \\[1em] \Rightarrow x = \frac{144000}{64} \\[1em] \Rightarrow x = 2250\end{aligned}$$

Hence, the total strength of the school is 2250.

Given:

Percentage of salary saved by Geeta = 18%

Amount spent by Geeta per month = ₹10,250

Percentage of salary spent by Geeta = (100 - 18)% = 82%

Let the monthly salary of Geeta be ₹$x$.

According to the given condition,

82% of $x$ = 10250

$$\begin{aligned}\Rightarrow \frac{82}{100} \times x = 10250 \\[1em] \Rightarrow \frac{82x}{100} = 10250 \\[1em] \Rightarrow x = \frac{10250 \times 100}{82} \\[1em] \Rightarrow x = \frac{1025000}{82} \\[1em] \Rightarrow x = 12500\end{aligned}$$

Hence, the monthly salary of Geeta is ₹12,500.

Given:

Percentage of marks required to pass = 40%

Marks secured by Rahul = 178

Marks by which Rahul fails = 32

Passing marks = Marks secured + Marks by which he fails

= 178 + 32

= 210

Let the maximum marks be $x$.

According to the given condition,

40% of $x$ = 210

$$\begin{aligned}\Rightarrow \frac{40}{100} \times x = 210 \\[1em] \Rightarrow \frac{40x}{100} = 210 \\[1em] \Rightarrow x = \frac{210 \times 100}{40} \\[1em] \Rightarrow x = \frac{21000}{40} \\[1em] \Rightarrow x = 525\end{aligned}$$

Hence, the maximum marks are 525.

Given:

Percentage of students absent = 8%

Number of students who attended the school = 1633

Percentage of students present = (100 - 8)% = 92%

Let the total number of students in the school be $x$.

According to the given condition,

92% of $x$ = 1633

$$\begin{aligned}\Rightarrow \frac{92}{100} \times x = 1633 \\[1em] \Rightarrow \frac{92x}{100} = 1633 \\[1em] \Rightarrow x = \frac{1633 \times 100}{92} \\[1em] \Rightarrow x = \frac{163300}{92} \\[1em] \Rightarrow x = 1775\end{aligned}$$

Total number of students in the school = 1775

Number of students who remained absent = 1775 - 1633 = 142

Hence, 142 students remained absent.

Given:

Increase in the price of the article = 14%

Increased price of the article = ₹1,995

Let the original price of the article be ₹$x$.

Increased price = $\Big(1 + \frac{14}{100}\Big)x$ = ₹$\frac{114x}{100}$

According to the given condition,

$$\begin{aligned}\frac{114x}{100} = 1995 \\[1em] \Rightarrow x = \frac{1995 \times 100}{114} \\[1em] \Rightarrow x = \frac{199500}{114} \\[1em] \Rightarrow x = 1750\end{aligned}$$

Hence, the original price of the article was ₹1,750.

Given:

Decrease in the price of the article = 6%

Decreased price of the article = ₹1,551

Let the original price of the article be ₹$x$.

Decreased price = $\Big(1 - \frac{6}{100}\Big)x$ = ₹$\frac{94x}{100}$

According to the given condition,

$$\begin{aligned}\frac{94x}{100} = 1551 \\[1em] \Rightarrow x = \frac{1551 \times 100}{94} \\[1em] \Rightarrow x = \frac{155100}{94} \\[1em] \Rightarrow x = 1650\end{aligned}$$

Hence, the original price of the article was ₹1,650.

Given:

Reduction in the weight of Reenu = 15%

Present weight of Reenu = 52.7 kg

Let the earlier weight of Reenu be $x$ kg.

Reduced weight = $\Big(1 - \frac{15}{100}\Big)x$ = $\frac{85x}{100}$ kg

According to the given condition,

$$\begin{aligned}\frac{85x}{100} = 52.7 \\[1em] \Rightarrow x = \frac{52.7 \times 100}{85} \\[1em] \Rightarrow x = \frac{5270}{85} \\[1em] \Rightarrow x = 62\end{aligned}$$

Hence, the earlier weight of Reenu was 62 kg.

Given:

Percentage of votes secured by the winning candidate = 58%

Margin of votes by which he won = 2560

Percentage of votes secured by the other candidate = (100 - 58)% = 42%

Difference in the percentage of votes = (58 - 42)% = 16%

Let the total number of votes polled be $x$.

According to the given condition,

16% of $x$ = 2560

$$\begin{aligned}\Rightarrow \frac{16}{100} \times x = 2560 \\[1em] \Rightarrow \frac{16x}{100} = 2560 \\[1em] \Rightarrow x = \frac{2560 \times 100}{16} \\[1em] \Rightarrow x = \frac{256000}{16} \\[1em] \Rightarrow x = 16000\end{aligned}$$

Hence, 16000 votes were polled in all.

To compare her performance in the four subjects, we express the marks scored in each subject as a percentage of the maximum marks of that subject.

Percentage of marks in Science = $\frac{60}{75} \times 100$%

= $\frac{6000}{75}$%

= 80%

Percentage of marks in Mathematics = $\frac{84}{100} \times 100$%

= $\frac{8400}{100}$%

= 84%

Percentage of marks in Hindi = $\frac{36}{50} \times 100$%

= $\frac{3600}{50}$%

= 72%

Percentage of marks in English = $\frac{30}{45} \times 100$%

= $\frac{3000}{45}$%

= $\frac{200}{3}$%

= $66\frac{2}{3}$%

Arranging the percentages in ascending order, we get

$66\frac{2}{3}$% < 72% < 80% < 84%

(i) The lowest percentage is $66\frac{2}{3}$%, which is in English.

Hence, her performance is the worst in English.

(ii) The highest percentage is 84%, which is in Mathematics.

Hence, her performance is the best in Mathematics.

(iii) Total marks scored by Preeti = 60 + 84 + 36 + 30 = 210

Total maximum marks = 75 + 100 + 50 + 45 = 270

Aggregate percentage of marks = $\frac{210}{270} \times 100$%

= $\frac{21000}{270}$%

= $\frac{700}{9}$%

= $77\frac{7}{9}$%

Hence, her aggregate percentage of marks is $\mathbf{77\frac{7}{9}}$%.

Let the original price of the article be ₹100.

Increase in the price = 25%

Increased price = $\Big(1 + \frac{25}{100}\Big) \times 100$

$= \frac{125}{100} \times 100$

= ₹125

To restore the price to its former value, the decrease required = ₹(125 - 100) = ₹25

This decrease is to be calculated on the new value, that is, on ₹125.

Required percentage decrease = $\frac{25}{125} \times 100$%

= $\frac{2500}{125}$%

= 20%

Hence, the new value must be decreased by 20%.

Let the original price of the article be ₹100.

Reduction in the price = 10%

Reduced price = $\Big(1 - \frac{10}{100}\Big) \times 100$

$= \frac{90}{100} \times 100$

= ₹90

To restore the price to its former value, the increase required = ₹(100 - 90) = ₹10

This increase is to be calculated on the new value, that is, on ₹90.

Required percentage increase = $\frac{10}{90} \times 100$%

= $\frac{1000}{90}$%

= $\frac{100}{9}$%

= $11\frac{1}{9}$%

Hence, the value must be increased by $\mathbf{11\frac{1}{9}}$%.

Let the earlier consumption of tea be $x$ units and let the price of $x$ units be ₹100.

Increase in the price = 20%

New price of $x$ units = $\Big(1 + \frac{20}{100}\Big) \times 100$

$= \frac{120}{100} \times 100$

= ₹120

So that the expenditure does not increase, the housewife can still spend only ₹100.

Now, ₹120 will fetch $x$ units.

∴ ₹100 will fetch $\frac{x}{120} \times 100$ units = $\frac{5x}{6}$ units

Reduction in consumption = $\Big(x - \frac{5x}{6}\Big)$ units = $\frac{x}{6}$ units

Reduction percentage = $\frac{\frac{x}{6}}{x} \times 100$%

= $\frac{x}{6} \times \frac{1}{x} \times 100$%

= $\frac{100}{6}$%

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

= $16\frac{2}{3}$%

Hence, the housewife should reduce the consumption of tea by $\mathbf{16\frac{2}{3}}$%.

Let the total money the man had be ₹$x$.

Money given to the elder son = 35% of $x$

= $\frac{35}{100} \times x$

= ₹$\frac{7x}{20}$

Remainder = $\Big(x - \frac{7x}{20}\Big)$

$$\begin{aligned}= \frac{20x - 7x}{20} \\[1em] = ₹\frac{13x}{20}\end{aligned}$$

Money given to the younger son = 40% of $\frac{13x}{20}$

$$\begin{aligned}= \frac{40}{100} \times \frac{13x}{20} \\[1em] = \frac{2}{5} \times \frac{13x}{20} \\[1em] = ₹\frac{13x}{50}\end{aligned}$$

Money left with the man = $\Big(\frac{13x}{20} - \frac{13x}{50}\Big)$

$$\begin{aligned}= \frac{65x - 26x}{100} \\[1em] = ₹\frac{39x}{100}\end{aligned}$$

According to the given condition,

$$\begin{aligned}\frac{39x}{100} = 11700 \\[1em] \Rightarrow x = \frac{11700 \times 100}{39} \\[1em] \Rightarrow x = \frac{1170000}{39} \\[1em] \Rightarrow x = 30000\end{aligned}$$

Hence, the man had ₹30,000.

Let the population of the town at the beginning be $x$.

Percentage of population killed in the earthquake = 5%

Population left after the earthquake = $\Big(1 - \frac{5}{100}\Big)x$

$$\begin{aligned}= \frac{95}{100} \times x \\[1em] = \frac{19x}{20}\end{aligned}$$

Percentage of the remainder that left the town = 8%

Population of the town now = $\Big(1 - \frac{8}{100}\Big) \times \frac{19x}{20}$

$$\begin{aligned}= \frac{92}{100} \times \frac{19x}{20} \\[1em] = \frac{23}{25} \times \frac{19x}{20} \\[1em] = \frac{437x}{500}\end{aligned}$$

According to the given condition,

$$\begin{aligned}\frac{437x}{500} = 43700 \\[1em] \Rightarrow x = \frac{43700 \times 500}{437} \\[1em] \Rightarrow x = \frac{21850000}{437} \\[1em] \Rightarrow x = 50000\end{aligned}$$

Hence, the population of the town at the beginning was 50000.

Let the height of tower B be 100 units.

The height of tower A is 20% less than that of B.

Height of tower A = $\Big(1 - \frac{20}{100}\Big) \times 100$

$$\begin{aligned}= \frac{80}{100} \times 100 \\[1em] = 80 \text{ units}\end{aligned}$$

Difference in the heights = (100 - 80) units = 20 units

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

Required percentage = $\frac{20}{80} \times 100$%

= $\frac{2000}{80}$%

= 25%

Hence, B's height is 25% more than that of A.

Given:

Percentage of candidates who failed in English = 30%

Percentage of candidates who failed in G.K. = 35%

Percentage of candidates who failed in both the subjects = 27%

Percentage of candidates who failed in English only = (30 - 27)% = 3%

Percentage of candidates who failed in G.K. only = (35 - 27)% = 8%

Percentage of candidates who failed in one or both the subjects = (3 + 27 + 8)% = 38%

Percentage of candidates who passed in both the subjects = (100 - 38)% = 62%

Let the number of candidates who appeared in the examination be $x$.

According to the given condition,

62% of $x$ = 310

$$\begin{aligned}\Rightarrow \frac{62}{100} \times x = 310 \\[1em] \Rightarrow \frac{62x}{100} = 310 \\[1em] \Rightarrow x = \frac{310 \times 100}{62} \\[1em] \Rightarrow x = \frac{31000}{62} \\[1em] \Rightarrow x = 500\end{aligned}$$

Hence, 500 candidates appeared in the examination.

Given:

Present value of the car = ₹6,50,000

Rate of annual depreciation = 10%

As the value depreciates by 10% every year,

Value after 1 year = $\Big(1 - \frac{10}{100}\Big) \times$ Present value

Value after 2 years = $\Big(1 - \frac{10}{100}\Big)^2 \times$ Present value

$$\begin{aligned}\text{Value after 2 years} = \Big(1 - \frac{10}{100}\Big)^2 \times 650000 \\[1em] = \frac{90}{100} \times \frac{90}{100} \times 650000 \\[1em] = \frac{9}{10} \times \frac{9}{10} \times 650000 \\[1em] = \frac{52650000}{100} \\[1em] = ₹5{,}26{,}500\end{aligned}$$

Hence, the value of the car after 2 years is ₹5,26,500.

Given:

Present population of the village = 80000

Rate of annual increase = 7%

As the population increases by 7% every year,

Population after 1 year = $\Big(1 + \frac{7}{100}\Big) \times$ Present population

Population after 2 years = $\Big(1 + \frac{7}{100}\Big)^2 \times$ Present population

$$\begin{aligned}\text{Population after 2 years} = \Big(1 + \frac{7}{100}\Big)^2 \times 80000 \\[1em] = \frac{107}{100} \times \frac{107}{100} \times 80000 \\[1em] = \frac{11449}{10000} \times 80000 \\[1em] = \frac{915920000}{10000} \\[1em] = 91592\end{aligned}$$

Hence, the population of the village after two years will be 91592.

Let the number be $x$.

Actual value = $\frac{5}{3} \times x$ = $\frac{5x}{3}$

Value obtained by the student = $\frac{3}{5} \times x$ = $\frac{3x}{5}$

As $\frac{5x}{3} \gt \frac{3x}{5}$,

Error = Actual value - Value obtained

$$\begin{aligned}= \frac{5x}{3} - \frac{3x}{5} \\[1em] = \frac{25x - 9x}{15} \\[1em] = \frac{16x}{15}\end{aligned}$$

Percentage error = $\frac{\text{Error}}{\text{Actual Value}} \times 100$%

= $\frac{\frac{16x}{15}}{\frac{5x}{3}} \times 100$%

= $\frac{16x}{15} \times \frac{3}{5x} \times 100$%

= $\frac{16}{25} \times 100$%

= $\frac{1600}{25}$%

= 64%

Hence, the percentage error in the calculation is 64%.

Given:

Percentage of voters who did not cast their vote = 10%

Percentage of the votes polled that were invalid = 10%

Percentage of valid votes got by the successful candidate = 54%

Majority by which he won = 1620

Percentage of valid votes got by the other candidate = (100 - 54)% = 46%

Let the number of valid votes be $x$.

According to the given condition,

(54 - 46)% of $x$ = 1620

$$\begin{aligned}\Rightarrow \frac{8}{100} \times x = 1620 \\[1em] \Rightarrow \frac{8x}{100} = 1620 \\[1em] \Rightarrow x = \frac{1620 \times 100}{8} \\[1em] \Rightarrow x = \frac{162000}{8} \\[1em] \Rightarrow x = 20250\end{aligned}$$

Number of valid votes = 20250

As 10% of the votes polled were invalid, the valid votes are 90% of the votes polled.

Let the number of votes polled be $y$.

$$\begin{aligned}\frac{90}{100} \times y = 20250 \\[1em] \Rightarrow y = \frac{20250 \times 100}{90} \\[1em] \Rightarrow y = \frac{2025000}{90} \\[1em] \Rightarrow y = 22500\end{aligned}$$

Number of votes polled = 22500

As 10% of the voters did not cast their vote, the votes polled are 90% of the voters enrolled.

Let the number of voters enrolled be $z$.

$$\begin{aligned}\frac{90}{100} \times z = 22500 \\[1em] \Rightarrow z = \frac{22500 \times 100}{90} \\[1em] \Rightarrow z = \frac{2250000}{90} \\[1em] \Rightarrow z = 25000\end{aligned}$$

Hence, 25000 voters were enrolled on the voters list.

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 = $\frac{18}{7200} \times 100$%

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

= 0.25%

Hence, option 3 is the correct option.

Let the required number be $x$.

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

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

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

Hence, option 3 is the correct option.

Let the required number be $x$.

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

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

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

Hence, option 2 is the correct option.

Let the required number be $x$.

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

According to the given condition,

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

$$\begin{aligned}\Rightarrow \frac{75}{2} \times \frac{1}{100} \times x = 900 \\[1em] \Rightarrow \frac{75x}{200} = 900 \\[1em] \Rightarrow x = \frac{900 \times 200}{75} \\[1em] \Rightarrow x = \frac{180000}{75} \\[1em] \Rightarrow x = 2400\end{aligned}$$

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

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

$$\begin{aligned}= \frac{125 \times 2400}{200} \\[1em] = \frac{300000}{200} \\[1em] = 1500\end{aligned}$$

Hence, option 4 is the correct option.

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

= $\frac{280}{5}$

= 56

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

= $\frac{560}{7}$

= 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 = $\frac{24}{80} \times 100$%

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

= 30%

Hence, option 2 is the correct option.

Let the required number be $x$.

According to the given condition,

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

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

Hence, option 3 is the correct option.

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

According to the given condition,

96% of $x$ = 23040

$$\begin{aligned}\Rightarrow \frac{96}{100} \times x = 23040 \\[1em] \Rightarrow \frac{96x}{100} = 23040 \\[1em] \Rightarrow x = \frac{23040 \times 100}{96} \\[1em] \Rightarrow x = \frac{2304000}{96} \\[1em] \Rightarrow x = 24000\end{aligned}$$

Hence, option 1 is the correct option.

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

$$\begin{aligned}\Rightarrow \frac{35}{100} \times x = 168 \\[1em] \Rightarrow \frac{35x}{100} = 168 \\[1em] \Rightarrow x = \frac{168 \times 100}{35} \\[1em] \Rightarrow x = \frac{16800}{35} \\[1em] \Rightarrow x = 480\end{aligned}$$

Total number of examinees = 480

Number of examinees who passed = 480 - 168 = 312

Hence, option 2 is the correct option.

Let the original price of the item be ₹100.

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

$$\begin{aligned}= \frac{120}{100} \times 100 \\[1em] = ₹120\end{aligned}$$

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

$$\begin{aligned}= \frac{80}{100} \times 120 \\[1em] = \frac{9600}{100} \\[1em] = ₹96\end{aligned}$$

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

Decrease in price = ₹100 - ₹96 = ₹4

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

Hence, option 3 is the correct option.

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 $y$

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

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

Number of goods required by the customer = $x$

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

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

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

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

Hence, option 4 is the correct option.

Let the original wages of the man be ₹100.

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

$$\begin{aligned}= \frac{50}{100} \times 100 \\[1em] = ₹50\end{aligned}$$

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

$$\begin{aligned}= \frac{150}{100} \times 50 \\[1em] = \frac{7500}{100} \\[1em] = ₹75\end{aligned}$$

As ₹75 < ₹100, the man has a loss.

Loss = ₹100 - ₹75 = ₹25

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

Hence, option 4 is the correct option.

Let B's income be ₹100.

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

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

$$\begin{aligned}= \frac{130}{100} \times 100 \\[1em] = ₹130\end{aligned}$$

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 = $\frac{30}{130} \times 100$%

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

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

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

Hence, option 1 is the correct option.

Let B's income be ₹100.

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

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

$$\begin{aligned}= \frac{70}{100} \times 100 \\[1em] = ₹70\end{aligned}$$

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 = $\frac{30}{70} \times 100$%

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

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

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

Hence, option 3 is the correct option.

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

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

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

According to the given condition,

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

Hence, option 3 is the correct option.

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

$$\begin{aligned}\Rightarrow \frac{36}{100} \times x = 198 \\[1em] \Rightarrow \frac{36x}{100} = 198 \\[1em] \Rightarrow x = \frac{198 \times 100}{36} \\[1em] \Rightarrow x = \frac{19800}{36} \\[1em] \Rightarrow x = 550\end{aligned}$$

Hence, option 2 is the correct option.

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

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

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

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

$$\begin{aligned}2b = 112 \\[1em] \Rightarrow b = \frac{112}{2} \\[1em] \Rightarrow b = 56\end{aligned}$$

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

$$\begin{aligned}56 + g = 100 \\[1em] \Rightarrow g = 100 - 56 \\[1em] \Rightarrow g = 44\end{aligned}$$

Ratio of boys to girls = 56 : 44

= $\frac{56}{44}$

= $\frac{14}{11}$

= 14 : 11

Hence, option 2 is the correct option.

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

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

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

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

= 8%

Hence, 240 ml is 8% of 3 l.

(ii) $x$% of 35 = 42

$$\begin{aligned}\Rightarrow \frac{x}{100} \times 35 = 42 \\[1em] \Rightarrow \frac{35x}{100} = 42 \\[1em] \Rightarrow x = \frac{42 \times 100}{35} \\[1em] \Rightarrow x = \frac{4200}{35} \\[1em] \Rightarrow x = 120\end{aligned}$$

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

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

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

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 = $\frac{270}{1440} \times 100$%

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

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

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

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

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

According to the given condition,

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

$$\begin{aligned}\Rightarrow x - \frac{20}{100} \times x = 40 \\[1em] \Rightarrow \frac{100x - 20x}{100} = 40 \\[1em] \Rightarrow \frac{80x}{100} = 40 \\[1em] \Rightarrow x = \frac{40 \times 100}{80} \\[1em] \Rightarrow x = \frac{4000}{80} \\[1em] \Rightarrow x = 50\end{aligned}$$

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 = $\frac{22}{24} \times 100$%

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

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

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

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

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

$$\begin{aligned}\Rightarrow \frac{50}{3} \times \frac{1}{100} \\[1em] \Rightarrow \frac{50}{300} \\[1em] \Rightarrow \frac{1}{6} \\[1em] \Rightarrow 0.1666.....\end{aligned}$$

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 = $\frac{5}{1000} \times 100$%

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

= 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 = $\frac{x}{x} \times 100$% = 100%

Hence, the given statement is true.

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

$$\begin{aligned}\Rightarrow \frac{9}{25} \\[1em] \Rightarrow 9 : 25\end{aligned}$$

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

Hence, the given statement is false.

(v) 26% of $x$ = 65

$$\begin{aligned}\Rightarrow \frac{26}{100} \times x = 65 \\[1em] \Rightarrow \frac{26x}{100} = 65 \\[1em] \Rightarrow x = \frac{65 \times 100}{26} \\[1em] \Rightarrow x = \frac{6500}{26} \\[1em] \Rightarrow x = 250\end{aligned}$$

Hence, the given statement is true.

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 = $\frac{56}{80} \times 100$%

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

= 70%

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

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

= 72%

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

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

= 68%

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

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

= 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 = $\frac{51}{75} \times 100$% = 68%

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

Increase in his marks = 12% of 75

$$\begin{aligned}= \frac{12}{100} \times 75 \\[1em] = \frac{900}{100} \\[1em] = 9\end{aligned}$$

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

$$\begin{aligned}= \frac{40}{100} \times 80 \\[1em] = \frac{3200}{100} \\[1em] = 32\end{aligned}$$

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 = $\frac{24}{32} \times 100$%

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

= 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 = $\frac{42}{120} \times 100$%

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

= 35%

Hence, option (b) is the correct option.

Total population of the village = 8000

Number of literate people = 80% of 8000

$$\begin{aligned}= \frac{80}{100} \times 8000 \\[1em] = \frac{640000}{100} \\[1em] = 6400\end{aligned}$$

Number of literate women = 40% of 6400

$$\begin{aligned}= \frac{40}{100} \times 6400 \\[1em] = \frac{256000}{100} \\[1em] = 2560\end{aligned}$$

Number of illiterate people = 8000 - 6400 = 1600

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

= $\frac{2560}{1600}$

= $\frac{8}{5}$

= 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.

On increasing 240 by 5%, we get

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

So, Assertion (A) is true.

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

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.

$x$% of $y$ = 1000

$$\begin{aligned}\Rightarrow \frac{x}{100} \times y = 1000 \\[1em] \Rightarrow xy = 1000 \times 100 \\[1em] \Rightarrow xy = 100000\end{aligned}$$

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

$y$% of $z$ = 2000

$$\begin{aligned}\Rightarrow \frac{y}{100} \times z = 2000 \\[1em] \Rightarrow yz = 2000 \times 100 \\[1em] \Rightarrow yz = 200000\end{aligned}$$

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

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

$$\begin{aligned}\frac{xy}{yz} = \frac{100000}{200000} \\[1em] \Rightarrow \frac{x}{z} = \frac{1}{2} \\[1em] \Rightarrow x = \frac{z}{2}\end{aligned}$$

Hence, option 3 is the correct option.

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.

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

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

$$\begin{aligned}= \frac{625}{10000} \\[1em] = 0.0625\end{aligned}$$

Hence, option 3 is the correct option.

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

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

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

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

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

= 2% of $xy$

Hence, option 4 is the correct option.

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

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

$$\begin{aligned}= \frac{100x - 5x}{100} \\[1em] = \frac{95x}{100} \\[1em] = 0.95x\end{aligned}$$

Hence, option 2 is the correct option.

Let the required number be $x$.

According to the given condition,

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

$$\begin{aligned}\Rightarrow \frac{75}{100} \times x + 75 = x \\[1em] \Rightarrow \frac{75x}{100} + 75 = x \\[1em] \Rightarrow 75 = x - \frac{75x}{100} \\[1em] \Rightarrow 75 = \frac{100x - 75x}{100} \\[1em] \Rightarrow 75 = \frac{25x}{100} \\[1em] \Rightarrow x = \frac{75 \times 100}{25} \\[1em] \Rightarrow x = \frac{7500}{25} \\[1em] \Rightarrow x = 300\end{aligned}$$

Hence, option 3 is the correct option.

Quantity of the mixture = 15 litres

Strength of milk in the mixture = 20%

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

$$\begin{aligned}= \frac{20}{100} \times 15 \\[1em] = \frac{300}{100} \\[1em] = 3 \text{ litres}\end{aligned}$$

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 = $\frac{3}{18} \times 100$%

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

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

Hence, option 2 is the correct option.

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

On increasing $y$ by 10%,

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

On increasing $z$ by 10%,

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

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

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

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

$$\begin{aligned}= \frac{11x - 10x}{10} \\[1em] = \frac{x}{10}\end{aligned}$$

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

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

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

= 10%

Hence, option 1 is the correct option.

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

According to the given condition,

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

$$\begin{aligned}\Rightarrow \frac{35}{100} \times a = \frac{25}{100} \times b \\[1em] \Rightarrow 35a = 25b \\[1em] \Rightarrow \frac{a}{b} = \frac{25}{35} \\[1em] \Rightarrow \frac{a}{b} = \frac{5}{7}\end{aligned}$$

Ratio of their incomes = 5 : 7

Hence, option 3 is the correct option.

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 : $x$

$$\begin{aligned}\frac{2}{x} = \frac{70}{30} \\[1em] \Rightarrow \frac{2}{x} = \frac{7}{3} \\[1em] \Rightarrow 7x = 2 \times 3 \\[1em] \Rightarrow 7x = 6 \\[1em] \Rightarrow x = \frac{6}{7}\end{aligned}$$

Hence, option 3 is the correct option.