Percentage and its Applications

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

(i) 25%

$$\begin{aligned}\Rightarrow 25 \times \frac{1}{100}\\[1em] \Rightarrow \frac{25}{100}\\[1em] \Rightarrow \frac{1}{4}\end{aligned}$$

Hence, 25% = $\frac{1}{4}$.

(ii) 150%

$$\begin{aligned}\Rightarrow 150 \times \frac{1}{100}\\[1em] \Rightarrow \frac{150}{100}\\[1em] \Rightarrow \frac{3}{2}\end{aligned}$$

Hence, 150% = $\frac{3}{2}$.

(iii) $7\frac{1}{2}$%

$$\begin{aligned}\Rightarrow \frac{15}{2}\\[1em] \Rightarrow \frac{15}{2} \times \frac{1}{100}\\[1em] \Rightarrow \frac{15}{200}\\[1em] \Rightarrow \frac{3}{40}\end{aligned}$$

Hence, $7\frac{1}{2}$% $= \frac{3}{40}$.

(iv) $33\frac{1}{3}$%

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

Hence, $33\frac{1}{3}$% $= \frac{1}{3}$.

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

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

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

$\Rightarrow \left(\frac{1}{8} \times 100\right)$%

$\Rightarrow 12.5$%

Hence, $\frac{1}{8} = 12.5$%.

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

$\Rightarrow \left(\frac{5}{4} \times 100\right)$%

$\Rightarrow \frac{500}{4}$%

$\Rightarrow 125$%

Hence, $\frac{5}{4} = 125$%.

(iii) $\frac{9}{16}$

$\Rightarrow \left(\frac{9}{16} \times 100\right)$%

$\Rightarrow \frac{900}{16}$%

$\Rightarrow \frac{225}{4}$%

$\Rightarrow 56\frac{1}{4}$%

Hence, $\frac{9}{16} = 56\frac{1}{4}$%.

(iv) $\frac{3}{7}$

$\Rightarrow \left(\frac{3}{7} \times 100\right)$%

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

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

Hence, $\frac{3}{7} = 42\frac{6}{7}$%.

(v) $\frac{11}{15}$

$\Rightarrow \left(\frac{11}{15} \times 100\right)$%

$\Rightarrow \frac{1100}{15}$%

$\Rightarrow \frac{220}{3}$%

$\Rightarrow 73\frac{1}{3}$%

Hence, $\frac{11}{15} = 73\frac{1}{3}$%.

(vi) $1\frac{3}{8}$

$\Rightarrow \frac{11}{8}$

$\Rightarrow \left(\frac{11}{8} \times 100\right)$%

$\Rightarrow \frac{1100}{8}$%

$\Rightarrow 137.5$%

$\Rightarrow 137\frac{1}{2}$%

Hence, $1\frac{3}{8} = 137\frac{1}{2}$%.

Number of students absent = 6 and total number of students = 40.

Percentage of students absent = $\left(\frac{6}{40} \times 100\right)$%

$\Rightarrow \frac{600}{40}$%

$\Rightarrow 15$%

Hence, the percentage of students absent = 15%.

Marks secured by Antony = 384 out of 500.

Percentage of marks = $\left(\frac{384}{500} \times 100\right)$%

$\Rightarrow \frac{38400}{500}$%

$\Rightarrow 76.8$%

Hence, the percentage of marks secured by Antony = 76.8%.

Number of defective shirts = 15 and total number of shirts = 500.

Percentage of defective shirts = $\left(\frac{15}{500} \times 100\right)$%

$\Rightarrow \frac{1500}{500}$%

$\Rightarrow 3$%

Hence, the percentage of defective shirts = 3%.

Number of gold bangles = 20 and number of silver bangles = 10.

Total number of bangles = 20 + 10 = 30.

Percentage of gold bangles = $\left(\frac{20}{30} \times 100\right)$%

$\Rightarrow \frac{2000}{30}$%

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

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

Percentage of silver bangles = $\left(\frac{10}{30} \times 100\right)$%

$\Rightarrow \frac{1000}{30}$%

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

$\Rightarrow 33\frac{1}{3}$%

Hence, gold bangles = $66\frac{2}{3}$% and silver bangles = $33\frac{1}{3}$%.

Total number of voters = 120 and number who voted = 90.

Number who did not vote = 120 − 90 = 30.

Percentage who did not vote = $\left(\frac{30}{120} \times 100\right)$%

$\Rightarrow \frac{3000}{120}$%

$\Rightarrow 25$%

Hence, the percentage of voters who did not vote = 25%.

(i) From the figure, shaded part = $\frac{3}{4}$.

Percentage of shaded part = $\left(\frac{3}{4} \times 100\right)$% = 75%.

Hence, shaded part = $\frac{3}{4}$ and percentage of shaded part = 75%.

(ii) From the figure, shaded part = $\frac{2}{6} = \frac{1}{3}$.

Percentage of shaded part = $\left(\frac{1}{3} \times 100\right)$% $= 33\frac{1}{3}$%.

Hence, shaded part = $\frac{1}{3}$ and percentage of shaded part = $33\frac{1}{3}$%.

(iii) From the figure, shaded part = $\frac{5}{8}$.

Percentage of shaded part = $\left(\frac{5}{8} \times 100\right)$% = 62.5%.

Hence, shaded part = $\frac{5}{8}$ and percentage of shaded part = 62.5%.

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

(i) 14%

$$\begin{aligned}\Rightarrow \frac{14}{100}\\[1em] \Rightarrow \frac{7}{50}\\[1em] \Rightarrow 7 : 50\end{aligned}$$

Hence, 14% = 7 : 50.

(ii) $1\frac{3}{4}$%

$\Rightarrow \frac{7}{4}$%

$\Rightarrow \frac{7}{4} \times \frac{1}{100}$

$\Rightarrow \frac{7}{400}$

$\Rightarrow 7 : 400$

Hence, $1\frac{3}{4}$% = 7 : 400.

(iii) $33\frac{1}{3}$%

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

$\Rightarrow \frac{100}{3} \times \frac{1}{100}$

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

$\Rightarrow 1 : 3$

Hence, $33\frac{1}{3}$% = 1 : 3.

(iv) 37.5%

$$\begin{aligned}\Rightarrow \frac{37.5}{100}\\[1em] \Rightarrow \frac{375}{1000}\\[1em] \Rightarrow \frac{3}{8}\\[1em] \Rightarrow 3 : 8\end{aligned}$$

Hence, 37.5% = 3 : 8.

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

(i) 5 : 4

$\Rightarrow \frac{5}{4}$

$\Rightarrow \left(\frac{5}{4} \times 100\right)$%

$\Rightarrow 125$%

Hence, 5 : 4 = 125%.

(ii) 1 : 1

$\Rightarrow \frac{1}{1}$

$\Rightarrow \left(\frac{1}{1} \times 100\right)$%

$\Rightarrow 100$%

Hence, 1 : 1 = 100%.

(iii) 2 : 3

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

$\Rightarrow \left(\frac{2}{3} \times 100\right)$%

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

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

Hence, 2 : 3 = $66\frac{2}{3}$%.

(iv) 9 : 16

$\Rightarrow \frac{9}{16}$

$\Rightarrow \left(\frac{9}{16} \times 100\right)$%

$\Rightarrow \frac{900}{16}$%

$\Rightarrow 56\frac{1}{4}$%

Hence, 9 : 16 = $56\frac{1}{4}$%.

Parts of zinc = 7 and parts of copper = 33.

Total number of parts = 7 + 33 = 40.

Percentage of copper = $\left(\frac{33}{40} \times 100\right)$%

$\Rightarrow \frac{3300}{40}$%

$\Rightarrow 82.5$%

Hence, the percentage of copper in the alloy = 82.5%.

Ratio of calcium : carbon : sand = 12 : 3 : 10.

Total number of parts = 12 + 3 + 10 = 25.

Percentage of carbon = $\left(\frac{3}{25} \times 100\right)$%

$\Rightarrow \frac{300}{25}$%

$\Rightarrow 12$%

Hence, the percentage of carbon in the chalk = 12%.

Total amount = ₹ 2500.

Ravi : Raju : Roy = 2 : 3 : 5.

Total number of parts = 2 + 3 + 5 = 10.

Ravi's share = $\frac{2}{10} \times 2500 = ₹\ 500$.

Raju's share = $\frac{3}{10} \times 2500 = ₹\ 750$.

Roy's share = $\frac{5}{10} \times 2500 = ₹\ 1250$.

Ravi's percentage = $\left(\frac{2}{10} \times 100\right)$% = 20%.

Raju's percentage = $\left(\frac{3}{10} \times 100\right)$% = 30%.

Roy's percentage = $\left(\frac{5}{10} \times 100\right)$% = 50%.

Hence, Ravi gets ₹ 500 (20%), Raju gets ₹ 750 (30%) and Roy gets ₹ 1250 (50%).

To convert a percentage into a decimal, replace the % sign with $\frac{1}{100}$ and then express the fraction as a decimal.

(i) 28%

$\Rightarrow \frac{28}{100}$

$\Rightarrow 0.28$

Hence, 28% = 0.28.

(ii) 3%

$\Rightarrow \frac{3}{100}$

$\Rightarrow 0.03$

Hence, 3% = 0.03.

(iii) 0.44%

$\Rightarrow \frac{0.44}{100}$

$\Rightarrow 0.0044$

Hence, 0.44% = 0.0044.

(iv) $37\frac{1}{2}$%

$\Rightarrow \frac{75}{2}$%

$\Rightarrow \frac{75}{2} \times \frac{1}{100}$

$\Rightarrow \frac{75}{200}$

$\Rightarrow 0.375$

Hence, $37\frac{1}{2}$% = 0.375.

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

(i) 0.65

$\Rightarrow (0.65 \times 100)$%

$\Rightarrow 65$%

Hence, 0.65 = 65%.

(ii) 0.9

$\Rightarrow (0.9 \times 100)$%

$\Rightarrow 90$%

Hence, 0.9 = 90%.

(iii) 2.1

$\Rightarrow (2.1 \times 100)$%

$\Rightarrow 210$%

Hence, 2.1 = 210%.

(iv) 0.02

$\Rightarrow (0.02 \times 100)$%

$\Rightarrow 2$%

Hence, 0.02 = 2%.

15% of 250

$$\begin{aligned}\Rightarrow \frac{15}{100} \times 250\\[1em] \Rightarrow \frac{3750}{100}\\[1em] \Rightarrow 37.5\end{aligned}$$

Hence, 15% of 250 = 37.5.

25% of 120 litres

$$\begin{aligned}\Rightarrow \frac{25}{100} \times 120\\[1em] \Rightarrow \frac{3000}{100}\\[1em] \Rightarrow 30 \text{ litres}\end{aligned}$$

Hence, 25% of 120 litres = 30 litres.

1% of 1 hour

We know, 1 hour = 60 minutes.

$$\begin{aligned}\Rightarrow \frac{1}{100} \times 60 \text{ minutes}\\[1em] \Rightarrow \frac{60}{100} \text{ minutes}\\[1em] \Rightarrow 0.6 \text{ minutes}\\[1em] \Rightarrow 0.6 \times 60 \text{ seconds}\\[1em] \Rightarrow 36 \text{ seconds}\end{aligned}$$

Hence, 1% of 1 hour = 36 seconds.

75% of 1 kg

We know, 1 kg = 1000 g.

$$\begin{aligned}\Rightarrow \frac{75}{100} \times 1000\\[1em] \Rightarrow \frac{75000}{100}\\[1em] \Rightarrow 750 \text{ g}\end{aligned}$$

Hence, 75% of 1 kg = 750 g.

120% of ₹ 250

$$\begin{aligned}\Rightarrow \frac{120}{100} \times 250\\[1em] \Rightarrow \frac{30000}{100}\\[1em] \Rightarrow ₹\ 300\end{aligned}$$

Hence, 120% of ₹ 250 = ₹ 300.

0.6% of 2 km

We know, 1 km = 1000 m, so 2 km = 2000 m.

$$\begin{aligned}\Rightarrow \frac{0.6}{100} \times 2000\\[1em] \Rightarrow \frac{1200}{100}\\[1em] \Rightarrow 12 \text{ m}\end{aligned}$$

Hence, 0.6% of 2 km = 12 m.

Total number of children = 25.

Number of children who like getting wet in the rain = 8% of 25

$$\begin{aligned}\Rightarrow \frac{8}{100} \times 25\\[1em] \Rightarrow \frac{200}{100}\\[1em] \Rightarrow 2\end{aligned}$$

Hence, 2 children like getting wet in the rain.

Number of ice cream cups eaten = 3 out of 20.

Required percentage = $\left(\frac{3}{20} \times 100\right)$%

$= \frac{300}{20}$%

$= 15$%

Hence, Vasundara ate 15% of the ice cream cups.

To express one quantity as a percentage of another quantity, use percentage = $\left(\frac{\text{one quantity}}{\text{second quantity}} \times 100\right)$%, taking both quantities in the same units.

(i) 20 as a percentage of 50

$\Rightarrow \left(\frac{20}{50} \times 100\right)$%

$\Rightarrow \frac{2000}{50}$%

$\Rightarrow 40$%

Hence, 20 is 40% of 50.

(ii) 60 litres as a percentage of 40 litres

$\Rightarrow \left(\frac{60}{40} \times 100\right)$%

$\Rightarrow \frac{6000}{40}$%

$\Rightarrow 150$%

Hence, 60 litres is 150% of 40 litres.

(iii) 90 cm as a percentage of 4.5 m

We know, 1 m = 100 cm, so 4.5 m = 450 cm.

$\Rightarrow \left(\frac{90}{450} \times 100\right)$%

$\Rightarrow \frac{9000}{450}$%

$\Rightarrow 20$%

Hence, 90 cm is 20% of 4.5 m.

(iv) 350 g as a percentage of 5.6 kg

We know, 1 kg = 1000 g, so 5.6 kg = 5600 g.

$\Rightarrow \left(\frac{350}{5600} \times 100\right)$%

$\Rightarrow \frac{35000}{5600}$%

$\Rightarrow \frac{25}{4}$%

$\Rightarrow 6\frac{1}{4}$%

Hence, 350 g is $6\frac{1}{4}$% of 5.6 kg.

(i) 12 of 80

$\Rightarrow \left(\frac{12}{80} \times 100\right)$%

$\Rightarrow \frac{1200}{80}$%

$\Rightarrow 15$%

Hence, 12 is 15% of 80.

(ii) 25 paise of 4 rupees

We know, 1 rupee = 100 paise, so 4 rupees = 400 paise.

$\Rightarrow \left(\frac{25}{400} \times 100\right)$%

$\Rightarrow \frac{2500}{400}$%

$\Rightarrow \frac{25}{4}$%

$\Rightarrow 6\frac{1}{4}$%

Hence, 25 paise is $6\frac{1}{4}$% of 4 rupees.

(iii) 300 g of 2 kg

We know, 1 kg = 1000 g, so 2 kg = 2000 g.

$\Rightarrow \left(\frac{300}{2000} \times 100\right)$%

$\Rightarrow \frac{30000}{2000}$%

$\Rightarrow 15$%

Hence, 300 g is 15% of 2 kg.

Games won last year = 4 and games won this year = 6.

Increase in games won = 6 − 4 = 2.

Percentage increase = $\left(\frac{\text{increase in value}}{\text{original value}} \times 100\right)$%

$\Rightarrow \left(\frac{2}{4} \times 100\right)$%

$\Rightarrow \frac{200}{4}$%

$\Rightarrow 50$%

Hence, the percent increase = 50%.

Original price = ₹ 80 and new price = ₹ 60.

Decrease in price = 80 − 60 = ₹ 20.

Percentage decrease = $\left(\frac{\text{decrease in value}}{\text{original value}} \times 100\right)$%

$\Rightarrow \left(\frac{20}{80} \times 100\right)$%

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

$\Rightarrow 25$%

Hence, the percentage decrease in price = 25%.

Original price = ₹ 1 per litre and present price = ₹ 95 per litre.

Increase in price = 95 − 1 = ₹ 94.

Percentage increase = $\left(\frac{94}{1} \times 100\right)$%

= 9400%

Hence, the price of petrol has gone up by 9400%.

Last year's price = ₹ 40 per kg.

Increase in price = 20% of 40 = $\frac{20}{100} \times 40 = ₹\ 8$.

Price this year = 40 + 8 = ₹ 48.

Hence, the price of tomatoes this year = ₹ 48 per kg.

Total number of students = 300.

Percentage of students who failed = 28%.

Percentage of students who passed = (100 − 28)% = 72%.

Number of students who passed = 72% of 300

$$\begin{aligned}\Rightarrow \frac{72}{100} \times 300\\[1em] \Rightarrow \frac{21600}{100}\\[1em] \Rightarrow 216\end{aligned}$$

Hence, 216 students passed the exam.

Total number of voters = 15000.

Percentage of voters who voted = 60%.

Percentage of voters who did not vote = (100 − 60)% = 40%.

Number of voters who did not vote = 40% of 15000

$$\begin{aligned}\Rightarrow \frac{40}{100} \times 15000\\[1em] \Rightarrow \frac{600000}{100}\\[1em] \Rightarrow 6000\end{aligned}$$

Hence, 6000 voters did not vote.

Length of the flagpole = 18 m.

Percentage painted green = 20% and percentage painted yellow = 45%.

Percentage painted red = 100% − (20% + 45%) = 100% − 65% = 35%.

Length painted red = 35% of 18 m

$$\begin{aligned}\Rightarrow \frac{35}{100} \times 18\\[1em] \Rightarrow \frac{630}{100}\\[1em] \Rightarrow 6.3 \text{ m}\end{aligned}$$

Hence, the length painted red = 6.3 m.

Total quantity of chalk = $2\frac{1}{2}$ kg = 2.5 kg = 2500 g.

Amount of carbon = 3% of 2500 g

$$\begin{aligned}\Rightarrow \frac{3}{100} \times 2500\\[1em] \Rightarrow 75 \text{ g}\end{aligned}$$

Amount of calcium = 10% of 2500 g

$$\begin{aligned}\Rightarrow \frac{10}{100} \times 2500\\[1em] \Rightarrow 250 \text{ g}\end{aligned}$$

Percentage of sand = 100% − (10% + 3% + 12%) = 100% − 25% = 75%.

Amount of sand = 75% of 2500 g

$$\begin{aligned}\Rightarrow \frac{75}{100} \times 2500\\[1em] \Rightarrow 1875 \text{ g}\\[1em] \Rightarrow 1.875 \text{ kg}\end{aligned}$$

Hence, carbon = 75 g, calcium = 250 g and sand = 1.875 kg.

(i) 25% of it is 9

Let the whole quantity be x.

Given,

25% of x = 9

$$\begin{aligned}\Rightarrow \frac{25}{100} \times x = 9 \\[1em] \Rightarrow x = \frac{9 \times 100}{25} \\[1em] \Rightarrow x = 36.\end{aligned}$$

Hence, the whole quantity = 36.

(ii) 75% of it is 15

Let the whole quantity be x.

$$\begin{aligned}\Rightarrow \frac{75}{100} \times x = 15\\[1em] \Rightarrow x = \frac{15 \times 100}{75}\\[1em] \Rightarrow x = 20\end{aligned}$$

Hence, the whole quantity = 20.

(iii) 12% of it is ₹ 1080

Let the whole quantity be ₹ x.

$$\begin{aligned}\Rightarrow \frac{12}{100} \times x = 1080\\[1em] \Rightarrow x = \frac{1080 \times 100}{12}\\[1em] \Rightarrow x = 9000\end{aligned}$$

Hence, the whole quantity = ₹ 9000.

(iv) 8% of it is 40 litres

Let the whole quantity be x litres.

$$\begin{aligned}\Rightarrow \frac{8}{100} \times x = 40\\[1em] \Rightarrow x = \frac{40 \times 100}{8}\\[1em] \Rightarrow x = 500\end{aligned}$$

Hence, the whole quantity = 500 litres.

Let Mohini's salary be ₹ x.

Given, 10% of her salary = ₹ 4000.

$$\begin{aligned}\Rightarrow \frac{10}{100} \times x = 4000\\[1em] \Rightarrow x = \frac{4000 \times 100}{10}\\[1em] \Rightarrow x = 40000\end{aligned}$$

Hence, Mohini's salary = ₹ 40,000.

Percentage of apples that go bad = 16%.

Percentage of good apples = (100 − 16)% = 84%.

Let the total number of apples be x.

84% of x = 42

$$\begin{aligned}\Rightarrow \frac{84}{100} \times x = 42 \\[1em] \Rightarrow x = \frac{42 \times 100}{84} \\[1em] \Rightarrow x = 50\end{aligned}$$

Hence, the total number of apples in the basket = 50.

Marks scored by Varun = 251 and marks needed to pass = 19 more.

Pass marks = 251 + 19 = 270.

Let the maximum marks be x. Since 45% of the maximum marks are needed to pass,

45% of x = 270

$$\begin{aligned}\Rightarrow \frac{45}{100} \times x = 270\\[1em] \Rightarrow x = \frac{270 \times 100}{45}\\[1em] \Rightarrow x = 600\end{aligned}$$

Hence, the maximum marks = 600.

Percentage of students present = 94%.

Percentage of students absent = (100 − 94)% = 6%.

Let the total strength of the school be x.

6% of x = 174

$$\begin{aligned}\Rightarrow \frac{6}{100} \times x = 174\\[1em] \Rightarrow x = \frac{174 \times 100}{6}\\[1em] \Rightarrow x = 2900\end{aligned}$$

Hence, the total strength of the school = 2900.

Percentage of men = 40% and percentage of women = 39%.

Percentage of children = 100% − (40% + 39%) = 100% − 79% = 21%.

Let the total population be x.

21% of x = 12600

$$\begin{aligned}\Rightarrow \frac{21}{100} \times x = 12600\\[1em] \Rightarrow x = \frac{12600 \times 100}{21}\\[1em] \Rightarrow x = 60000\end{aligned}$$

Number of men = 40% of 60000

$$\begin{aligned}\Rightarrow \frac{40}{100} \times 60000\\[1em] \Rightarrow 24000\end{aligned}$$

Hence, the number of men = 24000.

Let the earlier price of the watch be ₹ x.

Given, 15% of the price = ₹ 90.

$$\begin{aligned}\Rightarrow \frac{15}{100} \times x = 90\\[1em] \Rightarrow x = \frac{90 \times 100}{15}\\[1em] \Rightarrow x = 600\end{aligned}$$

Hence, the earlier price of the watch = ₹ 600.

(i) Let the number be x.

When increased by 30%, the number becomes 130% of x.

130% of x = 39

$$\begin{aligned}\Rightarrow \frac{130}{100} \times x = 39\\[1em] \Rightarrow x = \frac{39 \times 100}{130}\\[1em] \Rightarrow x = 30\end{aligned}$$

Hence, the required number = 30.

(ii) Let the number be x.

When decreased by 8%, the number becomes (100 − 8)% = 92% of x.

92% of x = 506

$$\begin{aligned}\Rightarrow \frac{92}{100} \times x = 506\\[1em] \Rightarrow x = \frac{506 \times 100}{92}\\[1em] \Rightarrow x = 550\end{aligned}$$

Hence, the required number = 550.

Let the original price of the shirt be ₹ x.

When reduced by 7%, the price becomes (100 − 7)% = 93% of x.

93% of x = 465

$$\begin{aligned}\Rightarrow \frac{93}{100} \times x = 465\\[1em] \Rightarrow x = \frac{465 \times 100}{93}\\[1em] \Rightarrow x = 500\end{aligned}$$

Hence, the original price of the shirt = ₹ 500.

15% of 60 = $\frac{15}{100} \times 60 = 9$.

Let the number be x.

Given, 15% of 60 is greater than 25% of x by 3.

$\Rightarrow 9 = 25$% of x + 3

$\Rightarrow 25$% of x = 9 - 3

$\Rightarrow \frac{25}{100} \times x = 6$

$\Rightarrow x = \frac{6 \times 100}{25}$

$\Rightarrow x = 24$

Hence, the required number = 24.

Petrol in the full tank = 60 litres.

Petrol used = 30% of 60 = $\frac{30}{100} \times 60 = 18$ litres.

Petrol poured into the 50 litre tank = 60 − 18 = 42 litres.

(i) Percentage of 50 litre tank filled = $\left(\frac{42}{50} \times 100\right)$%

$\Rightarrow \frac{4200}{50}$%

$\Rightarrow 84$%

Hence, 84% of the 50 litre tank was filled with petrol.

(ii) Petrol used in 10 days = 2.8 × 10 = 28 litres.

Petrol present in the 50 litre tank = 42 litres.

Percentage of petrol used = $\left(\frac{28}{42} \times 100\right)$%

$\Rightarrow \frac{2800}{42}$%

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

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

Hence, $66\frac{2}{3}$% of the petrol in the 50 litre tank would be used in 10 days.

C.P. of the calculator = ₹ 760 and S.P. of the calculator = ₹ 874.

Since S.P. > C.P., there is a profit.

Profit = S.P. − C.P. = 874 − 760 = ₹ 114.

Profit percentage = $\left(\frac{\text{profit}}{\text{cost price}} \times 100\right)$%

$\Rightarrow \left(\frac{114}{760} \times 100\right)$%

$\Rightarrow \frac{11400}{760}$%

$\Rightarrow 15$%

Hence, profit = ₹ 114 and profit percentage = 15%.

C.P. of the saree = ₹ 2500 and S.P. of the saree = ₹ 2300.

Since C.P. > S.P., there is a loss.

Loss = C.P. − S.P. = 2500 − 2300 = ₹ 200.

Loss percentage = $\left(\frac{\text{loss}}{\text{cost price}} \times 100\right)$%

$\Rightarrow \left(\frac{200}{2500} \times 100\right)$%

$\Rightarrow \frac{20000}{2500}$%

$\Rightarrow 8$%

Hence, loss = ₹ 200 and loss percent = 8%.

(i) C.P. of the shears = ₹ 250 and S.P. of the shears = ₹ 325.

Since S.P. > C.P., there is a profit.

Profit = S.P. − C.P. = 325 − 250 = ₹ 75.

Profit percentage = $\left(\frac{75}{250} \times 100\right)$%

$\Rightarrow \frac{7500}{250}$%

$\Rightarrow 30$%

Hence, profit = ₹ 75 and profit percent = 30%.

(ii) C.P. of the shirt = ₹ 250 and S.P. of the shirt = ₹ 150.

Since C.P. > S.P., there is a loss.

Loss = C.P. − S.P. = 250 − 150 = ₹ 100.

Loss percentage = $\left(\frac{100}{250} \times 100\right)$%

$\Rightarrow \frac{10000}{250}$%

$\Rightarrow 40$%

Hence, loss = ₹ 100 and loss percent = 40%.

First almirah:

C.P. = ₹ 4800 and gain = $13\frac{1}{3}$% = $\frac{40}{3}$%.

Gain = $\frac{40}{3}$% of 4800 = $\frac{40}{3 \times 100} \times 4800 = ₹\ 640$.

S.P. of first almirah = 4800 + 640 = ₹ 5440.

Second almirah:

C.P. = ₹ 3640 and loss = 15%.

Loss = 15% of 3640 = $\frac{15}{100} \times 3640 = ₹\ 546$.

S.P. of second almirah = 3640 − 546 = ₹ 3094.

Whole deal:

Total C.P. = 4800 + 3640 = ₹ 8440.

Total S.P. = 5440 + 3094 = ₹ 8534.

Since total S.P. > total C.P., there is a gain.

Gain = 8534 − 8440 = ₹ 94.

Hence, Rajinder gained ₹ 94 in the whole deal.

Total C.P. = 24 × 4500 = ₹ 108000.

S.P. of 16 tables = 16 × 6000 = ₹ 96000.

Remaining tables = 24 − 16 = 8.

S.P. of 8 tables = 8 × 4000 = ₹ 32000.

Total S.P. = 96000 + 32000 = ₹ 128000.

Since total S.P. > total C.P., there is a gain.

Gain = 128000 − 108000 = ₹ 20000.

Gain percentage = $\left(\frac{20000}{108000} \times 100\right)$%

$\Rightarrow \frac{2000000}{108000}$%

$\Rightarrow \frac{500}{27}$%

$\Rightarrow 18\frac{14}{27}$%

Hence, the shopkeeper has a gain of $18\frac{14}{27}$%.

S.P. of the lamp = ₹ 810 and profit = ₹ 60.

C.P. = S.P. − profit = 810 − 60 = ₹ 750.

Profit percentage = $\left(\frac{60}{750} \times 100\right)$%

$\Rightarrow \frac{6000}{750}$%

$\Rightarrow 8$%

Hence, the cost price of the lamp = ₹ 750 and profit percent = 8%.

S.P. of the jacket = ₹ 3906 and loss = ₹ 294.

C.P. = S.P. + loss = 3906 + 294 = ₹ 4200.

Loss percentage = $\left(\frac{294}{4200} \times 100\right)$%

$\Rightarrow \frac{29400}{4200}$%

$\Rightarrow 7$%

Hence, the cost price of the jacket = ₹ 4200 and loss percentage = 7%.

C.P. of the vase = ₹ 120 and loss = 10%.

Loss = 10% of 120 = $\frac{10}{100} \times 120 = ₹\ 12$.

S.P. = C.P. − loss = 120 − 12 = ₹ 108.

Hence, the vase was sold for ₹ 108.

C.P. of the T.V. = ₹ 10000 and profit = 20%.

Profit = 20% of 10000 = $\frac{20}{100} \times 10000 = ₹\ 2000$.

S.P. = C.P. + profit = 10000 + 2000 = ₹ 12000.

Hence, I get ₹ 12000 for the T.V.

S.P. of the article = ₹ 300 and profit = 20%.

Let the C.P. be ₹ x.

S.P. = C.P. + 20% of C.P. = (100 + 20)% of C.P. = 120% of C.P.

$$\begin{aligned}\Rightarrow \frac{120}{100} \times x = 300\\[1em] \Rightarrow x = \frac{300 \times 100}{120}\\[1em] \Rightarrow x = 250\end{aligned}$$

Hence, the cost price of the article = ₹ 250.

S.P. of the article = ₹ 320 and loss = 20%.

Let the C.P. be ₹ x.

S.P. = C.P. − 20% of C.P. = (100 − 20)% of C.P. = 80% of C.P.

$$\begin{aligned}\Rightarrow \frac{80}{100} \times x = 320\\[1em] \Rightarrow x = \frac{320 \times 100}{80}\\[1em] \Rightarrow x = 400\end{aligned}$$

Hence, the cost price of the article = ₹ 400.

S.P. of the chair = ₹ 522 and profit = 16%.

Let the C.P. be ₹ x.

S.P. = (100 + 16)% of C.P. = 116% of C.P.

$$\begin{aligned}\Rightarrow \frac{116}{100} \times x = 522\\[1em] \Rightarrow x = \frac{522 \times 100}{116}\\[1em] \Rightarrow x = 450\end{aligned}$$

Hence, the cost price of the chair = ₹ 450.

S.P. of the garments = ₹ 7360 and loss = 8%.

Let the C.P. be ₹ x.

S.P. = (100 − 8)% of C.P. = 92% of C.P.

$$\begin{aligned}\Rightarrow \frac{92}{100} \times x = 7360\\[1em] \Rightarrow x = \frac{7360 \times 100}{92}\\[1em] \Rightarrow x = 8000\end{aligned}$$

Hence, the cost price of the garments = ₹ 8000.

S.P. of the table = ₹ 3168 and loss = 12%.

Let the C.P. be ₹ x.

S.P. = (100 − 12)% of C.P. = 88% of C.P.

$$\begin{aligned}\Rightarrow \frac{88}{100} \times x = 3168\\[1em] \Rightarrow x = \frac{3168 \times 100}{88}\\[1em] \Rightarrow x = 3600\end{aligned}$$

So, the cost price of the table = ₹ 3600.

Now, if the table is sold for ₹ 3870:

Since 3870 > 3600, there is a gain.

Gain = 3870 − 3600 = ₹ 270.

Gain percentage = $\left(\frac{270}{3600} \times 100\right)$%

$\Rightarrow \frac{27000}{3600}$%

$\Rightarrow 7.5$%

Hence, the cost price = ₹ 3600 and he would gain 7.5% by selling it for ₹ 3870.

S.P. of the article = ₹ 4550 and loss = 9%.

Let the C.P. be ₹ x.

S.P. = (100 − 9)% of C.P. = 91% of C.P.

$$\begin{aligned}\Rightarrow \frac{91}{100} \times x = 4550\\[1em] \Rightarrow x = \frac{4550 \times 100}{91}\\[1em] \Rightarrow x = 5000\end{aligned}$$

So, the cost price of the article = ₹ 5000.

Now, if the article is sold for ₹ 4825:

Since 5000 > 4825, there is a loss.

Loss = 5000 − 4825 = ₹ 175.

Loss percentage = $\left(\frac{175}{5000} \times 100\right)$%

$\Rightarrow \frac{17500}{5000}$%

$\Rightarrow 3.5$%

Hence, he would lose 3.5% by selling it for ₹ 4825.

Cost of the car = ₹ 80000.

Amount spent on repairs = 12.5% of 80000 = $\frac{12.5}{100} \times 80000 = ₹\ 10000$.

Total cost price = 80000 + 10000 = ₹ 90000.

To make a profit of 15%:

S.P. = (100 + 15)% of C.P. = 115% of 90000

$$\begin{aligned}\Rightarrow \frac{115}{100} \times 90000\\[1em] \Rightarrow \frac{10350000}{100}\\[1em] \Rightarrow 103500\end{aligned}$$

Hence, Arif should sell the car for ₹ 103500.

Simple interest is given by the formula $I = \frac{P \times R \times T}{100}$ and amount = principal + interest.

(i) Here, P = ₹ 350, R = 11% per annum and T = 2 years.

$$\begin{aligned}\Rightarrow I = \frac{350 \times 11 \times 2}{100}\\[1em] \Rightarrow I = \frac{7700}{100}\\[1em] \Rightarrow I = ₹\ 77\end{aligned}$$

Amount = P + I = 350 + 77 = ₹ 427.

Hence, simple interest = ₹ 77 and amount = ₹ 427.

(ii) Here, P = ₹ 20000, R = $8\frac{1}{2} = \frac{17}{2}$% per annum and T = $4\frac{1}{2} = \frac{9}{2}$ years.

$$\begin{aligned}\Rightarrow I = \frac{20000 \times \frac{17}{2} \times \frac{9}{2}}{100}\\[1em] \Rightarrow I = \frac{20000 \times 17 \times 9}{2 \times 2 \times 100}\\[1em] \Rightarrow I = \frac{3060000}{400}\\[1em] \Rightarrow I = ₹\ 7650\end{aligned}$$

Amount = P + I = 20000 + 7650 = ₹ 27650.

Hence, simple interest = ₹ 7650 and amount = ₹ 27650.

(iii) Here, P = ₹ 648, R = $16\frac{2}{3} = \frac{50}{3}$% per annum and T = 8 months = $\frac{8}{12} = \frac{2}{3}$ year.

$$\begin{aligned}\Rightarrow I = \frac{648 \times \frac{50}{3} \times \frac{2}{3}}{100}\\[1em] \Rightarrow I = \frac{648 \times 50 \times 2}{3 \times 3 \times 100}\\[1em] \Rightarrow I = \frac{64800}{900}\\[1em] \Rightarrow I = ₹\ 72\end{aligned}$$

Amount = P + I = 648 + 72 = ₹ 720.

Hence, simple interest = ₹ 72 and amount = ₹ 720.

Using $I = \frac{P \times R \times T}{100}$, we get $T = \frac{I \times 100}{P \times R}$.

(i) Here, P = ₹ 2500, R = 4% per annum and I = ₹ 200.

$$\begin{aligned}\Rightarrow T = \frac{200 \times 100}{2500 \times 4}\\[1em] \Rightarrow T = \frac{20000}{10000}\\[1em] \Rightarrow T = 2 \text{ years}\end{aligned}$$

Hence, the time = 2 years.

(ii) Here, P = ₹ 12000, R = $6\frac{1}{2} = \frac{13}{2}$% per annum and I = ₹ 2730.

$$\begin{aligned}\Rightarrow T = \frac{2730 \times 100}{12000 \times \frac{13}{2}}\\[1em] \Rightarrow T = \frac{2730 \times 100 \times 2}{12000 \times 13}\\[1em] \Rightarrow T = \frac{546000}{156000}\\[1em] \Rightarrow T = \frac{7}{2}\\[1em] \Rightarrow T = 3\frac{1}{2} \text{ years}\end{aligned}$$

Hence, the time = $3\frac{1}{2}$ years.

Using $I = \frac{P \times R \times T}{100}$, we get $R = \frac{I \times 100}{P \times T}$.

(i) Here, P = ₹ 1560, T = 3 years and I = ₹ 585.

$\Rightarrow R = \frac{585 \times 100}{1560 \times 3}$%

$\Rightarrow R = \frac{58500}{4680}$%

$\Rightarrow R = \frac{25}{2}$%

$\Rightarrow R = 12\frac{1}{2}$%

Hence, the rate of interest = $12\frac{1}{2}$% per annum.

(ii) Here, P = ₹ 1625, T = $2\frac{1}{2} = \frac{5}{2}$ years and I = ₹ 325.

$\Rightarrow R = \frac{325 \times 100}{1625 \times \frac{5}{2}}$%

$\Rightarrow R = \frac{325 \times 100 \times 2}{1625 \times 5}$%

$\Rightarrow R = \frac{65000}{8125}$%

$\Rightarrow R = 8$%

Hence, the rate of interest = 8% per annum.

Using $I = \frac{P \times R \times T}{100}$, we get $P = \frac{I \times 100}{R \times T}$.

(i) Here,

    R = 16% per annum,

    T = $2\frac{1}{2} = \frac{5}{2}$ years and

     I = ₹ 3840.

$$\begin{aligned}\Rightarrow P = \frac{3840 \times 100}{16 \times \frac{5}{2}}\\[1em] \Rightarrow P = \frac{3840 \times 100 \times 2}{16 \times 5}\\[1em] \Rightarrow P = \frac{768000}{80}\\[1em] \Rightarrow P = ₹\ 9600\end{aligned}$$

Hence, the principal = ₹ 9600.

(ii) Here,

     R = $7\frac{1}{2} = \frac{15}{2}$% per annum,

     T = 2 years 4 months = $2\frac{4}{12} = \frac{7}{3}$ years and

     I = ₹ 2730.

$$\begin{aligned}\Rightarrow P = \frac{2730 \times 100}{\frac{15}{2} \times \frac{7}{3}}\\[1em] \Rightarrow P = \frac{2730 \times 100 \times 2 \times 3}{15 \times 7}\\[1em] \Rightarrow P = \frac{1638000}{105}\\[1em] \Rightarrow P = ₹\ 15600\end{aligned}$$

Hence, the principal = ₹ 15600.

(i) Here, P = ₹ 1200, A = ₹ 1320 and T = 2 years.

I = A − P = 1320 − 1200 = ₹ 120.

$\Rightarrow R = \frac{I \times 100}{P \times T}$%

$\Rightarrow R = \frac{120 \times 100}{1200 \times 2}$%

$\Rightarrow R = \frac{12000}{2400}$%

$\Rightarrow R = 5$%

Hence, the rate of interest = 5% per annum.

(ii) Here, P = ₹ 300, A = ₹ 400 and T = 2 years.

I = A − P = 400 − 300 = ₹ 100.

$\Rightarrow R = \frac{100 \times 100}{300 \times 2}$%

$\Rightarrow R = \frac{10000}{600}$%

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

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

Hence, the rate of interest = $16\frac{2}{3}$% per annum.

(i) Here,

P = ₹ 1250, A = ₹ 1950 and R = 16% per annum.

I = A − P = 1950 − 1250 = ₹ 700.

$$\begin{aligned}\Rightarrow T = \frac{I \times 100}{P \times R}\\[1em] \Rightarrow T = \frac{700 \times 100}{1250 \times 16}\\[1em] \Rightarrow T = \frac{70000}{20000}\\[1em] \Rightarrow T = \frac{7}{2}\\[1em] \Rightarrow T = 3\frac{1}{2} \text{ years}\end{aligned}$$

Hence, the time = $3\frac{1}{2}$ years.

(ii) Here,

P = ₹ 6540, A = ₹ 8447.50 and R = $12\frac{1}{2} = \frac{25}{2}$% per annum.

I = A − P = 8447.50 − 6540 = ₹ 1907.50.

$$\begin{aligned}\Rightarrow T = \frac{1907.50 \times 100}{6540 \times \frac{25}{2}}\\[1em] \Rightarrow T = \frac{1907.50 \times 100 \times 2}{6540 \times 25}\\[1em] \Rightarrow T = \frac{381500}{163500}\\[1em] \Rightarrow T = \frac{7}{3}\\[1em] \Rightarrow T = 2\frac{1}{3} \text{ years} = 2 \text{ years } 4 \text{ months}\end{aligned}$$

Hence, the time = 2 years 4 months.

Here,

P = ₹ 14000, A = ₹ 16240 and R = 4% per annum.

I = A − P = 16240 − 14000 = ₹ 2240.

$$\begin{aligned}\Rightarrow T = \frac{I \times 100}{P \times R}\\[1em] \Rightarrow T = \frac{2240 \times 100}{14000 \times 4}\\[1em] \Rightarrow T = \frac{224000}{56000}\\[1em] \Rightarrow T = 4 \text{ years}\end{aligned}$$

Hence, it will take 4 years.

Let the principal be ₹ P.

Since the money trebled, amount A = 3P.

Interest I = A − P = 3P − P = 2P.

Here, T = 6 years.

$\Rightarrow R = \frac{I \times 100}{P \times T}$%

$\Rightarrow R = \frac{2P \times 100}{P \times 6}$%

$\Rightarrow R = \frac{200}{6}$%

$\Rightarrow R = \frac{100}{3}$%

$\Rightarrow R = 33\frac{1}{3}$%

Hence, the rate of interest = $33\frac{1}{3}$% per annum.

(i) Here,

A = ₹ 4500, R = 20% per annum and T = 5 years.

Let the principal be ₹ P.

$$\begin{aligned}\Rightarrow I = \frac{P \times 20 \times 5}{100} = P\\[1em] \Rightarrow A = P + I = P + P = 2P\\[1em] \Rightarrow 2P = 4500\\[1em] \Rightarrow P = \frac{4500}{2}\\[1em] \Rightarrow P = ₹\ 2250\end{aligned}$$

Hence, the principal = ₹ 2250.

(ii) Here,

A = ₹ 2420, R = 4% per annum and T = $2\frac{1}{2} = \frac{5}{2}$ years.

Let the principal be ₹ P.

$$\begin{aligned}\Rightarrow I = \frac{P \times 4 \times \frac{5}{2}}{100} = \frac{P \times 10}{100} = \frac{P}{10}\\[1em] \Rightarrow A = P + I = P + \frac{P}{10} = \frac{11P}{10}\\[1em] \Rightarrow \frac{11P}{10} = 2420\\[1em] \Rightarrow P = \frac{2420 \times 10}{11}\\[1em] \Rightarrow P = ₹\ 2200\end{aligned}$$

Hence, the principal = ₹ 2200.

Let the sum (principal) be ₹ P and T = 3 years.

Given, simple interest I = $\frac{3}{10}$ of P = $\frac{3P}{10}$.

$\Rightarrow R = \frac{I \times 100}{P \times T}$%

$\Rightarrow R = \frac{\frac{3P}{10} \times 100}{P \times 3}$%

$\Rightarrow R = \frac{3P \times 100}{10 \times P \times 3}$%

$\Rightarrow R = \frac{300}{30}$%

$\Rightarrow R = 10$%

Hence, the rate of interest = 10% per annum.

Here,

A = ₹ 2760, T = 3 years and R = 5% per annum.

Let the sum (principal) be ₹ P.

$$\begin{aligned}\Rightarrow I = \frac{P \times 5 \times 3}{100} = \frac{15P}{100} = \frac{3P}{20}\\[1em] \Rightarrow A = P + I = P + \frac{3P}{20} = \frac{23P}{20}\\[1em] \Rightarrow \frac{23P}{20} = 2760\\[1em] \Rightarrow P = \frac{2760 \times 20}{23}\\[1em] \Rightarrow P = ₹\ 2400\end{aligned}$$

Hence, the required sum = ₹ 2400.

Here,

P = ₹ 6000, A = ₹ 6900 and T = 3 years.

I = A − P = 6900 − 6000 = ₹ 900.

$\Rightarrow R = \frac{I \times 100}{P \times T}$%

$\Rightarrow R = \frac{900 \times 100}{6000 \times 3}$%

$\Rightarrow R = \frac{90000}{18000}$%

$\Rightarrow R = 5$%

New rate of interest = 5% + 2% = 7% per annum.

New interest = $\frac{6000 \times 7 \times 3}{100} = \frac{126000}{100} = ₹\ 1260$.

New amount = P + new interest = 6000 + 1260 = ₹ 7260.

Hence, the sum will amount to ₹ 7260.

(i) 6% of ₹ 50 = $\frac{6}{100} \times 50 = ₹\ 3$.

Hence, 6% of ₹ 50 = ₹ 3.

(ii) Let the number be x. Then 25% of x = 12, so $x = \frac{12 \times 100}{25} = 48$.

Hence, the number is 48.

(iii) $1\frac{3}{4} = \frac{7}{4} = \left(\frac{7}{4} \times 100\right)$% = 175%.

Hence, $1\frac{3}{4}$ = 175%.

(iv) Increase = 28 − 20 = 8. Increase percentage = $\left(\frac{8}{20} \times 100\right)$% = 40%.

Hence, the increase percentage is 40%.

(v) Loss = 15% of 400 = $\frac{15}{100} \times 400 = ₹\ 60$. Selling price = 400 − 60 = ₹ 340.

Hence, the selling price is ₹ 340.

(vi) Hence, the profit or loss percentage is always calculated on cost price (C.P.).

(vii) S.I. = $\frac{5600 \times 8 \times 1}{100} = ₹\ 448$.

Hence, the simple interest is ₹ 448.

(viii) 135% = $\frac{135}{100} = 1.35$.

Hence, 135% = 1.35.

(ix) 50% of 60 = 30. So the number = 60 + 30 = 90.

Hence, 90 is 50% more than 60.

(x) 5 litres = 5000 mL. Required percentage = $\left(\frac{25}{5000} \times 100\right)$% = 0.5%.

Hence, 25 mL is 0.5 percent of 5 litres.

(i) 20% of 30 = 6, so 20% more than 30 = 30 + 6 = 36. The statement is correct.

Hence, the statement is True.

(ii) 2 : 5 = $\frac{2}{5} = \left(\frac{2}{5} \times 100\right)$% = 40%, not 60%. The statement is incorrect.

Hence, the statement is False.

(iii) Solving,

$\Rightarrow 6\frac{1}{4}$%

$\Rightarrow \frac{25}{4}$%

$\Rightarrow \frac{25}{4} \times \frac{1}{100} = \frac{25}{400} = \frac{1}{16}$.

Thus, the statement is correct.

Hence, the statement is True.

(iv) 80% of 450 m = $\frac{80}{100} \times 450 = 360$ m. The statement is correct.

Hence, the statement is True.

(v) Decrease = 20 − 15 = 5. Decrease percentage = $\left(\frac{5}{20} \times 100\right)$% = 25%. The statement is correct.

Hence, the statement is True.

(vi) Percentage of marks = $\left(\frac{336}{600} \times 100\right)$% = 56%, not 33.6. The statement is incorrect.

Hence, the statement is False.

(vii) 0.018 = (0.018 × 100)% = 1.8%, not 8%. The statement is incorrect.

Hence, the statement is False.

(viii) 1 km = 100000 cm. Required percentage = $\left(\frac{250}{100000} \times 100\right)$% = 0.25%, not 4%. The statement is incorrect.

Hence, the statement is False.

(ix) C.P. = S.P. + loss = 540 + 40 = ₹ 580, not ₹ 500. The statement is incorrect.

Hence, the statement is False.

(x) S.P. = ₹ 500 at a loss of 10% means S.P. = 90% of C.P. So C.P. = $\frac{500 \times 100}{90} = ₹\ 555.56$ (approximately), not ₹ 600. The statement is incorrect.

Hence, the statement is False.

Income : saving = 4 : 1.

Percentage of money saved = $\left(\frac{\text{saving}}{\text{income}} \times 100\right)$%

= $\left(\frac{1}{4} \times 100\right)$%

= 25%.

Hence, Option 2 is the correct option.

225% = $\frac{225}{100} = \frac{9}{4} = 9 : 4$.

Hence, Option 4 is the correct option.

30% of x = 72.

$$\begin{aligned}\Rightarrow \frac{30}{100} \times x = 72\\[1em] \Rightarrow x = \frac{72 \times 100}{30}\\[1em] \Rightarrow x = 240\end{aligned}$$

Hence, Option 2 is the correct option.

x% of 80 = 12.

$$\begin{aligned}\Rightarrow \frac{x}{100} \times 80 = 12\\[1em] \Rightarrow x = \frac{12 \times 100}{80}\\[1em] \Rightarrow x = 15\end{aligned}$$

Hence, Option 1 is the correct option.

0.025 = (0.025 × 100)% = 2.5%.

Hence, Option 4 is the correct answer.

Percentage of girls = 45%, so percentage of boys = (100 − 45)% = 55%.

Let the total number of students be x.

55% of x = 22

$\Rightarrow \frac{55}{100} \times x = 22 \Rightarrow x = \frac{22 \times 100}{55} \Rightarrow x = 40$

Hence, Option 3 is the correct option.

C.P. = ₹ 80 and S.P. = ₹ 100.

Gain = 100 − 80 = ₹ 20.

Gain percentage = $\left(\frac{20}{80} \times 100\right)$% = 25%.

Hence, Option 2 is the correct option.

C.P. = ₹ 120 and S.P. = ₹ 100.

Loss = 120 − 100 = ₹ 20.

Loss percentage = $\left(\frac{20}{120} \times 100\right)$%

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

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

Hence, Option 4 is the correct answer.

Increase = 25% of 24000 = $\frac{25}{100} \times 24000 = ₹\ 6000$.

New salary = 24000 + 6000 = ₹ 30000.

Hence, Option 3 is the correct option.

S.P. = ₹ 100 and gain = ₹ 20.

C.P. = S.P. − gain = 100 − 20 = ₹ 80.

Gain percentage = $\left(\frac{20}{80} \times 100\right)$% = 25%.

Hence, Option 1 is the correct option.

S.I. = $\frac{P \times R \times T}{100} = \frac{6000 \times 8 \times 1}{100} = ₹\ 480$.

Hence, Option 2 is the correct option.

S.I. = $\frac{P \times R \times T}{100} = \frac{4800 \times 5 \times 2}{100} = ₹\ 480$.

Amount = P + S.I. = 4800 + 480 = ₹ 5280.

Hence, Option 3 is the correct option.

Statement I: 125% = $\frac{125}{100} = \frac{5}{4} = \frac{10}{8} = 10 : 8$. So Statement I is true.

Statement II: 10 : 8 = 5 : 4, while 4 : 5 = 4 : 5. Since 5 : 4 ≠ 4 : 5, they are not equal ratios. So Statement II is false.

Hence, Option 1 is the correct option.

Statement I: C.P. = ₹ 4000, S.P. = ₹ 4400. Gain = 4400 − 4000 = ₹ 400. Gain percentage = $\left(\frac{400}{4000} \times 100\right)$% = 10%. So Statement I is true.

Statement II: Profit and loss percentage is always calculated on the cost price. So Statement II is true.

Hence, Option 3 is the correct option.

Statement I: S.I. = $\frac{10000 \times 6 \times 3}{100} = ₹\ 1800$. Amount = 10000 + 1800 = ₹ 11800, not ₹ 13000. So Statement I is false.

Statement II: The formula for simple interest is $\frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100}$, which is correct. So Statement II is true.

Hence, Option 2 is the correct option.

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

(i) $12\frac{1}{2}$%

$\Rightarrow \frac{25}{2}$%

$\Rightarrow \frac{25}{2} \times \frac{1}{100}$

$\Rightarrow \frac{25}{200}$

$\Rightarrow \frac{1}{8}$

Hence, $12\frac{1}{2}$% = $\frac{1}{8}$.

(ii) $66\frac{2}{3}$%

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

$\Rightarrow \frac{200}{3} \times \frac{1}{100}$

$\Rightarrow \frac{200}{300}$

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

Hence, $66\frac{2}{3}$% = $\frac{2}{3}$.

(iii) $8\frac{1}{3}$%

$\Rightarrow \frac{25}{3}$%

$\Rightarrow \frac{25}{3} \times \frac{1}{100}$

$\Rightarrow \frac{25}{300}$

$\Rightarrow \frac{1}{12}$

Hence, $8\frac{1}{3}$% = $\frac{1}{12}$.

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

(i) $\frac{5}{8}$

$\Rightarrow \left(\frac{5}{8} \times 100\right)$%

$\Rightarrow \frac{500}{8}$%

$\Rightarrow 62.5$%

$\Rightarrow 62\frac{1}{2}$%

Hence, $\frac{5}{8} = 62\frac{1}{2}$%.

(ii) $\frac{13}{40}$

$\Rightarrow \left(\frac{13}{40} \times 100\right)$%

$\Rightarrow \frac{1300}{40}$%

$\Rightarrow 32.5$%

$\Rightarrow 32\frac{1}{2}$%

Hence, $\frac{13}{40} = 32\frac{1}{2}$%.

(iii) $\frac{7}{6}$

$\Rightarrow \left(\frac{7}{6} \times 100\right)$%

$\Rightarrow \frac{700}{6}$%

$\Rightarrow \frac{350}{3}$%

$\Rightarrow 116\frac{2}{3}$%

Hence, $\frac{7}{6} = 116\frac{2}{3}$%.

To convert a percentage into a decimal, replace the % sign with $\frac{1}{100}$ and express as a decimal.

(i) 122%

$$\begin{aligned}\Rightarrow \frac{122}{100}\\[1em] \Rightarrow 1.22\end{aligned}$$

Hence, 122% = 1.22.

(ii) 2.2%

$$\begin{aligned}\Rightarrow \frac{2.2}{100}\\[1em] \Rightarrow 0.022\end{aligned}$$

Hence, 2.2% = 0.022.

(iii) $3\frac{1}{8}$%

$\Rightarrow \frac{25}{8}$%

$\Rightarrow \frac{25}{8} \times \frac{1}{100}$

$\Rightarrow \frac{25}{800}$

$\Rightarrow 0.03125$

Hence, $3\frac{1}{8}$% = 0.03125.

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

⇒ (0.0345 × 100)%

⇒ 3.45%

Hence, 0.0345 = 3.45%.

The given ratio is 5 : 6 : 9.

Total number of parts = 5 + 6 + 9 = 20.

Percentage of first part = $\left(\frac{5}{20} \times 100\right)$% = 25%.

Percentage of second part = $\left(\frac{6}{20} \times 100\right)$% = 30%.

Percentage of third part = $\left(\frac{9}{20} \times 100\right)$% = 45%.

Hence, the three parts are 25%, 30% and 45%.

(i) We know, 1 day = 24 hours = 24 × 60 = 1440 minutes.

Half an hour = 30 minutes.

Required percentage = $\left(\frac{30}{1440} \times 100\right)$%

$\Rightarrow \frac{3000}{1440}$%

$\Rightarrow \frac{25}{12}$%

$\Rightarrow 2\frac{1}{12}$%

Hence, half an hour is $2\frac{1}{12}$% of a day.

(ii) Here, $4\frac{1}{2}$ m = $\frac{9}{2}$ m.

Required percentage = $\left(\frac{\frac{3}{4}}{\frac{9}{2}} \times 100\right)$%

$\Rightarrow \left(\frac{3}{4} \times \frac{2}{9} \times 100\right)$%

$\Rightarrow \left(\frac{6}{36} \times 100\right)$%

$\Rightarrow \left(\frac{1}{6} \times 100\right)$%

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

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

Hence, $\frac{3}{4}$ metres is $16\frac{2}{3}$% of $4\frac{1}{2}$ metres.

Original population = 25000 and new population = 24500.

Decrease in population = 25000 − 24500 = 500.

Percentage decrease = $\left(\frac{500}{25000} \times 100\right)$%

$\Rightarrow \frac{50000}{25000}$%

$\Rightarrow 2$%

Hence, the percentage decrease = 2%.

Original price = ₹ 350000 and new price = ₹ 370000.

Increase in price = 370000 − 350000 = ₹ 20000.

Percentage increase = $\left(\frac{20000}{350000} \times 100\right)$%

$\Rightarrow \frac{2000000}{350000}$%

$\Rightarrow \frac{40}{7}$%

$\Rightarrow 5\frac{5}{7}$%

Hence, the percentage increase in price = $5\frac{5}{7}$%.

Original population = 3650.

Decrease = 6% of 3650 = $\frac{6}{100} \times 3650 = 219$.

Population after decrease = 3650 − 219 = 3431.

Hence, the population after decrease = 3431.

Percentage of girls = 43%, so percentage of boys = (100 − 43)% = 57%.

(i) Let the total strength of the school be x.

57% of x = 1482

$\Rightarrow \frac{57}{100} \times x = 1482$

$\Rightarrow x = \frac{1482 \times 100}{57}$

$\Rightarrow x = 2600$

Hence, the total strength of the school = 2600.

(ii) Number of girls = total strength − number of boys = 2600 − 1482 = 1118.

Hence, the number of girls in the school = 1118.

S.P. of the article = ₹ 1027 and loss = ₹ 273.

C.P. = S.P. + loss = 1027 + 273 = ₹ 1300.

Loss percentage = $\left(\frac{273}{1300} \times 100\right)$%

$\Rightarrow \frac{27300}{1300}$%

$\Rightarrow 21$%

Hence, the loss percentage = 21%.

S.P. of the lamp = ₹ 710 and loss = ₹ 40.

C.P. = S.P. + loss = 710 + 40 = ₹ 750.

To gain 10%:

Gain = 10% of 750 = $\frac{10}{100} \times 750 = ₹\ 75$.

New S.P. = C.P. + gain = 750 + 75 = ₹ 825.

Hence, the cost price of the lamp = ₹ 750 and to gain 10% it should be sold for ₹ 825.

Here, P = ₹ 6000, R = 6.5% per annum and T = 3 years.

$$\begin{aligned}\Rightarrow I = \frac{P \times R \times T}{100}\\[1em] \Rightarrow I = \frac{6000 \times 6.5 \times 3}{100}\\[1em] \Rightarrow I = \frac{117000}{100}\\[1em] \Rightarrow I = ₹\ 1170\end{aligned}$$

Amount = P + I = ₹ 6000 + ₹ 1170 = ₹ 7170.

Hence, the interest = ₹ 1170 and the amount = ₹ 7170.

Here, P = ₹ 1860, A = ₹ 2449 and R = 9.5% per annum.

I = A − P = 2449 − 1860 = ₹ 589.

$$\begin{aligned}\Rightarrow T = \frac{I \times 100}{P \times R}\\[1em] \Rightarrow T = \frac{589 \times 100}{1860 \times 9.5}\\[1em] \Rightarrow T = \frac{58900}{17670}\\[1em] \Rightarrow T = \frac{10}{3}\\[1em] \Rightarrow T = 3\frac{1}{3} \text{ years} = 3 \text{ years } 4 \text{ months}\end{aligned}$$

Hence, it will take 3 years 4 months.

Here, P = ₹ 7200, I = ₹ 3024 and T = 4 years.

$\Rightarrow R = \frac{I \times 100}{P \times T}$%

$\Rightarrow R = \frac{3024 \times 100}{7200 \times 4}$%

$\Rightarrow R = \frac{302400}{28800}$%

$\Rightarrow R = 10.5$%

Hence, the rate of interest = 10.5% p.a.

Here, I = ₹ 1155, R = 11% per annum and T = 3 years 6 months = $3\frac{1}{2} = \frac{7}{2}$ years.

$$\begin{aligned}\Rightarrow P = \frac{I \times 100}{R \times T}\\[1em] \Rightarrow P = \frac{1155 \times 100}{11 \times \frac{7}{2}}\\[1em] \Rightarrow P = \frac{1155 \times 100 \times 2}{11 \times 7}\\[1em] \Rightarrow P = \frac{231000}{77}\\[1em] \Rightarrow P = ₹\ 3000\end{aligned}$$

Hence, the required sum = ₹ 3000.

Let the money Medha originally had be ₹ x.

Money deposited in the bank = 20% of x = $\frac{20}{100} \times x = \frac{x}{5}$.

Remainder = $x - \frac{x}{5} = \frac{4x}{5}$.

Amount spent = 20% of the remainder = $\frac{20}{100} \times \frac{4x}{5} = \frac{4x}{25}$.

Money left = $\frac{4x}{5} - \frac{4x}{25}$

$$\begin{aligned}\Rightarrow \frac{20x}{25} - \frac{4x}{25}\\[1em] \Rightarrow \frac{16x}{25}\end{aligned}$$

Given, money left = ₹ 48000.

$$\begin{aligned}\Rightarrow \frac{16x}{25} = 48000\\[1em] \Rightarrow x = \frac{48000 \times 25}{16}\\[1em] \Rightarrow x = 75000\end{aligned}$$

Hence, Medha originally had ₹ 75000.

Let Raman's income be ₹ 100.

Mohan's income = 100 + 25% of 100 = 100 + 25 = ₹ 125.

Raman's income is less than Mohan's income by = 125 − 100 = ₹ 25.

Required percentage = $\left(\frac{25}{125} \times 100\right)$%

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

$\Rightarrow 20$%

Hence, Raman's income is 20% less than Mohan's income.

Alcohol present in 400 mL of 15% solution = 15% of 400 = $\frac{15}{100} \times 400 = 60$ mL.

Let the quantity of pure alcohol to be added be x mL.

New total quantity of solution = (400 + x) mL.

New quantity of alcohol = (60 + x) mL.

For a 32% alcohol solution:

$$\begin{aligned}\Rightarrow \frac{60 + x}{400 + x} = \frac{32}{100}\\[1em] \Rightarrow 100(60 + x) = 32(400 + x)\\[1em] \Rightarrow 6000 + 100x = 12800 + 32x\\[1em] \Rightarrow 100x - 32x = 12800 - 6000\\[1em] \Rightarrow 68x = 6800\\[1em] \Rightarrow x = 100\end{aligned}$$

Hence, he should mix 100 mL of pure alcohol.

The shopkeeper sold the item for ₹ 594 at a profit of 20%.

Shopkeeper's C.P. = $\frac{594 \times 100}{100 + 20} = \frac{594 \times 100}{120} = ₹\ 495$.

This ₹ 495 is the agency's selling price (at 10% profit).

Agency's C.P. = $\frac{495 \times 100}{100 + 10} = \frac{495 \times 100}{110} = ₹\ 450$.

This ₹ 450 is the manufacturer's selling price (at 25% profit).

Manufacturing price = $\frac{450 \times 100}{100 + 25} = \frac{450 \times 100}{125} = ₹\ 360$.

Hence, the manufacturing price = ₹ 360.