Percent and Percentage

⇒ $\frac{3}{4}$

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

⇒ $\left(\frac{3}{4} \times 100\right)$%

⇒ $\frac{300}{4}$%

⇒ 75%

Hence, $\frac{3}{4} = 75$%.

⇒ $\frac{2}{3}$

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

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

⇒ $\frac{200}{3}$%

⇒ $66\frac{2}{3}$%

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

⇒ 0.025

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

⇒ 0.025 × 100%

⇒ 2.5%

Hence, 0.025 = 2.5%.

⇒ 0.125

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

⇒ 0.125 × 100%

⇒ 12.5%

Hence, 0.125 = 12.5%.

⇒ $7\frac{1}{2}$% = $\frac{15}{2}$%

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

⇒ $\frac{15}{2}$%

$$\begin{aligned}\Rightarrow \frac{15}{2} \times \frac{1}{100}\\[1em] \Rightarrow \frac{15}{200}\\[1em] \Rightarrow \frac{3}{40} \\[1em] \Rightarrow 0.075\end{aligned}$$

Hence, $7\frac{1}{2}$% = $\frac{3}{40}$ = 0.075.

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

⇒ 2.50%

$$\begin{aligned}\Rightarrow 2.50 \times \frac{1}{100}\\[1em] \Rightarrow \frac{2.5}{100}\\[1em] \Rightarrow \frac{25}{1000}\\[1em] \Rightarrow \frac{1}{40}\end{aligned}$$

As a decimal, $\frac{1}{40} = 0.025$.

Hence, 2.50% $= \frac{1}{40} = 0.025$.

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

⇒ 0.02%

$$\begin{aligned}\Rightarrow 0.02 \times \frac{1}{100}\\[1em] \Rightarrow \frac{0.02}{100}\\[1em] \Rightarrow \frac{2}{10000}\\[1em] \Rightarrow \frac{1}{5000}\end{aligned}$$

As a decimal, $\frac{1}{5000} = 0.0002$.

Hence, 0.02% $= \frac{1}{5000} = 0.0002$.

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

⇒ 175%

$$\begin{aligned}\Rightarrow 175 \times \frac{1}{100}\\[1em] \Rightarrow \frac{175}{100}\\[1em] \Rightarrow \frac{7}{4}\\[1em] \Rightarrow 1\frac{3}{4}\end{aligned}$$

As a decimal, $\frac{7}{4} = 1\frac{3}{4}= 1.75$.

Hence, 175% $= 1\frac{3}{4} = 1.75$.

To express one quantity as a percent of another, both must be in the same units; then percentage = $\left(\frac{\text{one quantity}}{\text{other quantity}} \times 100\right)$%.

⇒ 16 hours of 2 days

We know, 1 day = 24 hours, so 2 days = 48 hours.

$\Rightarrow \left(\frac{16}{48} \times 100\right)$%

$\Rightarrow \frac{1600}{48}$%

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

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

Hence, 16 hours is $33\frac{1}{3}$% of 2 days.

To express one quantity as a percent of another, both must be in the same units; then percentage = $\left(\frac{\text{one quantity}}{\text{other quantity}} \times 100\right)$%.

⇒ 40 paisa of ₹ 2

We know, ₹ 1 = 100 paisa, so ₹ 2 = 200 paisa.

$\Rightarrow \left(\frac{40}{200} \times 100\right)$%

$\Rightarrow \frac{4000}{200}$%

$\Rightarrow 20$%

Hence, 40 paisa is 20% of ₹ 2.

To express one quantity as a percent of another, both must be in the same units; then percentage = $\left(\frac{\text{one quantity}}{\text{other quantity}} \times 100\right)$%.

⇒ 25 cm of 4 metres

We know, 1 m = 100 cm, so 4 m = 400 cm.

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

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

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

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

Hence, 25 cm is $6\frac{1}{4}$% of 4 metres.

To express one quantity as a percent of another, both must be in the same units; then percentage = $\left(\frac{\text{one quantity}}{\text{other quantity}} \times 100\right)$%.

⇒ 600 g of 5 kg

We know, 1 kg = 1000 g, so 5 kg = 5000 g.

$\Rightarrow \left(\frac{600}{5000} \times 100\right)$%

$\Rightarrow \frac{60000}{5000}$%

⇒ 12%

Hence, 600 g is 12% of 5 kg.

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

⇒ 5% of ₹ 350

$$\begin{aligned}\Rightarrow \frac{5}{100} \times 350\\[1em] \Rightarrow \frac{1750}{100}\\[1em] \Rightarrow ₹\ 17.50\end{aligned}$$

Hence, 5% of ₹ 350 = ₹ 17.50.

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

⇒ 10% of ₹ 400.40

$$\begin{aligned}\Rightarrow \frac{10}{100} \times 400.40\\[1em] \Rightarrow \frac{4004}{100}\\[1em] \Rightarrow ₹\ 40.04\end{aligned}$$

Hence, 10% of ₹ 400.40 = ₹ 40.04.

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

⇒ 1% of ₹ 500

$\Rightarrow \frac{1}{100} \times 500$

⇒ ₹ 5

Hence, 1% of ₹ 500 = ₹ 5.

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

As, $12\frac{1}{2}$% = $\frac{25}{2}$%

⇒ $\frac{25}{2}$% of 80 kg

$$\begin{aligned}\Rightarrow \frac{25}{2 \times 100} \times 80\\[1em] \Rightarrow \frac{25 \times 80}{200}\\[1em] \Rightarrow \frac{2000}{200}\\[1em] \Rightarrow 10 \text{ kg}\end{aligned}$$

Hence, $12\frac{1}{2}$% of 80 kg = 10 kg.

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

⇒ $\frac{5}{8}$% of ₹ 600

$$\begin{aligned}\Rightarrow \frac{5}{8 \times 100} \times 600\\[1em] \Rightarrow \frac{5 \times 600}{800}\\[1em] \Rightarrow \frac{3000}{800}\\[1em] \Rightarrow ₹\ 3.75\end{aligned}$$

Hence, $\frac{5}{8}$% of ₹ 600 = ₹ 3.75.

⇒ $33\frac{1}{3}$% = $\frac{100}{3}$%

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

⇒ $\frac{100}{3}$% of 27 m

$$\begin{aligned}\Rightarrow \frac{100}{3 \times 100} \times 27\\[1em] \Rightarrow \frac{27}{3}\\[1em] \Rightarrow 9 \text{ m}\end{aligned}$$

Hence, $33\frac{1}{3}$% of 27 m = 9 m.

Total number of children = 60.

Percentage of girls = 30%.

Percentage of boys = (100 − 30)% = 70%.

Number of boys = 70% of 60

$$\begin{aligned}= \frac{70}{100} \times 60\\[1em] = \frac{4200}{100}\\[1em] = 42\end{aligned}$$

Hence, there are 42 boys in the class.

Total votes polled = 8000.

Votes got by A = 60% of 8000

$= \frac{60}{100} \times 8000$

= 4800

Votes got by B = 8000 − 4800 = 3200.

Hence, A got 4800 votes and B got 3200 votes.

Total salary = ₹ 2,500.

Percentage saved = 12%.

Percentage of expenditure = (100 − 12)% = 88%.

Expenditure = 88% of 2,500

$$\begin{aligned}= \frac{88}{100} \times 2,500\\[1em] = \frac{220000}{100}\\[1em] = ₹\ 2,200\end{aligned}$$

Hence, his expenditure = ₹ 2,200.

Total marks = 800.

Percentage of marks got = 75%.

Percentage of marks lost = (100 − 75)% = 25%.

Marks lost = 25% of 800

$= \frac{25}{100} \times 800$

= 200

Hence, Seeta lost 200 marks.

Value of the shop = ₹ 25,000.

Amount insured = 95% of 25,000

$$\begin{aligned}= \frac{95}{100} \times 25,000\\[1em] = \frac{2375000}{100}\\[1em] = ₹\ 23,750\end{aligned}$$

Hence, the owner would get ₹ 23,750.

Number of boys = 30 and number of girls = 25.

Total number of students = 30 + 25 = 55.

Percentage of boys = $\left(\frac{30}{55} \times 100\right)$%

$= \frac{3000}{55}$%

$= \frac{600}{11}$%

$= 54\frac{6}{11}$%

Hence, the percentage of boys in the class = $54\frac{6}{11}$%.

$\Rightarrow 3\frac{2}{5} = \frac{17}{5}$

$\Rightarrow \frac{17}{5}$ as percent

$\Rightarrow \left(\frac{17}{5} \times 100\right)$%

$\Rightarrow \frac{1700}{5}$%

⇒ 340%

Hence, $3\frac{2}{5}$ = 340%.

⇒ 0.0075 as percent

⇒ 0.0075 x 100%

⇒ 0.75%

Hence, 0.0075 = 0.75%.

⇒ 3 : 20 = $\frac{3}{20}$

$\Rightarrow \frac{3}{20}$ as percent

$\Rightarrow \left(\frac{3}{20} \times 100\right)$%

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

⇒ 15%

Hence, 3 : 20 = 15%.

⇒ 60 cm as percent of 1 m 25 cm

We know, 1 m 25 cm = 100 cm + 25 cm = 125 cm.

$\Rightarrow \left(\frac{60}{125} \times 100\right)$%

$\Rightarrow \frac{6000}{125}$%

⇒ 48%

Hence, 60 cm is 48% of 1 m 25 cm.

⇒ 9 hours as percent of 4 days

We know, 1 day = 24 hours, so 4 days = 96 hours.

$\Rightarrow \left(\frac{9}{96} \times 100\right)$%

$\Rightarrow \frac{900}{96}$%

$\Rightarrow \frac{75}{8}$%

$\Rightarrow 9\frac{3}{8}$%

⇒ 9.375 %

Hence, 9 hours is $9\frac{3}{8}$% = 9.375 % of 4 days.

We know, 2 hours 30 min = (2 × 60 + 30) min = 150 min.

$$\begin{aligned}\Rightarrow \frac{2}{100} \times 150\\[1em] \Rightarrow \frac{300}{100}\\[1em] \Rightarrow 3 \text{ min}\end{aligned}$$

Hence, 2% of 2 hours 30 min = 3 minutes.

We know, 1 kg = 1000 g, so 12 kg = 12000 g.

$\Rightarrow \left(\frac{725}{12000} \times 100\right)$%

$\Rightarrow \frac{72500}{12000}$%

$\Rightarrow \frac{145}{24}$%

$\Rightarrow 6\frac{1}{24}$%

Hence, 725 g is $6\frac{1}{24}$% of 12 kg.

Total number of mangoes = 250.

Number of rotten mangoes = 12% of 250

$= \frac{12}{100} \times 250$

= 30

Remaining mangoes = 250 − 30 = 220.

Number of crushed mangoes = 10% of 220

$= \frac{10}{100} \times 220$

= 22

Mangoes in good condition = 220 − 22 = 198.

Hence, 198 mangoes were in good condition.

Total number of questions = 60.

Chandra's correct answers = 90% of 60

$= \frac{90}{100} \times 60$

= 54

Ram's correct answers = 80% of 60

$= \frac{80}{100} \times 60$

= 48

Ram's correct answers as a percent of Chandra's = $\left(\frac{48}{54} \times 100\right)$%

$= \frac{4800}{54}$%

$= \frac{800}{9}$%

$= 88\frac{8}{9}$%

Hence, Chandra got 54 correct answers, Ram got 48 correct answers, and Ram's correct answers are $88\frac{8}{9}$% of Chandra's.

Maximum marks = 900.

Marks obtained by the student = 33% of 900

$= \frac{33}{100} \times 900$

= 297

Since the student fails by 45 marks,

Passing marks = 297 + 45 = 342.

Passing percentage = $\left(\frac{342}{900} \times 100\right)$%

$= \frac{34200}{900}$%

= 38%

Hence, the passing marks = 342 and the passing percentage = 38%.

Percentage travelling in first class = 15%.

Percentage travelling in second class = 35%.

Percentage travelling in A.C. class = 100% − (15% + 35%)

= 100% − 50%

= 50%

Hence, the percentage of A.C. class travellers = 50%.

Percentage of cake eaten = 25%.

Percentage of cake given away = 35%.

Percentage of cake left = 100% − (25% + 35%)

= 100% − 60%

= 40%

Hence, 40% of the cake is still left with him.

Total number of letters in the English alphabet = 26.

Number of vowels (a, e, i, o, u) = 5.

Percentage of vowels = $\left(\frac{5}{26} \times 100\right)$%

$= \frac{500}{26}$%

$= \frac{250}{13}$%

$= 19\frac{3}{13}$%

Hence, the percentage of vowels in the English alphabet = $19\frac{3}{13}$%.

$6\frac{1}{4}$% = $\frac{25}{4}$%

Let the number be x whose $\frac{25}{4}$% = 375.

$\Rightarrow \frac{25}{4}$% of x = 375

$$\begin{aligned}\Rightarrow \frac{25}{4}\times \frac{1}{100} \times x = 375\\[1em] \Rightarrow x = \frac{375 \times 400}{25}\\[1em] \Rightarrow x = 6000\end{aligned}$$

Hence, the required number = 6000.

Let the number be x whose 0.2% = 5.

⇒ 0.2% of x = 5

$$\begin{aligned}\Rightarrow \frac{0.2}{100} \times x = 5\\[1em] \Rightarrow x = \frac{5 \times 100}{0.2}\\[1em] \Rightarrow x = \frac{500}{0.2}\\[1em] \Rightarrow x = 2500\end{aligned}$$

Hence, the required number = 2500.

$16\frac{2}{3}$% = $\frac{50}{3}$%

Let the number be x whose $\frac{50}{3}$% = 30.

$\Rightarrow \frac{50}{3}$% of x = 30

$$\begin{aligned}\Rightarrow \frac{50}{3} \times \frac{1}{100} \times x = 30\\[1em] \Rightarrow \frac{50}{300} \times x = 30\\[1em] \Rightarrow x = \frac{30 \times 300}{50}\\[1em] \Rightarrow x = 180\end{aligned}$$

Hence, the required number = 180.

Let the cost (value) of the house be ₹ x.

Given, 1% of its value = ₹ 5,000.

$$\begin{aligned}\Rightarrow \frac{1}{100} \times x = 5,000\\[1em] \Rightarrow x = 5,000 \times 100\\[1em] \Rightarrow x = 5,00,000\end{aligned}$$

Hence, the cost of the house = ₹ 5,00,000.

Total number of students = 300.

Number of girls = 70% of 300 = $\frac{70}{100} \times 300 = 210$.

Number of boys = 30% of 300 = $\frac{30}{100} \times 300 = 90$.

When 30 girls leave:

New number of girls = 210 − 30 = 180.

New total number of students = 300 − 30 = 270.

New percentage of girls = $\left(\frac{180}{270} \times 100\right)$%

$= \frac{18000}{270}$%

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

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

Hence, the new percentage of girls in the school = $66\frac{2}{3}$%.

Cost of the transistor = ₹ 960.

$12\frac{1}{2}$% = $\frac{25}{2}$%

Cash money paid = $\frac{25}{2}$% of 960

$$\begin{aligned}= \frac{25}{2} \times \frac{1}{100} \times 960\\[1em] = \frac{25 \times 960}{200}\\[1em] = ₹\ 120\end{aligned}$$

Remaining amount = 960 − 120 = ₹ 840.

Amount paid each month = $\frac{840}{12}$ = ₹ 70.

Hence, Kumar will pay ₹ 70 each month.

Let the quantity of ore required be x kg.

Given, 20% of the ore = 45 kg of zinc.

$$\begin{aligned}\Rightarrow \frac{20}{100} \times x = 45\\[1em] \Rightarrow x = \frac{45 \times 100}{20}\\[1em] \Rightarrow x = 225\end{aligned}$$

Hence, 225 kg of ore will be required.

Original salary = ₹ 600 and new salary = ₹ 850.

Increase in salary = 850 − 600 = ₹ 250.

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

$= \left(\frac{250}{600} \times 100\right)$%

$= \frac{25000}{600}$%

$= \frac{125}{3}$%

$= 41\frac{2}{3}$%

Hence, the increase in salary = $41\frac{2}{3}$%.

Increase = 5% of 60 = $\frac{5}{100} \times 60 = 3$.

Increased value = 60 + 3 = 63.

Hence, 60 increased by 5% = 63.

Increase = 15% of 20 = $\frac{15}{100} \times 20 = 3$.

Increased value = 20 + 3 = 23.

Hence, 20 increased by 15% = 23.

Increase = $12\frac{1}{2}$% of 48 = $\frac{25}{2 \times 100} \times 48 = 6$.

Increased value = 48 + 6 = 54.

Hence, 48 increased by $12\frac{1}{2}$% = 54.

Increase = 140% of 80 = $\frac{140}{100} \times 80 = 112$.

Increased value = 80 + 112 = 192.

Hence, 80 increased by 140% = 192.

Increase = 3.5% of 1000 = $\frac{3.5}{100} \times 1000 = 35$.

Increased value = 1000 + 35 = 1035.

Hence, 1000 increased by 3.5% = 1035.

Decrease = 20% of 80 = $\frac{20}{100} \times 80 = 16$.

Decreased value = 80 − 16 = 64.

Hence, 80 decreased by 20% = 64.

Decrease = 10% of 300 = $\frac{10}{100} \times 300 = 30$.

Decreased value = 300 − 30 = 270.

Hence, 300 decreased by 10% = 270.

Decrease = 12.5% of 50 = $\frac{12.5}{100} \times 50 = 6.25$.

Decreased value = 50 − 6.25 = 43.75.

Hence, 50 decreased by 12.5% = 43.75.

Let the number be x. When increased by 10%, it becomes 110% of x.

$$\begin{aligned}\Rightarrow \frac{110}{100} \times x = 88\\[1em] \Rightarrow x = \frac{88 \times 100}{110}\\[1em] \Rightarrow x = 80\end{aligned}$$

Hence, the required number = 80.

Let the number be x. When increased by 15%, it becomes 115% of x.

$$\begin{aligned}\Rightarrow \frac{115}{100} \times x = 230\\[1em] \Rightarrow x = \frac{230 \times 100}{115}\\[1em] \Rightarrow x = 200\end{aligned}$$

Hence, the required number = 200.

Let the number be x. When decreased by 15%, it becomes 85% of x.

$$\begin{aligned}\Rightarrow \frac{85}{100} \times x = 170\\[1em] \Rightarrow x = \frac{170 \times 100}{85}\\[1em] \Rightarrow x = 200\end{aligned}$$

Hence, the required number = 200.

Let the number be x. When decreased by 40%, it becomes 60% of x.

$$\begin{aligned}\Rightarrow \frac{60}{100} \times x = 480\\[1em] \Rightarrow x = \frac{480 \times 100}{60}\\[1em] \Rightarrow x = 800\end{aligned}$$

Hence, the required number = 800.

Let the number be x. When increased by 100%, it becomes 200% of x.

$$\begin{aligned}\Rightarrow \frac{200}{100} \times x = 100\\[1em] \Rightarrow x = \frac{100 \times 100}{200}\\[1em] \Rightarrow x = 50\end{aligned}$$

Hence, the required number = 50.

Let the number be x. When decreased by 50%, it becomes 50% of x.

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

Hence, the required number = 100.

Let the original price be ₹ x. When lowered by 20%, it becomes 80% of x.

$$\begin{aligned}\Rightarrow \frac{80}{100} \times x = 40,000\\[1em] \Rightarrow x = \frac{40,000 \times 100}{80}\\[1em] \Rightarrow x = 50,000\end{aligned}$$

Reduction in price = 50,000 − 40,000 = ₹ 10,000.

Hence, the original price = ₹ 50,000 and the reduction in price = ₹ 10,000.

Let the original price be ₹ x.

Given, 25% of x = ₹ 10.

$$\begin{aligned}\Rightarrow \frac{25}{100} \times x = 10\\[1em] \Rightarrow x = \frac{10 \times 100}{25}\\[1em] \Rightarrow x = 40\end{aligned}$$

New price = original price + increase = 40 + 10 = ₹ 50.

Hence, the new price = ₹ 50.

Let the old price be ₹ x.

Given, 10% of x = ₹ 40.

$$\begin{aligned}\Rightarrow \frac{10}{100} \times x = 40\\[1em] \Rightarrow x = \frac{40 \times 100}{10}\\[1em] \Rightarrow x = 400\end{aligned}$$

Hence, the old price = ₹ 400.

Let the old price be ₹ 100.

Change in price (reduction) = 25% of 100 = ₹ 25.

New price = 100 − 25 = ₹ 75.

(i) Change in price : old price = 25 : 100 = 1 : 4.

Hence, the ratio of the change in price to the old price = 1 : 4.

(ii) Old price : new price = 100 : 75 = 4 : 3.

Hence, the ratio of the old price to the new price = 4 : 3.

Since x is 20% less than y,

$$\begin{aligned}\Rightarrow x = y - 20% \text{ of } y\\[1em] \Rightarrow x = y - \frac{20}{100}y\\[1em] \Rightarrow x = y - \frac{y}{5}\\[1em] \Rightarrow x = \frac{4y}{5}\end{aligned}$$

$x = \frac{4y}{5}$

(i) $\frac{x}{y}$

$$\begin{aligned}\frac{x}{y} = \frac{\frac{4y}{5}}{y}\\[1em] = \frac{4y}{5} \times \frac{1}{y}\\[1em] = \frac{4}{5}\end{aligned}$$

Hence, $\frac{x}{y} = \frac{4}{5}$.

(ii) $\frac{y - x}{y}$

$$\begin{aligned}\frac{y - x}{y} = \frac{y - \frac{4y}{5}}{y}\\[1em] = \frac{\frac{y}{5}}{y}\\[1em] = \frac{y}{5} \times \frac{1}{y}\\[1em] = \frac{1}{5}\end{aligned}$$

Hence, $\frac{y - x}{y} = \frac{1}{5}$.

(iii) $\frac{x}{y - x}$

$$\begin{aligned}\frac{x}{y - x} = \frac{\frac{4y}{5}}{y - \frac{4y}{5}}\\[1em] = \frac{\frac{4y}{5}}{\frac{y}{5}}\\[1em] = \frac{4y}{5} \times \frac{5}{y}\\[1em] = 4\end{aligned}$$

Hence, $\frac{x}{y - x} = 4$.

Since x is 30% more than y,

$$\begin{aligned}\Rightarrow x = y + 30% \text{ of } y\\[1em] \Rightarrow x = y + \frac{30}{100}y\\[1em] \Rightarrow x = y + \frac{3y}{10}\\[1em] \Rightarrow x = \frac{13y}{10}\end{aligned}$$

x = $\frac{13y}{10}$

(i) $\frac{x}{y}$

$$\begin{aligned}\frac{x}{y} = \frac{\frac{13y}{10}}{y}\\[1em] = \frac{13y}{10} \times \frac{1}{y}\\[1em] = \frac{13}{10}\end{aligned}$$

Hence, $\frac{x}{y} = \frac{13}{10}$.

(ii) $\frac{y + x}{x}$

$$\begin{aligned}\frac{y + x}{x} = \frac{y + \frac{13y}{10}}{\frac{13y}{10}}\\[1em] = \frac{\frac{23y}{10}}{\frac{13y}{10}}\\[1em] = \frac{23y}{10} \times \frac{10}{13y}\\[1em] = \frac{23}{13}\end{aligned}$$

Hence, $\frac{y + x}{x} = \frac{23}{13}$.

(iii) $\frac{y}{y - x}$

$$\begin{aligned}\frac{y}{y - x} = \frac{y}{y - \frac{13y}{10}}\\[1em] = \frac{y}{-\frac{3y}{10}}\\[1em] = y \times \left(-\frac{10}{3y}\right)\\[1em] = -\frac{10}{3}\end{aligned}$$

Hence, $\frac{y}{y - x} = -\frac{10}{3}$.

Actual weight = 40 kg and weight measured = 40.8 kg.

Error = 40.8 − 40 = 0.8 kg.

Error percent = $\left(\frac{\text{error}}{\text{actual weight}} \times 100\right)$%

$= \left(\frac{0.8}{40} \times 100\right)$%

$= \frac{80}{40}$%

= 2%

Hence, the error percent = 2%.

Total petrol = 450 litres.

Petrol lost = 8% of 450 = $\frac{8}{100} \times 450 = 36$ litres.

Petrol left = 450 − 36 = 414 litres.

Hence, 414 litres of petrol were left in the cask.

Let the weight corresponding to each part be x kg.

So, copper = 13x, zinc = 7x and nickel = 5x.

Total weight of the alloy = 13x + 7x + 5x = 25x.

Percentage of copper = $\left(\frac{13x}{25x} \times 100\right)$%

$= \frac{1300}{25}$%

= 52%

Percentage of zinc = $\left(\frac{7x}{25x} \times 100\right)$%

$= \frac{700}{25}$%

= 28%

Percentage of nickel = $\left(\frac{5x}{25x} \times 100\right)$%

$= \frac{500}{25}$%

= 20%

Hence, copper = 52%, zinc = 28% and nickel = 20%.

Marks secured by the student = 538.

Since the student misses the first division by 2 marks,

Marks needed for first division = 538 + 2 = 540.

Let the total marks be x. Since 60% of the total marks are needed for first division,

$$\begin{aligned}\Rightarrow \frac{60}{100} \times x = 540\\[1em] \Rightarrow x = \frac{540 \times 100}{60}\\[1em] \Rightarrow x = 900\end{aligned}$$

Hence, the total marks of the examination = 900.

Total number of pupils = 1200.

Number of boys = 900 and number of girls = 1200 − 900 = 300.

Boys wearing spectacles = 20% of 900 = $\frac{20}{100} \times 900 = 180$.

Girls wearing spectacles = 30% of 300 = $\frac{30}{100} \times 300 = 90$.

(i) Total pupils wearing spectacles = 180 + 90 = 270.

Hence, 270 pupils wear spectacles.

(ii) Percentage wearing spectacles = $\left(\frac{270}{1200} \times 100\right)$%

$= \frac{27000}{1200}$%

= 22.5%

Hence, 22.5% of the total number of pupils wear spectacles.

Total number of bulbs = 25.

Number of red bulbs = 17 and number of black bulbs = 3.

Number of yellow bulbs = 25 − (17 + 3) = 25 − 20 = 5.

Difference between red and yellow bulbs = 17 − 5 = 12.

Difference as a percent of total = $\left(\frac{12}{25} \times 100\right)$%

$= \frac{1200}{25}$%

= 48%

Hence, the difference is 12 bulbs, which is 48% of the total.

Let the number be 100.

After an increase of 20%:

$\Rightarrow 100 + \frac{20}{100} \times 100$

⇒ 100 + 20

⇒ 120

After a decrease of 20% on 120:

$\Rightarrow 120 - \frac{20}{100} \times 120$

⇒ 120 - 24

⇒ 96

Net change = 96 − 100 = −4, i.e. a decrease of 4.

Percentage change = $\left(\frac{4}{100} \times 100\right)$% = 4% decrease.

Hence, there is a 4% decrease on the whole.

Let the number be 100.

After a decrease of 40%:

$\Rightarrow 100 - \frac{40}{100} \times 100$

⇒ 100 - 40

⇒ 60

After a decrease of 60% on 60:

$\Rightarrow 60 - \frac{60}{100} \times 60$

⇒ 60 - 36

⇒ 24

Net change = 24 − 100 = −76, i.e. a decrease of 76.

Percentage change = $\left(\frac{76}{100} \times 100\right)$% = 76% decrease.

Hence, there is a 76% decrease on the whole.

Let the number be x.

$$\begin{aligned}\Rightarrow \frac{150}{100} \times x = 750\\[1em] \Rightarrow x = \frac{750 \times 100}{150}\\[1em] \Rightarrow x = 500\end{aligned}$$

Now, 60% of 500 = $\frac{60}{100} \times 500 = 300$.

Hence, 60% of the number = 300.

Let the number be x.

$$\begin{aligned}\Rightarrow \frac{28}{100} \times x = 84\\[1em] \Rightarrow x = \frac{84 \times 100}{28}\\[1em] \Rightarrow x = 300\end{aligned}$$

Hence, the required number = 300.

Percentage of income spent = 72%.

Percentage of income saved = (100 − 72)% = 28%.

(i) Let his monthly income be ₹ x. Given, 28% of x = ₹ 12,600.

$$\begin{aligned}\Rightarrow \frac{28}{100} \times x = 12,600\\[1em] \Rightarrow x = \frac{12,600 \times 100}{28}\\[1em] \Rightarrow x = 45,000\end{aligned}$$

So, his monthly income = ₹ 45,000.

(ii) Monthly expenses = income − savings = 45,000 − 12,600 = ₹ 32,400.

Hence, (i) his monthly income = ₹ 45,000 and (ii) his monthly expenses = ₹ 32,400.

Number of boys = 1800 and number of girls = 900.

Total number of students = 1800 + 900 = 2700.

(i) Number of boys passed = 42% of 1800

$= \frac{42}{100} \times 1800$

= 756

Hence, the number of boys passed is 756.

(ii) Number of girls passed = 30% of 900

$= \frac{30}{100} \times 900$

= 270

Hence, the number of girls passed is 270.

(iii) Total number of students passed = 756 + 270 = 1026.

Hence, the total number of students passed is 1026.

(iv) Number of students failed = 2700 − 1026 = 1674.

Hence, the number of students failed is 1674.

(v) Percentage of students failed = $\left(\frac{1674}{2700} \times 100\right)$%

$= \frac{167400}{2700}$%

= 62%

Hence, the percentage of students failed is 62%.

Let the weight be x kg.

As, $6\frac{1}{4}$% = $\frac{25}{4}$%

$$\begin{aligned}\Rightarrow \frac{25}{4 \times 100} \times x = 0.25\\[1em] \Rightarrow \frac{25}{400} \times x = 0.25\\[1em] \Rightarrow x = \frac{0.25 \times 400}{25}\\[1em] \Rightarrow x = 4\end{aligned}$$

So, the weight = 4 kg.

Now, 45% of 4 kg = $\frac{45}{100} \times 4 = 1.8$ kg.

Hence, 45% of the weight = 1.8 kg.

Let the weight corresponding to each part be x kg.

So, copper = 13x, zinc = 7x and nickel = 5x.

Total weight of the alloy = 13x + 7x + 5x = 25x.

Percentage of copper = $\left(\frac{13x}{25x} \times 100\right)$%

$= \frac{1300}{25}$%

= 52%

Hence, the percentage of copper in the alloy = 52%.

Let the quantity of ore required be x kg.

Given, 15% of the ore = 36 kg of iron.

$$\begin{aligned}\Rightarrow \frac{15}{100} \times x = 36\\[1em] \Rightarrow x = \frac{36 \times 100}{15}\\[1em] \Rightarrow x = 240\end{aligned}$$

Hence, 240 kg of ore will be required.

Let the number be x. When increased by 6%, it becomes 106% of x.

$$\begin{aligned}\Rightarrow \frac{106}{100} \times x = 424\\[1em] \Rightarrow x = \frac{424 \times 100}{106}\\[1em] \Rightarrow x = 400\end{aligned}$$

Hence, the required number = 400.

Let the number be x. When decreased by 15%, it becomes 85% of x.

$$\begin{aligned}\Rightarrow \frac{85}{100} \times x = 1360\\[1em] \Rightarrow x = \frac{1360 \times 100}{85}\\[1em] \Rightarrow x = 1600\end{aligned}$$

Hence, the required number = 1600.

Original cost = ₹ 17,000 and new cost = ₹ 15,980.

Decrease in cost = 17000 − 15980 = ₹ 1020.

Percentage decrease = $\left(\frac{1020}{17000} \times 100\right)$%

$= \frac{102000}{17000}$%

= 6%

Hence, the percentage decrease = 6%.

Actual length = 22.5 m and measured length = 21.6 m.

Error = 22.5 − 21.6 = 0.9 m.

Percentage error = $\left(\frac{\text{error}}{\text{actual length}} \times 100\right)$%

$= \left(\frac{0.9}{22.5} \times 100\right)$%

$= \frac{90}{22.5}$%

= 4%

Hence, the percentage error = 4%.

Ratio = 1 : 2 : 5, so the whole = 1 + 2 + 5 = 8 and the middle term = 2.

Required percentage = $\left(\frac{2}{8} \times 100\right)$%

$= \frac{200}{8}$%

= 25%

Hence, option 3 is the correct option.

Solving,

$\Rightarrow \frac{10}{100} \times 10 + \frac{20}{100} \times 20 - \frac{30}{100} \times 30$

⇒ 1 + 4 - 9

⇒ -4

Since −4 is not among the given values,

Hence, option 4 is the correct option.

Original cost = ₹ 60 and present cost = ₹ 40.

Since the cost has decreased, decrease = 60 − 40 = ₹ 20.

Percentage change = $\left(\frac{20}{60} \times 100\right)$%

$= \frac{2000}{60}$%

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

$= 33\frac{1}{3}$% decrease

Hence, option 2 is the correct option.

Given, A = 80% of B = $\frac{80}{100}B = \frac{4}{5}B$.

$$\begin{aligned}\Rightarrow B = \frac{5}{4}A\\[1em] \Rightarrow B = \left(\frac{5}{4} \times 100\right)% \text{ of } A\\[1em] \Rightarrow B = 125% \text{ of } A\end{aligned}$$

Hence, option 3 is the correct option.

Let the number be x.

$$\begin{aligned}\Rightarrow \frac{15}{100} \times x = 30\\[1em] \Rightarrow x = \frac{30 \times 100}{15}\\[1em] \Rightarrow x = 200\end{aligned}$$

Now, 20% of 200 = $\frac{20}{100} \times 200 = 40$.

Hence, option 2 is the correct option.

Marks scored = 140 and she failed by 10 marks.

Passing marks = 140 + 10 = 150.

Let the maximum marks be x. Since 30% of the maximum marks are needed to pass,

$$\begin{aligned}\Rightarrow \frac{30}{100} \times x = 150\\[1em] \Rightarrow x = \frac{150 \times 100}{30}\\[1em] \Rightarrow x = 500\end{aligned}$$

Hence, option 3 is the correct option.

Percentage of boys = 40%.

Percentage of girls = (100 − 40)% = 60%.

Let the total strength be x. Then 60% of x = 180.

$$\begin{aligned}\Rightarrow \frac{60}{100} \times x = 180\\[1em] \Rightarrow x = \frac{180 \times 100}{60}\\[1em] \Rightarrow x = 300\end{aligned}$$

Hence, option 3 is the correct option.

Let the number be x. When increased by 20%, it becomes 120% of x.

$$\begin{aligned}\Rightarrow \frac{120}{100} \times x = 720\\[1em] \Rightarrow x = \frac{720 \times 100}{120}\\[1em] \Rightarrow x = 600\end{aligned}$$

Hence, option 2 is the correct option.

Let the marked price be ₹ x.

Given, 20% of x = ₹ 40 (saving).

$$\begin{aligned}\Rightarrow \frac{20}{100} \times x = 40\\[1em] \Rightarrow x = \frac{40 \times 100}{20}\\[1em] \Rightarrow x = 200\end{aligned}$$

Price after discount = 200 − 40 = ₹ 160.

Hence, option 3 is the correct option.

Let the number be x.

$$\begin{aligned}\Rightarrow 35% \text{ of } x - 20% \text{ of } x = 300\\[1em] \Rightarrow (35 - 20)% \text{ of } x = 300\\[1em] \Rightarrow \frac{15}{100} \times x = 300\\[1em] \Rightarrow x = \frac{300 \times 100}{15}\\[1em] \Rightarrow x = 2000\end{aligned}$$

Hence, option 1 is the correct option.

Let Raj's salary be ₹ x.

After losing 20%, remainder = (100 − 20)% of x = 80% of x = $\frac{4x}{5}$.

After spending 25% of the remainder, money left = (100 − 25)% of the remainder = 75% of $\frac{4x}{5}$.

$$\begin{aligned}\Rightarrow \frac{75}{100} \times \frac{4x}{5} = 2400\\[1em] \Rightarrow \frac{3}{4} \times \frac{4x}{5} = 2400\\[1em] \Rightarrow \frac{3x}{5} = 2400\\[1em] \Rightarrow x = \frac{2400 \times 5}{3}\\[1em] \Rightarrow x = 4000\end{aligned}$$

Hence, option 2 is the correct option.

Total number of students = 100 + 100 = 200.

Boys who leave = 10% of 100 = 10.

Girls who leave = 20% of 100 = 20.

Total students who leave = 10 + 20 = 30.

Students left = 200 − 30 = 170.

Percentage of students left = $\left(\frac{170}{200} \times 100\right)$%

$= \frac{17000}{200}$%

= 85%

Hence, option 2 is the correct option.

Total money = ₹ 400.

Money lost = 15% of 400 = $\frac{15}{100} \times 400 = ₹\ 60$.

Money spent = one fifth of 400 = $\frac{1}{5} \times 400 = ₹\ 80$.

Money left = ₹ (400 − 60 − 80) = ₹ 260.

Hence, option 4 is the correct option.

Actual number = 45 and misread number = 54.

Error = 54 − 45 = 9.

Percentage error = $\left(\frac{9}{45} \times 100\right)$%

$= \frac{900}{45}$%

= 20%

Hence, option 3 is the correct option.

Let the number be x. When reduced by 25%, it becomes 75% of x.

$$\begin{aligned}\Rightarrow \frac{75}{100} \times x = 150\\[1em] \Rightarrow x = \frac{150 \times 100}{75}\\[1em] \Rightarrow x = 200\end{aligned}$$

To become 250, increase needed = 250 − 200 = 50.

Percentage increase = $\left(\frac{50}{200} \times 100\right)$%

$= \frac{5000}{200}$%

= 25%

Hence, option 2 is the correct option.

Let y = 100.

x is 20% more than y

$\Rightarrow x = 100 + \frac{20}{100} \times 100$

= 100 + 20

= 120

Now, y is less than x by 120 − 100 = 20.

Percentage by which y is less than x

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

$= \frac{2000}{120}%$

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

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

Since y is $16\frac{2}{3}$% less than x and not 20% less than x, Statement 1 is false.

We know that,

% change = $\left(\frac{\text{Decrease (or increase) in the value}}{\text{Original value}} \times 100\right)$%

This is the correct formula. So, Statement 2 is true.

Statement 1 is false and Statement 2 is true.

Hence, option 4 is the correct option.

We know, 1 hour = 60 × 60 seconds = 3600 seconds.

4% of 1 hour

$$\begin{aligned}= \frac{4}{100} \times 3600 \text{ seconds}\\[1em] = \frac{14400}{100} \text{ seconds}\\[1em] = 144 \text{ seconds}\end{aligned}$$

Now, 144 seconds = (120 + 24) seconds = 2 min 24 seconds.

Since 4% of 1 hour = 2 min 24 seconds, Statement 1 is true.

We know that,

While comparing two quantities, they must be of the same kind and in the same units, and

required % = $\left(\frac{\text{Any Quantity}}{\text{Another Quantity}} \times 100\right)$%

This is correct. So, Statement 2 is true.

Both Statement 1 and Statement 2 are true.

Hence, option 1 is the correct option.

7% of 12 km

$$\begin{aligned}= \frac{7}{100} \times 12 \text{ km}\\[1em] = \frac{84}{100} \text{ km}\\[1em] = 0.84 \text{ km}\end{aligned}$$

We know, 1 km = 1000 m.

⇒ 0.84 km = 0.84 × 1000 m = 840 m

Since 7% of 12 km = 840 m, Assertion (A) is true.

We know that,

To find the percentage of a given quantity, we convert the percentage into a fraction and then multiply it by the given quantity, and not divide it by the given quantity. So, Reason (R) is false.

Assertion (A) is true and Reason (R) is false.

Hence, option 1 is the correct option.

$2\frac{3}{4} = \frac{11}{4}$

Converting $\frac{11}{4}$ into percent,

$$\begin{aligned}= \left(\frac{11}{4} \times 100\right)%\\[1em] = \frac{1100}{4}%\\[1em] = 275%\end{aligned}$$

Since $2\frac{3}{4}$ as percent is 275% and not 265%, Assertion (A) is false.

We know that,

For converting a fraction or a decimal into a percentage, we multiply it by 100 and put the % sign. So, Reason (R) is true.

Assertion (A) is false and Reason (R) is true.

Hence, option 2 is the correct option.

Original attendance = 300 and reduced attendance = 210.

Decrease in attendance = 300 − 210 = 90

Percentage decrease:

$= \left(\frac{\text{Decrease}}{\text{Original value}} \times 100\right)%$

$= \left(\frac{90}{300} \times 100\right)%$

$= \frac{9000}{300}%$

$= 30%$

Since the decrease percent is 30%, Assertion (A) is true.

We know that,

% decrease = $\left(\frac{\text{Decrease}}{\text{Original value}} \times 100\right)$%. So, Reason (R) is true.

Both Assertion (A) and Reason (R) are true.

Hence, option 3 is the correct option.