Percentage and its Applications — Check Your Progress

Convert $\dfrac{7}{24}$ into a percentage correct to two decimal places.

Answer

Converting,

$\Rightarrow \dfrac{7}{24} \times 100$%

$\Rightarrow \dfrac{700}{24}$%

$\Rightarrow \dfrac{175}{6}$%

⇒ 29.1666.....%

Correct to two decimal places,

$\dfrac{7}{24}$ = 29.17%

Hence, $\mathbf{\dfrac{7}{24}}$ = 29.17%.

Express 750 mL as a percentage of 4 litres.

Answer

Given:

Quantity = 750 mL

Total quantity = 4 litres

Both the quantities must be in the same units.

4 litres = (4 x 1000) mL = 4000 mL

Required percentage = $\dfrac{750}{4000} \times 100$%

= $\dfrac{75000}{4000}$%

= 18.75%

Hence, 750 mL is 18.75% of 4 litres.

A football team won 9 matches out of the total number of matches they played. If their win percentage was 45, then how many matches did they play in all?

Answer

Given:

Number of matches won = 9

Win percentage = 45%

Let the total number of matches played be $x$.

According to the given condition,

45% of $x$ = 9

$\Rightarrow \dfrac{45}{100} \times x = 9 \\[1em] \Rightarrow \dfrac{45x}{100} = 9 \\[1em] \Rightarrow x = \dfrac{9 \times 100}{45} \\[1em] \Rightarrow x = \dfrac{900}{45} \\[1em] \Rightarrow x = 20$

Hence, the football team played 20 matches in all.

In a straight student election, Rahul got 66% of the votes polled. If the defeated student got 187 votes, find the total number of votes polled.

Answer

Given:

Percentage of votes polled by Rahul = 66%

Votes got by the defeated student = 187

There are only two candidates in a straight election.

Percentage of votes polled by the defeated student = (100 - 66)% = 34%

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

According to the given condition,

34% of $x$ = 187

$\Rightarrow \dfrac{34}{100} \times x = 187 \\[1em] \Rightarrow \dfrac{34x}{100} = 187 \\[1em] \Rightarrow x = \dfrac{187 \times 100}{34} \\[1em] \Rightarrow x = \dfrac{18700}{34} \\[1em] \Rightarrow x = 550$

Hence, the total number of votes polled is 550.

36 kg of gunpowder contains 27 kg nitre, 5.4 kg charcoal and rest sulphur. Find the percentage content of sulphur in the gunpowder.

Answer

Given:

Total quantity of gunpowder = 36 kg

Quantity of nitre = 27 kg

Quantity of charcoal = 5.4 kg

Quantity of sulphur = Total gunpowder - (Nitre + Charcoal)

= 36 - (27 + 5.4)

= 36 - 32.4

= 3.6 kg

Percentage content of sulphur = $\dfrac{\text{Quantity of sulphur}}{\text{Total quantity of gunpowder}} \times 100$%

= $\dfrac{3.6}{36} \times 100$%

= $\dfrac{360}{36}$%

= 10%

Hence, the percentage content of sulphur in the gunpowder is 10%.

75% of the students in a class passed an exam. If 2 more students had passed the exam, 80% would have been successful. How many students are there in the class?

Answer

Given:

Percentage of students who passed = 75%

If 2 more students had passed, the pass percentage would be 80%.

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

Number of students who passed = 75% of $x$ = $\dfrac{75x}{100}$

If 2 more students had passed, then

$\Rightarrow \dfrac{75}{100} \times x + 2 = \dfrac{80}{100} \times x \\[1em] \Rightarrow \dfrac{80x}{100} - \dfrac{75x}{100} = 2 \\[1em] \Rightarrow \dfrac{5x}{100} = 2 \\[1em] \Rightarrow \dfrac{x}{20} = 2 \\[1em] \Rightarrow x = 2 \times 20 \\[1em] \Rightarrow x = 40$

Hence, there are 40 students in the class.

The monthly salary of a factory worker in 2022 was ₹24,000. It increased by 10% in 2023 and again by 10% in 2024. What is his salary in 2024?

Answer

Given:

Monthly salary in 2022 = ₹24,000

Increase in 2023 = 10%

Increase in 2024 = 10%

New salary = $\Big(1 + \dfrac{\text{Increase Percentage}}{100}\Big) \times$ Original salary

$\text{Salary in 2023} = \Big(1 + \dfrac{10}{100}\Big) \times 24000 \\[1em] = \dfrac{110}{100} \times 24000 \\[1em] = \dfrac{2640000}{100} \\[1em] = ₹26{,}400\\[1em] \text{Salary in 2024} = \Big(1 + \dfrac{10}{100}\Big) \times 26400 \\[1em] = \dfrac{110}{100} \times 26400 \\[1em] = \dfrac{2904000}{100} \\[1em] = ₹29{,}040$

Hence, his salary in 2024 is ₹29,040.

Price of a commodity decreased by 10% last year and increased by 20% this year. Find the percentage change in two years.

Answer

Given:

Decrease in price last year = 10%

Increase in price this year = 20%

Let the original price of the commodity be ₹100.

New price = $\Big(1 - \dfrac{\text{Decrease Percentage}}{100}\Big) \times$ Original price

Price after a decrease of 10% = $\Big(1 - \dfrac{10}{100}\Big) \times 100$

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

= ₹90

New price = $\Big(1 + \dfrac{\text{Increase Percentage}}{100}\Big) \times$ Original price

Price after an increase of 20% = $\Big(1 + \dfrac{20}{100}\Big) \times 90$

$= \dfrac{120}{100} \times 90 \\[1em] = \dfrac{10800}{100} \\[1em] = ₹108$

Change in price = Final price - Original price

= ₹108 - ₹100

= ₹8

As the change is positive, the price has increased.

Percentage change = $\dfrac{\text{Change in price}}{\text{Original price}} \times 100$%

= $\dfrac{8}{100} \times 100$%

= 8%

Hence, the price of the commodity increased by 8% in two years.

A shop gives 15% discount. What would the sale price of each of these be?

(i) A cap marked at ₹150

(ii) A pair of shoes marked at ₹800.

Answer

Given:

Rate of discount = 15%

(i) M.P. of the cap = ₹150

S.P. = $\Big(1 - \dfrac{\text{Discount Percentage}}{100}\Big) \times$ M.P.

$\text{Sale price of the cap} = \Big(1 - \dfrac{15}{100}\Big)\text{ of M.P.} \\[1em] = \dfrac{85}{100} \times 150 \\[1em] = \dfrac{12750}{100} \\[1em] = ₹127.50$

Hence, the sale price of the cap is ₹127.50.

(ii) M.P. of the pair of shoes = ₹800

S.P. = $\Big(1 - \dfrac{\text{Discount Percentage}}{100}\Big) \times$ M.P.

$\text{Sale price of the pair of shoes} = \Big(1 - \dfrac{15}{100}\Big)\text{ of M.P.} \\[1em] = \dfrac{85}{100} \times 800 \\[1em] = \dfrac{68000}{100} \\[1em] = ₹680$

Hence, the sale price of the pair of shoes is ₹680.

A washing machine marked at ₹18,000 is available for ₹17,100. Find the discount given and the discount percent.

Answer

Given:

M.P. of the washing machine = ₹18,000

S.P. of the washing machine = ₹17,100

Discount = M.P. - S.P.

= ₹18,000 - ₹17,100

= ₹900

Discount percentage = $\dfrac{\text{Discount}}{\text{M.P.}} \times 100$%

= $\dfrac{900}{18000} \times 100$%

= $\dfrac{90000}{18000}$%

= 5%

Hence, the discount given is ₹900 and the discount percent is 5%.

Ramesh bought two table fans for ₹1,800 each. He sold one at a loss of 8% and the other at a profit of 12%. Find the selling price of each. Also find out the total profit or loss.

Answer

Given:

C.P. of each table fan = ₹1,800

Loss on the first table fan = 8%

Profit on the second table fan = 12%

S.P. = $\Big(1 - \dfrac{\text{Loss Percentage}}{100}\Big) \times$ C.P.

$\text{S.P. of the first table fan} = \Big(1 - \dfrac{8}{100}\Big) \times 1800 \\[1em] = \dfrac{92}{100} \times 1800 \\[1em] = \dfrac{165600}{100} \\[1em] = ₹1{,}656$

S.P. = $\Big(1 + \dfrac{\text{Profit Percentage}}{100}\Big) \times$ C.P.

$\text{S.P. of the second table fan} = \Big(1 + \dfrac{12}{100}\Big) \times 1800 \\[1em] = \dfrac{112}{100} \times 1800 \\[1em] = \dfrac{201600}{100} \\[1em] = ₹2{,}016$

Total C.P. of both the table fans = ₹1,800 + ₹1,800 = ₹3,600

Total S.P. of both the table fans = ₹1,656 + ₹2,016 = ₹3,672

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

Profit = S.P. - C.P.

= ₹3,672 - ₹3,600

= ₹72

Hence, the selling prices are ₹1,656 and ₹2,016, and the total profit is ₹72.

Ajit bought an old car for ₹78,000. He spent ₹2,000 on repairs and repainting the car. He sold the car to Anuj at a gain of 15%. Anuj sold it to Anthony at a loss of 5%. What did the car cost to Anthony?

Answer

Given:

Cost of the old car = ₹78,000

Overhead expenses on repairs and repainting = ₹2,000

Gain of Ajit = 15%

Loss of Anuj = 5%

Total C.P. of the car for Ajit = Cost + Overheads

= ₹78,000 + ₹2,000

= ₹80,000

S.P. = $\Big(1 + \dfrac{\text{Gain Percentage}}{100}\Big) \times$ C.P.

$\text{S.P. of the car for Ajit} = \Big(1 + \dfrac{15}{100}\Big) \times 80000 \\[1em] = \dfrac{115}{100} \times 80000 \\[1em] = \dfrac{9200000}{100} \\[1em] = ₹92{,}000$

This becomes the C.P. of the car for Anuj.

S.P. = $\Big(1 - \dfrac{\text{Loss Percentage}}{100}\Big) \times$ C.P.

$\text{S.P. of the car for Anuj} = \Big(1 - \dfrac{5}{100}\Big) \times 92000 \\[1em] = \dfrac{95}{100} \times 92000 \\[1em] = \dfrac{8740000}{100} \\[1em] = ₹87{,}400$

Hence, the car cost ₹87,400 to Anthony.

Sai sold an article at a profit of 12%. Had it been sold for ₹16 more, the profit would have been 20%. Find the C.P. of the article.

Answer

Given:

Profit on selling the article = 12%

On selling it for ₹16 more, profit = 20%

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

S.P. = $\Big(1 + \dfrac{\text{Profit Percentage}}{100}\Big) \times$ C.P.

S.P. of the article at 12% profit = $\Big(1 + \dfrac{12}{100}\Big)$ of ₹$x$

$= \dfrac{112}{100} \times x \\[1em] = ₹\dfrac{28x}{25}$

To obtain 20% profit,

$\text{S.P. of the article} = \Big(1 + \dfrac{20}{100}\Big)\text{ of ₹}x \\[1em] = \dfrac{120}{100} \times x \\[1em] = ₹\dfrac{6x}{5}$

According to the given information,

$\Rightarrow \dfrac{6x}{5} = \dfrac{28x}{25} + 16 \\[1em] \Rightarrow \dfrac{6x}{5} - \dfrac{28x}{25} = 16 \\[1em] \Rightarrow \dfrac{30x - 28x}{25} = 16 \\[1em] \Rightarrow \dfrac{2x}{25} = 16 \\[1em] \Rightarrow x = \dfrac{16 \times 25}{2} \\[1em] \Rightarrow x = \dfrac{400}{2} \\[1em] \Rightarrow x = 200$

Hence, the C.P. of the article is ₹200.

A shopkeeper allows 8% discount on her goods and still earns a profit of 15%. If an article is sold for ₹460, find

(i) the marked price

(ii) the cost price.

Answer

Given:

S.P. of the article = ₹460

Rate of discount = 8%

Profit = 15%

(i) S.P. = $\Big(1 - \dfrac{\text{Discount Percentage}}{100}\Big) \times$ M.P.

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

$\text{S.P.} = \Big(1 - \dfrac{8}{100}\Big)\text{ of M.P.} \\[1em] \Rightarrow 460 = \dfrac{92}{100} \times x \\[1em] \Rightarrow x = \dfrac{460 \times 100}{92} \\[1em] \Rightarrow x = \dfrac{46000}{92} \\[1em] \Rightarrow x = 500$

Hence, the marked price of the article is ₹500.

(ii) S.P. of the article = ₹460, profit = 15%

S.P. = $\Big(1 + \dfrac{\text{Profit Percentage}}{100}\Big) \times$ C.P.

$\Rightarrow 460 = \Big(1 + \dfrac{15}{100}\Big)\text{ of C.P.} \\[1em] \Rightarrow 460 = \dfrac{115}{100}\text{ of C.P.} \\[1em] \Rightarrow \text{C.P.} = \dfrac{460 \times 100}{115} \\[1em] \Rightarrow \text{C.P.} = \dfrac{46000}{115} \\[1em] \Rightarrow \text{C.P.} = ₹400$

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

Rajesh bought a laptop for ₹41,300 including 18% GST. Find the price of laptop before GST was added.

Answer

Given:

Buying price of the laptop = ₹41,300 including 18% GST

Let the price of the laptop before GST be ₹$x$.

GST = 18% of ₹$x$ = ₹$\dfrac{18x}{100}$

Buying price = Price before GST + GST

$\Rightarrow 41300 = x + \dfrac{18x}{100} \\[1em] \Rightarrow 41300 = \dfrac{100x + 18x}{100} \\[1em] \Rightarrow 41300 = \dfrac{118x}{100} \\[1em] \Rightarrow x = \dfrac{41300 \times 100}{118} \\[1em] \Rightarrow x = \dfrac{4130000}{118} \\[1em] \Rightarrow x = 35000$

Hence, the price of the laptop before GST was added is ₹35,000.

A gaming mouse is marked for ₹3,000. The shopkeeper offers 10% discount on it. If GST is received 18%, find the amount paid by the customer.

Answer

Given:

M.P. of the gaming mouse = ₹3,000

Rate of discount = 10%

Rate of GST = 18%

Discount = 10% of ₹3,000

$= \dfrac{10}{100} \times 3000 \\[1em] = \dfrac{30000}{100} \\[1em] = ₹300$

S.P. of the gaming mouse = M.P. - Discount

= ₹3,000 - ₹300

= ₹2,700

Now GST = 18% of ₹2,700

$= \dfrac{18}{100} \times 2700 \\[1em] = \dfrac{48600}{100} \\[1em] = ₹486$

Amount paid by the customer = S.P. + GST

= ₹2,700 + ₹486

= ₹3,186

Hence, the amount paid by the customer is ₹3,186.

An OTT platform offers monthly subscription plan for ₹200 per month or an Annual subscription plan for ₹2,000 per annum. How much discount percentage do you get on Annual subscription plan compared to monthly subscription plan?

Answer

Given:

Monthly subscription charge = ₹200 per month

Annual subscription charge = ₹2,000 per annum

Cost of the monthly plan for one year = 12 x ₹200 = ₹2,400

Cost of the annual plan for one year = ₹2,000

Discount = ₹2,400 - ₹2,000 = ₹400

Discount percentage = $\dfrac{\text{Discount}}{\text{Cost of monthly plan for one year}} \times 100$%

= $\dfrac{400}{2400} \times 100$%

= $\dfrac{40000}{2400}$%

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

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

Hence, the discount on the Annual subscription plan is $\mathbf{16\dfrac{2}{3}}$%.

The sum of two numbers is $\dfrac{31}{25}$ of the first number. What percent is the second number of the first number?

Answer

Given:

Sum of the two numbers = $\dfrac{31}{25}$ of the first number

Let the first number be $x$ and the second number be $y$.

According to the given condition,

$\Rightarrow x + y = \dfrac{31}{25} \times x \\[1em] \Rightarrow y = \dfrac{31x}{25} - x \\[1em] \Rightarrow y = \dfrac{31x - 25x}{25} \\[1em] \Rightarrow y = \dfrac{6x}{25}$

Required percentage = $\dfrac{y}{x} \times 100$%

= $\dfrac{\dfrac{6x}{25}}{x} \times 100$%

= $\dfrac{6x}{25} \times \dfrac{1}{x} \times 100$%

= $\dfrac{600}{25}$%

= 24%

Hence, the second number is 24% of the first number.

The boys and girls in a college are in the ratio 3 : 2. If 20% of the boys and 25% of the girls wear spectacles, find the percentage of students who do not wear spectacles.

Answer

Given:

Ratio of boys and girls = 3 : 2

Percentage of boys who wear spectacles = 20%

Percentage of girls who wear spectacles = 25%

Let the number of boys be $3x$ and the number of girls be $2x$.

Total number of students = $3x + 2x = 5x$

Number of boys who wear spectacles = 20% of $3x$

$= \dfrac{20}{100} \times 3x \\[1em] = \dfrac{60x}{100} \\[1em] = \dfrac{3x}{5}$

Number of girls who wear spectacles = 25% of $2x$

$= \dfrac{25}{100} \times 2x \\[1em] = \dfrac{50x}{100} \\[1em] = \dfrac{x}{2}\\[1em] \text{Total number of students who wear spectacles} = \dfrac{3x}{5} + \dfrac{x}{2} \\[1em] = \dfrac{6x + 5x}{10} \\[1em] = \dfrac{11x}{10}\\[1em] \text{Number of students who do not wear spectacles} = 5x - \dfrac{11x}{10} \\[1em] = \dfrac{50x - 11x}{10} \\[1em] = \dfrac{39x}{10}$

Percentage of students who do not wear spectacles = $\dfrac{\dfrac{39x}{10}}{5x} \times 100$%

= $\dfrac{39x}{10} \times \dfrac{1}{5x} \times 100$%

= $\dfrac{3900}{50}$%

= 78%

Hence, 78% of the students do not wear spectacles.

Some candies are bought at the rate of 11 for ₹10 and the same number of candies at the rate of 9 for ₹10. If the whole lot is sold at one rupee per candy, find the gain or loss percent in the whole transaction.

Answer

Given:

First lot bought at the rate of 11 candies for ₹10

Second lot bought at the rate of 9 candies for ₹10

Selling rate of the whole lot = ₹1 per candy

As the LCM of 11 and 9 is 99, let 99 candies of each type be bought.

C.P. of 99 candies of the first type = $\dfrac{99}{11} \times 10$

= 9 x 10

= ₹90

C.P. of 99 candies of the second type = $\dfrac{99}{9} \times 10$

= 11 x 10

= ₹110

Total number of candies bought = 99 + 99 = 198

Total C.P. of 198 candies = ₹90 + ₹110 = ₹200

S.P. of 198 candies at ₹1 per candy = 198 x ₹1 = ₹198

As S.P. < C.P., there is a loss.

Loss = C.P. - S.P.

= ₹200 - ₹198

= ₹2

Loss percentage = $\dfrac{\text{Loss}}{\text{C.P.}} \times 100$%

= $\dfrac{2}{200} \times 100$%

= $\dfrac{200}{200}$%

= 1%

Hence, there is a loss of 1% in the whole transaction.