Percentage — Exercise 6(B)

36% of the students in a school are girls. If the number of boys is 1440, find the total strength of the school.

Answer

Given:

Percentage of girls in the school = 36%

Number of boys in the school = 1440

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

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

According to the given condition,

64% of $x$ = 1440

$\Rightarrow \dfrac{64}{100} \times x = 1440 \\[1em] \Rightarrow \dfrac{64x}{100} = 1440 \\[1em] \Rightarrow x = \dfrac{1440 \times 100}{64} \\[1em] \Rightarrow x = \dfrac{144000}{64} \\[1em] \Rightarrow x = 2250$

Hence, the total strength of the school is 2250.

Geeta saves 18% of her monthly salary. If she spends ₹10250 per month, what is her monthly salary?

Answer

Given:

Percentage of salary saved by Geeta = 18%

Amount spent by Geeta per month = ₹10,250

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

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

According to the given condition,

82% of $x$ = 10250

$\Rightarrow \dfrac{82}{100} \times x = 10250 \\[1em] \Rightarrow \dfrac{82x}{100} = 10250 \\[1em] \Rightarrow x = \dfrac{10250 \times 100}{82} \\[1em] \Rightarrow x = \dfrac{1025000}{82} \\[1em] \Rightarrow x = 12500$

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

In an examination, a student has to secure 40% marks to pass. Rahul gets 178 marks and fails by 32 marks. What are the maximum marks?

Answer

Given:

Percentage of marks required to pass = 40%

Marks secured by Rahul = 178

Marks by which Rahul fails = 32

Passing marks = Marks secured + Marks by which he fails

= 178 + 32

= 210

Let the maximum marks be $x$.

According to the given condition,

40% of $x$ = 210

$\Rightarrow \dfrac{40}{100} \times x = 210 \\[1em] \Rightarrow \dfrac{40x}{100} = 210 \\[1em] \Rightarrow x = \dfrac{210 \times 100}{40} \\[1em] \Rightarrow x = \dfrac{21000}{40} \\[1em] \Rightarrow x = 525$

Hence, the maximum marks are 525.

8% of the students in a school remained absent on a day. If 1633 attended the school on that day, how many remained absent?

Answer

Given:

Percentage of students absent = 8%

Number of students who attended the school = 1633

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

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

According to the given condition,

92% of $x$ = 1633

$\Rightarrow \dfrac{92}{100} \times x = 1633 \\[1em] \Rightarrow \dfrac{92x}{100} = 1633 \\[1em] \Rightarrow x = \dfrac{1633 \times 100}{92} \\[1em] \Rightarrow x = \dfrac{163300}{92} \\[1em] \Rightarrow x = 1775$

Total number of students in the school = 1775

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

Hence, 142 students remained absent.

On increasing the price of an article by 14%, it becomes ₹ 1995. What was its original price?

Answer

Given:

Increase in the price of the article = 14%

Increased price of the article = ₹1,995

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

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

According to the given condition,

$\dfrac{114x}{100} = 1995 \\[1em] \Rightarrow x = \dfrac{1995 \times 100}{114} \\[1em] \Rightarrow x = \dfrac{199500}{114} \\[1em] \Rightarrow x = 1750$

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

On decreasing the price of an article by 6%, it becomes ₹ 1551. What was its original price?

Answer

Given:

Decrease in the price of the article = 6%

Decreased price of the article = ₹1,551

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

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

According to the given condition,

$\dfrac{94x}{100} = 1551 \\[1em] \Rightarrow x = \dfrac{1551 \times 100}{94} \\[1em] \Rightarrow x = \dfrac{155100}{94} \\[1em] \Rightarrow x = 1650$

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

Reenu reduced her weight by 15%. If now she weighs 52.7 kg, what was her earlier weight?

Answer

Given:

Reduction in the weight of Reenu = 15%

Present weight of Reenu = 52.7 kg

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

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

According to the given condition,

$\dfrac{85x}{100} = 52.7 \\[1em] \Rightarrow x = \dfrac{52.7 \times 100}{85} \\[1em] \Rightarrow x = \dfrac{5270}{85} \\[1em] \Rightarrow x = 62$

Hence, the earlier weight of Reenu was 62 kg.

Two candidates contested an election. One of them secured 58% votes and won the election by a margin of 2560 votes. How many votes were polled in all?

Answer

Given:

Percentage of votes secured by the winning candidate = 58%

Margin of votes by which he won = 2560

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

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

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

According to the given condition,

16% of $x$ = 2560

$\Rightarrow \dfrac{16}{100} \times x = 2560 \\[1em] \Rightarrow \dfrac{16x}{100} = 2560 \\[1em] \Rightarrow x = \dfrac{2560 \times 100}{16} \\[1em] \Rightarrow x = \dfrac{256000}{16} \\[1em] \Rightarrow x = 16000$

Hence, 16000 votes were polled in all.

In an examination, Preeti scored 60 out of 75 in Science, 84 out of 100 in Mathematics, 36 out of 50 in Hindi and 30 out of 45 in English.

(i) In which subject her performance is worst?

(ii) In which subject her performance is the best?

(iii) What is her aggregate percentage of marks?

Answer

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

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

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

= 80%

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

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

= 84%

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

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

= 72%

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

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

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

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

Arranging the percentages in ascending order, we get

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

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

Hence, her performance is the worst in English.

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

Hence, her performance is the best in Mathematics.

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

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

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

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

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

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

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

The price of an article is increased by 25%. By how much per cent must this new value be decreased to restore it to its former value?

Answer

Let the original price of the article be ₹100.

Increase in the price = 25%

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

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

= ₹125

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

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

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

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

= 20%

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

The price of an article is reduced by 10%. By how much per cent this value be increased to restore it to its former value?

Answer

Let the original price of the article be ₹100.

Reduction in the price = 10%

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

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

= ₹90

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

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

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

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

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

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

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

The price of tea is increased by 20%. By how much per cent a housewife should reduce the consumption of tea so as not to increase the expenditure on tea?

Answer

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

Increase in the price = 20%

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

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

= ₹120

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

Now, ₹120 will fetch $x$ units.

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

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

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

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

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

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

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

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

A man gave 35% of his money to his elder son and 40% of the remainder to the younger son. Now, he is left with ₹ 11700. How much money had he?

Answer

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

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

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

= ₹$\dfrac{7x}{20}$

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

$= \dfrac{20x - 7x}{20} \\[1em] = ₹\dfrac{13x}{20}$

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

$= \dfrac{40}{100} \times \dfrac{13x}{20} \\[1em] = \dfrac{2}{5} \times \dfrac{13x}{20} \\[1em] = ₹\dfrac{13x}{50}$

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

$= \dfrac{65x - 26x}{100} \\[1em] = ₹\dfrac{39x}{100}$

According to the given condition,

$\dfrac{39x}{100} = 11700 \\[1em] \Rightarrow x = \dfrac{11700 \times 100}{39} \\[1em] \Rightarrow x = \dfrac{1170000}{39} \\[1em] \Rightarrow x = 30000$

Hence, the man had ₹30,000.

5% of the population of a town were killed in an earthquake and 8% of the remainder left the town. If the population of the town now is 43700, what was its population at the beginning?

Answer

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

Percentage of population killed in the earthquake = 5%

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

$= \dfrac{95}{100} \times x \\[1em] = \dfrac{19x}{20}$

Percentage of the remainder that left the town = 8%

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

$= \dfrac{92}{100} \times \dfrac{19x}{20} \\[1em] = \dfrac{23}{25} \times \dfrac{19x}{20} \\[1em] = \dfrac{437x}{500}$

According to the given condition,

$\dfrac{437x}{500} = 43700 \\[1em] \Rightarrow x = \dfrac{43700 \times 500}{437} \\[1em] \Rightarrow x = \dfrac{21850000}{437} \\[1em] \Rightarrow x = 50000$

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

A and B are two towers. The height of tower A is 20% less than that of B. How much per cent is B's height more than that of A?

Answer

Let the height of tower B be 100 units.

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

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

$= \dfrac{80}{100} \times 100 \\[1em] = 80 \text{ units}$

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

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

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

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

= 25%

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

In an examination, 30% of the candidates failed in English, 35% failed in G.K. and 27% failed in both the subjects. If 310 candidates passed in both, how many candidates appeared in the examination?

Answer

Given:

Percentage of candidates who failed in English = 30%

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

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

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

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

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

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

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

According to the given condition,

62% of $x$ = 310

$\Rightarrow \dfrac{62}{100} \times x = 310 \\[1em] \Rightarrow \dfrac{62x}{100} = 310 \\[1em] \Rightarrow x = \dfrac{310 \times 100}{62} \\[1em] \Rightarrow x = \dfrac{31000}{62} \\[1em] \Rightarrow x = 500$

Hence, 500 candidates appeared in the examination.

The value of a car depreciates annually by 10%. If the present value of the car be ₹ 6,50,000, find its value after 2 years.

Answer

Given:

Present value of the car = ₹6,50,000

Rate of annual depreciation = 10%

As the value depreciates by 10% every year,

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

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

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

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

The population of a village increases by 7% every year. If the present population is 80000, then what will be the population after two years?

Answer

Given:

Present population of the village = 80000

Rate of annual increase = 7%

As the population increases by 7% every year,

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

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

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

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

A student was asked to multiply a number by $\dfrac{5}{3}$. He multiplied it by $\dfrac{3}{5}$ instead. Find the percentage error in the calculation.

Answer

Let the number be $x$.

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

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

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

Error = Actual value - Value obtained

$= \dfrac{5x}{3} - \dfrac{3x}{5} \\[1em] = \dfrac{25x - 9x}{15} \\[1em] = \dfrac{16x}{15}$

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

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

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

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

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

= 64%

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

In an election between two candidates, 10% of the voters did not cast their vote. 10% of the votes polled were found invalid. The successful candidate got 54% of the valid votes and won by a majority of 1620 votes. Find the number of voters enrolled on the voters list.

Answer

Given:

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

Percentage of the votes polled that were invalid = 10%

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

Majority by which he won = 1620

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

Let the number of valid votes be $x$.

According to the given condition,

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

$\Rightarrow \dfrac{8}{100} \times x = 1620 \\[1em] \Rightarrow \dfrac{8x}{100} = 1620 \\[1em] \Rightarrow x = \dfrac{1620 \times 100}{8} \\[1em] \Rightarrow x = \dfrac{162000}{8} \\[1em] \Rightarrow x = 20250$

Number of valid votes = 20250

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

Let the number of votes polled be $y$.

$\dfrac{90}{100} \times y = 20250 \\[1em] \Rightarrow y = \dfrac{20250 \times 100}{90} \\[1em] \Rightarrow y = \dfrac{2025000}{90} \\[1em] \Rightarrow y = 22500$

Number of votes polled = 22500

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

Let the number of voters enrolled be $z$.

$\dfrac{90}{100} \times z = 22500 \\[1em] \Rightarrow z = \dfrac{22500 \times 100}{90} \\[1em] \Rightarrow z = \dfrac{2250000}{90} \\[1em] \Rightarrow z = 25000$

Hence, 25000 voters were enrolled on the voters list.