Selina Concise Solutions for ICSE Class 8 Mathematics Chapter 7 Percent and Percentage

Exercise 7(A)

Question 1. Evaluate :
(i) 55% of 160 + 24% of 50 - 36% of 150
(ii) 9.3% of 500 - 4.8% of 250 - 2.5% of 240
Answer:
(i) To solve this, we change each percentage into a fraction and multiply:
\( \frac{55}{100} \times 160 = 88 \)
\( \frac{24}{100} \times 50 = 12 \)
\( \frac{36}{100} \times 150 = 54 \)
Now, combine the results as given in the problem:
\( 88 + 12 - 54 = 46 \)

(ii) We calculate each part separately:
\( \frac{9.3}{100} \times 500 = 9.3 \times 5 = 46.5 \)
\( \frac{4.8}{100} \times 250 = 4.8 \times 2.5 = 12 \)
\( \frac{2.5}{100} \times 240 = 2.5 \times 2.4 = 6 \)
Now, we put these values together:
\( 46.5 - 12 - 6 = 28.5 \)
In simple words: Turn the percentages into fractions by dividing by 100. Multiply them by the numbers, and then add or subtract to find the final answer.

Exam Tip: Simplify your math by canceling zeros in the fractions before you multiply the numbers together.

Question 2. (i) A number is increased from 125 to 150 ; find the percentage increase.
(ii) A number is decreased from 125 to 100 ; find the percentage decrease.
Answer:
(i) The starting number is 125 and the new value is 150.
The amount of increase is:
\( 150 - 125 = 25 \)
Next, we calculate the percentage of this rise compared to the start value:
\( \text{Percentage Increase} = \frac{25}{125} \times 100 = 20\% \)

(ii) The starting number is 125 and the new value is 100.
The amount of decrease is:
\( 125 - 100 = 25 \)
Next, we find the percentage of this drop compared to the start value:
\( \text{Percentage Decrease} = \frac{25}{125} \times 100 = 20\% \)
In simple words: Find the difference between the two numbers first. Then divide that difference by the original starting number and multiply by 100.

Exam Tip: Remember to always divide by the original starting number, not the new number, when working out percentage change.

Question 3. Find :
(i) 45 is what percent of 54 ?
(ii) 2.7 is what percent of 18 ?
Answer:
(i) Let the percentage be \( x\% \).
So, we write:
\( 54 \times \frac{x}{100} = 45 \)
\( x = \frac{45 \times 100}{54} \)
\( x = \frac{5 \times 100}{6} = \frac{250}{3} = 83\frac{1}{3}\% \)
So, 45 is \( 83\frac{1}{3}\% \) of 54.

(ii) Let the percentage be \( x\% \).
So, we have:
\( 18 \times \frac{x}{100} = 2.7 \)
\( x = \frac{2.7 \times 100}{18} \)
\( x = \frac{270}{18} = 15\% \)
So, 2.7 is \( 15\% \) of 18.
In simple words: Write the two numbers as a fraction with the main number on the bottom. Multiply by 100 to change it to a percent.

Exam Tip: Write down your fraction steps clearly and simplify them before multiplying by 100 to avoid simple math mistakes.

Question 4. (i) 252 is 35% of a certain number, find the number.
(ii) If 14% of a number is 315 ; find the number.
Answer:
(i) Let us call the unknown number \( x \).
Based on the details given:
\( \frac{x \times 35}{100} = 252 \)
Solve for \( x \):
\( x = \frac{252 \times 100}{35} \)
\( x = \frac{252 \times 20}{7} = 36 \times 20 = 720 \)
So, the number is 720.

(ii) Let us call the unknown number \( x \).
Based on the details given:
\( \frac{x \times 14}{100} = 315 \)
Solve for \( x \):
\( x = \frac{315 \times 100}{14} \)
\( x = \frac{45 \times 100}{2} = 45 \times 50 = 2250 \)
So, the number is 2250.
In simple words: To find the whole number when you know a percentage of it, multiply the value by 100 and divide by the percent.

Exam Tip: Always state clearly what your variable represents (like letting \( x \) be the number) at the very start of your answer.

Question 5. Find the percentage change, when a number is changed from :
(i) 80 to 100
(ii) 100 to 80
(iii) 6.25 to 7.50
Answer:
(i) Original value is 80 and the new value is 100.
Increase is:
\( 100 - 80 = 20 \)
Percentage change is:
\( \frac{20}{80} \times 100 = 25\% \) (increase)

(ii) Original value is 100 and the new value is 80.
Decrease is:
\( 100 - 80 = 20 \)
Percentage change is:
\( \frac{20}{100} \times 100 = 20\% \) (decrease)

(iii) Original value is 6.25 and the new value is 7.50.
Increase is:
\( 7.50 - 6.25 = 1.25 \)
Percentage change is:
\( \frac{1.25}{6.25} \times 100 = 20\% \) (increase)
In simple words: Find the change between the two numbers. Divide that change by the first number, then multiply by 100 to get the percent.

Exam Tip: Always write whether the percentage is an "increase" or a "decrease" to get full marks on these questions.

Question 6. An auctioneer charges 8% for selling a house. If a house is sold for Rs. 2,30,500; find the charges of the auctioneer.
Answer:
The price of the house is Rs. 2,30,500.
The rate of fee charged by the auctioneer is 8%.
We calculate the final charge like this:
\( 8\% \text{ of } \text{Rs. } 2,30,500 \)
\( = \frac{8}{100} \times 2,30,500 \)
\( = 8 \times 2305 = \text{Rs. } 18,440 \)
In simple words: Work out 8% of the selling price to find the total money the auctioneer gets for the sale.

Exam Tip: Be sure to write the units, like Rs., in your final answer so you do not lose any basic marks.

Question 7. Out of 800 oranges, 50 are rotten. Find the percentage of good oranges.
Answer:
Total number of oranges is 800.
Rotten oranges are 50.
So, the number of fresh oranges is:
\( 800 - 50 = 750 \)
Now, we find the percentage of these fresh oranges:
\( \text{Percentage} = \frac{750}{800} \times 100 \)
\( = \frac{750}{8} = \frac{375}{4} = 93\frac{3}{4}\% \)
In simple words: Subtract the bad oranges from the total to find the good ones. Then find what percentage the good ones are out of the total.

Exam Tip: It is best to write fractional percentages as mixed numbers like \( 93\frac{3}{4}\% \) or as a decimal like 93.75%.

Question 8. A cistern contains 5 thousand litres of water. If 6% water is leaked. Find how many litres of water are left in the cistern.
Answer:
The total water in the cistern is 5000 litres.
Amount of water that leaked out is:
\( 6\% \text{ of } 5000 \text{ litres} \)
\( = \frac{6}{100} \times 5000 = 300 \text{ litres} \)
So, the amount of water left is:
\( 5000 - 300 = 4700 \text{ litres} \)
In simple words: Calculate 6% of 5,000 to find the leaked water, which is 300 litres. Subtract that from the total to get the water left.

Exam Tip: Read the words carefully. "5 thousand" must be written out as the number 5,000 before you start your calculation.

Question 9. A man spends 87% of his salary. If he saves Rs. 325 ; find his salary.
Answer:
Let us assume the man's total salary is Rs. \( x \).
Since he spends 87%, his savings percentage is:
\( 100\% - 87\% = 13\% \)
We are given that his actual savings are Rs. 325.
So, we write the equation:
\( 13\% \text{ of } x = 325 \)
\( \frac{13}{100} \times x = 325 \)
Solve for \( x \):
\( x = \frac{325 \times 100}{13} \)
\( x = 25 \times 100 = 2500 \)
So, his salary is Rs. 2,500.
In simple words: Since he spends 87%, he must save 13%. If that 13% is Rs. 325, his total salary is Rs. 2,500.

Exam Tip: Finding the savings percentage first is a very fast way to solve this type of word problem.

Question 10. (i) A number 3.625 is wrongly read as 3.265; find the percentage error.
(ii) A number \( 5.78 \times 10^3 \) is wrongly written as \( 5.87 \times 10^3 \) ; find the percentage error.
Answer:
(i) Correct value is 3.625.
Wrongly read value is 3.265.
The error is:
\( 3.625 - 3.265 = 0.360 \)
Now, calculate the percentage error:
\( \text{Percentage Error} = \frac{0.360}{3.625} \times 100 \)
\( = \frac{36000}{3625} \approx 9.93\% \)

(ii) Correct value is \( 5.78 \times 10^3 \).
Wrongly written value is \( 5.87 \times 10^3 \).
The error is:
\( 5.87 \times 10^3 - 5.78 \times 10^3 = 0.09 \times 10^3 \)
Now, calculate the percentage error:
\( \text{Percentage Error} = \frac{0.09 \times 10^3}{5.78 \times 10^3} \times 100 \)
\( = \frac{0.09}{5.78} \times 100 = \frac{900}{578} \approx 1.56\% \)
In simple words: Find the difference between the two numbers. Divide that difference by the correct number, and multiply by 100 to get the error percentage.

Exam Tip: Always use the correct original number in the denominator when calculating the percentage error.

Question 11. In an election between two candidates, one candidate secured 58% of the votes polled and won the election by 18,336 votes. Find the total number of votes polled and the votes secured by each candidate.
Answer:
The winner received 58% of the votes.
This means the loser received:
\( 100\% - 58\% = 42\% \text{ of the votes} \)
The difference between their vote percentages is:
\( 58\% - 42\% = 16\% \)
We are told this 16% difference equals 18,336 votes.
Let us assume the total votes polled is \( x \).
\( \frac{16}{100} \times x = 18,336 \)
Solve for \( x \):
\( x = \frac{18,336 \times 100}{16} = 1,14,600 \)
So, the total votes polled are 1,14,600.

Now, we find the votes for each candidate:
Votes for the winner:
\( \frac{58}{100} \times 1,14,600 = 66,468 \)
Votes for the loser:
\( \frac{42}{100} \times 1,14,600 = 48,132 \)
In simple words: One candidate got 58% and the other got 42%. The difference is 16%, which is 18,336 votes. We use this to find the total votes, then calculate each candidate's share.

Exam Tip: You can double-check your final answer by subtracting the loser's votes from the winner's votes to see if you get the original vote difference of 18,336.

Question 12. In an election between two candidates, one candidate secured 47% of votes polled and lost the election by 12, 366 votes. Find the total votes polled and die votes secured by the winning candidate.
Answer:
The losing candidate received 47% of the votes.
This means the winning candidate received:
\( 100\% - 47\% = 53\% \text{ of the votes} \)
The difference between their vote percentages is:
\( 53\% - 47\% = 6\% \)
We are told this 6% difference is equal to 12,366 votes.
Let us assume the total votes polled is \( x \).
\( \frac{6}{100} \times x = 12,366 \)
Solve for \( x \):
\( x = \frac{12,366 \times 100}{6} = 2,06,100 \)
So, the total votes polled are 2,06,100.

Now, we find the votes for the winning candidate:
\( \frac{53}{100} \times 2,06,100 = 1,09,233 \)
In simple words: The loser got 47% and the winner got 53%. The winning margin is 6% of the total, which is 12,366 votes. We use this to calculate the total votes and the winner's votes.

Exam Tip: Make sure to read carefully. This question only asks for the total votes and the winner's votes, not the loser's votes.

Question 13. The cost of a scooter depreciates every year by 15% of its value at the beginning of the year. If the present cost of the scooter is Rs. 8,000; find its cost:
(i) after one year
(ii) after 2 years
Answer:
The initial price of the scooter is Rs. 8,000.
The price drops by 15% each year, which means it retains 85% of its value.

(i) Value after one year:
\( \frac{85}{100} \times 8000 = \text{Rs. } 6,800 \)

(ii) Value after two years (using the value at the end of the first year):
\( \frac{85}{100} \times 6800 = \text{Rs. } 5,780 \)
In simple words: Since the price drops by 15% every year, we find 85% of Rs. 8,000 for the first year. For the second year, we find 85% of that new price.

Exam Tip: Do not just subtract 30% for two years. Depreciation must be worked out step-by-step on the new value at the start of each year.

Question 14. In an examination, the pass mark is 40%. If a candidate gets 65 marks and fails by 3 marks ; find the maximum marks.
Answer:
The candidate got 65 marks and needed 3 more marks to pass.
So, the minimum passing score is:
\( 65 + 3 = 68 \)
We are told that the passing score is 40% of the total marks.
Let us assume the total maximum marks is \( x \).
\( \frac{40}{100} \times x = 68 \)
Solve for \( x \):
\( x = 68 \times \frac{100}{40} = 170 \)
So, the maximum marks are 170.
In simple words: The passing mark is 68. Since 68 is 40% of the total, we use this to find that the maximum marks are 170.

Exam Tip: Always find the actual passing marks first to make setting up your equation much easier.

Question 15. In an examination, a candidate secured 125 marks and failed by 15 marks. If the pass percentage was 35% ; find the maximum marks.
Answer:
The candidate got 125 marks and was 15 marks short of passing.
So, the passing marks are:
\( 125 + 15 = 140 \)
We know the passing percentage is 35%.
Let us assume the maximum marks is \( x \).
\( \frac{35}{100} \times x = 140 \)
Solve for \( x \):
\( x = \frac{140 \times 100}{35} = 4 \times 100 = 400 \)
So, the maximum marks are 400.
In simple words: The passing score is 140 marks. Since 140 is 35% of the total score, the full marks for the exam are 400.

Exam Tip: Practice dividing numbers by common values like 35 to speed up your calculations during exams.

Question 16. In an objective type paper of 150 questions; John got 80% correct answers and Mohan got 64% correct answers.
(i) How many correct answers did each get?
(ii) What percent is Mohan’s correct answers to John’s correct answers ?
Answer:
The total number of questions is 150.

(i) John's correct answers:
\( \frac{80}{100} \times 150 = 120 \)
Mohan's correct answers:
\( \frac{64}{100} \times 150 = 96 \)

(ii) Percentage of Mohan's correct answers compared to John's correct answers:
\( \text{Percentage} = \frac{96}{120} \times 100 = 80\% \)
In simple words: John got 120 correct answers and Mohan got 96. Comparing Mohan's score of 96 to John's 120 gives a percentage of 80%.

Exam Tip: In part (ii), be careful to divide by John's correct answers, not the total number of questions.

Question 17. The number 8,000 is first increased by 20% and then decreased by 20%. Find the resulting number.
Answer:
The original number is 8,000.
First, apply the 20% increase:
\( \text{Value} = 8000 \times \left(1 + \frac{20}{100}\right) = 8000 \times \frac{120}{100} = 9600 \)

Next, apply the 20% decrease to this new value of 9,600:
\( \text{Final Value} = 9600 \times \left(1 - \frac{20}{100}\right) = 9600 \times \frac{80}{100} = 7680 \)
So, the final number is 7,680.
In simple words: Increasing 8,000 by 20% gives 9,600. Then, lowering 9,600 by 20% leaves you with a final answer of 7,680.

Exam Tip: Increasing a number and then decreasing it by the same percentage always results in a final value that is less than the original number.

Question 18. The number 12,000 is first decreased by 25% and then increased by 25%. Find the resulting number.
Answer:
The starting number is 12,000.
First, apply the 25% decrease:
\( \text{Value} = 12000 \times \left(1 - \frac{25}{100}\right) = 12000 \times \frac{75}{100} = 9000 \)

Next, apply the 25% increase to this new value of 9,000:
\( \text{Final Value} = 9000 \times \left(1 + \frac{25}{100}\right) = 9000 \times \frac{125}{100} = 11,250 \)
So, the final number is 11,250.
In simple words: Lowering 12,000 by 25% gives 9,000. Then, raising 9,000 by 25% gives a final answer of 11,250.

Exam Tip: Be sure to calculate the second percentage change on the updated value, not on the original starting number.

Question 19. The cost of an article is first increased by 20% and then decreased by 30%, find the percentage change in the cost of the article.
Answer:
Let us assume the starting cost of the article is Rs. 100.
After the 20% increase:
\( \text{Cost} = 100 + 20 = \text{Rs. } 120 \)

Next, apply the 30% decrease on this new cost of Rs. 120:
\( \text{Decrease} = \frac{30}{100} \times 120 = \text{Rs. } 36 \)
\( \text{Final Cost} = 120 - 36 = \text{Rs. } 84 \)

The total change from the original Rs. 100 is:
\( 100 - 84 = \text{Rs. } 16 \)
So, the percentage change is:
\( \frac{16}{100} \times 100 = 16\% \text{ decrease} \)
In simple words: If you start with Rs. 100, raising it by 20% makes it Rs. 120. Dropping it by 30% from there makes it Rs. 84, which is a 16% drop overall.

Exam Tip: Using Rs. 100 as a starting value is an easy way to solve percentage change questions when no actual price is given.

Question 20. The cost of an article is first decreased by 25% and then further decreased by 40%. Find the percentage change in the cost of the article.
Answer:
Let us assume the starting cost of the article is Rs. 100.
First, apply the 25% decrease:
\( \text{Cost} = 100 - 25 = \text{Rs. } 75 \)

Next, apply the 40% decrease on this new cost of Rs. 75:
\( \text{Decrease} = \frac{40}{100} \times 75 = \text{Rs. } 30 \)
\( \text{Final Cost} = 75 - 30 = \text{Rs. } 45 \)

The total change from the original Rs. 100 is:
\( 100 - 45 = \text{Rs. } 55 \)
So, the percentage change is:
\( \frac{55}{100} \times 100 = 55\% \text{ decrease} \)
In simple words: Starting with Rs. 100, the first drop makes it Rs. 75. The next drop of 40% takes away Rs. 30, leaving you at Rs. 45. This is a total reduction of 55%.

Exam Tip: Do not just add the two percentages (25% + 40% = 65%). The second drop is on the reduced price, making the total change 55%.

Exercise 7(B)

Question 1. A man bought a certain number of oranges ; out of which 13 percent were found rotten. He gave 75% of the remaining in charity and still has 522 oranges left. Find how many had he bought?
Answer:
Let us assume the total number of oranges bought was 100.
Rotten oranges = 13
Remaining good oranges:
\( 100 - 13 = 87 \)
He gave 75% of these 87 oranges in charity:
\( \text{Charity} = \frac{75}{100} \times 87 = \frac{261}{4} \)
Oranges remaining with him:
\( 87 - \frac{261}{4} = \frac{348 - 261}{4} = \frac{87}{4} \)

Using the unitary method, if he is left with \( \frac{87}{4} \) oranges, the original count is 100.
If he is left with 1 orange, the original count is:
\( 100 \times \frac{4}{87} \)
Given that he actually has 522 oranges left, the original count is:
\( 100 \times \frac{4}{87} \times 522 = 100 \times 4 \times 6 = 2400 \)
So, the man bought 2,400 oranges.
In simple words: We assume he starts with 100 oranges. After removing the rotten ones and those given to charity, he has 21.75 left. Since he actually has 522 left, we calculate that he started with 2,400.

Exam Tip: Be very careful when calculating with fractions. Do not round off numbers mid-step, as it can change your final answer.

Question 2. 5% pupil in a town died due to some diseases and 3% of the remaining left the town. If 2, 76, 450 pupil are still in the town; find the original number of pupil in the town.
Answer:
Let us assume the starting number of pupils in the town was 100.
Number of pupils who died:
\( 5\% \text{ of } 100 = 5 \)
Remaining pupils:
\( 100 - 5 = 95 \)
Number of pupils who left the town:
\( \frac{3}{100} \times 95 = \frac{57}{20} \)
Actual remaining pupils:
\( 95 - \frac{57}{20} = \frac{1900 - 57}{20} = \frac{1843}{20} \)

Using the unitary method, if the remaining is \( \frac{1843}{20} \), the original count is 100.
If the remaining is 1, the original count is:
\( 100 \times \frac{20}{1843} \)
Given that the actual remaining is 2,76,450, the original count is:
\( 100 \times \frac{20}{1843} \times 2,76,450 \)
\( = 100 \times 20 \times 150 = 3,00,000 \)
So, the original number of pupils was 3,00,000.
In simple words: We assume there are 100 pupils. After 5 die, 95 remain. Then 3% of 95 leave, which leaves 92.15 pupils. Comparing this to the actual count of 2,76,450 shows that the town had 3,00,000 pupils to start with.

Exam Tip: Write down your steps for the unitary method clearly to make it easy for the examiner to follow your work.

Question 3. In a combined test in English and Physics ; 36% candidates failed in English ; 28% failed in Physics and 12% in both ; find:
(i) the percentage of passed candidates
(ii) the total number of candidates appeared, if 208 candidates have failed.
Answer:
Percentage of candidates who failed only in English:
\( 36\% - 12\% = 24\% \)
Percentage of candidates who failed only in Physics:
\( 28\% - 12\% = 16\% \)
Percentage of candidates who failed in both subjects = 12%
Total percentage of candidates who failed:
\( 24\% + 16\% + 12\% = 52\% \)

(i) Percentage of candidates who passed:
\( 100\% - 52\% = 48\% \)

(ii) We are given that 52% of the candidates failed, which is equal to 208 candidates.
Let the total number of candidates be \( y \).
\( \frac{52}{100} \times y = 208 \)
Solve for \( y \):
\( y = \frac{208 \times 100}{52} = 4 \times 100 = 400 \)
So, the total number of candidates who appeared is 400.
In simple words: We find that 52% of the students failed in at least one subject. This means the remaining 48% passed. Since the 52% who failed equals 208 students, the total number of students must be 400.

Exam Tip: Drawing a quick Venn diagram can help you understand set-based percent questions and avoid double-counting.

Question 4. In a combined test in Maths and Chemistry; 84% candidates passsed in Maths; 76% in Chemistry and 8% failed in both. Find :
(i) the percentage of failed candidates ;
(ii) if 340 candidates passed in the test ; then how many appeared ?
Answer:
(i) First, let us find the percentage of students who failed. We know that 84% of candidates passed in Mathematics.
This means the percentage of those who failed Mathematics is:
\( 100\% - 84\% = 16\% \)

Similarly, 76% of candidates passed in Chemistry.
So, the percentage of students who failed Chemistry is:
\( 100\% - 76\% = 24\% \)

We are given that 8% of the candidates failed in both subjects.

To find the percentage of candidates who failed in Mathematics only, we subtract those who failed both:
\( 16\% - 8\% = 8\% \)

To find the percentage of candidates who failed in Chemistry only, we subtract:
\( 24\% - 8\% = 16\% \)

We get the total percentage of failed candidates by adding these groups:
\( 8\% + 16\% + 8\% = 32\% \)
Thus, the percentage of failed candidates is 32%.

(ii) The percentage of candidates who passed the test is:
\( 100\% - 32\% = 68\% \)

We are told that 340 candidates passed.
If 68% of the candidates corresponds to 340, we find the total number of candidates who appeared:
Total candidates = \( \frac{100}{68} \times 340 = 5 \times 100 = 500 \).
Therefore, 500 candidates appeared for the test.
In simple words: We find the percentage of students who failed in only one subject and those who failed in both to get the total fail percentage. Then, we use the pass percentage to calculate the total number of students who took the test.

Exam Tip: Remember to subtract the percentage of students who failed both subjects from the individual fail percentages to avoid double-counting.

Question 5. A’s income is 25% more than B’s. Find, B’s income is how much percent less than A’s.
Answer: Let us assume B's income is Rs. 100.
Since A's income is 25% higher, A earns:
\( 100 + 25 = \text{Rs. } 125 \)

The difference in their earnings is:
\( 125 - 100 = \text{Rs. } 25 \)

To find this difference as a percentage of A's income, we calculate:
\( \frac{25}{125} \times 100\% = \frac{1}{5} \times 100\% = 20\% \)

Therefore, B's income is 20% less than A's income.
In simple words: When comparing, the base value changes. A's income is larger than B's, so the same Rs. 25 difference is a smaller percentage when compared to A's larger income of Rs. 125.

Exam Tip: Always use the person whose income is being compared *against* as the denominator. Here, we compare with A's income, so A's income (125) goes in the denominator.

Question 6. Mona is 20% younger than Neetu. How much percent is Neetu older than Mona ?
Answer: Let Neetu's age be 100 years.
Since Mona is 20% younger, her age is:
\( 100 - 20 = 80 \text{ years} \)

The difference between their ages is:
\( 100 - 80 = 20 \text{ years} \)

To find how much older Neetu is compared to Mona as a percentage, we calculate:
\( \frac{20}{80} \times 100\% = \frac{1}{4} \times 100\% = 25\% \)

Thus, Neetu is 25% older than Mona.
In simple words: Mona's age is smaller, so the same 20-year difference represents a larger percentage (25%) when we compare it to her younger age of 80.

Exam Tip: For 'older than Mona' questions, Mona's age must be in the denominator because we are comparing Neetu's extra age to Mona's base age.

Question 7. If the price of sugar is increased by 25% today; by what percent should it be decreased tomorrow to bring the price back to the original ?
Answer: Let the initial price of sugar be Rs. 100.
After a 25% increase, today's price becomes:
\( 100 + 25 = \text{Rs. } 125 \)

To return to the original price of Rs. 100, we must lower the price by:
\( 125 - 100 = \text{Rs. } 25 \)

We calculate the percentage decrease based on the new price of Rs. 125:
\( \frac{25}{125} \times 100\% = \frac{1}{5} \times 100\% = 20\% \)

Therefore, the price must be decreased by 20% tomorrow to restore the original price.
In simple words: When the price goes up, the new price is higher. To drop back down to the old price, the reduction is figured on this new higher price, making the percentage drop smaller.

Exam Tip: To find the reduction percentage, use the formula: \( \frac{\text{Increase}}{\text{New Price}} \times 100\% \).

Question 8. A number increased by 15% becomes 391. Find the number.
Answer: Let the unknown number be \( x \).
Since increasing the number by 15% makes it 391, we can write the equation:
\( x + 15\% \text{ of } x = 391 \)

\( x + \frac{15}{100}x = 391 \)

\( x \left( 1 + \frac{15}{100} \right) = 391 \)

\( x \times \frac{115}{100} = 391 \)

Solving for \( x \):
\( x = \frac{391 \times 100}{115} \)

Simplifying the fraction by dividing both 391 and 115 by 23:
\( x = \frac{17 \times 100}{5} \)

\( x = 17 \times 20 = 340 \)

Hence, the original number is 340.
In simple words: An increase of 15% means the new number is 115% of the original. We can find the starting number by solving this relationship.

Exam Tip: To simplify calculations, divide 391 and 115 by their common factor, which is 23 (since \( 115 \div 23 = 5 \) and \( 391 \div 23 = 17 \)).

Question 9. A number decreased by 23 % becomes 539. Find the number.
Answer: Let the starting number be \( x \).
According to the problem, the number is reduced by 23%:
\( x - 23\% \text{ of } x = 539 \)

\( x - \frac{23}{100}x = 539 \)

\( x \left( 1 - \frac{23}{100} \right) = 539 \)

\( x \left( \frac{100 - 23}{100} \right) = 539 \)

\( x \left( \frac{77}{100} \right) = 539 \)

Rearranging to solve for \( x \):
\( x = \frac{539 \times 100}{77} \)

Since \( 539 \div 77 = 7 \):
\( x = 7 \times 100 = 700 \).

Therefore, the required number is 700.
In simple words: When we decrease a number by 23%, it becomes 77% of what it was. Knowing that 77% is 539 helps us find that the whole number is 700.

Exam Tip: Whenever a number decreases, think of it as \( (100 - \text{decrease percentage})\% \) of the original value to set up your equation quickly.

Question 10. Two numbers are respectively 20 percent and 50 percent more than a third number. What percent is the second of the first ?
Answer: Let us denote the third number as \( x \).
The first number is 20% more than the third:
\( \text{First number} = x + \frac{20}{100}x = \frac{120x}{100} \)

The second number is 50% more than the third:
\( \text{Second number} = x + \frac{50}{100}x = \frac{150x}{100} \)

Now, we find what percent the second number is of the first:
\( \text{Required percentage} = \frac{\frac{150x}{100}}{\frac{120x}{100}} \times 100\% \)

Simplifying this expression:
\( \text{Required percentage} = \frac{150}{120} \times 100\% = \frac{5}{4} \times 100\% = 125\% \).

So, the second number is 125% of the first.
In simple words: We compare the two numbers by writing them both in terms of a third number. Then, we find the ratio of the second number to the first and convert it to a percentage.

Exam Tip: To make this even simpler, you can assume the third number is 100. Then the first is 120 and the second is 150, which makes the ratio \( \frac{150}{120} \times 100 \) very easy to compute.

Question 11. Two numbers are respectively 20 percent and 50 percent of a third number. What percent is the second of the first ?
Answer: Let the third number be assumed as 100.
The first number is 20% of this third number:
\( 20\% \text{ of } 100 = \frac{20}{100} \times 100 = 20 \)

The second number is 50% of this third number:
\( 50\% \text{ of } 100 = \frac{50}{100} \times 100 = 50 \)

Now, we calculate the percentage of the second number relative to the first:
\( \text{Required percentage} = \frac{50}{20} \times 100\% = 250\% \).

Therefore, the second number is 250% of the first.
In simple words: We can use 100 for the third number. This makes the first number 20 and the second number 50. Comparing 50 to 20 gives us 250%.

Exam Tip: Be careful with the wording: this question asks for "percent of a third number" rather than "percent more than a third number". Always read the terms carefully!

Question 12. Two numbers are respectively 30 percent and 40 percent less than a third number. What percent is the second of the first ?
Answer: Let the third number be \( x \).
The first number is 30% less than the third:
\( \text{First number} = x - \frac{30}{100}x = \frac{70x}{100} = \frac{7x}{10} \)

The second number is 40% less than the third:
\( \text{Second number} = x - \frac{40}{100}x = \frac{60x}{100} = \frac{6x}{10} \)

We find the percentage of the second number compared to the first:
\( \text{Required percentage} = \frac{\frac{6x}{10}}{\frac{7x}{10}} \times 100\% = \frac{6}{7} \times 100\% = \frac{600}{7}\% = 85\frac{5}{7}\% \).

Thus, the second number is \( 85\frac{5}{7}\% \) of the first.
In simple words: We find how big each number is compared to the third one. Since they are smaller, we subtract from 100%. Then we divide the second number by the first and write it as a mixed fraction percentage.

Exam Tip: When converting an improper fraction like \( \frac{600}{7} \) to a mixed number, divide 600 by 7 to get the quotient 85 and the remainder 5, which gives \( 85\frac{5}{7}\% \).

Exercise 7(C)

Question 1. A bag contains 8 red balls, 11 blue balls and 6 green balls. Find the percentage of blue balls in the bag.
Answer: The total number of balls in the bag is:
\( 8 + 11 + 6 = 25 \)

The count of blue balls is 11.

To find their percentage, we divide the number of blue balls by the total balls and multiply by 100:
\( \text{Percentage} = \frac{11}{25} \times 100\% = 11 \times 4\% = 44\% \).

So, 44% of the balls in the bag are blue.
In simple words: We add all the balls together to see how many there are in total. Then, we find what portion of this total is made up of blue balls and turn it into a percentage.

Exam Tip: Always find the correct total sum first before calculating any percentage, as any error in addition will make the percentage wrong.

Question 2. Mohan gets Rs. 1, 350 from Geeta and Rs. 650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit. what percent does he get from Rohit ?
Answer: First, calculate the total sum of money Mohan receives:
\( \text{Total money} = \text{Rs. } (1350 + 650) = \text{Rs. } 2000 \)

The amount he received from Rohit is Rs. 650.

Now, we find what percentage this is of the total amount:
\( \text{Percentage} = \frac{650}{2000} \times 100\% = \frac{65}{2}\% = 32.5\% \).

So, Mohan gets 32.5% of his total money from Rohit.
In simple words: We add the two amounts of money together to find the total sum. Next, we divide the amount from Rohit by this total and multiply by 100 to get the percentage.

Exam Tip: You can easily simplify the fraction \( \frac{650}{2000} \times 100 \) by canceling out the zeros first to get \( \frac{65}{2} \).

Question 3. The monthly income of a man is Rs. 16, 000. 15 percent of it is paid as income-tax and 75% of the remainder is spent on rent, food, clothing, etc. How much money is still left with the man?
Answer: The monthly earnings of the man are Rs. 16,000.
The income tax paid by him is 15% of his total earnings:
\( \text{Income Tax} = \frac{16000 \times 15}{100} = \text{Rs. } 2400 \)

The remaining income after paying tax is:
\( \text{Remaining Income} = 16000 - 2400 = \text{Rs. } 13600 \)

He spends 75% of this remainder on living expenses:
\( \text{Expenses} = \frac{13600 \times 75}{100} = 13600 \times \frac{3}{4} = 3400 \times 3 = \text{Rs. } 10200 \)

The money that is left with him now is:
\( \text{Balance left} = 13600 - 10200 = \text{Rs. } 3400 \).

Therefore, Rs. 3,400 is still left with the man.
In simple words: First, calculate the tax amount and subtract it from the total income. Then, calculate 75% of the leftover money and subtract that too to find the final remaining amount.

Exam Tip: Be careful with the word "remainder" or "remaining". Always calculate the second expense on the balance money, not on the starting income.

Question 4. A number is first increased by 20% and the resulting number is then decreased by 10%. Find the overall change in the number as percent.
Answer: Let us assume the starting number is 100.
First, we increase it by 20%:
\( \text{New value} = 100 + 20 = 120 \)

Now, this new value is decreased by 10%:
\( \text{Decrease} = \frac{120 \times 10}{100} = 12 \)

So, the final value becomes:
\( 120 - 12 = 108 \)

The total net change from the starting value of 100 is:
\( 108 - 100 = 8 \)

As a percentage of the original number:
\( \text{Percentage change} = \frac{8}{100} \times 100\% = 8\% \text{ (increase)} \).

Therefore, there is an overall increase of 8%.
In simple words: We start with 100. Adding 20% makes it 120. Subtracting 10% of 120 (which is 12) leaves us with 108. Since 108 is 8 more than 100, the overall increase is 8%.

Exam Tip: Do not just subtract percentages directly (like 20% - 10% = 10%). The second change is calculated on the already changed number, not the starting number.

Question 5. A number is increased by 10% and the resulting number is again increased by 20%. What is the overall percentage increase in the number ?
Answer: Let the original number be 100.
After the first increase of 10%, the number becomes:
\( 100 + 10 = 110 \)

Now, this number is increased again by 20%:
\( \text{Second increase} = \frac{110 \times 20}{100} = 22 \)

The final value of the number is:
\( 110 + 22 = 132 \)

The net increase from our starting value of 100 is:
\( 132 - 100 = 32 \)

As a percentage of the original number:
\( \text{Percentage increase} = \frac{32}{100} \times 100\% = 32\% \).

So, the overall percentage increase is 32%.
In simple words: Starting with 100, we add 10 to get 110. Then we add 20% of 110 (which is 22) to get 132. The total rise is 32, which is 32%.

Exam Tip: Consecutive increases can also be solved using the successive percentage formula: \( A + B + \frac{AB}{100} \), which gives \( 10 + 20 + \frac{10 \times 20}{100} = 32\% \).

Question 6. During 2003, the production of a factory decreased by 25%. But, during 2004, it (production) increased by 40% of what it was at the beginning of2004. Calculate the resulting change (increase or decrease) in production during these two years.
Answer: Let the factory production at the start of 2003 be 100 units.
During 2003, there was a decrease of 25%:
\( \text{Production at the end of 2003} = 100 - 25 = 75 \)

At the start of 2004, the production was 75 units, which then increased by 40% during the year:
\( \text{Increase amount} = \frac{75 \times 40}{100} = 30 \)

So, the new production level is:
\( 75 + 30 = 105 \)

The net change over the two-year period is:
\( 105 - 100 = 5 \text{ (increase)} \)

As a percentage of the initial production:
\( \text{Percentage change} = \frac{5}{100} \times 100\% = 5\% \).

Thus, the net change is a 5% increase in production.
In simple words: If we start with 100 units, a 25% drop leaves us with 75. Then, a 40% rise on 75 adds 30 units, bringing us to 105. Since 105 is 5% more than 100, we have an overall increase of 5%.

Exam Tip: Make sure to calculate the second year's increase based on the reduced production (75 units) from the end of the first year.

Question 7. Last year, oranges were available at Rs. 24 per dozen ; but this year, they are available at Rs. 50 per score. Find the percentage change in the price of oranges.
Answer: Last year, the price of one orange was:
\( \text{Price last year} = \text{Rs. } \frac{24}{12} = \text{Rs. } 2 \)

This year, they cost Rs. 50 per score (1 score contains 20 items):
\( \text{Price this year} = \text{Rs. } \frac{50}{20} = \text{Rs. } 2.50 \)

The increase in the price of a single orange is:
\( \text{Price increase} = \text{Rs. } 2.50 - \text{Rs. } 2.00 = \text{Rs. } 0.50 \)

The percentage change in price is calculated based on the original price:
\( \text{Percentage change} = \frac{0.50}{2} \times 100\% = 0.25 \times 100\% = 25\% \).

Therefore, there is a 25% increase in the price of oranges.
In simple words: First, find the price of a single orange for both years (Rs. 2 last year and Rs. 2.50 this year). Then, find the price increase (Rs. 0.50) and see what percent it is of last year's price.

Exam Tip: Remember that 1 dozen is 12 units and 1 score is 20 units. Finding the unit price makes it easy to compare different packaging sizes.

Question 8. In an examination, Kavita scored 120 out of 150 in Maths, 136 out of 200 in English and 108 out of 150 in Science. Find her percentage score in each subject and also on the whole (aggregate).
Answer: Her percentage score in Mathematics is:
\( \text{Maths Percentage} = \frac{120}{150} \times 100\% = \frac{4}{5} \times 100\% = 80\% \)

Her percentage score in English is:
\( \text{English Percentage} = \frac{136}{200} \times 100\% = \frac{136}{2}\% = 68\% \)

Her percentage score in Science is:
\( \text{Science Percentage} = \frac{108}{150} \times 100\% = \frac{108 \times 2}{3}\% = 36 \times 2\% = 72\% \)

To find her overall score, we first sum up her marks across all subjects:
\( \text{Total marks scored} = 120 + 136 + 108 = 364 \)

We also sum up the maximum marks possible:
\( \text{Total maximum marks} = 150 + 200 + 150 = 500 \)

Thus, her aggregate percentage score is:
\( \text{Aggregate Percentage} = \frac{364}{500} \times 100\% = \frac{364}{5}\% = 72.8\% \).
In simple words: To find individual percentages, divide the marks obtained in each subject by its maximum marks and multiply by 100. For the aggregate, divide the sum of all scored marks by the sum of all maximum marks and multiply by 100.

Exam Tip: Never average the individual percentages directly (like \( \frac{80+68+72}{3} \)) unless the maximum marks for all subjects are equal. Here, they are different, so you must use the total marks method.

Question 9. A is 25% older than B. By what percent is B younger than A ?
Answer: Let B's age be 100 years.
Since A is 25% older, A's age is:
\( 100 + 100 \times \frac{25}{100} = 125 \text{ years} \)

The age difference between them is:
\( 125 - 100 = 25 \text{ years} \)

To find how much younger B is compared to A, we calculate:
\( \text{Percentage} = \frac{25}{125} \times 100\% = \frac{1}{5} \times 100\% = 20\% \).

Therefore, B is 20% younger than A.
In simple words: Since A is older, his age (125) is larger. When we find how much younger B is, we compare the 25-year difference to A's larger age, which gives us 20%.

Exam Tip: When the question asks "younger than A", the denominator must be A's age because A is the base of comparison.

Question 10.
(i) Increase 180 by 25%.
(ii) Decrease 140 by 18%.
Answer:
(i) To increase 180 by 25%:
\( \text{New value} = 180 + \frac{180 \times 25}{100} = 180 + 45 = 225 \)

(ii) To decrease 140 by 18%:
\( \text{New value} = 140 - \frac{140 \times 18}{100} = 140 - \frac{14 \times 18}{10} = 140 - \frac{252}{10} = 140 - 25.2 = 114.8 \)
In simple words: To increase a number, calculate the specified percentage of that number and add it. To decrease a number, calculate the percentage and subtract it.

Exam Tip: You can also use multipliers: to increase by 25%, multiply by 1.25. To decrease by 18%, multiply by 0.82 (since \( 100\% - 18\% = 82\% \)).

Question 11. In an election, three candidates contested and secured 29200, 58800 and 72000 votes. Find the percentage of votes scored by winning candidate.
Answer: First, let's find the total number of votes cast:
\( \text{Total votes} = 29200 + 58800 + 72000 = 160000 \)

The candidate who won has the highest number of votes, which is 72,000. We find their percentage of the total votes as:
\( \text{Winning percentage} = \frac{72000}{160000} \times 100\% = \frac{72 \times 10}{16}\% = \frac{9 \times 10}{2}\% = 45\% \).

So, the winning candidate got 45% of the votes.
In simple words: We add all the votes together to find the total vote count. Then, we take the highest vote count (72,000) and divide it by this total, multiplying by 100 to get the percentage.

Exam Tip: The winning candidate is simply the one with the highest number of votes. Always confirm which number is the largest before calculating the percentage.

Question 12.
(i) A number when increased by 23% becomes 861 ; find the number.
(ii) A number when decreased by 16% becomes 798 ; find the number.
Answer:
(i) Let the unknown number be \( x \).
According to the problem:
\( x + \frac{x \times 23}{100} = 861 \)

\( \frac{100x + 23x}{100} = 861 \)

\( \frac{123}{100}x = 861 \)

\( x = \frac{861 \times 100}{123} \)
Since \( 861 \div 123 = 7 \):
\( x = 7 \times 100 = 700 \).
So, the number is 700.

(ii) Let the number be \( x \).
According to the problem:
\( x - \frac{x \times 16}{100} = 798 \)

\( \frac{100x - 16x}{100} = 798 \)

\( \frac{84}{100}x = 798 \)

\( x = \frac{798 \times 100}{84} \)
Reducing the fraction:
\( x = \frac{114 \times 100}{12} = \frac{114 \times 25}{3} \)
Since \( 114 \div 3 = 38 \):
\( x = 38 \times 25 = 950 \).
So, the number is 950.
In simple words: To find the starting number, we set up equations. An increase of 23% makes the number 123% of its original value. A decrease of 16% makes it 84% of its original value.

Exam Tip: Look for easy division steps when solving these equations. For example, in part (i), notice that \( 123 \times 7 = 861 \), which simplifies the calculation significantly.

Question 13. The price of sugar is increased by 20%. By what percent must the consumption of sugar be decreased so that the expenditure on sugar may remain the same ?
Answer: Let us assume the price of \( x \) kg of sugar is Rs. 100.
When the price increases by 20%, the new price becomes:
\( 100 + 20 = \text{Rs. } 120 \)

This means Rs. 120 now buys \( x \) kg of sugar.
Since the family wants to keep their spending at Rs. 100, they can now only buy:
\( \text{New consumption} = \frac{x}{120} \times 100 = \frac{5x}{6} \text{ kg} \)

The starting consumption was \( x \) kg, and the new consumption is \( \frac{5x}{6} \) kg.
The reduction in consumption is:
\( \text{Decrease} = x - \frac{5x}{6} = \frac{x}{6} \text{ kg} \)

We find this decrease as a percentage of the original consumption:
\( \text{Required percentage decrease} = \frac{\frac{x}{6}}{x} \times 100\% = \frac{100}{6}\% = \frac{50}{3}\% = 16\frac{2}{3}\% \).
Therefore, consumption must be decreased by \( 16\frac{2}{3}\% \).
In simple words: If the price goes up, we can buy less sugar with the same amount of money. We calculate the new amount of sugar we can buy, find the difference from the original amount, and write it as a percentage.

Exam Tip: Use the shortcut formula for consumption reduction when price increases by \( r\% \): \( \frac{r}{100 + r} \times 100\% \). Here, \( \frac{20}{120} \times 100 = \frac{100}{6} = 16\frac{2}{3}\% \).