Percentage

Q1: 36% of the students in a school are girls. If the number of boys is 1440, find the total strength of the school.

Step 1: Let the total number of students in the school be \(x\)
Then, number of girls = 36% of \(x = \frac{36}{100} \times x = \frac{36x}{100}\)
Step 2: Number of boys = Total – Number of girls
So, boys = \(x – \frac{36x}{100} = \frac{64x}{100}\)
Step 3: We are given that number of boys = 1440
\(\frac{64x}{100} = 1440\)
Step 4: Solve for \(x\)
\(x = \frac{1440 \times 100}{64} = \frac{144000}{64} = 2250\)
Answer: Total strength = 2250 students

Q2: Geeta saves 18% of her monthly salary. If she spends ₹10250 per month, what is her monthly salary?

Step 1: Let Geeta’s monthly salary be ₹\(x\)
She saves 18% of her salary ⇒ She spends the remaining 82%
Step 2: Spending = 82% of salary
\(\frac{82}{100} \times x = 10250\)
Step 3: Solve for \(x\)
\(x = \frac{10250 \times 100}{82} = \frac{1025000}{82} = 12500\)
Answer: Monthly salary = ₹12,500

Q3: 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?

Step 1: Let the maximum marks be \(x\)
To pass, a student must secure 40% of \(x\)
Step 2: Rahul failed by 32 marks ⇒ passing marks = 178 + 32 = 210
Step 3: Form the equation:
\(\frac{40}{100} \times x = 210\)
Step 4: Solve for \(x\)
\(x = \frac{210 \times 100}{40} = \frac{21000}{40} = 525\)
Answer: Maximum Marks = 525

Q4: 8% of the students in a school remained absent on a day. If 1633 attended the school on that day, how many remained absent?

Step 1: Let the total number of students in the school be \(x\)
If 8% were absent, then 92% were present
Step 2: 92% of total = 1633
\(\frac{92}{100} \times x = 1633\)
Step 3: Solve for \(x\)
\(x = \frac{1633 \times 100}{92} = \frac{163300}{92} = 1775\)
Step 4: Find number of absent students
Absent = Total − Present = 1775 − 1633 = 142
Answer: 142 students were absent

Q5: On increasing the price of an article by 14%, it becomes ₹1995. What was its original price?

Step 1: Let the original price be ₹\(x\)
After a 14% increase, the price becomes ₹1995
Step 2: Use the formula:
Increased Price = \(x + 14\%\text{ of }x = x \times \left(1 + \frac{14}{100}\right) = x \times \frac{114}{100}\)
Step 3: Set up the equation:
\(\frac{114}{100} \times x = 1995\)
Step 4: Solve for \(x\)
\(x = \frac{1995 \times 100}{114} = \frac{199500}{114} = 1750\)
Answer: Original Price = ₹1750

Q6: On decreasing the price of an article by 6%, it becomes ₹1551. What Was its original price?

Step 1: Let the original price be ₹\(x\)
After decreasing by 6%, the price becomes ₹1551
Step 2: Use the formula:
Decreased Price = \(x \times \left(1 – \frac{6}{100}\right) = x \times \frac{94}{100}\)
Step 3: Set up the equation:
\(\frac{94}{100} \times x = 1551\)
Step 4: Solve for \(x\)
\(x = \frac{1551 \times 100}{94} = \frac{155100}{94} = 1650\)
Answer: Original Price = ₹1650

Q7: Reenu reduced her weight by 15%. If now she weighs 52.7 kg, what was her earlier weight?

Step 1: Let Reenu’s earlier weight be \(x\) kg
After reducing her weight by 15%, her current weight = 52.7 kg
Step 2: Use the formula:
Reduced Weight = \(x \times \left(1 – \frac{15}{100}\right) = x \times \frac{85}{100}\)
Step 3: Form the equation:
\(\frac{85}{100} \times x = 52.7\)
Step 4: Solve for \(x\)
\(x = \frac{52.7 \times 100}{85} = \frac{5270}{85} = 62\)
Answer: Earlier weight = 62 kg

Q8: 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?

Step 1: Let the total number of votes polled be \(x\)
One candidate got 58% of the votes, the other got the remaining 42%
Step 2: Difference in their votes = 58% − 42% = 16%
Step 3: This 16% vote difference = 2560 votes
So, \(\frac{16}{100} \times x = 2560\)
Step 4: Solve for \(x\)
\(x = \frac{2560 \times 100}{16} = \frac{256000}{16} = 16000\)
Answer: Total votes polled = 16,000

Q9: 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?

Step 1: Calculate percentage in each subject:
Science = \(\frac{60}{75} \times 100 = 80\%\)
Mathematics = \(\frac{84}{100} \times 100 = 84\%\)
Hindi = \(\frac{36}{50} \times 100 = 72\%\)
English = \(\frac{30}{45} \times 100 = \frac{2}{3} \times 100 = 66.67\%\)
Step 2: Lowest percentage = English (66.67%)
Answer: English

ii. In which subject her performance is the best?

Step 1: Highest percentage = Mathematics (84%)
Answer: Mathematics

iii. What is her aggregate percentage of marks?

Step 1: Add total marks obtained:
60 (Science) + 84 (Maths) + 36 (Hindi) + 30 (English) = 210
Step 2: Add maximum marks:
75 + 100 + 50 + 45 = 270
Step 3: Calculate aggregate percentage:
\(\frac{210}{270} \times 100 = \frac{7000}{9} = 77 \frac{7}{9}\%\)
Answer: Aggregate percentage = \(77 \frac{7}{9}\)%

Q10: 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?

Step 1: Assume the original price = ₹100
Increased by 25% ⇒ New price = ₹\(100 + 25 = 125\)
Step 2: Let the required % decrease on ₹125 = \(x\)
\(\frac{x}{100} \times 125 = 25\) ← we want to reduce ₹125 back to ₹100
Step 3: Solve for \(x\)
\(\frac{125x}{100} = 25\)
\(\Rightarrow 125x = 2500\)
\(\Rightarrow x = \frac{2500}{125} = 20\)
Answer: 20% decrease required

Q11: The price of an article is reduced by 10%. By how much cent this value be increased to restore it to its former value?

Step 1: Assume the original price = ₹100
10% reduction ⇒ New price = ₹\(100 – 10 = 90\)
Step 2: Let the required % increase on ₹90 be \(x\)
We want: ₹90 + \(x\%\) of ₹90 = ₹100
\(\Rightarrow \frac{x}{100} \times 90 = 10\)
Step 3: Solve for \(x\)
\(\frac{90x}{100} = 10\)
\(\Rightarrow 90x = 1000\)
\(\Rightarrow x = \frac{1000}{90} = \frac{100}{9}\%\)
\(\Rightarrow x = 11 \frac{1}{9}\%\)
Answer: \( 11 \frac{1}{9}\)% increase required

Q12: 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?

Step 1: Let original price of tea = ₹1 per unit
Let original consumption = 100 units
Then, original expenditure = ₹1 × 100 = ₹100
Step 2: Price increases by 20% ⇒ New price = ₹1.20 per unit
Step 3: To keep expenditure same (₹100):
New quantity consumed = ₹100 ÷ ₹1.20 = \(\frac{100}{1.20} = 83.33\) units
Step 4: Find % reduction in consumption:
Reduction = \(100 – 83.33 = 16.67\) units
% reduction = \(\frac{16.67}{100} \times 100 = 16.67\%\)
Answer: She should reduce consumption by 16.67%

Q13: 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?

Step 1: Let the total money be ₹\(x\)
He gave 35% to elder son ⇒ ₹\(\frac{35}{100} \times x = \frac{35x}{100}\)
Step 2: Remaining = \(x – \frac{35x}{100} = \frac{65x}{100}\)
Step 3: He gave 40% of this remainder to the younger son: \[ \frac{40}{100} \times \frac{65x}{100} = \frac{2600x}{10000} = \frac{13x}{50} \]Step 4: Money left = Total − (Elder’s share + Younger’s share) \[ x – \left(\frac{35x}{100} + \frac{13x}{50}\right) \] First, convert both to like denominators: \[ \frac{35x}{100} + \frac{13x}{50} = \frac{35x}{100} + \frac{26x}{100} = \frac{61x}{100} \] So, money left = \(\frac{39x}{100}\)
Step 5: Set up the equation:
\[ \frac{39x}{100} = 11700 \\ x = \frac{11700 \times 100}{39} = \frac{1170000}{39} = 30000 \]Answer: Total money = ₹30,000

Q14: 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?

Step 1: Let the original population be ₹\(x\)
5% were killed ⇒ Remaining = \(x – \frac{5x}{100} = \frac{95x}{100}\)
Step 2: 8% of the remaining people left the town: \[ \text{Left the town} = \frac{8}{100} \times \frac{95x}{100} = \frac{760x}{10000} = \frac{19x}{250} \]Step 3: Population now = Remaining people after both reductions \[ \text{Now} = \frac{95x}{100} – \frac{19x}{250} \]Take LCM = 500: \[ \frac{475x}{500} – \frac{38x}{500} = \frac{437x}{500} \]Step 4: This remaining population = 43700 \[ \frac{437x}{500} = 43700 \] Multiply both sides by 500: \[ 437x = 43700 \times 500 = 21850000 \\ x = \frac{21850000}{437} = 50000 \]Answer: Original population = 50,000

Q15: 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?

Step 1: Let the height of tower B = 100 m (assumed)
A is 20% less than B ⇒ A = 100 − 20 = 80 m
Step 2: Now, B is how much % more than A?
Difference in height = 100 − 80 = 20 m
Step 3: % increase = \(\frac{20}{80} \times 100 = 25\%\)
Answer: B is 25% more than A

Q16: 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?

Step 1: Let total number of candidates = \(x\)
Failed in English = 30% of \(x = \frac{30x}{100}\)
Failed in G.K. = 35% of \(x = \frac{35x}{100}\)
Failed in both = 27% of \(x = \frac{27x}{100}\)
Step 2: Use the formula for union of two sets:
Failed in at least one = Failed in English + Failed in G.K. − Failed in both \[ = \frac{30x}{100} + \frac{35x}{100} – \frac{27x}{100} = \frac{38x}{100} \]Step 3: Passed in both = Total − Failed in at least one \[ \text{Passed in both} = x – \frac{38x}{100} = \frac{62x}{100} \] We are given: \(\frac{62x}{100} = 310\)
Step 4: Solve for \(x\): \[ x = \frac{310 \times 100}{62} = \frac{31000}{62} = 500 \]Answer: Total candidates appeared = 500

Q17: The value of a car depreciates annually by 10%. If the present value of the car be ₹650000, find its value after 2 years.

Step 1: Use the compound depreciation formula: \[ \text{Value after } n \text{ years} = P \times \left(1 – \frac{r}{100}\right)^n \] Here, \(P = ₹650000\), \(r = 10\), \(n = 2\)
Step 2: Apply values into the formula: \[ \text{Value after 2 years} = 650000 \times \left(1 – \frac{10}{100}\right)^2 = 650000 \times (0.9)^2 = 650000 \times 0.81 \]Step 3: Multiply: \[ 650000 \times 0.81 = ₹526500 \]Answer: Value after 2 years = ₹5,26,500

Q18: 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?

Step 1: Use the compound growth formula: \[ \text{Population after } n \text{ years} = P \times \left(1 + \frac{r}{100}\right)^n \] Here, \(P = 80000\), \(r = 7\), \(n = 2\)
Step 2: Apply values: \[ \text{Population after 2 years} = 80000 \times \left(1 + \frac{7}{100}\right)^2 = 80000 \times (1.07)^2 = 80000 \times 1.1449 \]Step 3: Multiply: \[ 80000 \times 1.1449 = 91592 \]Answer: Population after 2 years = 91,592

Q19: A student was asked to multiply a number by \(\frac{5}{3}\). He multiplied it by \(\frac{3}{5}\) instead. Find the percentage error in the calculation.

Step 1: Let the number be 1 (for simplicity)
Correct result = \(1 \times \frac{5}{3} = \frac{5}{3}\)
Wrong result = \(1 \times \frac{3}{5} = \frac{3}{5}\)
Step 2: Error = Correct − Incorrect = \(\frac{5}{3} – \frac{3}{5}\)
Take LCM of 3 and 5 = 15: \[ \frac{5}{3} = \frac{25}{15}, \quad \frac{3}{5} = \frac{9}{15} \\ \text{Error} = \frac{25}{15} – \frac{9}{15} = \frac{16}{15} \]Step 3: Percentage error = \(\frac{\text{Error}}{\text{Correct value}} \times 100\) \[ \frac{\frac{16}{15}}{\frac{25}{15}} \times 100 = \frac{16}{25} \times 100 = 64\% \]Answer: Percentage error = 64%

Q20: 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.

Step 1: Let the total number of voters be \( x \)
10% did not vote ⇒ Votes polled = \( 90\% \) of \( x = \frac{90x}{100} \)
Step 2: 10% of the votes polled were invalid ⇒ Valid votes = \( 90\% \) of \( \frac{90x}{100} \) \[ = \frac{90}{100} \times \frac{90x}{100} = \frac{8100x}{10000} = \frac{81x}{100} \]Step 3: The winner got 54% of valid votes, the loser got 46%
Majority = \( 54\% – 46\% = 8\% \) of valid votes \[ \text{8% of valid votes} = \frac{8}{100} \times \frac{81x}{100} = \frac{648x}{10000} \]Step 4: This is given as 1620 votes: \[ \frac{648x}{10000} = 1620 \] Multiply both sides by 10000: \[ 648x = 1620 \times 10000 = 16200000 \\ x = \frac{16200000}{648} = 25000 \]Answer: Total number of enrolled voters = 25,000