Percent and Percentage

Q1. Multiple Choice Type:

i. Out of two students A and B, A does 10 questions and B does 30 questions in the same time. The percentage of number of questions done by B to the number of questions by A is:

Step 1: Number of questions done by A = 10
Number of questions done by B = 30
Step 2: Percentage of questions done by B to A = (B / A) × 100
= (30 / 10) × 100 = 3 × 100 = 300%
Answer: b. 300%

ii. In an election, there are only two candidates A and B. A gets 60% of the total votes polled and wins the election by 960 votes. What is the number of total votes polled?

Step 1: Let total votes be x.
Votes secured by A = 60% of x = 0.60x
Votes secured by B = Remaining votes = 40% of x = 0.40x
Step 2: Difference between A and B = 0.60x − 0.40x = 0.20x
Given difference = 960 votes
Step 3: Solve for x:
0.20x = 960
x = 960 ÷ 0.20 = 4800
Answer: d. 4800 votes

iii. If A is 20% less than B, then B is:

Step 1: Assume B = 100 units.
Then A = 100 − 20% of 100 = 100 − 20 = 80 units.
Step 2: Find how much % B is more than A:
Difference = B − A = 100 − 80 = 20
Percentage more than A = (Difference / A) × 100 = (20 / 80) × 100 = 25%
Answer: d. 25% more than A

iv. A student has to obtain 35% of the total marks to pass. He got 25% of the total marks and failed by 80 marks. The total of marks is:

Step 1: Let total marks be x.
Pass marks = 35% of x = 0.35x
Obtained marks = 25% of x = 0.25x
Step 2: Failed by = Pass marks − Obtained marks = 0.35x − 0.25x = 0.10x
Given failed by 80 marks
Step 3: Solve for x:
0.10x = 80
x = 80 ÷ 0.10 = 800
Answer: b. 800

v. A mixture of milk and water contains 4 parts of milk and 1 part of water. The percentage of milk in the mixture is:

Step 1: Total parts = 4 + 1 = 5 parts
Milk parts = 4
Step 2: Percentage of milk = (Milk parts / Total parts) × 100 = (4 / 5) × 100 = 80%
Answer: d. 80%

Q2: A man bought a certain number of oranges; out of which 13% were found rotten. He gave 75% of the remaining in charity and still had 522 oranges. Find how many oranges had he bought?

Step 1: Let total number of oranges bought = x.
Step 2: Rotten oranges = 13% of x = (13 / 100) × x = 0.13x
Step 3: Remaining good oranges = x − 0.13x = 0.87x
Step 4: Oranges given in charity = 75% of remaining = 0.75 × 0.87x = 0.6525x
Step 5: Oranges left after charity = Remaining − Given in charity
= 0.87x − 0.6525x = 0.2175x
Step 6: Given that oranges left = 522
So, 0.2175x = 522
Step 7: Solve for x:
x = 522 ÷ 0.2175 = 2400
Answer: The man bought 2400 oranges.

Q3: 5% pupil in a town died due to some disease 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.

Step 1: Let the original number of pupils be x.
Step 2: Pupils affected by disease = 5% of x = 0.05x
Step 3: Remaining pupils = x − 0.05x = 0.95x
Step 4: Pupils who left town = 3% of remaining = 0.03 × 0.95x = 0.0285x
Step 5: Pupils still in town = Remaining − Left = 0.95x − 0.0285x = 0.9215x
Step 6: Given pupils still in town = 276,450
So, 0.9215x = 276,450
Step 7: Solve for x:
x = 276,450 ÷ 0.9215 ≈ 300,000
Answer: The original number of pupils in the town = 3,00,000

Q4: In a combined test in English and Physics;
36% candidates failed in English;
28% failed in Physics and 12% failed in both;
find:

i. The percentage of passed candidates.

Step 1: Percentage failed in English (E) = 36%
Percentage failed in Physics (P) = 28%
Percentage failed in both (E ∩ P) = 12%
Step 2: Use formula for union:
Percentage failed in English or Physics = E + P − (E ∩ P)
= 36% + 28% − 12% = 52%
Step 3: Percentage passed = 100% − Percentage failed in either
= 100% − 52% = 48%
Answer: 48%

ii. The total number of candidates appeared, if 208 candidates have failed.

Step 1: Let total candidates appeared = x
Step 2: Number of failed candidates = 52% of x = (52/100) × x = 0.52x
Step 3: Given failed candidates = 208
So, 0.52x = 208
Step 4: Solve for x:
x = 208 ÷ 0.52 = 400
Answer: 400 candidates

Q5: In a combined test in Maths and Chemistry, 84% candidates passed in Maths, 76% in Chemistry and 8% failed in both. Find:

i. The percentage of failed candidates.

Step 1: Percentage passed in Maths (M) = 84%
Percentage passed in Chemistry (C) = 76%
Percentage failed in both = 8%
Step 2: Percentage failed in Maths = 100% − 84% = 16%
Percentage failed in Chemistry = 100% − 76% = 24%
Step 3: Use formula for union of failed candidates:
Failed in Maths or Chemistry = Failed in Maths + Failed in Chemistry − Failed in both
= 16% + 24% − 8% = 32%
Answer: 32%

ii. if 340 candidates passed in the test, then, how many candidates had appeared in test?

Step 1: Percentage passed in test = 100% − Percentage failed
= 100% − 32% = 68%
Step 2: Let total candidates appeared = x
Number of candidates passed = 68% of x = (68/100) × x = 0.68x
Step 3: Given number passed = 340
So, 0.68x = 340
Step 4: Solve for x:
x = 340 ÷ 0.68 = 500
Answer: 500 candidates

Q6: A’s income is 25% more than B’s. Find out by how much percent is B’s income less than A’s?

Step 1: Let B’s income = 100 units.
Step 2: A’s income = 100 + 25% of 100 = 100 + 25 = 125 units.
Step 3: Difference between A’s and B’s income = 125 − 100 = 25 units.
Step 4: Percentage by which B’s income is less than A’s:
= (Difference / A’s income) × 100 = (25 / 125) × 100 = 20%
Answer: B’s income is 20% less than A’s.

Q7: Mona is 20% younger than Neetu. By how much percent is Neetu older than Mona?

Step 1: Let Neetu’s age = 100 units.
Step 2: Mona is 20% younger than Neetu, so Mona’s age = 100 − 20 = 80 units.
Step 3: Difference in age = Neetu − Mona = 100 − 80 = 20 units.
Step 4: Percentage by which Neetu is older than Mona:
= (Difference / Mona’s age) × 100 = (20 / 80) × 100 = 25%
Answer: Neetu is 25% older than Mona.

Q8: 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?

Step 1: Let the original price be 100 units.
Step 2: Price after 25% increase = 100 + 25% of 100 = 100 + 25 = 125 units.
Step 3: Let the required percentage decrease be x%.
Step 4: Price after decrease = 125 − x% of 125 = 125 × (1 − x/100)
Step 5: To bring price back to original:
125 × (1 − x/100) = 100
Step 6: Solve for x:
1 − x/100 = 100 / 125 = 0.8
x/100 = 1 − 0.8 = 0.2
x = 0.2 × 100 = 20%
Answer: The price should be decreased by 20% tomorrow.

Q9: A number increased by 15% becomes 391. Find the number.

Step 1: Let the original number be x.
Step 2: Number after 15% increase = x + 15% of x = 1.15x
Step 3: Given 1.15x = 391
Step 4: Solve for x:
x = 391 ÷ 1.15 = 340
Answer: The original number is 340.

Q10: A number decreased by 23% becomes 539. Find the number.

Step 1: Let the original number be x.
Step 2: Number after 23% decrease = x − 23% of x = (1 − 0.23)x = 0.77x
Step 3: Given 0.77x = 539
Step 4: Solve for x:
x = 539 ÷ 0.77 = 700
Answer: The original number is 700.

Q11: Two numbers are respectively 20 percent and 50 percent more than a third number. What percent of the first number is the second number?

Step 1: Let the third number be \( x \).
Step 2: First number = \( x + 20\% \text{ of } x = \frac{120}{100}x = 1.20x \)
Second number = \( x + 50\% \text{ of } x = \frac{150}{100}x = 1.50x \)
Step 3: Find what percent of the first number is the second number: \[ = \left(\frac{\text{Second number}}{\text{First number}}\right) \times 100 = \left(\frac{1.50x}{1.20x}\right) \times 100 = \frac{150}{120} \times 100 = \frac{5}{4} \times 100 = 125\% \]Answer: The second number is \( 125\% \) of the first number.

Q12: Two numbers are respectively 20 percent and 50 percent of a third number. What percent of the first number is the second number?

Step 1: Let the third number be \( x \).
Step 2: First number = \( 20\% \text{ of } x = \frac{20}{100} x = 0.20x \)
Second number = \( 50\% \text{ of } x = \frac{50}{100} x = 0.50x \)
Step 3: Find what percent of the first number is the second number: \[ = \left(\frac{\text{Second number}}{\text{First number}}\right) \times 100 = \left(\frac{0.50x}{0.20x}\right) \times 100 = \frac{50}{20} \times 100 = \frac{5}{2} \times 100 = 250\% \]Answer: The second number is \( 250\% \) of the first number.

Q13: Two numbers are respectively 30 percent 40 percent less than a third number. What percent of the first number is the second number?

Step 1: Let the third number be \( x \).
Step 2: First number = \( x – 30\% \text{ of } x = \frac{70}{100} x = 0.70x \)
Second number = \( x – 40\% \text{ of } x = \frac{60}{100} x = 0.60x \)
Step 3: Find what percent of first number is the second number: \[ = \left(\frac{\text{Second number}}{\text{First number}}\right) \times 100 = \left(\frac{0.60x}{0.70x}\right) \times 100 = \frac{60}{70} \times 100 = \frac{6}{7} \times 100 = \frac{600}{7} = 85 \frac{5}{7}\% \]Answer: The second number is \( 85 \frac{5}{7}\% \) of the first number.

Q14: Mohan gets ₹1,350 from Geeta and ₹650 from Rohit. Out of the total money that Mohan gets from Geeta and Rohit, what percent does he get from Rohit?

Step 1: Money received from Geeta = ₹1,350
Money received from Rohit = ₹650
Step 2: Total money received by Mohan = ₹1,350 + ₹650 = ₹2,000
Step 3: Required percentage = \[ \left( \frac{650}{2000} \right) \times 100 = \frac{13}{40} \times 100 = 32.5\% \]Answer: Rohit’s contribution is \( 32.5\% \) of the total money received by Mohan.

Q15: The monthly income of a man is ₹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?

Step 1: Monthly income = ₹16,000
Step 2: Income-tax paid = \[ \frac{15}{100} \times 16000 = ₹2,400 \]Step 3: Remaining after tax = ₹16,000 − ₹2,400 = ₹13,600
Step 4: Amount spent on rent, food, clothing, etc. = \[ \frac{75}{100} \times 13,600 = ₹10,200 \]Step 5: Money still left = ₹13,600 − ₹10,200 = ₹3,400
Answer: ₹3,400 is still left with the man after all expenses.

Q16: 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 of 2004. Calculate the resulting change (increase or decrease) in production during these two years.

Step 1: Let the initial production at the beginning of 2003 be \( x \).
Step 2: Production at end of 2003 after 25% decrease = \[ x – \frac{25}{100}x = \frac{75}{100}x = \frac{3}{4}x \]Step 3: This becomes the starting production for 2004.
Now 40% increase in 2004 means:
Increase = \[ \frac{40}{100} \times \frac{3}{4}x = \frac{3}{10}x \]Step 4: Final production at end of 2004 = \[ \frac{3}{4}x + \frac{3}{10}x = \left( \frac{15}{20} + \frac{6}{20} \right)x = \frac{21}{20}x \]Step 5: Compare final production with original production \( x \): \[ \text{Change} = \frac{21}{20}x – x = \frac{1}{20}x \]Step 6: Percentage change = \[ \frac{\frac{1}{20}x}{x} \times 100 = \frac{1}{20} \times 100 = 5\% \]Answer: There is a net increase of 5% in production over the two years.

Q17: Last year, oranges were available at ₹24 per dozen; but this year, they are available at ₹50 per score. Find the percentage change in the price of oranges.

Step 1: Last year’s price: ₹24 per dozen = 12 oranges \[ \text{Price per orange last year} = \frac{24}{12} = ₹2 \]Step 2: This year’s price: ₹50 per score = 20 oranges \[ \text{Price per orange this year} = \frac{50}{20} = ₹2.50 \]Step 3: Find percentage change in price: \[ \text{Increase in price} = ₹2.50 – ₹2 = ₹0.50 \\ \text{Percentage change} = \left( \frac{0.50}{2} \right) \times 100 = \frac{1}{4} \times 100 = 25\% \]Answer: There is a 25% increase in the price of oranges.

Q18:

i. Increase 180 by 25%

Step 1: 25% of 180 = \[ \frac{25}{100} \times 180 = \frac{1}{4} \times 180 = 45 \]Step 2: Increased value = \[ 180 + 45 = 225 \]Answer: The number after 25% increase is 225.

ii. Decrease 140 by 18%

Step 1: 18% of 140 = \[ \frac{18}{100} \times 140 = \frac{2520}{100} = 25.2 \]Step 2: Decreased value = \[ 140 – 25.2 = 114.8 \]Answer: The number after 18% decrease is 114.8.

Q19:

i. A number when increased by 23% becomes 861; find the number.

Step 1: Let the original number be \( x \).
Step 2: Increase of 23% means: \[ x + \frac{23}{100}x = \frac{123}{100}x = 861 \]Step 3: Solve for \( x \): \[ \frac{123}{100}x = 861 \Rightarrow x = \frac{861 \times 100}{123} = \frac{86100}{123} = 700 \]Answer: The original number is 700.

ii. A number when decreased by 16% becomes 798; find the number.

Step 1: Let the original number be \( x \).
Step 2: Decrease of 16% means: \[ x – \frac{16}{100}x = \frac{84}{100}x = 798 \]Step 3: Solve for \( x \): \[ \frac{84}{100}x = 798 \Rightarrow x = \frac{798 \times 100}{84} = \frac{79800}{84} = 950 \]Answer: The original number is 950.

Q20: 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?

Step 1: Let the original price of sugar per kg = ₹1 (for simplicity).
Let the original consumption = 100 kg.
So, original expenditure = ₹1 × 100 = ₹100
Step 2: New price after 20% increase = \[ ₹1 + \frac{20}{100} × ₹1 = ₹1.20 \]Let the new consumption be \( x \) kg.
To keep the expenditure the same: \[ ₹1.20 × x = ₹100 \Rightarrow x = \frac{100}{1.20} = \frac{1000}{12} = 83.33 \text{ kg} \]Step 3: Decrease in consumption = \[ 100 – 83.33 = 16.67 \text{ kg} \]Step 4: Percentage decrease = \[ \frac{16.67}{100} × 100 = 16.67\% \]Answer: The consumption must be decreased by \( \frac{1}{6} × 100 = 16\frac{2}{3}\% \).