Percentages: Solved Examples

Q1. A student obtained 62.5 % marks in a certain examination. If the maximum marks are 800. Find the total marks obtained by her.

Sol: Option D
Solution: 800 * 62.5/100 = 500

Q2. The annual sales of company B were Rs 96,000 in fiscal year 95-96 and Rs 1,08,000 in fiscal 96-97. What was the % increase in turnover?

Sol: Option D
Solution: Increase = 96,000 – 1,08, 000 = 12,000
∴ % Increase = 12000 * 100/96000 = 100 * 1/8 = 12.5 %

Q3. The price of a Bajaj Scooter is Rs 25,000 which is 20 % lesser than a LML Scooter. What is the price of a LML Scooter?

Sol: Option B
Solution: Here note that the % given is defined in terms of the price of LML Scooter and not Bajaj Scooter. ∴ Computing the price of SMS as 1.2 times Bajaj's price will give incorrect answer. The data given is:
LML * (100 – 20)/100 = Bajaj = 25,000
∴ LML =25,000 × 100/80 = Rs 31,250
Or you can simply do = 25,000/0.8 = 31250
(20 % less means it is 0.80 times of the other one)

Q4. The price of rice increases by 30 %. In order to keep the expenses on rice constant as before, by what percentage should a person cut down his consumption?

Sol: Option C
Solution: Applying the formula 100 × R/ (100 + R), Since the price has increased, the consumption should be reduced by:
(100 × 30)/ (100+30) = 23 1/13 % = 300/13 %. Hence option C

Q5. Tom’s income is 20 % less than Jerry’s. How much is Jerry’s income more than Tom’s?

Sol: Option A
Solution: Applying the formula 100 * R/(100 - R)
As , income is 20% less
Jerry’s income is more by = 100 * 20/80 = 25 %

Q6. A traveling salesman carried 75 % of his money in traveler’s cheque and 25 % in cash. During one of his journeys, he lost his entire cash and spent from his traveler’s cheque. On completion of the journey, he returned 30% of the traveler’s cheque, which amounted to Rs 180. What was the total money that he carried?

Sol: Option D
Solution: 30 % of Traveler’s Cheque = Rs 180
∴ 100 % of Traveler's Cheque = 180 * 100/30 = Rs 600
Since Traveler's Cheque accounted for 75% of the total money that he had carried, the total money that he carried is: 600 * 100/75 = Rs 800

Q7. A number is increased by 20%, and then it is decreased by 30%, what is the net change in the number?

Sol: Option B
Solution: Using the formula: X + Y + XY/100
⇒ 20 + (-30) + 20 *-30 /100 = -16
This means there is a decrease of 16 %

Q8. A square is converted into a rectangle by increasing one of its sides by 5 % and reducing the other by 5 %. What will be the % change in the area of the two figures?

Sol: Option D
Solution: Let the side of the square = a
∴ Its area = a²; When the square is being converted to a rectangle, the length becomes 1.05a and the width becomes 0.95a. ∴ New area = 1.05a × 0.95a = 0.9975a²
Change in area = a decrease of 0.0025a²
∴ % decrease in area =0.0025a² * 100/a²= 0.25 %

Q9. In an election there were two candidates P and S. The poll turnout was only 90 %. 500 of the given votes were declared invalid. P won 440 of X = 400 % of the total votes. Find the total number of eligible voters and the number of votes received by each of the candidates.

Sol: Option A
Solution: Let the registered voters = X
P got 440/9 % of X = 4.4X/9
Since P got 400 votes more than S, S got 4.4X/9 – 400
Total votes registered = X = P + S + Invalid votes
X = 4.4 X/9 + (4.4 X/9 – 400)+ 500 ⇒ X = 8.8X/9 + 100
⇒ X = 4500. Since the poll turnout was only 90%, the total number of eligible voters = 100/90 * 4500 = 5000
Number of votes P got = 4.4/9 * 4500 = 2200
Number of votes S got = 2200 – 400 = 1800.

Q10. If the population of a town is 926100 and it has been growing annually at 5 %, what was the population 3 years ago?

Sol: Option A
Let the population be X
X * (100 + 5)³ / (100) ³ =926100; X * (105/ 100) * (105/100) * (105/100) = 926100
⇒ X * (21/20) * (21/20) * (21/20) = 926100
⇒ X = 926100 * (20/21) * (20/21) * (20/21) = 800,000