Types of Percentage Questions in MAH MBA CET
Percentage questions in the MAH MBA CET exam include basic percentage calculations, percentage increase/decrease, successive percentage changes and many others. Some questions are mentioned below:
1.3.A man's annual income has increased by Rs. 5 lakh, but the tax on income that he has to pay has reduced from 12% to 10%. He now pays Rs. 10,000 more income tax. What is his increased income (in Rs. lakhs)?
Options:
1.20
2.25
3.15
4.10
Solution:
Let the previous income of man be Rs. $x$.
Man's new income after increment = Rs. $(x + 500000)$
According to the given condition,
⇒ $(x+500000)×\frac{10}{100}-x×\frac{12}{100}=10000$
⇒ $\frac{10x}{100}+50000-\frac{12x}{100}=10000$
⇒ $\frac{2x}{100}$ = 50000–10000
⇒ $x=\frac{4000000}{2}$
⇒ $x$ = Rs. 20,00,000
His increased income = 20,00,000 + 5,00,000 = Rs. 25,00,000 = Rs. 25 lakhs
Answer:
Hence, the correct answer is 25.
2. Amit donated 20% of his income to a school and deposited 20% of the remainder in his bank. If he has Rs. 12800 now, what is the income (in Rs.) of Amit?
Options:
1.18000
2.20000
3.24000
4.32000
Solution:
Let the income of Amit be Rs. $100x$.
After donating 20% of income to school, the remaining amount = $80x$
After depositing 20% of his remaining income to the bank, he has = $80x×\frac{80}{100}=64x$
Now, as per the question,
$64x = 12800$
$\therefore x=\frac{12800}{64} = 200$
Income of Amit $=100x = 100 × 200 = 20000$
Answer:
Hence, the correct answer is 20000.
3. If A's salary is 30% more than that of B, then by how much percent is B's salary less than that of A?
Options:
1.13.1%
2.13.07%
3.23.07%
4.23.01%
Solution:
Given: A's salary is 30% more than that of B.
Let the salary of B be 100 units.
So, salary of A will be = 100 + (100 × $\frac{30}{100}$) = 130 units
The percentage B is less than A
= $\frac{A-B}{A}\times 100$
= $\frac{(130-100)}{130}× 100$
= $\frac{30}{130} × 100$
= $23.07\%$
Answer:
Hence, the correct answer is 23.07%.
4. A worker suffers a 20% cut in his wages. He may regain his original wages by obtaining a rise of:
Options:
1.27.5%
2.25%
3.22.5%
4.20%
Solution:
After reduction increased percentage, so that no effect = $\frac{R}{100-R}\times 100$
Required percentage = $\frac{20}{80}\times 100$ = 25%
Answer:
Hence, the correct answer is the option (2).
5. The allowances of an employee constitute 165% of his basic pay. If he receives Rs 11925 as gross salary, then his basic pay is (in Rs.)
Options:
1.4000
2.5000
3.4500
4.5500
Solution:
Let's assume basic salary = 100 units
Then the allowances = 165 units
Gross Salary = 265 units = Rs. 11925
So, basic salary (100 units) = Rs. $\frac{11925}{265}$ × 100
= Rs. 4500
Answer:
Hence, the correct answer is option (3).
6. If the value of a company's stock drops from Rs. 25 per share to Rs. 21 per share, the percentage decrease per share is:
Options:
1.4
2.8
3.12
4.16
Solution:
The initial value of the stock is Rs. $25$ per share.
The final value of the stock is Rs. $21$ per share.
Decrease in percentage = $\frac{\text{Initial value –Final value}}{\text{Initial value}}\times100$
= $\frac{25 –21}{ 25}×100$
= $\frac{400}{ 25}$
= 16%
Answer:
Hence, the correct answer is 16.
7. A man spends $7\frac{1}{2}$% of his money, and after spending 75% of the remaining, he has Rs. 370 left. How much money did he initially have?
Options:
1.Rs. 1200
2.Rs. 1600
3.Rs. 1500
4.Rs. 1400
Solution:
Given:
A man spends $7\frac{1}{2}$% of his money and after spending 75% of the remaining has Rs. 370 left.
Now, $7\frac{1}{2}$% = $\frac{3}{40}$ and $75$% = $\frac{3}{4}$
Let the income of the man be Rs. $x$.
At first, the man spent $\frac{3}{40}$ of $x$ money.
So, the money left = $(1 - \frac{3}{40})x = (\frac{37}{40})x$,
and then he spent $\frac{3}{4}$th of the remaining money.
So, the left amount of money will be $(1 - \frac{3}{4}) = \frac{1}{4}$th of the remaining money.
According to the given information,
$x × (\frac{37}{40}) × (\frac{1}{4}) = 370$
⇒ $x = 370 × (\frac{4}{1}) × (\frac{40}{37}$)
So, $x = 1600$
Answer:
Hence, the correct answer is Rs. 1600.
8. In a class, the average score of girls in an examination is 73 and that of boys is 71. The average score for the whole class is 71.8. Find the percentage of girls:
Options:
1.40%
2.50%
3.55%
4.60%
Solution:
Boys = (73 – 71.8) = 1.2, Girls = (71.8 – 71) = 0.8
So the ratio of Boys : Girls = 1.2 : 0.8 = 3 : 2
$\therefore$ Percentage of Girls $=\frac{2}{5}×100 = 40$%
Answer:
Hence, the correct answer is 40%.
8.0.06% of 250% of 1600 is:
Options:
1.24
2.0.24
3.0.024
4.2.4
Solution:
Given: 0.06% of 250% of 1600
= $\frac{(1600×250×0.06)}{(100×100)}$
= $\frac{(16×6)}{40}$
= $\frac{24}{10}$
= 2.4
Answer:
Hence, the correct answer is 2.4.
9. If a number is increased by 84, it becomes 107% of itself. What is the number?
Options:
1.600
2.900
3.1500
4.1200
Solution:
Given: If a number is increased by 84, it becomes 107% of itself.
Let the number be y.
According to the question,
y + 84 = 107% of y
⇒ y + 84 = 1.07y
⇒ 84 = 1.07y – y
⇒ 84 = 0.07y
⇒ y = 1200
Answer:
Hence, the correct answer is 1200.
10. If P is 25% less than Q, then Q is how much percent more than P?
Options:
1.20
2.16.66
3.33.33
4.12.5
Solution:
Let Q be 100
Then P = 25% less than Q = 75% of Q = 75
So, Q - P = 100 - 75 = 25
Required percentage = $\frac{\text{Q - P}}{\text{Q}}$ × 100
= $\frac{25}{75}$ × 100
= 33.33%
Answer:
Hence, the correct answer is 33.33.
11.$x$ is 5 times longer than $y$. The percentage by which $y$ is less than $x$ is:
Options:
1.50%
2.40%
3.80%
4.70%
Solution:
Let the length of $y$ be $l$.
Then, the length of $x$ is $5l$.
Required percentage = $\frac{x-y}{x}\times 100 = \frac{5l-l}{5l}×100 = 80$%
Answer:
Hence, the correct answer is 80%.
12. What percentage of a day is 36 minutes?
Options:
1.25%
2.2.5%
3.3.6%
4.0.25%
Solution:
1 day = 24 hours = (24 × 60) minutes = 1440 minutes
So, the required percentage = $\frac{\text{36 }}{\text{1440}}$ × 100 = 2.5%
Answer:
Hence, the correct answer is 2.5%.
13.If 60% of A = 30% of B, B = 40% of C, and C = $x$% of A, the value of $x$ is:
Options:
1.200
2.500
3.800
4.300
Solution:
Given: 60% of A = 30% of B, B = 40% of C, and C = $x$% of A
Here, C = $x$% of A ⇒ C = $\frac{x×A}{100}$
and B = 40% of C ⇒ B = $\frac{40×C}{100}$ ⇒ B = $\frac{40×x×A}{100×100}$ = $\frac{2×x×A}{500}$
Now, 60% of A = 30% of B
⇒ $\frac{60×A}{100}$ = $\frac{30×B}{100}$
⇒ 2A = B
⇒ 2A = $\frac{2×x×A}{500}$
⇒ $x$ = 500
So, the value of $x$ is 500.
Answer:
Hence, the correct answer is 500.
14. The value of equipment depreciates by 20% each year. How much less will the value of the equipment be after 3 years?
Options:
1.48.8%
2.51.2%
3.54%
4.60%
Solution:
Assume the initial value as $P$ and calculate the depreciation value by using the formula $P(1-\frac{R}{100})^n$ where $R$ is the rate and $n$ is the time.
So, the value after 3 years
= $P(1-\frac{20}{100})^3$
= $P(1-\frac{1}{5})^3$
= $P(\frac{4}{5})^3$
= $P(0.8)^3$
= $0.512P$
So, the required percentage $=\frac{P - 0.512P}{P} ×100 = 48.8$%
The value of the equipment will be 48.8% less than the initial value after 3 years.
Answer:
Hence, the correct answer is 48.8%.
15.If $X$ is 20% less than $Y$, then find the values of$\frac{Y–X}{Y}$ and $\frac{X}{X–Y}$.
Options:
1.$\frac{1}{5}$ and $-4$
2.$5$ and $-\frac{1}{4}$
3.$\frac{2}{5}$ and $-\frac{5}{2}$
4.$\frac{3}{5}$ and $-\frac{5}{3}$
Solution:
Let $Y$ = 100
Then, $X$ = 100 × 80% = 80
The value of $\frac{Y–X}{Y}$ is $\frac{100–80}{100}=\frac{20}{100}=\frac{1}{5}$
The value of $\frac{X}{X–Y}$ is $\frac{80}{80–100} = –\frac{80}{20}=–4$
Answer:
Hence, the correct answer is $\frac{1}{5}$ and $-4$.