Percentage — BSSC Quantitative Aptitude

50 Questions • 40 Minutes • Quantitative Aptitude Mock Test in Hindi and English

Sample Questions from this Test

Question 1:

If $A$'s income is $25\%$ more than $B$'s income, by what percentage is $B$'s income less than $A$'s income?
A.$20\%$
B.$25\%$
C.$16\frac{2}{3}\%$
D.$30\%$

Formula: Required Percentage Less = $\left(\frac{R}{100 + R}\right) \times 100\%$ Step-by-Step Solution:

• Let $B$'s income be $\text{Rs. } 100$.

• Then $A$'s income = $100 + 25 = \text{Rs. } 125$.

• Difference = $125 - 100 = \text{Rs. } 25$.

• Percentage by which $B$ is less than $A$ = $\left(\frac{25}{125}\right) \times 100 = 20\%$. Exam Trick: Using fraction conversion: $25\% = \frac{1}{4}$. Since $A = 4 + 1 = 5$ parts and $B = 4$ parts, $B$ is $1$ part less out of $5$ parts. Required percentage = $\frac{1}{5} \times 100 = 20\%$.

Question 2:

A price of an item is increased by $15\%$ and then decreased by $15\%$. What is the net percentage change in the price?
A.No change
B.$2.25\%$ increase
C.$2.25\%$ decrease
D.$1.5\%$ decrease

Formula: When a value is increased by $x\%$ and then decreased by $x\%$, there is always a net decrease given by $\frac{x^2}{100}\%$. Step-by-Step Solution:

• Here $x = 15$.

• Net percentage change = $-\frac{15^2}{100}\% = -\frac{225}{100}\% = -2.25\%$.

• The negative sign indicates a decrease of $2.25\%$. Exam Trick: Net percentage formula $a + b + \frac{ab}{100}$: Here $a = +15$ and $b = -15$. Net change = $15 - 15 + \frac{(15)(-15)}{100} = -2.25\%$, which means a $2.25\%$ decrease.

Question 3:

If the price of sugar increases by $20\%$, by what percentage must a family reduce its consumption of sugar so that the expenditure on sugar increases by only $8\%$?
A.$10\%$
B.$12\%$
C.$15\%$
D.$8\%$

Formula: $\text{Expenditure} = \text{Price} \times \text{Consumption}$ Step-by-Step Solution:

• Let initial Price = $100$ and initial Consumption = $100$.

• Initial Expenditure = $100 \times 100 = 10000$.

• New Price = $120$.

• New Expenditure = $10000 + 8\% \text{ of } 10000 = 10800$.

• New Consumption = $\frac{10800}{120} = 90$.

• Percentage reduction in consumption = $\frac{100 - 90}{100} \times 100 = 10\%$. Exam Trick: Ratio method: Initial price $100 \rightarrow$ New price $120$. Target expenditure $108$. Reduction needed in price ratio = $120 \rightarrow 108$, which is a decrease of $12$ on $120$. Percentage reduction = $\frac{12}{120} \times 100 = 10\%$.

Question 4:

A person saves $20\%$ of his income. If his income increases by $15\%$ and his expenditure increases by $10\%$, then by what percentage do his savings increase?
A.$25\%$
B.$30\%$
C.$35\%$
D.$40\%$

Formula: $\text{Income} = \text{Expenditure} + \text{Savings}$ Step-by-Step Solution:

• Let initial Income = $100$.

• Initial Savings = $20$, initial Expenditure = $80$.

• New Income = $100 + 15\% = 115$.

• New Expenditure = $80 + 10\% \text{ of } 80 = 80 + 8 = 88$.

• New Savings = $115 - 88 = 27$.

• Increase in savings = $27 - 20 = 7$.

• Percentage increase in savings = $\frac{7}{20} \times 100 = 35\%$. Exam Trick: Alligation / Weighted Average Method: Savings change = $S\%$, Expenditure change = $10\%$, Total Income change = $15\%$. Ratio of Expenditure : Savings = $80 : 20 = 4 : 1$. Weighted change: $15 = \frac{4 \times 10 + 1 \times S}{4 + 1} \implies 15 \times 5 = 40 + S \implies S = 35\%$.

Question 5:

In an election between two candidates, $10\%$ of the voters did not cast their vote and $10\%$ of the votes cast were found invalid. The winning candidate got $54\%$ of the valid votes and won by a majority of $1620$ votes. Find the total number of voters enrolled in the voters list.
A.$20000$
B.$25000$
C.$22500$
D.$18000$

Formula: $\text{Winning Margin} = (\%\text{ Winner} - \%\text{ Loser}) \times \text{Valid Votes}$ Step-by-Step Solution:

• Let total enrolled voters = $x$.

• Votes cast = $0.90x$.

• Valid votes = $0.90 \times 0.90x = 0.81x$.

• Winner gets $54\%$ of valid votes, Loser gets $46\%$ of valid votes.

• Difference = $54\% - 46\% = 8\%$ of valid votes.

• $8\% \text{ of } 0.81x = 1620$.

• $\frac{8}{100} \times 0.81x = 1620 \implies 0.0648x = 1620$.

• $x = \frac{1620}{0.0648} = 25000$. Exam Trick: Chain rule: Total voters $x \times \frac{9}{10} \times \frac{9}{10} \times \frac{8}{100} = 1620$. $x \times \frac{648}{100000} = 1620 \implies x = \frac{1620 \times 100000}{648} = 25000$.

Question 6:

A candidate scoring $30\%$ marks fails by $15$ marks, while another candidate scoring $40\%$ marks gets $35$ marks more than the minimum passing marks. Find the maximum marks of the examination and the passing percentage.
A.$500$ marks, $33\%$
B.$500$ marks, $35\%$
C.$400$ marks, $35\%$
D.$600$ marks, $30\%$

Formula: $\text{Difference in Percentage} = \frac{\text{Sum of deficit and surplus}}{\text{Maximum Marks}} \times 100\%$ Step-by-Step Solution:

• Difference in marks percentage = $40\% - 30\% = 10\%$.

• Difference in actual marks = $15 + 35 = 50$ marks.

• $10\%$ of Maximum Marks = $50$.

• Maximum Marks = $\frac{50}{10} \times 100 = 500$.

• Passing marks = $30\% \text{ of } 500 + 15 = 150 + 15 = 165$.

• Passing percentage = $\frac{165}{500} \times 100 = 35\%$. Exam Trick: $10\% \rightarrow 50 \implies 100\% = 500$ (Max Marks). Since $50$ marks = $10\%$, $15$ marks = $\frac{15}{50} \times 10\% = 3\%$. Passing percentage = $30\% + 3\% = 35\%$.

Question 7:

The population of a town increases by $10\%$ in the first year, decreases by $20\%$ in the second year, and increases by $25\%$ in the third year. If the current population is $1,65,000$, what was the population $3$ years ago?
A.$1,50,000$
B.$1,60,000$
C.$1,40,000$
D.$1,25,000$

Formula: $P_{\text{present}} = P_{\text{initial}} \times \left(1 + \frac{r_1}{100}\right) \times \left(1 - \frac{r_2}{100}\right) \times \left(1 + \frac{r_3}{100}\right)$ Step-by-Step Solution:

• Let initial population = $P$.

• Growth factors: $10\% = \frac{11}{10}$, $-20\% = \frac{4}{5}$, $+25\% = \frac{5}{4}$.

• $P \times \frac{11}{10} \times \frac{4}{5} \times \frac{5}{4} = 1,65,000$.

• $P \times \frac{11}{10} = 1,65,000$.

• $P = 1,65,000 \times \frac{10}{11} = 15,000 \times 10 = 1,50,000$. Exam Trick: Cancel out common fractions: $\frac{4}{5} \times \frac{5}{4} = 1$. So the net multiplier over 3 years is simply $\frac{11}{10}$. $P \times \frac{11}{10} = 165000 \implies P = 150000$.

Question 8:

In a $60\text{ liter}$ solution of acid and water, the concentration of acid is $20\%$. How much pure acid (in liters) must be added to this solution to make the concentration of acid $40\%$?
A.$15\text{ liters}$
B.$20\text{ liters}$
C.$12\text{ liters}$
D.$18\text{ liters}$

Formula: Since water quantity remains constant: $\text{Initial Solution} \times \%\text{ Water}_{1} = \text{New Solution} \times \%\text{ Water}_{2}$ Step-by-Step Solution:

• Initial Water percentage = $100\% - 20\% = 80\%$.

• Initial quantity of water = $60 \times 80\% = 48\text{ liters}$.

• In the new solution, acid concentration = $40\% \implies$ Water concentration = $60\%$.

• Since water quantity does not change, $60\% \text{ of New Solution} = 48\text{ liters}$.

• New Solution = $\frac{48}{0.60} = 80\text{ liters}$.

• Acid added = New Solution - Initial Solution = $80 - 60 = 20\text{ liters}$. Exam Trick: Equating non-changing component (Water): $60 \times 80\% = N \times 60\% \implies N = 80\text{ liters}$. Added Acid = $80 - 60 = 20\text{ liters}$.

Question 9:

In an examination, $65\%$ of candidates passed in Mathematics, $55\%$ passed in English, and $20\%$ failed in both subjects. If $240$ candidates passed in both subjects, what is the total number of candidates who appeared in the examination?
A.$500$
B.$600$
C.$750$
D.$800$

Formula: $\text{Total Percentage} = n(A) + n(B) - n(A \cap B) + \text{Failed in Both}$ Step-by-Step Solution:

• Percentage of students failing in Math = $100\% - 65\% = 35\%$.

• Percentage of students failing in English = $100\% - 55\% = 45\%$.

• Failed in both = $20\%$.

• Percentage failing in at least one subject = $35\% + 45\% - 20\% = 60\%$.

• Percentage passing in both subjects = $100\% - 60\% = 40\%$.

• Given $40\% \text{ of Total} = 240 \implies \text{Total} = \frac{240}{40} \times 100 = 600$. Exam Trick: Passing in at least one subject = $100\% - 20\% = 80\%$. By Venn diagram formula: $65\% + 55\% - P(\text{Both}) = 80\% \implies P(\text{Both}) = 120\% - 80\% = 40\%$. $40\% \rightarrow 240 \implies 100\% \rightarrow 600$.

Question 10:

If the rate of income tax is increased by $19\%$, the net income is reduced by $1\%$. What is the initial rate of income tax?
A.$4\%$
B.$5\%$
C.$6\%$
D.$7.5\%$

Formula: $\text{Increase in Income Tax} = \text{Decrease in Net Income}$ $\text{Gross Income} = \text{Income Tax} + \text{Net Income}$ Step-by-Step Solution:

• Change in Tax = $19\% \text{ of Tax}$.

• Change in Net Income = $1\% \text{ of Net Income}$.

• $19\% \times \text{Tax} = 1\% \times \text{Net Income} \implies \frac{\text{Tax}}{\text{Net Income}} = \frac{1}{19}$.

• Gross Income = $\text{Tax} + \text{Net Income} = 1 + 19 = 20$.

• Initial Tax Rate = $\left(\frac{\text{Tax}}{\text{Gross Income}}\right) \times 100 = \frac{1}{20} \times 100 = 5\%$. Exam Trick: Direct Tax Rate Formula = $\frac{\text{Net Income Change}}{\text{Tax Change} + \text{Net Income Change}} \times 100\%$. $\text{Tax Rate} = \frac{1}{19 + 1} \times 100 = \frac{1}{20} \times 100 = 5\%$.

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