RRB JE Percentage Questions with Detailed Solutions

Given
Original Price (P) = 56000
Depreciation rate (R) = 5% per year
Time (T) = 3 years
Formula Used
Amount (A) = P ×  $$(1 - \frac{R}{100})^T$$​​
Solution
​$$A = 56000 × (1 - \frac{5}{100})^3 \\\ \\A = 56000 × (\frac{95}{100})^3 = 56000 × (\frac{19}{20})^3 \\\ \\A = 56000 × \frac{6859}{8000} \\\ \\A = 7 × 6859 \\\ \\A = 48013 \\\ \\$$​​
Final Answer
So the correct answer is (b)

Given:
Initial percentage reduction = 40%
Subsequent percentage raise = 24%
Formula Used:
Net Change% = $$x + y + \frac{x × y}{100}$$​​
Solution:
Let the initial salary of Chetan be 100 units.
Salary after a 40% reduction:
New Salary = 100 - 40 = 60 units.
Next, this salary is raised by 24%:
Final Salary = 60 + $$(60 × \frac{24}{100})$$​​
Final Salary = 60 + 14.4 = 74.4 units.
Now, calculate the overall reduction from the initial salary:
Reduction = 100 - 74.4 = 25.6 units.
Since the base is 100, the percentage reduction is exactly 25.6%.
Alternatively, using the net percentage change formula where x = -40 and y = +24:
Net Change = $$-40 + 24 + \frac{(-40) × 24}{100}$$​​
Net Change = -16 - 9.6 = -25.6%
The negative sign indicates a decrease of 25.6%.
Final Answer
So the correct answer is (c)

Given:
Marks scored by Girish = 664
Difference from 84% of maximum marks = 8 marks short
Marks scored by his friend = 472
Formula Used:
Percentage = $$\frac{Marks\ Obtained}{Maximum\ Marks}$$​ × 100
Solution:
Let the maximum marks of the examination be M.
According to the question, 84% of the maximum marks is 8 marks more than Girish's score.
0.84 × M = 664 + 8
0.84 × M = 672
M = $$\frac{672}{0.84}$$​= 800
The maximum marks for the examination is 800.
Now, calculate the percentage of marks scored by his friend:
Percentage =  $$\frac{472}{800}$$​ × 100 = 59%
Final Answer
So the correct answer is (a)

Given:
Present population (P) = 10 crores
Expected population after 3 years (A) = 17.28 crores
Time period (t) = 3 years
Formula Used:
A = $$P(1 + \frac{R}{100})^t$$​​
Solution:
Substitute the given values into the compound growth formula:
​$$17.28 = 10\left(1 + \frac{R}{100}\right)^3$$​​

​$$\frac{17.28}{10} = \left(1 + \frac{R}{100}\right)^3$$​​

​$$1.728 = \left(1 + \frac{R}{100}\right)^3$$​​

​$$\frac{1728}{1000} = \left(1 + \frac{R}{100}\right)^3$$​​

​$$\sqrt[3]{\frac{1728}{1000}} = 1 + \frac{R}{100}$$​​

​$$\frac{12}{10} = 1 + \frac{R}{100}$$​​

​$$1.2 = 1 + \frac{R}{100}$$​​

​$$\frac{R}{100} = 1.2 - 1 = 0.2$$​​
R = 0.2 × 100 = 20%
Final Answer
So the correct answer is (b)

Given
Successive changes: decrease of 35%, decrease of 10%, decrease of 50%, and an increase of 40%.
Formula Used
Final Value = Initial Value $$\times (1 \pm \frac{R_1}{100}) \times (1 \pm \frac{R_2}{100}) \dots$$​​
Solution
Let the initial price of the fuel be 100.
Price after first month =  $$100 \times (1 - 0.35) = 100 \times 0.65 = 65$$​​
Price after second month = $$65 \times (1 - 0.10) = 65 \times 0.90 = 58.5$$​​
Price after third month = $$58.5 \times (1 - 0.50) = 58.5 \times 0.50 = 29.25$$​​
Price after fourth month =  $$29.25 \times (1 + 0.40) = 29.25 \times 1.40 = 40.95$$​​
Total percentage decrease from initial price = $$100 - 40.95 = 59.05\%$$​​
Final Answer
So the correct answer is (c)

Given
English marks = 9/10
Life Sciences marks = 15/16
Ratio of % marks in English to Mathematics = 15:16
Total Mathematics marks = 20
Formula Used
Percentage = $$\left(\frac{\text{Obtained Marks}}{\text{Total Marks}}\right)\times 100$$​​
Solution
English percentage = $$\frac{9}{10} \times 100 = 90\%$$​​
Life Sciences percentage =$$\frac{15}{16} \times 100 = 93.75\%$$​​
Let Mathematics percentage be M.
Ratio of English to Mathematics percentage = 15:16
​$$\frac{90}{M} = \frac{15}{16} \\\ \\M = \frac{90 \times 16}{15} = 96\%$$​​
Difference between Mathematics and Life Sciences percentage = $$96\% - 93.75\% = 2.25\%$$​​

Final Answer
So the correct answer is (c)

Given:

A's salary is 31% more than B's salary.

We need to find the percentage by which B's salary is less than A's.

Formula Used:

​$$\text{Percentage decrease} = \left( \frac{\text{A's salary} - \text{B's salary}}{\text{A's salary}} \right) \times 100$$​

Solution:

Let B's salary be 100 units.
Then, A's salary = 100 + 31% of 100 = 131 units.

​$$\text{Percentage decrease} = \left( \frac{131 - 100}{131} \right) \times 100$$​

​$$= \left( \frac{31}{131} \right) \times 100 = 23.66\%$$​​

B's salary is 23.66% less than A's salary.

Given:

The number increases by 50% to give 2880.

Solution:

x + 50%  of  x = 2880

x + $$\frac{50}{100} \times$$​ x = 2880

x + $$\frac{x}{2}$$​ = 2880

​$$\frac{3x}{2}$$​ = 2880

3x = 5760

x = $$\frac{5760}{3}$$​ = 1920

Given:

A's salary is 72% more than B's salary.

Concept Used:

Percentage Decrease = $$\left( \frac{\text{Difference}}{\text{Original Value}} \right) \times$$​ 100

Percentage Decrease = $$\left( \frac{A - B}{A} \right) \times$$​ 100

Solution: 

Let B's salary be B.
Then A's salary = B + 72% of B = 1.72B.

$$= \left( \frac{1.72B - B}{1.72B} \right) \times 100\\ \ \\= \left( \frac{0.72B}{1.72B} \right) \times 100\\ \ \\= \left( \frac{0.72}{1.72} \right) \times 100\\ \ \\= 41.86\%$$​

Given:

Cost of a washing machine = 25% less than the cost of a TV.

Washing machine cost increases by 82%.

TV cost decreases by 16%.

Total items: 8 washing machines and 9 TVs.

Concept Used:
Percentage increase or decrease is applied as:

New Price = Original Price $$\times \left(1 + \frac{\text{Percentage Change}}{100}\right)$$​​

Solution:

Let the cost of a TV be ₹100.
Then, the cost of a washing machine = ₹75 (since 25% less than ₹100).

New cost of a washing machine: $$75 \times \left(1 + \frac{82}{100}\right) = 75 \times 1.82 = 136.5$$​​

New cost of a TV: $$100 \times \left(1 - \frac{16}{100}\right) = 100 \times 0.84 = 84$$​​

Old total cost: (8 × 75) + (9 × 100) = 600 + 900 = 1500

New total cost: (8 × 136.5) + (9 × 84) = 1092 + 756 = 1848

Change in cost = 1848 – 1500 = 348

Percentage change = $$\frac{348}{1500}\times 100 = 23.2\%$$​​

Given:

Let the original number be x
The number is increased by 50%, giving 3120.

Concept Used:

If a number is increased by P%, the new value becomes:

New Number = x + $$\frac{P}{100} \times$$​ x

New Number = x $$\times \left(1 + \frac{P}{100} \right)$$​

Solution:

x $$\times \left(1 + \frac{50}{100} \right)$$​ = 3120

x $$\times \left(1 + 0.5 \right)$$​ = 3120

x $$\times$$​ 1.5 = 3120

x = $$\frac{3120}{1.5}$$​

x = 2080

Given:

Population in 1974 = 222900

Increase per decade = 50%

Two decades: 1974 to 1994

Formula Used:

Final Population = Initial Population $$\times \left(1 + \frac{\text{Rate of Increase}}{100} \right)^n$$​

Solution:

Initial Population = 222900

Rate of Increase = 50%

Number of Decades = 2

P = 222900 $$\times \left(1 + \frac{50}{100}\right)^2$$​

P = 222900 $$\times \left(1.5\right)^2$$​

P = 222900$$\times$$​ 2.25

P = 501525

Given:
Initial population in 1927 = 232300
Increase per decade = 30%
Number of decades = 2
Formula Used:
Population Growth Formula:
​$$P_n = P_0 \times \left(1 + \frac{r}{100}\right)^n$$​​
where,
$$P_n$$​ = Population after n decades
$$P_0$$​ = Initial population
r = Growth rate per decade
n = Number of decades
Solution:
Population after the first decade (1937):
​$$P_{1937} = 232300 \times \left(1 + \frac{30}{100}\right)\\P_{1937} = 232300 \times 1.3\\P_{1937} = 301990\\$$​​
Population after the second decade (1947):
​$$P_{1947} = 301990 \times 1.3\\P_{1947} = 392587\\$$​​

Thus, the population of the town in 1947 was 392587. 

Alternate Solution:

$$\begin{array}{c|c}\textbf{Initial Population} & \textbf{Final Population} \\\hline10 & 13 \\10 & 13 \\\hline100 & 169 \\\end{array}\\\ \\100 \text{ unit} = 232300 \\1 \text{ unit} = \frac{232300}{100} = 2323 \\169 \text{ unit} = 2323 \times 169 = 392587$$​

Given:
A number when increased by 50 %', gives 2790
Solution:
Let the number = x
According to the question

x + $$x \times \frac{50}{100}$$  = 2790

$$\frac{3x}{2} = 2790$$

x = 1860

Given:

Percentage of salary spent on rent = 76%

Monthly expenditure on conveyance = ₹813

Monthly expenditure on grocery = ₹2508

Monthly savings = ₹687

Concept Used:

The total salary is the sum of all expenditures and savings.

The rent is 76% of the salary, while the remaining expenses (conveyance, grocery, and savings) make up the remaining 24%.

Solution:

Let the monthly salary be S.

Rent = 0.76S

Other expenses = Conveyance + Grocery + Savings = ₹813 + ₹2508 + ₹687 = ₹4008

The total salary is the sum of rent and other expenses:

S = 0.76S + 4008

S = 0.76S + 4008

S−0.76S = 4008

0.24S = 4008

S = $$\frac{4008}{0.24}$$​ = 16700

Thus, the man's monthly salary is ₹16,700.