RRB JE Maths Questions with Detailed Solutions

Given
Speed of boat in still water (v) = 4.5 km/h
Time taken to row upstream = 2 × Time taken to row downstream
Formula Used
Distance = Speed × Time
Upstream speed = v - u
Downstream speed = v + u
Solution
Let the speed of the stream be u km/h.
Let the distance covered be D.
Time taken downstream =$$\frac{D}{4.5 + u}$$​​

Time taken upstream = $$\frac{D}{4.5 - u}$$​​
According to the given condition:
Time upstream = 2 × Time downstream
​$$\frac{D}{4.5 - u} = 2 × \frac{D}{4.5 + u}$$​​
1(4.5 + u) = 2(4.5 - u)
4.5 + u = 9 - 2u
3u = 4.5
u = 1.5 km/h
Final Answer
So the correct answer is (c)

Given
Distance 1 = 425 km, Speed 1 = 5 km/hr
Distance 2 = 946 km, Speed 2 = 43 km/hr
Distance 3 = 340 km, Speed 3 = 34 km/hr
Formula Used
Time = $$\frac{\text{Distance}}{\text{Speed}}$$​​
Average Speed = $$\frac{\text{Total Distance}}{\text{Total Time}}$$​​
Solution
Total Distance = 425 + 946 + 340 = 1711 km
Time 1 = $$\frac{425}{5} = 85$$​ hours

Time 2 =$$\frac{946}{43} = 22$$​ hours

Time 3 =$$\frac{340}{34} = 10$$​ hours
Total Time = 85 + 22 + 10 = 117 hours
Average Speed = $$\frac{1711}{117}$$​ km/hr

Average Speed = $$14 \frac{73}{117}$$​km/hr
Final Answer
So the correct answer is (c)

Given
Time period (T) = 2 years
Rate of interest (R) = 10% per annum
Difference between Compound Interest (CI) and Simple Interest (SI) = ₹97
Formula Used
Difference between CI and SI for 2 years = $$P(\frac{R}{100})^2$$​​
Solution
Substitute the given values into the formula:
​$$97 = P × (\frac{10}{100})^2 \\\ \\97 = P × (\frac{1}{10})^2\\\ \\97 = P × \frac{1}{100}\\\ \\P = 97 × 100 = 9700$$​​
The sum invested in each scheme is ₹9700.
Final Answer
So the correct answer is (c)

Given
Time taken by A = 9 days
Time taken by B = 12 days
They work on alternate days starting with B.
Formula Used
Total Work = LCM of times taken
Efficiency = $$\frac{\text{Total Work}}{\text{Time Taken}}$$​​
Solution
Let the total work be the LCM of 9 and 12, which is 36 units.
Efficiency of A =$$\frac{36}{9} = 4$$​ units/day

Efficiency of B = $$\frac{36}{12} = 3$$​ units/day
In a 2-day cycle (B on day 1, A on day 2):
Work done = 3 + 4 = 7 units
In 5 such cycles (10 days):
Work done = 5 × 7 = 35 units
Remaining work = 36 - 35 = 1 unit
On the 11th day, it is B's turn.
Time taken by B to finish the remaining 1 unit =$$\frac{1}{3}$$​ day
Total time taken =$$10 + \frac{1}{3} = 10 \frac{1}{3}$$​ days
Final Answer
So the correct answer is (c)

Given
Ratio of ages of Amit and Bhanu 5 years ago = 2 : 3
Ratio of ages of Amit and Bhanu 15 years from now = 5 : 6
Solution
Let the present age of Amit be A and Bhanu be B.
Five years ago:
​$$\frac{A - 5}{B - 5} = \frac{2}{3}$$​​
3A - 15 = 2B - 10
3A - 2B = 5 (Equation 1)
Fifteen years from now:
​$$\frac{A + 15}{B + 15} = \frac{5}{6}$$​​
6A + 90 = 5B + 75
5B - 6A = 15 (Equation 2)
Multiply Equation 1 by 2:
6A - 4B = 10 (Equation 3)
Add Equation 2 and Equation 3:
(5B - 6A) + (6A - 4B) = 15 + 10
B = 25 years
Substitute B = 25 in Equation 1:
3A - 2(25) = 5
3A - 50 = 5
3A = 55
$$A = \frac{55}{3} = 18 \frac{1}{3}$$years =  18 years and 4 months old.
Sum of present ages = A + B = 18 years 4 months + 25 years = 43 years 4 months.
Final Answer
So the correct answer is (a)

Given
Data set: 13, 17, 23, 26, 31, 28, 15, 34, 17, 26
Number of terms (N) = 10
Formula Used
Mean (μ) = \frac{Sum of terms}{N}
Population Standard Deviation (σ) = $$\sqrt{\frac{Σ(x - μ)^2}{N}}$$​​
Solution
Sum of terms = 13 + 17 + 23 + 26 + 31 + 28 + 15 + 34 + 17 + 26 = 230
Mean (μ) = $$\frac{230}{10} = 23$$​​
Deviations from mean (x - μ): -10, -6, 0, 3, 8, 5, -8, 11, -6, 3
Squares of deviations $$(x - μ)^2$$​: 100, 36, 0, 9, 64, 25, 64, 121, 36, 9
Sum of squared deviations = 100 + 36 + 0 + 9 + 64 + 25 + 64 + 121 + 36 + 9 = 464
Variance = $$\frac{464}{10} = 46.4$$​​
Standard Deviation = √46.4 ≈ 6.8117
Rounded to two decimal places, the standard deviation is 6.81.
Final Answer
So the correct answer is (c)

Given
Assumed mean (A) = 55
Observed value (x) = 40
Formula Used
Deviation (d) = x - A
Solution
Deviation (d) = 40 - 55
Deviation (d) = -15
Final Answer
So the correct answer is (d)

Given
Proportion: 14.8 : x :: x : 3.7
Condition: x > 0
Formula Used
Product of extremes = Product of means
If a : b :: c : d, then a × d = b × c
Solution
​$$x × x = 14.8 × 3.7 \\x^2 = \frac{148}{10} × \frac{37}{10}\\\ \\x^2 = \frac{4 × 37 × 37}{100}\\\ \\x = \frac{2 × 37}{10}\\\ \\x = \frac{74}{10} = 7.4\\\ \\$$​​
Final Answer
So the correct answer is (a)

Given
Original numbers: 13, 19, 16, 23
Proportion equation: (13 + x) : (19 + x) :: (16 + x) : (23 + x)
Second Proportion: 5x : y :: y : (8x - 4), where y > 0
Formula Used
If a, b, c, d are in proportion, a × d = b × c
Solution
(13 + x) × (23 + x) = (19 + x) × (16 + x)
299 + 36x +$$x^2 = 304 + 35x + x^2$$​​
36x - 35x = 304 - 299
x = 5
Substitute x = 5 in the second proportion:
5(5) : y :: y : (8(5) - 4)
25 : y :: y : 36
​$$y^2 = 25 × 36 \\y = √(900) = 30 (since y > 0)$$​​
Final Answer
So the correct answer is (d)

Given
Cost of cheaper sugar (C) = ₹34/kg
Cost of dearer sugar (D) = ₹58/kg
Selling Price of mixture = ₹52.80/kg
Profit percentage = 32%
Formula Used
Cost Price = $$\frac{\text{Selling Price}}{1 + \frac{\text{Profit}}{100}}$$​​
Rule of Alligation: Ratio = $$\frac{\text{D - Mean Cost}}{\text{Mean Cost - C}}$$​​
Solution
Mean Cost Price of the mixture = $$\frac{52.80}{1.32} = 40 ₹/kg$$​​
Quantity of Cheaper : Quantity of Dearer = (58 - 40) : (40 - 34)
Ratio = 18 : 6
Ratio = 3 : 1
Final Answer
So the correct answer is (c)

Given
Total distance = 36 km
Time taken = 2 hours 5 minutes
Distance covered = $$\frac{6}{7}$$​ of the remaining distance
Formula Used
Speed = $$\frac{\text{Distance}}{\text{Time}}$$​​
Solution
Let the remaining distance be x.
Distance covered = $$\frac{6}{7}x$$​​
Total distance = Distance covered + Remaining distance
$$36 = \frac{6}{7}x + x \\\ \\36 = \frac{13x}{7} \\\ \\x = \frac{36 × 7}{13} = \frac{252}{13} km \\\ \\ \text{(Remaining distance)}$$
Distance covered = $$36 - \frac{252}{13} = \frac{468 - 252}{13} = \frac{216}{13} km$$​​

Time = $$2 + \frac{5}{60} = 2 + \frac{1}{12} = \frac{25}{12}$$​hours

Speed = $$\frac{\frac{216}{13}}{\frac{25}{12}} = \frac{216}{13} × \frac{12}{25} = \frac{2592}{325} \approx 7.97$$​km/h
The approximate speed is 8 km/h.
Final Answer
So the correct answer is (d)

Formula Used
​$$\frac{M_1 × D_1 × H_1}{W_1} = \frac{M_2 × D_2 × H_2}{W_2}$$​​
Solution
Substitute the values into the formula:
​$$\frac{25 × 5 × 10}{2} = \frac{15 × D_2 × 10}{18} \\\ \\\frac{1250}{2} = \frac{150 × D_2}{18}\\\ \\625 = \frac{25 × D_2}{3}\\\ \\D_2 = \frac{625 × 3}{25}\\\ \\D_2 = 25 × 3 = 75 days\\\ \\$$​​
Final Answer
So the correct answer is (d)

Given
Total amount = ₹22,576
$$\frac{5}{16}$$​ of Rashmi's amount = $$\frac{3}{7}$$​ of Varsha's amount
Formula Used
Solution
Let Rashmi's amount be R and Varsha's amount be V.
​$$\frac{5}{16} R = \frac{3}{7} V \\\ \\\frac{R}{V} = \frac{3}{7} × \frac{16}{5} = \frac{48}{35}$$​​
The ratio of Rashmi's amount to Varsha's amount is 48 : 35.
Total parts = 48 + 35 = 83 parts.
Total value = 22576
Value of 1 part = $$\frac{22576}{83} = 272$$​​
Rashmi's amount = 48 parts
Rashmi's amount = 48 × 272 = 13056

Final Answer
So the correct answer is (b)

Given
​$$x = \sqrt[3]{512} \\y = \sqrt[6]{64}\\z = 6^3 - 2 × 10^2 \\$$​​
Solution
​$$x = \sqrt[3]{512} = 8 \\y = \sqrt[6]{64} = 2\\ z = 216 - 2(100) = 216 - 200 = 16$$​​
Now, substitute x, y, and z into the expression $$\sqrt[3]{10(x^2 + y^3 + z + 12)}:$$​​
​$$= \sqrt[3]{10(8^2 + 2^3 + 16 + 12)} \\= \sqrt[3]{10(64 + 8 + 16 + 12)} \\= \sqrt[3]{10(100)} \\= \sqrt[3]{1000} \\= 10$$​​
Final Answer
So the correct answer is (b)

Given
Principal (P) = ₹4,40,000
Rate (R) = 6% per annum
Time (T) = 4 years
Formula Used
Simple Interest = $$\frac{P × R × T}{100}$$​​
Solution
Substitute the values into the formula:
Simple Interest = $$\frac{440000 × 6 × 4}{100}$$​​
Simple Interest = 4400 × 24
Simple Interest = 105600
Final Answer
So the correct answer is (b)