RRB JE Aptitude Questions with Detailed Solutions

Given:
Time period T = 3 years.
Rate of interest R = 20% per annum.
Difference between CI and SI = ₹256.
Formula Used:
For 3 years, the difference between compound interest and simple interest is:
​$$D = P × (\frac{R}{100})^{2} × (3 + \frac{R}{100})$$​​
where P is principal, R is rate, and D is difference.
Solution:
Substitute the given values D = 256 and R = 20 into the formula:
​$$256 = P × (\frac{20}{100})^{2} × (3 + \frac{20}{100})$$​​

​$$256 = P × (\frac{1}{5})^{2} × (3 + \frac{1}{5})$$​​

​$$256 = P × \frac{1}{25} × \frac{16}{5}$$​​

​$$256 = P × \frac{16}{125}$$​​

​$$P = \frac{256 × 125}{16}$$​​

P = 16 × 125 = 2000.
Therefore, the principal lent is ₹2,000.
Final Answer
So the correct answer is (b)

Given:
Time taken by P and Q together to fill cistern = 24 hours.
Time taken by P alone to fill cistern = 72 hours.
Fraction of cistern to be filled by Q alone = $$\frac{3}{4}.$$​​
Formula Used:
If total work is completed in T hours, work done in 1 hour is  $$\frac{1}{T}.$$​​
Solution:
Part of cistern filled by P and Q in 1 hour = $$\frac{1}{24}.$$​​

Part of cistern filled by P alone in 1 hour = $$\frac{1}{72}.$$​

Part of cistern filled by Q alone in 1 hour = $$\frac{1}{24} - \frac{1}{72}$$​​

​$$\frac{1}{Q} = \frac{3 - 1}{72} = \frac{2}{72} = \frac{1}{36}$$​​
Thus, Q alone can fill the complete cistern in 36 hours.
Time taken by Q alone to fill $$\frac{3}{4}$$​ of the cistern:
Time = $$\frac{3}{4}$$​× 36 = 3 × 9 = 27
Therefore, Q alone takes 27 hours.
Final Answer
So the correct answer is (c)

Given:
The numbers are 4, 8, and 16.
Formula Used:
Harmonic Mean of n observations $$x_{1}, x_{2}, ..., x_{n}$$​ is given by:
HM = $$\frac{n}{\frac{1}{x_{1}} + \frac{1}{x_{2}} + ... + \frac{1}{x_{n}}}$$​​
Solution:
Here n = 3.
Calculating the sum of reciprocals:
​$$\frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{4 + 2 + 1}{16} = \frac{7}{16}$$​​
Substituting into the harmonic mean formula:
HM = $$\frac{3}{\frac{7}{16}}$$​​

HM = $$\frac{3 × 16}{7} = \frac{48}{7}$$​​

HM = 6.857142...
Rounding off to four decimal places gives 6.8571.
Final Answer
So the correct answer is (d)

Given:
Height of observer = 1.8 m.
Height of platform = 1.0 m.
Distance from platform to building = $$25\sqrt{3}$$​ m 
Angle of elevation = $$30^{\circ}$$​ 
Formula Used:
In a right-angled triangle:
​$$\tan(θ) = \frac{p}{b}$$​​
where p is the perpendicular (opposite side) and b is the base (adjacent side).
Total height of building H = $$h + \text{eye level height}.$$​​
Solution:

Total height of the observer's eye level above the ground = 1.0 + 1.8 = 2.8 m.
Let the height of the building above the observer's eye level be h m.
The horizontal base distance b =$$25\sqrt{3}$$​ m.
Using the tangent ratio for the angle of elevation $$θ = 30^{\circ}:$$​​
​$$\tan(30^{\circ}) = \frac{h}{25\sqrt{3}}$$​​
Since $$\tan(30^{\circ}) = \frac{1}{\sqrt{3}}:$$​​
​$$\frac{1}{\sqrt{3}} = \frac{h}{25\sqrt{3}}$$​​
h = 25 m.
Total height of the building H = 25 + 2.8 = 27.8 m.
Final Answer
So the correct answer is (b)

Given:
∆PQR is an acute-angled triangle.
O is the orthocenter and C is the circumcenter.
​$$∠QCR = 138^{\circ}.$$​​
PM is perpendicular to QR.
​$$∠PQR = 52^{\circ}.$$​​
Formula Used:
Angle subtended by an arc at the circumcenter is double the angle subtended at any point on the remaining circumference:
∠QCR = 2∠QPR
In a right-angled triangle, sum of acute angles is $$90^{\circ}.$$​​
Solution:

Using the circumcenter property at vertex P:
​$$∠QPR = \frac{138^{\circ}}{2} = 69^{\circ}$$​​
In right-angled triangle PMQ, since PM ⊥ QR, ∠PMQ = $$90^{\circ}.$$​​
Thus, the sum of ∠PQM and ∠QPM is $$90^{\circ}:$$​​
​$$∠QPM = 90^{\circ} - 52^{\circ} = 38^{\circ}$$​​
From the figure, PM divides vertex angle QPR such that:
∠QPR = ∠QPM + ∠RPM
​$$69^{\circ} = 38^{\circ} + ∠RPM$$​​
∠RPM = $$69^{\circ} - 38^{\circ} = 31^{\circ}.$$​​
Therefore, the measure of ∠RPM is $$31^{\circ}.$$​​
Final Answer
So the correct answer is (d)

Given:
Length is 34 m more than breadth.
Area of rectangular field = 5040 $$m^{2}.$$​​
Formula Used:
Area of rectangle = l × b
where l is length and b is breadth.
Solution:
Let breadth be b m.
Length l = b + 34 m.
According to the area formula:
b × (b + 34) = 5040
$$b^{2}$$​ + 34b - 5040 = 0
By testing the given options for breadth b:
If b = 56:
56 × (56 + 34) = 56 × 90 = 5040.
This satisfies the area condition exactly.
Therefore, the breadth of the field is 56 m.
Final Answer
So the correct answer is (b)

Given:
Initial number of people = X.
Contract duration = 18 days.
Number of sick people who did not work = 2.
Actual days taken by remaining people = 27 days.
Formula Used:
Total work = $$M_{1} × D_{1} = M_{2} × D_{2}$$​​
where M represents men and D represents days.
Solution:
Total work planned = X × 18 man-days.
Actual workers present = X - 2.
Total work done = (X - 2) × 27 man-days.
Equating the total work:
18 × X = 27 × (X - 2)
2 × X = 3 × (X - 2)
2X = 3X - 6
3X - 2X = 6
X = 6.
Therefore, the initial number of people X was 6.
Final Answer
So the correct answer is (a)

Given:
Investment of A = 3 × Investment of B.
Total annual profit = ₹55,204.
Formula Used:
Ratio of profit shares equals ratio of investments.
Solution:
Let investment of B be 1 unit.
Investment of A = 3 units.
Ratio of profit shares A : B = 3 : 1.
Sum of ratio parts = 3 + 1 = 4.
Calculating share of A:
Share of A = $$\frac{3}{4}$$​ × 55204
Share of A = 3 × 13801 = 41403
Therefore, the share of A is ₹41,403
Final Answer
So the correct answer is (a)

Given:
Total profit = ₹14,500.
Ratio of share A to B = 16 : 3.
Ratio of share B to C = 6 : 20.
Formula Used:
If the combined ratio of shares is A : B : C, then:
Share of C = $$\frac{C}{A + B + C} × P$$​​
where P is the total profit.
Solution:
First, equate the common ratio term B in both ratios.
We have A : B = 16 : 3 = 32 : 6
And B : C = 6 : 20
Thus, the combined ratio is:
A : B : C = 32 : 6 : 20
Simplifying by dividing by 2:
A : B : C = 16 : 3 : 10
Sum of the ratio parts = 16 + 3 + 10 = 29
Calculating the share of C:
Share of C = $$\frac{10}{29} × 14500$$  = 10 × 500 = 5000

Therefore, the share of C is ₹5,000.
Final Answer
So the correct answer is (b)

Given:
Initial goals data in 10 matches: 2, 3, 1, 4, 2, 5, 2, 3, 2, 4.
Goals scored in the 11th match = 3.
Formula Used:
Mode is the value that appears most frequently in a data set.
Solution:
Frequencies in the initial 10 matches:
Goal 1: 1 time
Goal 2: 4 times
Goal 3: 2 times
Goal 4: 2 times
Goal 5: 1 time
The highest frequency is 4, which corresponds to goal 2.
Previous mode = 2.
After the 11th match, one more goal of 3 is added.
Updated frequencies:
Goal 2: 4 times
Goal 3: 2 + 1 = 3 times
The maximum frequency remains 4 (for goal 2).
New mode = 2.
Calculating the difference between previous mode and new mode:
Difference = 2 - 2 = 0.
Final Answer
So the correct answer is (d)

Given:
Principal borrowed P = ₹4,40,000.
Rate of interest R = 6% per annum.
Time period T = 2 years.
Formula Used:
Simple Interest I = $$\frac{P × R × T}{100}$$​​
Solution:
Substitute the given values into the simple interest formula:
​$$I = \frac{440000 × 6 × 2}{100}$$​​
I = 4400 × 12 = 52800.
Therefore, the simple interest paid is ₹52,800.
Final Answer
So the correct answer is (a)

Given:
Number of identical cones formed = 18.
Base radius of each cone r = 3 cm.
Height of each cone h = 18 cm.
Formula Used:
Volume of sphere $$V_{s} = \frac{4}{3}πR^{3}$$​​
Volume of cone $$V_{c} = \frac{1}{3}πr^{2}h$$​​
where R is radius of sphere, r is radius of cone, and h is height of cone.
Solution:
Since the sphere is melted and recast into 18 cones, the volumes are equal:
​$$\frac{4}{3}πR^{3} = 18 × \frac{1}{3}πr^{2}h$$​​

​$$4R^{3} = 18 × r^{2} × h$$​​
Substitute r = 3 and h = 18:
​$$4R^{3} = 18 × 3^{2} × 18$$​​
​$$4R^{3} = 18 × 9 × 18 = 2916$$​​
​$$R^{3} = \frac{2916}{4} = 729$$​​
R = $$\sqrt[3]{729}$$​ = 9.
Therefore, the radius of the sphere is 9 cm.
Final Answer
So the correct answer is (a)

Given:
Circumference of the circular base of the hemisphere = 35.2 cm.
Formula Used:
Circumference of base C = 2πr
Volume of hemisphere V = $$\frac{2}{3}πr^{3}$$​​
where r is the base radius.
Solution:
Using the base circumference formula to find radius r:
​$$2 × \frac{22}{7} × r = 35.2$$​​
​$$\frac{44}{7} × r = 35.2$$​​
​$$r = \frac{35.2 × 7}{44}$$​​
r = 0.8 × 7 = 5.6
The radius is r = 5.6 cm.
Now, calculate the volume of the hemisphere:
​$$V = \frac{2}{3} × \frac{22}{7} × (5.6)^{3}$$​​

​$$V = \frac{44}{21} × 175.616$$​

​$$V = \frac{7727.104}{21} = 367.9573...$$​​

Rounding off to two decimal places, Volume = 367.96 $$cm^{3}.$$​​
Final Answer
So the correct answer is (d)

Given:
Total distance of first race = 2000 m.
Start given by A to B = 500 m.
Start given by A to C = 300 m.
Total distance of second race = 3400 m.
Formula Used:
Ratio of speeds equals ratio of distances covered in the same time.
Solution:
In the first race of 2000 m, when A covers 2000 m:
Distance covered by B = 2000 - 500 = 1500 m.
Distance covered by C = 2000 - 300 = 1700 m.
Since they finish at the same time, the ratio of speeds of B and C is:
​$$\frac{S_{B}}{S_{C}} = \frac{1500}{1700} = \frac{15}{17}$$​​
In the second race of 3400 m, let C cover the entire distance of 3400 m.
In the same time, the distance covered by B will be:
​$$D_{B} = 3400 × \frac{15}{17}$$​​
= 200 × 15 = 3000 m.
The start that C can give to B so they finish together = 3400 - 3000 = 400 m.
Final Answer
So the correct answer is (c)

Given:
Total race distance = 2500 m.
Start given by A to B = 200 m.
Speed of A = 25 m/s.
Speed of B = 23 m/s.
Formula Used:
Time T = $$\frac{D}{S}$$​​
where D is distance and S is speed.
Solution:
Distance A has to run to reach the finish line = 2500 m.
Time taken by A to complete the race:
​$$T_{A} = \frac{2500}{25} = 100$$​​
A takes 100 seconds.
Since B receives a start of 200 m, the distance B has to run = 2500 - 200 = 2300 m.
Time taken by B to complete the race:
​$$T_{B} = \frac{2300}{23} = 100$$​​
B also takes 100 seconds.
Since $$T_{A} = T_{B}$$​ = 100 seconds, both A and B reach the finish line at the exact same time.
Therefore, they finish together.
Final Answer
So the correct answer is (a)