TCS Smart Hiring Aptitude Questions & Answers 2025 | TCS Smart Hiring Aptitude Ability Questions & Answers | TCS Smart Hiring Placement Papers | TCS Smart Hiring Aptitude Sample Questions | TCS Freshers Smart Hiring Aptitude Questions
TCS Smart Hiring Aptitude Questions & Answers 2025: Aspirants who pass in the Test will qualify for the TCS Smart Hiring Aptitude process. The coding test will be conducted for the process to check the candidate’s capability. The coding questions will be asked in any language. In the below section, we have provided the TCS Smart Hiring Aptitude Questions with Solutions. With the clear Explanation, we have provided all the details here. Solve the Frequently Asked Aptitude Test Questions to help with your study. Before entering the coding test, aspirants can start their preparation by downloading or viewing the practice questions. Here we have provided the study material for the TCS Smart Hiring Aptitude Test.
Clock Aptitude Questions
- An accurate clock shows 8 a.m. Through how many degrees will the hour hand rotate when the clock shows 2 p.m.?
- The reflex angle between the hands of a clock at 10.25 is:
- A watch that gains 5 seconds in 3 minutes was set right at 7 a.m. In the afternoon of the same day, when the watch indicated a quarter past 4 o’clock, the true time is:
- How much does a watch lose per day, if its hands coincide every 64 minutes?
- At what time between 5.30 and 6 will the hands of a clock be at right angles?
- The angle between the minute hand and the hour hand of a clock when the time is 4.20, is:
- At what angle the hands of a clock are inclined at 15 minutes past 5?
- At 3:40, the hour hand and the minute hand of a clock form an angle of:
TCS Smart Hiring Aptitude Ability Questions
- Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
- A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
- A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
- A train 110 metres long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
- A 270-meter-long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
Aptitude Ability Questions & Answers For TCS Smart Hiring
Problems on Trains
Q) The length of the bridge, which a train 130 meters long and traveling at 45 km/hr can cross in 30 seconds, is:
- 200 M
- 225 M
- 245 M
- 250 M
| Speed = |  |
45 x | 5 |  m/sec |
= | |
25 | m/sec. |
| 18 | 2 |
Time = 30 sec.
Let the length of the bridge be x meters.
| Then, | 130 + x | = | 25 |
| 30 | 2 |
2(130 + x) = 750
x = 245 m.
Q) Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
- 1 : 3
- 3 : 2
- 3 : 4
- None of these
Let the speeds of the two trains be x m/sec and y m/sec respectively.
Then, the length of the first train = 27x metres,
and length of the second train = 17y meters.
 |
27x + 17y | = 23 |
| x+ y |
27x + 17y = 23x + 23y
4x = 6y
|
x | = | 3 | . |
| y | 2 |
Q) A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
- 120 Metres
- 180 Metres
- 324 Metres
- 150 Metres
| Speed = | |
60 x | 5 | m/sec |
= | |
50 | m/sec. |
| 18 | 3 |
Length of the train = (Speed x Time).
| Length of the train = |
|
50 | x 9 | m = 150 m. |
| 3 |
Height and Distance
Q) The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6 m away from the wall. The length of the ladder is:
- 2.3 m
- 4.6 m
- 7.8 m
- 9.2 m
Let AB be the wall and BC be the ladder.
Then, 
ACB = 60° and AC = 4.6 m.
| AC | = cos 60° = | 1 |
| BC | 2 |
| BC |
= 2 x AC |
| = (2 x 4.6) m | |
| = 9.2 m. |
Q) An observer 1.6 m tall is 203 away from a tower. The angle of elevation from his eye to the top of the tower is 30°. The height of the tower is:
- 21.6 m
- 23.2 m
- 24.72 m
- None of these
Let AB be the observer and CD be the tower.
Draw BE 
CD.
Then, CE = AB = 1.6 m,
BE = AC = 203 m.
| DE | = tan 30° = | 1 |
| BE | 3 |
| DE = |
203 | m = 20 m. |
| 3 |
CD = CE + DE = (1.6 + 20) m = 21.6 m.
Q) From a point P on a level ground, the angle of elevation of the top tower is 30°. If the tower is 100 m high, the distance of point P from the foot of the tower is:
- 149 m
- 156 m
- 173 m
- 200 m
Let AB be the tower.
Then, APB = 30° and AB = 100 m.
| AB | = tan 30° = | 1 |
| AP | 3 |
| AP |
= (AB x 3) m |
| = 1003 m | |
| = (100 x 1.73) m | |
| = 173 m. |
Q) The angle of elevation of the sun, when the length of the shadow of a tree 3 times the height of the tree, is:
- 30°
- 45°
- 60°
- 90°
Let AB be the tree and AC be its shadow.
Let ACB = 
.
| Then, | AC | = | 3 cot = 3 |
| AB |
= 30°
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