| 1. Train X starts from A to B at 9:00 AM on Monday and takes 42 hours to reach B. At the same time, train Y starts from B to A and takes 56 hours to reach A. At what time will they cross each other if they travel at constant speeds. | Easy |
| A. 9:00 PM Monday B. 9:00 PM Tuesday C. 9:00 AM Tuesday D. None of these |
View Answer
Answer: Option C
Explanation:
Let the distance between A and B be d kms, and let the trains cross each other at point P, t hours after 9:00 am Monday. Then, distance travelled by train X will be AP, and that travelled by train Y will be BP.
Total distance covered = AP+PB = (d/42)t + (d/56)t = d
-> t/42+t/56 = 1
-> t = 24 hours
-> They cross each other at 9:00 AM on Tuesday.
| 2. Walking at 4/5th of his normal speed, Alex takes 14 minutes more than his normal time to reach a particular destination. What is the usual time taken by him to reach that destination? | Easy |
| A. 56 minutes B. 48 minutes C. 64 minutes D. None of these |
View Answer
Answer: Option A
Explanation:
If the distance is constant, speed is inversely proportional to time.
Thus, if Alex walks at 4/5 of his normal speed, he will take 5/4 of his normal time.
If his normal time to reach the destination is t minutes, then
(5/4)t – t = 14
-> t = 56 minutes
Thus, if Alex walks at 4/5 of his normal speed, he will take 5/4 of his normal time.
If his normal time to reach the destination is t minutes, then
(5/4)t – t = 14
-> t = 56 minutes
3.Alex reaches city Q from city P in 4 hours, driving at 45 km/hr for the first 2 hours and 55 km/hr for the next 2 hours. Brad follows the same route, from Q to P, and drives at three speeds: 40, 50 and 60 km/hr, covering an equal distance in each speed segment. The two cars are similar w.r.t petrol consumption characteristics, as given below:
How much more/less petrol is consumed by Brad for the journey compared to Alex? |
Difficult | ||||||||||||
| A. 0.67 litres less B. 0.37 litres more C. 0.17 litres more D. None of these |
View Answer
Answer: Option B
Explanation:
Distance covered by Alex @ 45km/hr = 45*2 = 90 km
Distance covered by Alex @ 555km/hr = 55*2 = 110 km
Petrol consumed by Alex = Distance/Milage = 90/16 + 110/20 = 11.125 litres
Distance travelled by Brad = 90+110 = 200 km
Brad travels this distance in 3 equal segments at different speeds.
Petrol consumed by Brad= Distance/Milage
= (200/3)/14 + (200/3)/18 + (200/3)/22 = 11.496 litres
-> Brad consumes (11.496-11.125 = 0.371 litres) more than Alex.
Distance covered by Alex @ 555km/hr = 55*2 = 110 km
Petrol consumed by Alex = Distance/Milage = 90/16 + 110/20 = 11.125 litres
Distance travelled by Brad = 90+110 = 200 km
Brad travels this distance in 3 equal segments at different speeds.
Petrol consumed by Brad= Distance/Milage
= (200/3)/14 + (200/3)/18 + (200/3)/22 = 11.496 litres
-> Brad consumes (11.496-11.125 = 0.371 litres) more than Alex.
| 4. Alex travelled by car from P to Q through R. The distance between P and R is 4/5 times the distance from R to Q. The average speed from P to R was 1/3 times more than the speed from R to Q. If the overall average speed for the entire journey was 81 km/hr, what was Alex’s average speed from R to Q? | Difficult |
| A. 48 km/hr B. 24 km/hr C. 72 km/hr D. None of these |
View Answer
Answer: Option C
Explanation:
Total distance travelled by Alex = PQ = PR+RQ.
Let RQ = 5x km -> PR = 4x kms
Let Alex’s average speed from R to Q be 3s km/hr
-> Average speed from P to R = 3s+(1/3)3s = 4s km/hr.
Time taken to travel from P to R = 4x/4s = x/s hrs.
Time taken to travel from R to Q = 5x/3s hrs.
-> Total time to travel from P to Q = x/s + (5/3)x/s = 8x/3s hrs. —– ❶
Also, time taken to travel from P to Q = total distance/ average speed = 9x/81—–❷
From ❶&❷, 8x/3s = 9x/81
-> s = 24 km/hr
-> 3s = 72 km/hr
Let RQ = 5x km -> PR = 4x kms
Let Alex’s average speed from R to Q be 3s km/hr
-> Average speed from P to R = 3s+(1/3)3s = 4s km/hr.
Time taken to travel from P to R = 4x/4s = x/s hrs.
Time taken to travel from R to Q = 5x/3s hrs.
-> Total time to travel from P to Q = x/s + (5/3)x/s = 8x/3s hrs. —– ❶
Also, time taken to travel from P to Q = total distance/ average speed = 9x/81—–❷
From ❶&❷, 8x/3s = 9x/81
-> s = 24 km/hr
-> 3s = 72 km/hr
| 5. Brad bicycled from P to Q at a speed of 15 km/hr and reached Q 10 mins late. Had he travelled at 20 km/hr, he would have reached Q 30 mins early. Find the distance between P and Q. | Easy |
| A. 40 km B. 20 km C. 30 km D. None of these |
View Answer
Answer: Option A
Explanation:
Let the distance between P and Q be d km.
-> d/15 – d/20 = 1/6 + 1/2 (converting 10 mins and 30 mins into 1/6 hr and 1/2 hr respectively).
-> d = 40 km
| 6. A vehicle travels at a 30 km/hr speed for 10 hours. It then increases its speed by 10 km/hr every hour for the next 5 hours and finally comes to rest. Find the average speed for the entire journey. | Easy |
| A. 50 km/hr B. 40 km/hr C. 60 km/hr D. None of these |
View Answer
Answer: Option B
Explanation:
Distance travelled at 30 km/hr for 10 hours = 300 km
Average Speed = (total distance travelled)/(total time taken) = (300+40+50+60+70+80)/(10+5) = 40
| 7. A train approaches a tunnel PQ from P. Inside the tunnel is a cat located at a point that is 1/7th of the distance PQ measured from P. When the train whistles the cat runs. If the cat runs towards P, then it just escapes the train entering the tunnel. If it runs towards Q, then the train just catches it at Q. What is the ratio of the speed of the train to that of the cat? | Difficult |
| A. 7:6 B. 6:7 C. 5:4 D. None of these |
View Answer
Answer: Option A
Explanation:
Let the length of the tunnel be 7x kms, and the cat be at point C.
Case I: The cat runs towards P
During the time the train enters the tunnel at point P, the cat travels a distance CP. Since CP is 1/7th of PQ, it is equal to x kms.
Case II: The cat runs towards Q
When the train enters the tunnel at point P, the cat would have already covered a distance of x kms towards Q (from case I above). Thus, during the time the cat covers 6x kms (CQ) to reach Q, the train would have to cover the entire distance PQ, which is 7x kms.
Now, if time is constant, distance is directly proportional to speed. If St and Sc are the respective speeds of the train and the cat, then ratio of speeds = St/Sc = 7x/6x = 7/6
| 8. One step of A is one and a half times as long as one step of B. If B takes twice as many steps per unit time as A does, then find the ratio of the speeds of A & B. | Medium |
| A. 3:4 B. 4:3 C. 2:3 D. None of these |
View Answer
Answer: Option A
Explanation:
| A | B | |
| Distance/Step | 3/2 | 1 |
| No. of steps/Unit time | 1 | 2 |
| Distance covered/Unit time | (3/2) *1 = 3/2 | 1*2 = 2 |
| Ratio of speeds | 3/2 : 2 = 3:4 | |
| 9. A and B leave point P for point Q at the same time. After reaching Q, both return to P immediately. A travels from P to Q @ 24 km/hr and returns @28 km/hr, while B travels @26 km/hr both ways. Who gets back to point P first? | Easy |
| A. A B. B C. Both reach at the same time D. Can’t be determined |
View Answer
Answer: Option B
Explanation:
Let distance between P and Q be d kms.
Time taken by A for the round trip = (d/24)+(d/28) = 13d/168
Time taken by B for the round trip = (d/26)+(d/26) = d/13
Now, d/13 < 13d/168
-> B reaches point P before A.
Time taken by A for the round trip = (d/24)+(d/28) = 13d/168
Time taken by B for the round trip = (d/26)+(d/26) = d/13
Now, d/13 < 13d/168
-> B reaches point P before A.
| 10. Alex and Brad start walking towards points P and Q, respectively. Alex reaches P in 25 minutes. Brad travels a distance which is 1/5th of the distance travelled by Alex, in order to reach Q. He cycles to Q at an average speed which is 1.25 times that of Alex’s speed. How long does Brad travel (in minutes)? | Easy |
| A. 5 minutes B. 3 minutes C. 4.5 minutes D. None of these |
View Answer
Answer: Option D
Explanation:
Time taken by Brad = Distance travelled by Brad/Brad’s speed
Let distance travelled by Alex be d kms
-> Distance travelled by Brad = d/5 kms
Now, Alex’s speed = d/25 km/min
-> Brad’s speed = 1.25×(d/25) = d/20 km/min
Therefore, time taken by Brad = (d/5)/( d/20) = 4 minutes