**Dear Bank Aspirants practice Quant Questions for upcoming SBI and other exams. Today's Topic is Time and Distance, very important for SBI clerk point of view.**

(a) 9

(b) 10

(c) 12

(d) 20

(e) None of these

2. Two men starting from the same place walk at the rate of 4 kmph and 4.5 kmph respectively. What time will they take to be 8.5 km apart, if they walk in the same direction?

(a) 1 hour

(b) 4 hrs 15 min

(c) 17 hrs

(d) 2 hrs. 55 min.

(e) None of these

3. Two cyclists start from the same place in opposite directions. One goes towards north at 18 kmph and the other goes towards south at 20 kmph. What time will they take to be 47.5 km apart?

(a) 2(1/4) hrs

(b) 1(1/4) hrs

(c) 2 hrs 23 min

(d) 23(1/4) hrs

(e) None of these

4. A man covers a certain distance on scooter. Had he moved 3 kmph faster, he would have taken 40 min. less. If he had moved 2 kmph slower, he would have taken 40 min. more. The distance (in km) is:

(a) 20

(b) 36

(c) 37.5

(d) 40

(e) None of these

5. If a train runs at 40 kmph, it reaches its destination late by 11 minutes but if it runs at 50 kmph, it is late by 5 minutes only. The correct time for the train to complete its journey is :

(a) 13 min.

(b) 15 min.

(c) 19 min.

(d) 21 min.

(e) None of these

6. Two buses travel to a place at 45 kmph and 60 kmph respectively. If the second bus takes 5(1/2) hours less than the first for the journey, the length of the journey is:

(a) 900 km

(b) 945 km

(c) 990 km

(d) 1350 km

(e) None of these

7. Walking 5/6 of his usual speed, a man is 10 minutes too late. The usual time taken by him to cover that distance is:

(a) 1 hours

(b) 50 min

(c) 12 min

(d) 8 min. 20 sec.

8. A car travels a distance of 840 km at a uniform speed. If the speed of the car is 10 kmph more, then it takes 2 hours less to cover the same distance. The original speed of the car (in kmph) was:

(a) 45

(b) 50

(c) 60

(d) 75

(e) None of these

9. Distance between two towns P and Q is 240 km. A motor cycle rider starts from P towards Q at 8 p.m. at a speed of 40 kmph. At the same time another motor cycle rider starts from Q towards P at 50 kmph. At what time will they meet?

(a) 9.45 p.m.

(b) 10.30 p.m.

(c) 10.40 p.m.

(d) 11 p.m.

(e) None of these

10. A train M leaves Meerut at 5 a.m. and reaches Delhi at 9 a.m. Another train N leaves Delhi at 7 a.m. and reaches Meerut at 10.30 a.m. At what time do the two trains cross one another?

(a) 8.26 a.m.

(b) 8 a.m.

(c) 7.36 a.m.

(d) 7.56 a.m.

(e) None of these

Answers

1. (b); Due to stoppages, it cover 9 km less.

Time taken to cover 9 km=[(9/54)X60)min=10 min

2. (c); To be 0.5 km apart, they take 1 hour

To be 8.5 km apart, they would take=[(1/0.5)X8.5]hrs=17 hrs.

3. (b); To be (18+20) km apart, they take 1 hour

To be 47.5 km apart, they would take=[(1/38)X47.5]hrs=1(1/4) hrs.

4. (d); Let distance=x km & usual rate=y km/hr.

(x/y)-x/(y+3)=40/60

2y(y+3)=9x…………….(i)

And, x/(y-2)-x/y=40/60

y(y-2)=3x……….(ii)

On dividing (i) by (ii), we get: (2(y+3))/(y-2)=3

y=12

Putting y=12 in (i), we get : x=40

5. (c); Let the correct time to complete the journey be x min.

Then, distance travelled in (x+11) min at 40 kmph

=distance covered in (x+5) min. at 50 kmph.

Therefore, (x+11)X(40/60)=(x+5)X(50/60)

10x=190

x=19 min.

6. (c); Let the length of journey be x km. Then,

(x/45)-(x/60)=11/2

X=990 km.

7. (b); With a speed of 5/6 th of the usual speed, the time taken is 6/5th of the usual time.

Therefore, (6/5 of usual time) – (usual time)=10 min.

Therefore, 1/5 of usual time=10 min. So, usual time=50 min.

8. (c); Let original speed=x kmph.

(840/x)-(840/(x+10))=2

840(x+10)-840x=2x(x+10)

x^2+10x-4200=0

(x+70)(x-60)=0

x=60

Therefore, Original speed=60 kmph.

9. (c); Suppose they meet x hrs after 8 p.m. Then,

Sum of distance covered by them in x hrs=240 km.

Therefore, 40x+50x=240

x=(240/90) hrs=2 hrs 40 min.

So, they will meet at 10.40 p.m.

10 . (d); Let distance between Meerut & Delhi be x km and let the train meet y hours after 7 a.m.

Clearly, M covers x km in 4 hrs. & N covers x km in 3(1/2) hr.

Therefore, Speed of M=(x/4)kmph & Speed of N=(2x/7) kmph.

[Distance covered by M in (y+2) hrs] [Distance covered by N in y hr]=x

(x(y+2))/4+(2xy/7)=x

((y+2)/4)+(2y/7)=1

y=(14/15) hrs=((14/15)X60) min=56 min.

So, the trains meet at 7.56 a.m.

Time taken to cover 9 km=[(9/54)X60)min=10 min

2. (c); To be 0.5 km apart, they take 1 hour

To be 8.5 km apart, they would take=[(1/0.5)X8.5]hrs=17 hrs.

3. (b); To be (18+20) km apart, they take 1 hour

To be 47.5 km apart, they would take=[(1/38)X47.5]hrs=1(1/4) hrs.

4. (d); Let distance=x km & usual rate=y km/hr.

(x/y)-x/(y+3)=40/60

2y(y+3)=9x…………….(i)

And, x/(y-2)-x/y=40/60

y(y-2)=3x……….(ii)

On dividing (i) by (ii), we get: (2(y+3))/(y-2)=3

y=12

Putting y=12 in (i), we get : x=40

5. (c); Let the correct time to complete the journey be x min.

Then, distance travelled in (x+11) min at 40 kmph

=distance covered in (x+5) min. at 50 kmph.

Therefore, (x+11)X(40/60)=(x+5)X(50/60)

10x=190

x=19 min.

6. (c); Let the length of journey be x km. Then,

(x/45)-(x/60)=11/2

X=990 km.

7. (b); With a speed of 5/6 th of the usual speed, the time taken is 6/5th of the usual time.

Therefore, (6/5 of usual time) – (usual time)=10 min.

Therefore, 1/5 of usual time=10 min. So, usual time=50 min.

8. (c); Let original speed=x kmph.

(840/x)-(840/(x+10))=2

840(x+10)-840x=2x(x+10)

x^2+10x-4200=0

(x+70)(x-60)=0

x=60

Therefore, Original speed=60 kmph.

9. (c); Suppose they meet x hrs after 8 p.m. Then,

Sum of distance covered by them in x hrs=240 km.

Therefore, 40x+50x=240

x=(240/90) hrs=2 hrs 40 min.

So, they will meet at 10.40 p.m.

10 . (d); Let distance between Meerut & Delhi be x km and let the train meet y hours after 7 a.m.

Clearly, M covers x km in 4 hrs. & N covers x km in 3(1/2) hr.

Therefore, Speed of M=(x/4)kmph & Speed of N=(2x/7) kmph.

[Distance covered by M in (y+2) hrs] [Distance covered by N in y hr]=x

(x(y+2))/4+(2xy/7)=x

((y+2)/4)+(2y/7)=1

y=(14/15) hrs=((14/15)X60) min=56 min.

So, the trains meet at 7.56 a.m.