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Quantitative Aptitude Quiz for SSC(Time, Speed and Distance): 23 March 2021

Updated Tue, 23 Mar 2021 03:44 PM IST
Source: Safalta.com
1. In a 100 m race, A runs at 8km per hour. If A gives B a start of 4 m and still beat him by 15 seconds, what is the speed of B?(approx.)

a) 6 km/hr
b) 3 km/hr
c) 5 km/hr
d) 4 km/hr
 

A runs at 8km/hr or 8 × (5/18) m/s = 2.22 m/s

He beats B by 15 seconds, ie, he covered 15 × 2.222 m = 33.33 m more than B if both were given the same time.

We can say that in the time A runs 100 m, B ran (100 - 33.33 - 4) m = 62.67 m

Time taken by A to run 100 m = 100/2.22 = 45.04 s ≈ 45 s

Hence in those 45 s, B runs 62.67 m, and his speed can be found by using

Speed = Distance/Time

⇒ 62.67/45

⇒ 1.393 m/s or 1.393 × (18/5) km/hr = 5.014 km/hr 

≈ 5 km/hr

2. A policeman starts chasing a thief 30 minutes after the thief had run from a spot. With an average speed of 20km per hour, he takes 2 hours to catch the thief. What is the average speed of the thief?

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a) 24 km /hr
b) 18 km /hr
c) 15 km /hr
d) 16 km /hr

⇒ 20 × 2 = Total Distance Travelled

⇒ Total Distance Travelled = 40 km

∵ The distance travelled by the thief = Distance Travelled by the policeman = 40 km

Time Taken by the thief = 2 + 0.5 = 2.5 hrs

⇒ Average Speed of thief = 40/2.5

= 16 km /hr

3. A car travels a certain distance taking 7 hrs in forward journey, during the return journey increases its speed by 12km/hr and takes the time 5 hrs. What is the distance travelled?

a) 280 kms
b) 140 kms
c) 196 kms
d) 210 kms

or Forward Journey:

Speed = S kmph

Time = 7 Hrs

Distance =  D kms

∵ Distance = (Speed ) ×  (Time)

⇒ D = 7S  ................... (i)

For Return Journey

Speed = ( S + 12 ) kmph

Time = 5 hrs

Distance = D kms

Distance = (Speed ) ×  (Time)

⇒ D = (S+12) × 5  ................... (ii)

Substituting value of "D" from Eq. (i)

⇒ 7S = 5S + 60

⇒ S = 30 kmph

⇒ Distance = D = 30 × 7 = 210 kms

4. A bus started from bus stand at 8.00a.m. and after 30 min staying at destination, it returned back to the bus stand. The destination is 27 miles from the bus stand. The speed of the bus while going to destination = 18 miles/hr, the speed of the bus while returning is 50 percent higher. At what time it returns to the bus stand.

a) 11 : 30 a.m 
b) 11 a.m 
c) 10 a.m 
d) 10 : 30 a.m 

Distance = 27 miles

Time = x hrs

Speed = 18 miles/hr

Distance = Speed × Time

⇒ Time = (Distance / Speed) = 27/18 = 1.5 hrs

Halt Time= 30minuts = (0.5) hr

For Destination to Bus Stand

Speed = 1.5 × 18 = 27 miles/hr

Time = y

Distance = 27 Miles

Distance = Speed × Time

⇒ Time = (Distance / Speed) = 27/27 = 1 hr

Total Time = 1.5 + 0.5 + 1 = 3 hrs

Reach time = 8:00 + 3 = 11:00 am

5. A bus starts at 6:00 pm. from starting point at the speed of 18m/s, reached its destination and waited for 40 minutes. And again returned back at the speed of 28m/s. If the time taken in forward journey is same as time taken in reverse and waiting time.
What will be the time when it reaches again at its starting point?

a) 9:44 PM
b) 9:54 PM
c) 10:04 PM
d) 9:34 PM

Say distance is d m.

Time taken = Distance/Speed

Time taken in forward journey = d/18 seconds.

Time taken in reverse journey = d/28 seconds.

Given that, d/18 = d/28 + (40 × 60)    [40 minutes = 40 × 60 seconds]

⇒ d × (5/252) = 2400

⇒ d = 2400 × 252/5 = 120960 m

∴ Total time taken = (d/18) + (d/28) + 2400 = 13440 seconds

⇒ Time taken = 13440/60 minutes = 224 minutes = 3 hours and 44 minutes

∴ The bus reaches back at 9:44 PM

6. If a person walks at 4/5th of his usual speed he reaches 40min late. If he walks at his usual speed how much time does he travel?

a) 120 minutes.
b) 180 minutes.
c) 140 minutes.
d) 160 minutes.

Let a man travels at speed of ‘m’ km/hr and takes ‘n’ minutes to reach his destination.

 Distance travelled = mn minutes.

Now we have the man walking at 4/5 of his speed ie, 0.8m km/hr

And then he taken n + 40 minutes.

Hence, distance travelled = 0.8m × (n + 40)

= 0.8mn + 32m

Distance in both the cases should be equal.

Hence

⇒ mn = 0.8mn + 32 m

⇒ mn – 0.8mn = 32 m

⇒ 0.2 mn = 32 m

⇒ 0.2n = 32

⇒n = 32/0.2

⇒ n = 160 minutes.

7. A man can row 30 km upstream and 44 km downstream in 10 h. Also, he can row 40 km upstream and 55 km downstream in 13 h. find the speed of the current.

a) 2 km/h.
b) 3 km/h.
c) 4 km/h.
d) 5 km/h.

Let speed in upstream = x km / hr and in the downstream = y km / hr.

Then, (40/x) + (55/y) = 13   ...(i)

And   (30/x) + (44/y) = 10       ...(ii)

Multiplying (ii) by 4 and (i) by 3 and subtracting, we get: 

11/y = 1                 

⇒ y = 11.

Substituting y = 11 in (i), we get : x = 5

Therefore, speed in still water = (Speed in upstream + speed in downstream)/2

= (11 + 5)/2 km/h = 8km/h.

Speed of current = (Speed in downstream - speed in upstream)/2

= (11 − 5)/2 km/h = 3km/h.

8. If Rachit drives a car four times a lap @10, 20, 30, 60 kmph. What is his average speed?

a) 15 km/hr
b) 30 km/hr
c) 25 km/hr
d) 20 km/hr

Let 60 km be the length of the lap.

Then time taken by Rachit to complete the lap at 10 km/hr = Distance/speed= 60/10

= 6 hours

Then time taken by Rachit to complete the lap at 20 km/hr = Distance/speed= 60/20

= 3 hours

Then time taken by Rachit to complete the lap at 30 km/hr = Distance/speed= 60/30

= 2 hours

Then time taken by Rachit to complete the lap at 60 km/hr = Distance/speed= 60/60

= 1 hour

Total distance covered = 60 + 60 + 60 + 60 = 240 kms

Total time taken = 6 + 3 + 2 + 1 = 12 hours

Average speed = Total distance/Total time

= 240/12

= 20 km/hr

9. A train passes a station platform in 35s and a man standing on the platform in 15s. If the speed of the train is 54 km/hr, find the length of the platform.

a) 400 m
b) 500 m
c) 600 m
d) 300 m

Given, speed of train = 54 km/hr. = 54 × (5/18) m/s = 15 m/s

Say T is the length of train and P is the length of platform.

Given, train crosses the man in 15s

⇒ Train covers T distance in 15 seconds

⇒ T = 15 × 15 = 225 m

[Distance = Speed × Time]

Given, train crosses the platform in 35s

⇒ Train covers (T + P) distance in 35 seconds

⇒ (225 + P) = 35 × 15 = 525

⇒ P = 525 – 225 = 300 m

∴ Length of platform = P = 300 m

10. Stephanie drove at an average rate of 50 miles per hour for two hours and then increased her average rate by 50% for the next 3 hours. Her average rate of speed for the 5 hours was t miles per hour. What is the value of t?

a) 75 mph
b) 70 mph
c) 65 mph
d) 60 mph

Distance travelled by Stephanie in first two hours at 50 miles per hour = 50 × 2 = 100 miles.

Stephanie increased speed by 50%

⇒ her new speed = 50 + 0.5× 50 = 75 miles per hour.

Distance travelled in next three hours at 75 miles per hour = 75 × 3 = 225 miles.

Total distance travelled in 5 hours = 100 + 225 = 325 miles.

Speed = Distance/Time

⇒ Average speed for 5 hours = t = 325/5 = 65 miles per hour.

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