High level Time and Distance Questions PDF for IBPS Clerk Mains 2021, CET and SBI PO Pre 2021 at Smartkeeda

Direction : Read the following questions carefully and choose the right answer.
1
Divya started from her house at 6:30 AM and travelled some distance till 7:30 AM. After covering 75% of that distance further, she found that she had covered 75% of total distance. Find the fraction of the total distance travelled by her by 8:30 AM.

(Given: She travels total distance with a constant speed)
» Explain it
C
Let the distance covered by Divya in 1st hour be 4x km. So the speed of Divya is 4x kmph.

4x + 3x = 75% of the total distance

i.e. 7x =  3  of total distance
4

Total distance =  28x  km
3

Distance travelled till 8:30 AM = 4x × 2 = 8x

Fraction of the total distance covered =  8x  =  6
(28x/3) 7

Hence, option C is correct.
2
Amit, Anil and Ajit ride from home to their common office with speeds in the ratio 5 : 4 : 3. If in total they take 94 minutes (sum of the individual time taken) to cover the individual distance(which is same for all), then find the time taken by 'Anil' to cover his distance.
 
» Explain it
A
Ratio of the speeds is given by = 5 : 4 : 3
 
For a given distance, speed and time are inversely proportional. 
 
So, their respective time will be in the ration of 1/5 : 1/4 : 1/3 Multiplying with their LCM which is 
 
60, we get the ratio as 12 : 15 : 20          
 
Give: 12x + 15x + 20x = 94 
 
x = 2
 
Time taken by Anil to cover the distance = 2 × 15 = 30 minutes
 
Hence, option A is correct.
3
Ashok and Bimal runs on a 450 m long circular track at a speed of 12 m/s and 18 m/s respectively. They start at same time but run in opposite direction. Find the number of times, they would have met when Bimal covers 3240 meters.
» Explain it
A
When Ashok and Bimal together cover 450 m distance, they meet and out of this Ashok covers :

12  × 450 = 180
12 + 18

Bimal covers 450 – 180 = 270 m.

So they meet, Bimal covers 270 m.

When Bimal covers 3240 meters, they would have meet

3240  = 12 times
270

Hence, option A is correct.
4
Pankaj was travelling to point B from point A with speed of 45 km/hr. After 1 hour, Pratik also started to travel from point A to point B. Pratik reached the point B 30 minutes before Pankaj. If Pratik would have decreased his speed by 6 km/hr then both would have reached point B at the same time. Find the original speed of Pratik. 
» Explain it
B
Let the original time taken by Pankaj to reach point B from point A = x hours

Then, original time taken by Pratik to reach point B from point A = (x – 1.5) hours

Also

Let original speed of Pratik = y km/h

nd, y × (x – 1.5) = (y – 6) × (x – 1)

So, 45 × x = y × (x – 1.5)

y × (x – 1.5) = (y – 6) × (x – 1)

yx – 1.5y = yx – 6x – y + 6

0.5y = 6x – 6

y = 12x – 12

Therefore, 45 × x = (12x – 12) × (x – 1.5)

45x = 12x2 – 12x – 18x + 18

12x2 – 75x + 18 = 0

4x2 – 25x + 6 = 0

4x2 – 24x – x + 6 = 0

4x (x – 6) – 1 (x – 6) = 0

(4x – 1) (x – 6) = 0

x =  1  , 6
4

Since, x =  1  is not possible
4

So, x = 6

Therefore, speed of Pratik = 12x – 12 = 60 km/hr

Hence, option B is correct.
5
Rahul travels every day from Powai to Juhu and back to Powai by train which takes 70 minutes for each journey. At a particular day the train starts 14 minutes late from Powai but reaches Juhu on time, while on return travel from Juhu due to some problem train reaches Powai 20 minutes late. If the average speed of the train on this particular day was 56km/h what is the general average speed of the train? 
» Explain it
B
Powai---------------------------------------------------------------Juhu
 
Let the general average speed = S and general time taken by train = T

Journey from Powai to Juhu

General time = 70 minutes, Time taken = 56 minutes,

So T1 =  4  × T
5

As distance remains same and time becomes 4/5 of the original time, speed will become 5/4 of the original speed.
 
So S1 =  5  × S
4

In return journey time taken = 90 min, T2 =  9  × T,
7

so the speed will become 7/9 of original speed.
 
So S2 =  7  × S
9

S average = 2 ×  S1 × S2
S1 + S2

56 = 2 ×  7S  ×  5S
9 4
      (7/9 × S + 5/4 × S)

56 =  70  × S
73

S = 58.4 km/h

Hence, option B is correct.