New Pattern Data Interpretation Based on Time and Distance and Download Trains Based DI PDF for SBI PO, IBPS PO, IBPS RRB Scale 1, IBPS Clerk, SBI Clerk, 2020 at Smartkeeda

Directions : Study the following line chart carefully and answer the questions given beside.

Below the pair of five trains are given. The sum of length of each pair and the time taken to cross each other when travelling in opposite direction of each pair is given in the line graph.    
 
Length (in decametre) and Time (in sec) taken to cross each other when travelling to opposite direction

1
Find the time taken by train B and train D to cross each other if both are travelling in opposite direction.
» Explain it
A
Let, length (in metres) of train A, train B, train C, train D, and train E be ‘a’, ’b’, ‘c’, ‘d’, and ‘e’, respectively.

And, speed (in m/s) of train A, train B, train C, train D, and train E be ‘p’, ‘q’, ‘r’, ‘s’, and ‘t’, respectively.

2 × (a + b + c + d + e) = 3380

⇒ 2 × (630 + 700 + e) = 3380

⇒ e = 360 m

a = 660 – 360 = 300 m

b = 630 – 300 = 330 m

c = 840 – 330 = 510 m

d = 700 – 510 = 190 m

Also,  a + b  = 14
p + q

⇒  630  = 14
p + q

⇒ p + q = 45
 
b + c  = 24
q + r

⇒  840  = 24
q + r

⇒ q + r = 35

Similarly,

r + s = 25

s + t = 40

p + t = 45

So, 2 × (p + q + r + s + t) = 190

⇒ p + q + r + s + t = 85

⇒ 45 + r + 40 = 95

⇒ r = 10

So, s = 15, t = 25, p = 20, q = 25
 
  Length
 (in metres)
Speed (in m/s)
Train A 300 m 20 m/s
Train B 330 m 25 m/s
Train C 510 m 10 m/s
Train D 190 m 15 m/s
Train E 360 m 25 m/s

Reqd. time =  330 + 190  = 13 sec.
25 + 15

Hence, option A is correct.
2
Find the time taken by train A to pass train C if they are travelling in the same direction.
» Explain it
C
Let, length (in metres) of train A, train B, train C, train D, and train E be ‘a’, ’b’, ‘c’, ‘d’, and ‘e’, respectively.

And, speed (in m/s) of train A, train B, train C, train D, and train E be ‘p’, ‘q’, ‘r’, ‘s’, and ‘t’, respectively.

2 × (a + b + c + d + e) = 3380

⇒2 × (630 + 700 + e) = 3380

⇒e = 360 m

a = 660 – 360 = 300 m

b = 630 – 300 = 330 m

c = 840 – 330 = 510 m

d = 700 – 510 = 190 m

Also,  a + b  = 14
p + q

⇒  630  = 14
p + q

⇒ p + q = 45
 
b + c  = 24
q + r

⇒  840  = 24
q + r

⇒ q + r = 35

Similarly,

r + s = 25

s + t = 40

p + t = 45

So, 2 × (p + q + r + s + t) = 190

⇒ p + q + r + s + t = 85

⇒ 45 + r + 40 = 95

⇒ r = 10

So, s = 15, t = 25, p = 20, q = 25
 
  Length (in metres) Speed (in m/s)
Train A 300 m 20 m/s
Train B 330 m 25 m/s
Train C 510 m 10 m/s
Train D 190 m 15 m/s
Train E 360 m 25 m/s

Reqd. time =  300 + 510  =  810  = 81 sec
20 – 10 10

Hence, option C is correct.
3
If train E crosses a platform of certain length in 49.6 seconds then find the time taken by train D to cross the same platform.
» Explain it
D
Let, length (in metres) of train A, train B, train C, train D, and train E be ‘a’, ’b’, ‘c’, ‘d’, and ‘e’, respectively.

And, speed (in m/s) of train A, train B, train C, train D, and train E be ‘p’, ‘q’, ‘r’, ‘s’, and ‘t’, respectively.

2 × (a + b + c + d + e) = 3380

⇒2 × (630 + 700 + e) = 3380

⇒e = 360 m

a = 660 – 360 = 300 m

b = 630 – 300 = 330 m

c = 840 – 330 = 510 m

d = 700 – 510 = 190 m

Also,  a + b  = 14
p + q

⇒  630  = 14
p + q

⇒ p + q = 45
 
b + c  = 24
q + r

⇒  840  = 24
q + r

⇒ q + r = 35

Similarly,

r + s = 25

s + t = 40

p + t = 45

So, 2 × (p + q + r + s + t) = 190

⇒ p + q + r + s + t = 85

⇒ 45 + r + 40 = 95

⇒ r = 10

So, s = 15, t = 25, p = 20, q = 25
 
  Length (in metres) Speed (in m/s)
Train A 300 m 20 m/s
Train B 330 m 25 m/s
Train C 510 m 10 m/s
Train D 190 m 15 m/s
Train E 360 m 25 m/s

Let, length of platform = ‘x’ m
 
So,  360 + x  = 49.6
25

⇒ 360 + x = 1240
 
⇒ x = 1240 – 360
 
⇒ x = 880 

So, time taken by D =  880 + 190  = 71.33 sec
15

Hence, option D is correct.
4
Train A was travelling from Patna to Delhi while train D was travelling from Delhi to Patna. Train A started after 2 hours of train D. If both trains meet at a distance of 405 km from Delhi then find the distance between Patna to Delhi. 
» Explain it
A
Let, length (in metres) of train A, train B, train C, train D, and train E be ‘a’, ’b’, ‘c’, ‘d’, and ‘e’, respectively.

And, speed (in m/s) of train A, train B, train C, train D, and train E be ‘p’, ‘q’, ‘r’, ‘s’, and ‘t’, respectively.

2 × (a + b + c + d + e) = 3380

⇒2 × (630 + 700 + e) = 3380

⇒e = 360 m

a = 660 – 360 = 300 m

b = 630 – 300 = 330 m

c = 840 – 330 = 510 m

d = 700 – 510 = 190 m

Also,  a + b  = 14
p + q

⇒  630  = 14
p + q

⇒ p + q = 45
 
b + c  = 24
q + r

⇒  840  = 24
q + r

⇒ q + r = 35

Similarly,

r + s = 25

s + t = 40

p + t = 45

So, 2 × (p + q + r + s + t) = 190

⇒p + q + r + s + t = 85

⇒45 + r + 40 = 95

⇒r = 10

So, s = 15, t = 25, p = 20, q = 25
 
  Length (in metres) Speed (in m/s)
Train A 300 m 20 m/s
Train B 330 m 25 m/s
Train C 510 m 10 m/s
Train D 190 m 15 m/s
Train E 360 m 25 m/s

Speed of train D = 15 ×  18  = 54 km/h
5

Speed of train A = 25 ×  18  = 90 km/h
5

Time taken by train D to travel 405 km =  405  = 7.5 hr
54

Distance travelled by train A to meet train D = 90  ×  5.5 = 495 km (We have multiplied by 5.5 as train A travelled 2 hrs less than train D)
 
Required distance = 405 + 495 = 900 km
 
Hence, option A is correct.
5
Train B and train C were travelling from station X to station Y which is 432 km apart. If both reached the station Y at the same time then find after how much time of train C, would train B leaved station X?  
» Explain it
E
Let, length (in metres) of train A, train B, train C, train D, and train E be ‘a’, ’b’, ‘c’, ‘d’, and ‘e’, respectively.

And, speed (in m/s) of train A, train B, train C, train D, and train E be ‘p’, ‘q’, ‘r’, ‘s’, and ‘t’, respectively.

2 × (a + b + c + d + e) = 3380

⇒ 2 × (630 + 700 + e) = 3380

⇒ e = 360 m

a = 660 – 360 = 300 m

b = 630 – 300 = 330 m

c = 840 – 330 = 510 m

d = 700 – 510 = 190 m

Also,  a + b  = 14
p + q

⇒  630  = 14
p + q

⇒ p + q = 45
 
b + c  = 24
q + r

⇒  840  = 24
q + r

⇒ q + r = 35

Similarly,

r + s = 25

s + t = 40

p + t = 45

So, 2 × (p + q + r + s + t) = 190

⇒p + q + r + s + t = 85

⇒45 + r + 40 = 95

⇒r = 10

So, s = 15, t = 25, p = 20, q = 25
 
  Length (in metres) Speed (in m/s)
Train A 300 m 20 m/s
Train B 330 m 25 m/s
Train C 510 m 10 m/s
Train D 190 m 15 m/s
Train E 360 m 25 m/s

Speed of train B = 25 ×  18  = 90 km/h
5

Speed of train C = 10 ×  18  = 36 km/h
5

Time taken by train C =  432  = 12 hours
36

Time taken by train B =  432  = 4.8 hours
90

So, train B would leave the station X after 7.2 hours of train C.

Hence, option E is correct.

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What is Data Interpretation and how its useful for SBI PO, SBI Clerk, LIC, NABARD etc

In our day-to-day life, data is reported in the newspapers, on television, and in magazines. The data involves business, government, sports and many other topics. It is represented in an organized fashion, making it easy to interpret what you hear and read in the media. Those set of problems is being asked in different kind of form like Pie Chart di, Bar Chart di, Table Chart di, line chart di and info chart etc.

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