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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

Important for :

1

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

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

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

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

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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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.

Now come back to the point data interpretation, Data has to be well organized for it to be useful. This process of interpreting and analysing data to extract meaningful information from it is Data Interpretation. Solving Data Interpretation problems involves the use of basic formulas and manipulation of numbers.

Data Interpretation is calculation-intensive. It consists of a myriad of graphs, charts and tables form which you have to glean and analyse data. The key to cracking this area is to quickly identify the key pieces of data that you require to work on the questions asked.

Problems in Data interpretation are probably the closest in resemblance to the kind of problems you will be dealing with as a manager. They test your decision-making ability and speed using the minimum possible data. They help you to draw conclusions from collected data, support decision making, and contribute to better process, product, and quality models.

Data Sufficiency problems involve testing your quantitative concepts. They usually take the form of a logical puzzle. Instead of solving a data sufficiency problem, all you need to do is determine if the given information is enough to solve the problem. To solve Data Interpretation Questions & Answers for Banking and Insurance Exams like SBI, IBPS, NABARD & RBI Grade B, NIACL, and LIC, you need to brush up your calculation skills; you need to perform calculations faster and accurately than others. The difficult Data interpretations should be solved within 10-12 minutes. The DI of moderate level should not consume more than 7-9 minutes of the time and easy DI's should be finished within 3.5-4 minutes.

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