Aptitude Discussion

Q. |
A train covers a distance in 50 min, if it runs at a speed of 48 kmph on an average. The speed at which the train must run to reduce the time of journey to 40 min will be. |

✖ A. |
45 kmph |

✔ B. |
60 kmph |

✖ C. |
75 kmph |

✖ D. |
None of these |

**Solution:**

Option(**B**) is correct

Time = \(\dfrac{50}{60}=\dfrac{5}{6}\) hr

Speed = 48 mph

Distance = \(S\times T=48\times \dfrac{5}{6}=40\) km

Time = \(\dfrac{40}{60} \) hr

New speed = \(40\times \dfrac{3}{2}=60\) kmph

**Veer**

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**Rini**

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no need of conversion needed?? minute and hour ryt?

**Pooja Vishwakarma**

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T1/T2 = V2 / V1

5/6 V2

---- = -----

4/6 48

V2 = 60 kmph.

T1/T2 = V2 / V1

$

5/6 V2

---- = -----

4/6 48

$

V2 = 60 kmph.

**Akshay**

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no need to find distance simple one liner solution inverse of ratio of speed as time and speed are inverse proportion .So simple 48 *5/4 =60

let me explain in detail using ratio method

given distance traveled is constant then speed indirect variation to time

the ratio of time is 5:4

ratio of speed is 4:5 if 4->48 then 5->?

cross mul then ==60