Question

If a person reduces his speed by 20%, he takes 20 min to reach his

destination. Find the time taken by the person (in min) to reach his

destination if he increases his speed by 14.28%.​

Answer 1

17Q:

A person takes 20 minutes more to cover a certain distance by decreasing his speed by 20%. What is the time taken to cover the distance at his original speed ?

A) 1hr

B) 1 hr 20 min

C) 1 hr 10 min

D) 50 min

Answer: B) 1 hr 20 min

Explanation:

Let the distance and original speed be 'd' km and 'k' kmph respectively.

d/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/3

=> (5d - 4d)/4k = 1/3 => d = 4/3 k

Time taken to cover the distance at original speed

= d/k = 4/3 hours = 1 hour 20 minutes.

Answer 2

The time taken will be 18.66 minutes.

Given,

He reduces hi speed by=20%

Time taken=20 minutes

He increases his speed by=14.28%.

To find,

the time taken by the person (in min) to reach his destination.

Solution:

Let the original speed be v.

He reduces hi speed by 20% so v1 will be,

$v1=v-20 percent of v\\v1=v-\frac{20}{100}v\\v1=\frac{80}{100}v .$

Time taken is 20 minutes, so the distance that he covers will be:

$Distance=SpeedXTime\\Distance=\frac{80}{100} v X20\\Distance=\frac{1600}{100} v\\Distance =16v m.$

He increases his speed by 14.28% so v2 will be:

$v2=v-14.28 percent of v\\v2=v-\frac{14.28}{100}v\\v2=\frac{85.72}{100}v .$

As the destination is same, the distance will also be same.

The time taken by the person will be:

$Time=\frac{Distance}{Speed} \\Time=\frac{16 v m}{\frac{85.72}{100}v } \\Time=16 vmX\frac{100}{85.72v} \\Time=18.66 minutes.$

Thus, the time taken is 18.66 minutes.

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