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When a person travels a distance with (3/2) times of his original speed then he reaches the destination 5 hours early. Find the original time taken by the person to travel the distance (in hours).
1. 12 hrs
2. 22 hrs
3. 15 hrs
4. 14 hrs
5.

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Correct Answer - Option 3 : 15 hrs

Given:

New speed = (3/2) × Original speed

New time = Old time - 5 hrs

Formula Used:

D = S × T 

where, D = Distance, S = Speed and T = Time

Calculation:

Let the original speed and time of the person be S1 and T1 respectively.

Similarly, let the new speed and new time of the person be S2 and T2 respectively.

Now, According to the question,

Distance travelled by the person in both the cases is the same.

Also, S2 = (3/2) × S1 and T2 = T1 - 5

So, 

S1 × T1 = S2 × T2 

⇒ S1 × T1 = (1.5 × S1) × (T1 - 5)

⇒ T1 = 1.5 × (T1 - 5)

⇒ T1 = 1.5 × T1 - 1.5 × 5

⇒ (1.5 - 1)T1 = 7.5

⇒ T1 = 7.5/0.5

⇒ T1 = 15 hrs

∴ The original time taken by the person to travel the distance is 15 hrs.

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