Lakshman Patel RJIT
asked
in Quantitative Aptitude
Mar 30, 2020
edited
Jun 29, 2020
by soujanyareddy13

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Mahira and Neelima started walking towards each other, simultaneously from Gandhi Nagar and Nehru Nagar respectively, which are $72$ miles apart. They met after $6$ hours. After their meeting, Mahira reduced her speed by $1$ mile/h and Neelima increased by $1$ mile/h. They arrived at Nehru Nagar and Gandhi Nagar respectively at the same time. Find their initial speeds :

- $6.5$ miles/h and $7.5$ miles/h
- $6$ miles/h and $7$ miles/h
- $6.5$ miles/h and $5.5$ miles/h
- $15.5$ miles/h and $9$ miles/h

1 vote

Best answer

Answer is **C**.

Detailed method of solving this problem.

Let Speed of Mahira be $S_{m}$ and Speed of Neelima be $S_{n}$

Relative Speed (${S_{m}} +S_{n}) = \frac{Dist\ travelled}{Time} = \frac{72}{6}=12 miles/hour$ → (1)

if the Dist traveled by Mahira be x, then the Dist traveled by Neelima be 72–x

**Case 1 **– When they meet each other, time taken by both are same

$\frac{Dist_{m}}{Speed_{m}} = \frac{Dist_{n}}{Speed_{n}}$

$\frac{x}{S_{m}} = \frac{72-x}{S_{n}}$ → (2)

**Case 2** – After they meet each other, time taken by both are same

After meeting each other, Dist need to travel by Mahira will be 72-x, and Dist need to travel by Neelima will be x

$\frac{72-x}{S_{m}} = \frac{x}{S_{n}}$ → (3)

Substitute ${S_{n}} = \frac{72-x}{x}*S_{m}$ from equation (2) and ${S_{n}}$ = 12 – ${S_{m}}$ from (1) in equation (3).

$\frac{S_{n}}{S_{m}} = \frac{S_{m}-1}{S_{n}+1}$

$\frac{12-x}{x} = \frac{x-1}{12-x+1}$

On solving x, we will get 6.5 miles/hour is the speed of Mahira and 5.5 miles/hour is the speed of Neelima