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Two circular tracks $\text{T}1$ and $\text{T}2$ of radii $100 \; \text{m}$ and $20 \; \text{m},$ respectively touch at a point $\text{A}.$ Starting from $\text{A}$ at the same time, Ram and Rahim are walking on track $\text{T}1$ and track $\text{T}2$ at speeds $15 \; \text{km/hr}$ and $5 \; \text{km/hr}$ respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is

1. $4$
2. $3$
3. $2$
4. $5$

Given that, radii of circular tracks $\text{T}_{1} = 100 \; \text{m},$ and radii of circular tracks $\text{T}_{2} = 20 \; \text{m}.$

We know that, $\text{Speed} = \frac{\text{Distance}}{ \text{Time}}$

$\Rightarrow$ Time taken by each of then to complete one round $= \frac{\text{Circumference of the circle}}{\text{Speed}}$

So, time taken by Ram to cover one round $= \frac{2 \pi (100)}{15 \times \frac{5}{18}} = 48 \pi$

Time taken by Rahim to cover one round $= \frac{2 \pi (20)}{5 \times \frac{5}{18}} = \frac{144 \pi}{5}$

Time taken by Ram and Rahim meet each other for the first time $= \text{LCM} (48 \pi , \frac{144 \pi}{5})$

$= 144 \pi$

$\therefore$ The number of full rounds that Ram makes befoe he meets Rahim for the first time $= \frac{144 \pi}{48 \pi} = 3$

Correct Answer$: \text{B}$

Second Method$:$

DIAGRAM

Speeds$:$

$\text{S}_{T_{1}} = 15 \; \text{km/hr} = 15 \times \frac{1000}{60 \times 60} = 15 \times \frac{5}{18} = \frac{25}{6} \; \text{m/sec.}$

$\text{S}_{T_{2}} = 5 \; \text{km/hr} = 5 \times \frac{1000}{60 \times 60} = 5 \times \frac{5}{18} = \frac{25}{18} \; \text{m/sec.}$

Distances$:$

$\text{D}_{T_{1}} = 2 \pi (100) = 200 \pi \; \text{m}$

$\text{D}_{T_{2}} = 2 \pi (20) = 40 \pi \; \text{m}$

$\text{Speed} = \frac{\text{Distance}}{ \text{Time}}$

$\text{Time} = \frac{\text{Distance}}{ \text{Speed}}$

$\text{Time}_{T_{1}} = \frac{\text{D}_{T_{1}}} {\text{S}_{{T_{1}}}} = \frac{200 \pi}{ \frac{25}{6}} = 48 \pi$

$\text{Time}_{T_{2}} = \frac{\text{D}_{T_{2}}} {\text{S}_{{T_{2}}}} = \frac{40 \pi}{\frac{25}{18}} = \frac{144}{5} \pi$

Ratio of the time taken by Ram : Rahim $= 48 \pi \frac{144 \pi}{5}$

$= 5 : 3$

Time taken by Ram and Rahim to meet each other first time $= \text{LCM} (5,3) = 15 \; \text{sec.}$

$\therefore$ Number of rounds made by Ram before he meets to Rahim for the first time $= \frac{15}{5} = 3.$
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