180 views

2 votes

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

- $4$
- $3$
- $2$
- $5$

2 votes

Best answer

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}} $

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 before he meets Rahim for the first time $ = \frac{144 \pi}{48 \pi} = 3.$

$\textbf{Second Method:}$

$\textbf{Speeds:}$

- $\text{S}_{T_{1}} = 15 \; \text{km/hr} = 15 \times \frac{1000}{60 \times 60} = 15 \times \frac{5}{18} = \frac{25}{6} \; \text{m/sec.}$
- $\text{S}_{T_{2}} = 5 \; \text{km/hr} = 5 \times \frac{1000}{60 \times 60} = 5 \times \frac{5}{18} = \frac{25}{18} \; \text{m/sec.}$

$\textbf{Distances:}$

- $\text{D}_{T_{1}} = 2 \pi (100) = 200 \pi \; \text{m} $
- $\text{D}_{T_{2}} = 2 \pi (20) = 40 \pi \; \text{m} $

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

$\textbf{Time:}$

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

Correct Answer$: \text{B}$