Question : Two friends, P and Q, start running around a circular track from the same point. They run in the same direction. P runs at 6 m/sec and Q runs at b m/sec. If they cross each other at exactly two points on the circular track and b is a natural number less than 6, how many values can b take?
Option 1: 2
Option 2: 1
Option 3: 4
Option 4: 3
Correct Answer: 1
Solution :
If the ratio of their speeds is $V_1:V_2$, the distinct meeting points will be $V_1-V_2$, with both are running in the same direction. Try different values of $b$ from 1 to 5 and check which satisfies $V_1-V_2 = 2$
$\frac{V_1}{V_2}=\frac{6}{b}$; number of points is 2
Given: b is a natural number less than 6
Checking for $b$ = 1: $\frac{V_1}{V_2}=\frac{6}{1}$, the difference is 6 – 1 = 5 (not satisfied).
Checking for $b$ = 2: $\frac{V_1}{V_2}=\frac{6}{2}$$=\frac{3}{1}$, the difference is 3 – 1 = 2 (satisfied).
Checking for $b$ = 3: $\frac{V_1}{V_2}=\frac{6}{3}$$=\frac{2}{1}$, the difference is 2 – 1 = 1 (not satisfied).
Checking for $b$ = 4: $\frac{V_1}{V_2}=\frac{6}{4}$ $=\frac{3}{2}$, the difference is 3 – 2 = 1 (not satisfied).
Checking for $b$ = 5: $\frac{V_1}{V_2}=\frac{6}{5}$, the difference is 6 – 5 = 1 (not satisfied).
The condition is satisfied only when b = 2. So, b can take only one value.
Hence, the correct answer is 1.
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