Run Race MCQ1( reach at the same time)
Run Race MCQ1: A runs 1 – 2/3 times as fast as B. If A gives B a start of 80 m, how far must the winning post be so that A and B might reach at the same time?
a) 200 m b) 300 m c) 270 m d) 160 m
Solution:
reach at the same time
Let B runs at speed ‘x’
By given condition,
A will run with speed \( 1\frac{2}{3}x=\frac{5}{3}x \)
We know the formula of distance;
\[ s=vt \]
Where,
S= distance
V= speed
t= time
From there we can say;
\[ t=\frac{s}{v} \]
or
\[ Time= \frac {Distance}{Speed} \]
In the statement “If A gives B a start of 80 m” means that B is 80 meter closer to the winning pole than A
Let the winning pole is at distance ‘y’ from B. Then by given condition it will be at distance ‘y+80’ from A.
Using above formula,
Time taken by B to reach winning pole;
\[ t=\frac{y}{x} \]
Time by A to reach winning pole;
\[ t= (y+80)\frac {3}{5x} \]
If bot A and B
then times taken by both should be equal
\[ (y+80)\frac {3}{5x}=\frac{y}{x} \]
or
\[ 5y=3(y+80) \]
\[ 2y=240 \]
\[ y=120 \]
So, the winning pole must be erected at distance of 120 from B
or
y+80 = 120+80=200 from A
Note:
This question has been solved on other websites as well — but in a confusing style and not in a proper and easy way
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