Run Race MCQ 2( In a 100 m race A can beat B by 25 m)

Race and Games 1

Run Race MCQ 2: In a 100 m race A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by________?

a) 21 m b) 26 m c) 28 m d) 29 m

Solution:

In a 100 m race, A can beat B by 25 m …..

Lets suppose that there is 100 meters run race in which A and B are participating.

So, by taking into consideration the statement of the question from given data we can say that in the race in which both A and B are taking part;

if A runs 100 m then B will be at 75 meters

Or ratio between running of A and B

i-e A:B=100:75

\[ \frac{A}{B}=\frac{100}{75}=\frac{4}{3} \] —-1

The same way ratio between B and C is expressible;

If B runs 100 meters then C will be at 96 meters

i-e

B:C=100:96

\[ \frac{B}{C}=\frac{100}{96}=\frac{25}{24} \] —–2

We have to find at what ration A beats C! or we have to find A:C=?

We can write;

\[ \frac{A}{C}=\frac{A}{B}\times \frac{B}{C} \]

Putting values from eq 1 & 2 above;

we get;

\[ \frac{A}{C}=\frac{25}{24}\times \frac{4}{3}= \frac {100}{72} \]

i-e

A:C= 100:72

In other words, If A runs 100 m then C will be at 72 m

i-e A will beat C by 100-72=28 m

So, the option C is the answer!

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