Question

A, B and C are three contestants in a 500 m race. If A can give B a start of 20 m and A can give C a start of 32 m, then how many metres start can B give to C?

• A. 12
• B. 14
• C. 12.5
• D. 13.5

Hint:

Use ratios of distances covered in same time to determine relative speeds and convert them to start distances.

Answer 1

The Correct Option is C

A gives B a start of 20 m in 500 m race $\Rightarrow$
This means when A runs 500 m, B runs 480 m
So, speed ratio of A : B = 500 : 480 = 25 : 24
A gives C a start of 32 m $\Rightarrow$ C runs 468 m
So, speed ratio of A : C = 500 : 468 = 125 : 117
Now, find B : C
A : B = 25 : 24
A : C = 125 : 117
To find B : C, divide A : C by A : B
$\dfrac{125}{117} \div \dfrac{25}{24} = \dfrac{125}{117} \times \dfrac{24}{25} = \dfrac{5 \times 25 \times 24}{5 \times 25 \times 117} = \dfrac{24}{117} = \dfrac{8}{39}$
Wait—error. Let's do correctly:
B : C = $\dfrac{A:C}{A:B} = \dfrac{125/117}{25/24} = \dfrac{125 \times 24}{117 \times 25} = \dfrac{5 \times 25 \times 24}{5 \times 25 \times 117} = \dfrac{24}{117} = \dfrac{24}{117} = \dfrac{8}{39}$
Error again. Let's re-calculate carefully:
$\dfrac{125}{117} \div \dfrac{25}{24} = \dfrac{125 \times 24}{117 \times 25} = \dfrac{3000}{2925} = \dfrac{120}{117} = 40 : 39$
So B : C = 40 : 39
Hence, when B runs 500 m, C runs $\dfrac{39}{40} \times 500 = 487.5$ m
So, B can give a start of $500 - 487.5 = \boxed{12.5}$ m to C