Problem Solving

A and B ran, at their respective rates, a race of 480m. In the first heat, A gives B a head start of 48m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144m and is beaten by 1/30th of a minute. What is B's speed in m/s?

A. 12
B. 14
C. 16
D. 18
E. 20

The OA is A.

I'm confused by this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

LUANDATO wrote:A and B ran, at their respective rates, a race of 480m. In the first heat, A gives B a head start of 48m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144m and is beaten by 1/30th of a minute. What is B's speed in m/s?

A. 12
B. 14
C. 16
D. 18
E. 20

The OA is A.

I'm confused by this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

Luandato, what are you confused about? Do let me know how you approached the question and I'd be happy to help you improve your logic. The question looks pretty similar to the other S = D/T questions you have posted before. I'll let you come back to me with your working/process and if you still need help I'll show you how I would approach the question.

Cheers,
Dan

when A gives B a head set of 48m, A ran 480m in, say, x seconds and
B ran (480-48)m =432m in
$$x+\frac{60}{10}=\left(x+6\right)\sec$$
$$B's\ rate\ can\ thus\ be\ \exp ressed\ as\ 432m\ in\ \left(x+6\right)s=\left(\frac{432}{x+6}\right)\ \frac{m}{s}$$
$$when\ A\ gives\ B\ a\ head\ start\ of\ 144m,$$
$$A\ ran\ 480m\ in\ the\ same\ x\ \sec onds,$$
$$and\ B\ ran\ 480-144m\ =\left(336m\right)\ in\ x-\frac{60}{30}=\left(x-2\right)\sec onds$$
$$B's\ rate\ can\ be\ \exp ressed\ as\ \left(\frac{336}{x-2}\right)\ \frac{m}{s}$$
$$Equating\ B's\ rate:\ i.e\ \frac{432}{x+6}=\frac{336}{x-2}$$
$$432\left(x-2\right)=336\left(x+6\right)\ \ ;\ dividing\ through\ by\ 48,\ we\ have$$
$$9\left(x-2\right)=7\left(x+6\right)=\ 9x-18=7x+42\ $$
$$or\ 9x-7x=60$$
$$x=\frac{60}{2}=30s$$
$$Therefore,\ B's\ rate=\frac{436}{30+6}\ \frac{m}{s}\ =12\ \frac{m}{s\ }\ \left[option\ A\right]$$