Catrina runs $100 \mathrm{~m}$ in 10 seconds. Sedra runs $400 \mathrm{~m}$ in 44 seconds. Maintaining these constant speeds, they participate in a $1 \mathrm{~km}$ race. How far ahead, to the nearest metre, is the winner as she crosses the finish line?
(A) $100 \mathrm{~m}$
(B) $110 \mathrm{~m}$
(C) $95 \mathrm{~m}$
(D) $90 \mathrm{~m}$
(E) $91 \mathrm{~m}$
Catrina runs $100 \mathrm{~m}$ in 10 seconds. Sedra runs $400 \mathrm{~m}$ in 44 seconds. Maintaining these constant speeds, they participate in a $1 \mathrm{~km}$ race. How far ahead, to the nearest metre, is the winner as she crosses the finish line?
(A) $100 \mathrm{~m}$
(B) $110 \mathrm{~m}$
(C) $95 \mathrm{~m}$
(D) $90 \mathrm{~m}$
(E) $91 \mathrm{~m}$

## Solution

Since Sedra runs $400 \mathrm{~m}$ in 44 seconds, then she can run $100 \mathrm{~m}$ in 11 seconds. So Catrina runs farther than Sedra.

If Catrina runs $100 \mathrm{~m}$ in 10 seconds, she will complete the race in $\frac{1000}{100} \times 10$ or 100 seconds. In 100 seconds, Sedra would have run only $\frac{400}{44} \times 100$ or 909.09 metres. Sedra would then be approximately 1000 - 909.09 or 90.91 metres behind as Catrina crossed the finish line.

ANswer: (E)