GCSE Foundation Maths question...?
- SamwiseLv 72 months ago
Car A completes 5 laps in 150 seconds.
The number of seconds car B takes to complete a lap must be a whole number
and a factor of 150 (so it completes a lap at 150 seconds after the start),
BUT it cannot be a factor of 30.
If it were, car B would complete a lap every time car A does.
Looking at the factorizations of 30 and 150, we can see that the only way
a factor of 150 cannot divide 30 is if it is a multiple of 25
so both the 5s in the factorization of 150 come into play.
So we are looking at possible factors of
2 * 3 * 5 * 5
that include the 5 * 5 part. These are
5 * 5 = 25 (Car B completes its sixth lap 150 seconds after the start.)
2 * 5 * 5 = 50 (Car B completes its third lap then.)
3 * 5 * 5 = 75 (Car B completes its second lap then.)
2 * 3 * 5 * 5 = 150 (Car B completes its first lap then.)