Question

ACT math question:

A tortoise moves x times as fast as a snail. A hare moves y times as fast as the tortoise. The hare moves how many times as fast as the snail?

a)y/x
b)x/y
c)xy
d)y-x
E)y+x

I tried to plug in numbers but it never seems to work?

Answer 1

@ganeshie8 @kirbykirby

Answer 2

Let's try a numerical example, and abstract it to numbers

Say the snail has a speed of 3 (say a=3)
Now, the tortoise moves x times as fast as the snail. Let's say it moves twice as fast. So, x=2. This means that the tortoise moves at a speed 2(3) = x(a) = xa

Now, the hare moves y times as fast as the tortoise. So, let's say he moves 4 times as fast (y=4) as the tortoise, meaning 4*(2(3)) = 4*2*3=y*x*a

Now... how fast is the hare moving in comparison to the snail?
The hare has a speed of 4*2*3 = y*x*a
The snail, a speed of 3 = a

So how much faster is the hare compared to the snail? can you see the result?

Answer 3

It's actually just a simple substitution problem

Answer 4

@kirbykirby thanks for replying give me a second to work it out :P

Answer 5

ok I see it now! I did not think of giving the snail a number.. I substituted the numbers really weirdly. thanks!

Answer 6

:)

Answer 7

Here's another way. It'll be less time consuming if you understand it, or you can continue solving by plugging numbers.

Tortoise moves x times as fast as snail
T= Sx

Hare moves y times as fast as tortoise.
H= Ty

Substitute the first equation into second equation.

H= (Sx)y
So
H= Sxy

So hare is xy times as fast as snail