How best can I approach the question below:

If Kwame had 3 cents more he would have twice as much as Ama. If he had 4 cents less, he would have the same amount.

How many cents does Kwame have?

a)4

b)7

c)11

d)14

e)none of the above

Printable View

- August 12th 2011, 05:58 AMKayPeeChallenging Money Math Question
How best can I approach the question below:

If Kwame had 3 cents more he would have twice as much as Ama. If he had 4 cents less, he would have the same amount.

How many cents does Kwame have?

a)4

b)7

c)11

d)14

e)none of the above - August 12th 2011, 07:24 AMCaptainBlackRe: Challenging Money Math Question
- August 12th 2011, 08:05 AMKayPeeRe: Challenging Money Math Question
Let

**a**be what Ama has and**b**be what Kwame has

b + 3 = 2a

b - 4= 2a

a = b + (3/2) = a

Substituting a into equation 2

b-4 = b + 3

I am stuck and the equation doesn’t sound ok

Where did I go wrong? - August 12th 2011, 11:14 AMCaptainBlackRe: Challenging Money Math Question
- September 7th 2011, 02:52 AMRoss137Re: Challenging Money Math Question
I solved it without using a (number thatAma has) in any equation. Here is how I did it (k being the number of Kwame's cents):

k + 3 = 2(k-4)

k + 3 = 2k - 8

k + 11 = 2k

11 = k

So Kwame has 11 cents. Using the simultaneous equations method you were using you can reach the same equation.

k + 3 = 2a

k - 4 = a

Using substituion and subbing the first into the second you can get to k + 3 = 2(k-4), as named earlier. Using elimination you can subtract the second from the first and get to 7 = a. Then sub the value for a into either equation to get k = 11. - September 7th 2011, 09:27 PMWilmerRe: Challenging Money Math Question
- September 8th 2011, 09:07 AMRoss137Re: Challenging Money Math Question
Yep - that's what I meant by elimination (Nod)