# Challenging Money Math Question

• August 12th 2011, 04:58 AM
KayPee
Challenging 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, 06:24 AM
CaptainBlack
Re: Challenging Money Math Question
Quote:

Originally Posted by KayPee
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

Let $a$ be the amount that Ama has, now rewrite what you are told in terms of $a$.

CB
• August 12th 2011, 07:05 AM
KayPee
Re: 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, 10:14 AM
CaptainBlack
Re: Challenging Money Math Question
Quote:

Originally Posted by KayPee
Let a be what Ama has and b be what Kwame has
b + 3 = 2a
b - 4= a

See the change in red above

CB
• September 7th 2011, 01:52 AM
Ross137
Re: 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, 08:27 PM
Wilmer
Re: Challenging Money Math Question
Quote:

Originally Posted by Ross137
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.

Simply subtract 2nd equation from 1st: k - k + 3 - (-4) = 2a - a
So 7 = a
• September 8th 2011, 08:07 AM
Ross137
Re: Challenging Money Math Question
Yep - that's what I meant by elimination (Nod)