Gnpth wrote:

A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse?

A. 26

B. 27

C. 28

D. 29

E. 30

I solved with two variables, and it works, but I have a question about my own method that applies to

vishalbalwani 's method, too.

Let A = the number of 5-cent coins

Let B = the number of 10- cent coins

In one scenario, the coins, in their respective unknown quantities, total $1.75

In a second scenario, the 5-cent coins and 10-cent coins exchange quantities exactly, and they total $2.15

5A + 10B = 175 (P)

10A + 5B = 215 (Q)

Multiply (Q) by two, and subtract (P)

20 A + 10B = 430

__5A + 10B = 17515A = 255

A = 17

There are 17 coins worth 5 cents.

Use (P) to find the number of 10-cent pieces

5(17) + 10B = 175

10B = 90

B = 9

There are 9 coins worth 19

0 cents.

A = 17, B = 9, total = 26 coins

Answer A

I have run this problem three ways, including

vishalbalwani 's, and working from answer choices. This combination (number of coins), is the only one that works.

Question: Is the method sound? The variables work, but they seem inconsistent.

I use A as a quantity for 5-cent coins. The coefficients of the variables -- 5 and 10 -- are the

values of the coins in cents.

But in the second equation, (Q) I do not think I have switched quantities, per the prompt. I think have switched values. In the second equation, (Q), the 5, a value, is in front of B -- which is supposed to be the

quantity of 10-cent coins.

vishalbalwani did the same.

What am I missing? We are supposed to be switching quantities. But switching values works.

I think

pushpitkc avoids the whole problem by immediately defining y in terms of x from one equation.

Is my method sound?

_________________

At the still point, there the dance is. -- T.S. Eliot

Formerly genxer123