# Math Help - coin problem

1. ## coin problem

I lost my coins! This morning I had 7 coins worth 53 cents. How many nickels (5 cent pieces) did I have?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5

2. Originally Posted by sri340
I lost my coins! This morning I had 7 coins worth 53 cents. How many nickels (5 cent pieces) did I have?

(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
So I'm assuming we have pennies, nickels, dimes, quarters.

Immediately we can see that there are three pennies. So now you have 4 coins worth 50 cents. Obviously there are no more pennies. Four dimes is not enough to make 50 cents so there must be at least one quarter. And there can't be two quarters because that would already be 50 cents.

So we have 3 coins worth 25 cents. Quick inspection reveals two dimes and one nickel.

3. Originally Posted by sri340
I lost my coins! This morning I had 7 coins worth 53 cents. How many nickels (5 cent pieces) did I have?
(A) 1 (B) 2 (C) 3 (D) 4 (E) 5
May I humbly suggest, Sir Sri340, that you show some effort.
We may soon get tired of doing YOUR work. Understand?

4. Hello, sri340!

I have an algebraic solution . . . sort of.

This morning I had 7 coins worth 53 cents.
How many nickels (5 cent pieces) did I have?

. . $(A)\;1\qquad (B)\;2 \qquad(C)\;3 \qquad (D)\; 4\qquad (E)\; 5$
Let: . $\begin{Bmatrix}Q &=&\text{no. of quarters} \\ D &=& \text{no. of dimes} \\ N &=& \text{no. of nickels} \\ P &=& \text{no. of pennies} \end{Bmatrix}$

$\begin{array}{ccccc}\text{Their value is 53 cents:} & 25Q + 10D + 5N + P &=& 53 & [1] \\
\text{There are 7 coins:} & Q + D + N + P &=& 7 & [2] \end{array}$

We see that $P = 3$

. . $\begin{array}{cccccccccc}25Q + 10D + 5N + 3 &=& 53 && \Rightarrow && 5Q + 2D + N &=& 10 & [3] \\
Q + D + N + 3 &=& 7 && \Rightarrow && Q + D + N &=& 4 & [4]\end{array}$

Subtract [1] - [2]: . $4Q + D \:=\:6$

We see that: . $Q \leq 1$, and we know that: . $D \leq 4$

. . If $Q = 0,\:D = 6$ . . . no

. . Hence: . $Q = 1,\;D = 2$ . . . and $N = 1$