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

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- May 19th 2010, 07:53 AMsri340coin 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 - May 19th 2010, 08:03 AMundefined
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.

So the answer is (A). - May 19th 2010, 08:30 AMWilmer
- May 19th 2010, 08:33 AMSoroban
Hello, sri340!

I have an algebraic solution . . . sort of.

Quote:

This morning I had 7 coins worth 53 cents.

How many nickels (5 cent pieces) did I have?

. . $\displaystyle (A)\;1\qquad (B)\;2 \qquad(C)\;3 \qquad (D)\; 4\qquad (E)\; 5$

$\displaystyle \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 $\displaystyle P = 3$

. . $\displaystyle \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]: .$\displaystyle 4Q + D \:=\:6 $

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

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

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

You had one nickel . . . answer (A).