In a jar, there are 50 coins with a total value of $\$ 5.00$. The coins are quarters (worth $\$ 0.25$ each), dimes (worth $\$ 0.10$ each), and nickels (worth $\$ 0.05$ each). The number of nickels in the jar is three times the number of quarters. The number of dimes is one more than the number of nickels. How many quarters are in the jar?
(A) 7
(B) 6
(C) 5
(D) 9
(E) 8
Solution 1

The number of dimes in the jar is one more than the number of nickels.

If we remove one dime from the jar, then the number of coins remaining in the jar is $50-1=49$, and the value of the coins remaining in the jar is $\$ 5.00-\$ 0.10=\$ 4.90$.

Also, the number of dimes remaining in the jar is now equal to the number of nickels remaining in the jar, and the number of nickels remaining in the jar is three times the number of quarters remaining in the jar.

That is, for every 1 quarter remaining in the jar, there are 3 nickels and 3 dimes.

Consider groups consisting of exactly 1 quarter, 3 nickels and 3 dimes.

In each of these groups, there are 7 coins whose total value is $\$ 0.25+3 \times \$ 0.05+3 \times \$ 0.10=\$ 0.25+\$ 0.15+\$ 0.30=\$ 0.70$.

Since there are 49 coins having a value of $\$ 4.90$ remaining in the jar, then there must be 7 such groups of 7 coins remaining in the jar (since $7 \times 7=49$ ).

(We may check that 7 such groups of coins, with each group having a value of $\$ 0.70$, has a total value of $7 \times \$ 0.70=\$ 4.90$, as required.)

Therefore, there are 7 quarters in the jar.

## Solution 2

To find the number of quarters in the jar, we need only focus on the total number of coins in the jar, 50 , or on the total value of the coins in the jar, $\$ 5.00$.

In the solution that follows, we consider both the number of coins in the jar as well as the value of the coins in the jar, to demonstrate that each approach leads to the same answer.

We use a trial and error approach.

Suppose that the number of quarters in the jar is 5 (the smallest of the possible answers given). The value of 5 quarters is $5 \times 25 \mathrm{c}=125 \mathrm{q}$.

Since the number of nickels in the jar is three times the number of quarters, there would be $3 \times 5=15$ nickels in the jar.

The value of 15 nickels is $15 \times 5 \mathrm{\phi}=75 \mathrm{\phi}$.

Since the number of dimes in the jar is one more than the number of nickels, there would be $15+1=16$ dimes in the jar.

The value of 16 dimes is $16 \times 10 \phi=160 \phi$.

If there were 5 quarters in the jar, then the total number of coins in the jar would be $5+15+16=36$, and so there must be more than 5 quarters in the jar.

Similarly, if there were 5 quarters in the jar, then the total value of the coins in the jar would be $125 \phi+75 \phi+160 \phi=360$.

Since the value of the coins in the jar is $\$ 5.00$ or 500 \& then the number of quarters in the jar is greater than 5 .

We summarize our next two trials in the table below.

| Number of  Quarters | Value of  Quarters | Number of  Nickels | Value of  Nickels | Number of  Dimes | Value of  Dimes | Total Value  of Coins |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 6 | $150 \phi$ | 18 | $90 \phi$ | 19 | $190 \phi$ | $430 \phi$ |
| 7 | $175 \phi$ | 21 | 105 | 22 | $220 థ$ | 500 |

When there are 7 quarters in the jar, there are $7+21+22=50$ coins in the jar, as required. When there are 7 quarters in the jar, the value of the coins in the jar is $175 \phi+105 \phi+220 \phi=500$ क or $\$ 5.00$, as required.

In either case, the number of quarters in the jar is 7 .

ANswer: (A)