0

# How many coins of each kind are there (full question below)?

A box contains \$6.70 in nickels, dimes, and quarters. There are 39 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

### 3 Answers by Expert Tutors

Jason S. | My goal is the success of my students. Knowledge-Patience-HonestyMy goal is the success of my students. K...
4.9 4.9 (115 lesson ratings) (115)
1
n = nickels
d = dimes
q = quarters

1)  n + d + q = 39

n+d = q - 3

Translate to:  2)   n + d - q = -3

.05n + .1d + .25q = 6.7   amount of each coin times the number of coins must add to 6.7
Multiply both sides by 20...

3)  1n + 2d + 5q = 134

Combine 1) & 2)
n + d + q = 39
n + d  - q = -3

2n + 2d = 36

Divide by 2:

4)  n + d = 18

Combine 1) and 3)

n + d + q = 39         Multiply by -5 =>  -5n -5d -5q = -195
n + 2d + 5q = 134                              n + 2d + 5q = 134

Add together->                                 5)   -4n - 3d  = -61

Combine 4) and 5)

n + d = 18   Multiply by 4 =>  4n + 4d = 72
-4n - 3d = 61                          -4n - 3d = -61

11 dimes...

Plug that into 4)

n + 11 = 18

n = 18-11 = 7 nickels.

Plug that into 1.

n + d + q = 39

7 + 11 + q = 39
18 + q = 39

q = 39-18 = 21 quarters

21 quarters, 7 nickels, and 11 dimes.

21(.25) + 7(.05) + 11(.1) = 5.25 + .35 + 1.1 = 5.6 + 1.1 = 6.7   Check.

Thank you so much Jason!
You are most welcome Katelyn.
Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (7 lesson ratings) (7)
0
Here's another way to solve.

n+d+q=39 and n+d=q-3
go to the first equation and replace n+d with q-3
q-3+q=39
2q-3=39
2q=42
q=21 quarters
we have \$6.70 so take away 21 quarters or \$5.25 to get \$1.45 left in nickels and dimes
39 coins minus 21 quarters gives us 18 coins in nickels and dimes
n+d=18 and 0.05n+0.10d=\$1.45
n=18-d
0.05(18-d)+0.10d=\$1.45
0.90-0.05d+0.10d=\$1.45
0.90+0.05d=\$1.45
0.05d=\$1.45-\$0.90
0.05d=\$0.55
5d=55
d=55/5=11 dimes and18-11=7 nickels
check:21+11+7=39
0.25(21)+0.10(11)+0.05(7)=\$5.25+\$1.10+\$0.35=\$6.70
Kirill Z. | Physics, math tutor with great knowledge and teaching skillsPhysics, math tutor with great knowledge...
4.9 4.9 (174 lesson ratings) (174)
0
Let x be number of nickels, y--number of dimes, and z--number of quarters. Then

5x+10y+25z=670
x+y+z=39
x+y+3=z

From the second and the third equation we obtain:

x+y=18

From the first and third we get:

30x+35y=595 or 6x+7y=119

Now we can easily determine x and y.

6x+6y=108
6x+7y=119

Thus y=119-108=11
x=18-y=7
z=x+y+3=21

So we have 7 nickels, 11 dimes, and 21 quarter.