Jason S. answered • 11/26/13

My goal is the success of my students. Knowledge-Patience-Honesty

Jason S.

11/26/13

Katelyn B.

asked • 11/26/13A 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?

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Jason S. answered • 11/26/13

Tutor

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(116)
My goal is the success of my students. Knowledge-Patience-Honesty

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

Add together

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

Add: d = 11

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.

Jason S.

You are most welcome Katelyn.

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11/26/13

Arthur D. answered • 11/26/13

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Forty Year Educator: Classroom, Summer School, Substitute, Tutor

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. answered • 11/26/13

Tutor

4.8
(222)
Physics, math tutor with great knowledge and teaching skills

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.

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Katelyn B.

11/26/13