SOLUTION: Micky and Yarberry saved nickels and dimes. They had a total of 34 coins whose value was $2.70. How many of each kind did they have?

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Question 1202342: Micky and Yarberry saved nickels and dimes. They had a total of 34 coins whose value was $2.70. How many of each kind did they have?
Found 4 solutions by math_tutor2020, josgarithmetic, MathTherapy, greenestamps:
Answer by math_tutor2020(3817) About Me  (Show Source):
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Answers:
14 nickels, 20 dimes


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Work Shown:

Method 1

n = number of nickels
34-n = number of dimes

5n = value of the nickels (cents)
10(34-n) = 340-10n = value of the dimes (cents)
5n+340-10n = -5n+340 = total value (cents)

$2.70 = 2.70*100 = 270 cents

-5n+340 = 270
-5n = 270-340
-5n = -70
n = -70/(-5)
n = 14
They have 14 nickels

And also
34-n = 34-14 = 20 dimes

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Method 2

Let's say we start with all 34 coins as nickels
34 nickels = 34*5 = 170 cents

The gap from 170 cents to the target goal of 270 cents is 270-170 = 100 cents.
To get to the target, we'll need to replace some nickels with dimes.

Each time we replace a nickel with a dime, the net value increase is 5 cents (because 10 cents - 5 cents = 5 cents)

Let x be the number of times we replace a nickel with a dime.
5x then represents the total net increase. We want this to equal the 100 cents calculated earlier.

5x = 100
x = 100/5
x = 20
We must replace 20 nickels with 20 dimes

34 nickels drops to 34-20 = 14 nickels
0 dimes increases to 0+20 = 20 dimes

------------------------

Check:

14 nickels + 20 dimes = 34 coins total
That validates the coin count requirement.

A = 14 nickels = 14*5 = 70 cents
B = 20 dimes = 20*10 = 200 cents
A+B = 70 cents + 200 cents = 270 cents = $2.70
That validates the total value requirement.

Both conditions have been met. The answers are confirmed.

Answer by josgarithmetic(39623) About Me  (Show Source):
You can put this solution on YOUR website!
n nickels
d dimes
n%2Bd=34
picking d as the single variable is a little more convenient than picking n.
n=34-d

10d%2B5%2834-d%29=270
-
10d%2B5%2A34-5d=270
10d-5d=270-5%2A34
5d=270-5%2A34
d=%28270-5%2A34%29%2F5
highlight%28d=20%29-----------and highlight%28n=14%29.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
Micky and Yarberry saved nickels and dimes. They had a total of 34 coins whose value was $2.70. How many of each kind did they have?

Let number of nickels be N
Then number of dimes = 34 - N
We then get: .05N + .1(34 - N) = 2.7
              .05N + 3.4 - .1N = 2.7
                    .05N - .1N = 2.7 - 3.4
                        - .05N = - .7
      Number of nickels, or 

      Number of dimes: 34 - 14 = 20

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


If a formal algebraic solution is not required, any problem like this can be solved quickly using logical reasoning and simple mental arithmetic. In this problem...

If all 34 coins were nickels, the total value would be 34(5) = 170 cents, or $1.70.
The actual total was $2.70, which is $1 or 100 cents more than $1.70.
The difference between the value of a nickel and a dime is 5 cents.
The number of dimes required to make that additional $1 of value is 100/5 = 20.

ANSWER: 20 dimes; so 34-10 = 14 nickels

CHECK: 20(10)+14(5) = 200+70 = 270