Question 409090: I need help with a word problem. This is what I have so far:
Joe has a collection of nickels and dimes that is worth $3.95. If the number of dimes was doubled and the number of nickels was increased by 21, the value of the coins would be $8.40. How many nickels and dimes does he have?
first part:
dimes value + nickles value = total amount
(value of a dime)(# of dimes) + (value of a nickle)(# of nickles) = total amount
(0.10)(# of dimes) + (0.05)(# of nickles) = 3.95
(0.10)(d) + (0.05)(n) = 3.95
so d = number of dimes we start with
n = number of nickles we start with
second part:
dimes value + nickles value = total amount
(value of a dime)(# of dimes) + (value of a nickle)(# of nickles) = total amount
(0.10)(2*d) + (0.05)(n + 21) = 8.40
because we're supposed to have double (*2 the dimes) and 21 more nickles
The two equations are then:
(0.10)(d) + (0.05)(n) = 3.95
(0.10)(2d) + (0.05)(n + 21) = 8.40
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
What you have done so far is absolutely spot on correct, but I'm going to show you a little trick that will get rid of all of the ugly decimal coefficients so that you are much less likely to make any arithmetic errors.
While it is true that a dime has a value of 0.10 dollars, it also can be said to have a value of 10 cents. Likewise a nickel is worth 5 cents. Furthermore, the total value of the coins in either situation can be said to be 395 cents or 840 cents.
Now we can write the equations as:
And
\ = 840)
Much tidier, no?
A little manipulation gets us:
Multiply the first equation by -1:
Then add the two equations, term by term:
And that leads us to
Meaning we started with 34 dimes, and it should follow directly from that fact that we started with 11 nickels accounting for the 55 cents remaining in the $3.95.
Checking our work: If we have twice the dimes, or 68 dimes, we have $6.80 in dimes and an additional 21 nickels would be 55 cents plus $1.05 or $1.60. Finally, $6.80 plus $1.60 is indeed $8.40. Ta Da!
John
My calculator said it, I believe it, that settles it
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