Question 1133457: A box contains 65 coins, only dimes and nickels. The amount of money in the box is $5.40, how many dimes and nickels are in the box? Found 2 solutions by ikleyn, ankor@dixie-net.com:Answer by ikleyn(52798) (Show Source):
Let's assume for a minute that all 34 coins are nickels - then the total would be 5*65 = 325 cents,
making the shortage of 540 - 325 = 215 cents.
Hence, we should replace some number of nickels by dimes to compensate the difference.
How much replacement we should do ?
- Obviously, = = 43 replacements diminishing the difference of 215 cents by 5 = 10-5 cents at any single replacement.
So, the answer is: the original collection has 43 dimes and the rest 65-43 = 22 coins are nickels.
Check. 43*10 + 22*5 = 540 cents. ! Correct !
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You can put this solution on YOUR website! A box contains 65 coins, only dimes and nickels.
d + n = 65
or for substitution the form is
n = (65-d)
The amount of money in the box is $5.40,
.10d + .05d = 5.40
replace n with (65-d)
.10d + .05(65-d) = 5.40
.10d + 3.25 - .05d = 5.40
.10d - .05d = 5.40 - 3.25
.05d = 2.15
d = 2.15/.05
d = 43 dimes
and
65 - 43 = 22 nickels
;
:
See if that checks out, find the actual amt of each
43(.10) = 4.30
22(.05) = 1.10
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Total $ = 5.40
