SOLUTION: two coin banks contain only dimes and quarters. the total value of the coins in the first bank is 15 the second bank, there are 1 more dimes then in the first bank and half as many
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Question 516037: two coin banks contain only dimes and quarters. the total value of the coins in the first bank is 15 the second bank, there are 1 more dimes then in the first bank and half as many quarters. the total value of the coins in the second bank is 10.25 find the number of dimes in the first bank. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! two coin banks contain only dimes and quarters.
the total value of the coins in the first bank is 15
the second bank, there are 1 more dimes than in the first bank and half as many quarters.
the total value of the coins in the second bank is 10.25 find the number of dimes in the first bank.
:
Let d = no. of dimes in the 1st bank
Let q = no. of quarters in the 1st bank
then
(d+1) = no. of dimes in the 2nd bank
.5q = no. of quarters in the 2nd bank
:
Write a equation for each bank
:
.10d + .25q = 15
and
.10(d+1) + .25(.5q) = 10.25
.10d + .10 + .125q = 10.25
.10d + .125q = 10.25 - .10
.10d + .125q = 10.15
:
Use these two equation for elimination
.10d + .25q = 15
.10d +.125q = 10.15
----------------------subtraction eliminates d, find q
0 + .125q = 4.85
q =
q = 38.8, I don't get an integer, is these something wrong with this problem???
:
:
If the amt in the 2nd bank was 10.35, I would get 38 quarters even, and could solve this (55 dimes) could you check on this???
