SOLUTION: A jar of dimes and quarters contained $10.95. There were only 93 coins in the jar. How many dimes were in the jar?

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Question 330520: A jar of dimes and quarters contained $10.95. There were only 93 coins in the jar. How many dimes were in the jar?
Found 2 solutions by Fombitz, vksarvepalli:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let D be the number of dimes, Q the number of quarters.
1.D%2BQ=93
2.10D%2B25Q=1095
From eq. 1,
D=93-Q
Substitute into eq. 2,
10%2893-Q%29%2B25Q=1095
930-10Q%2B25Q=1095
15Q=165
Q=11
From eq. 1,
D%2B11=93
D=82
There are 82 dimes and 11 quarters.

Answer by vksarvepalli(154) About Me  (Show Source):
You can put this solution on YOUR website!
let the no. of dimes be x
and no. of quarters be y
There were only 93 coins in the jar
so x+y=93 ----- eq. 1
also one dime = 10 cents
and one quarter = 25 cents
so 10x+25y=1095
10x+10y=930 (from equation 1)
=> 15y=165 (subtracting one eq. from another)
=> y=11
and x=82

so no. of dimes in the jar=82