PETE C. answered • 02/05/20
SAT MATH
Rylan, I believe you meant to say a total of $8.80.
With a total of $8.90, there is NO SOLUTION.
HOW MANY OF EACH COIN?
Let Q = number of QUARTERS
Let D = number of DIMES
.25Q = amount in quarters
.10D = amount in dimes
D + Q = 70 (There are 70 COINS.)
.10D + .25Q = $8.80 (Total of 70 coins is $8.80.)
Solve the top equation for D.
D + Q - Q = 70 - Q (Subtract Q from both sides.)
D = 70 - Q
Eliminate the decimals in the second equation by MULTIPLYING BY 100.
10D + 25Q = 880
Now, substitute 70 - Q into the equation above for D.
10(70 - Q) + 25Q = 880
700 - 10Q + 25Q = 880 (Distribute 10 times 70 and Q.)
700 + 15Q = 880 (Combine like terms.)
700 - 700 + 15Q = 880 - 700 (Subtract 700 from both sides.)
15Q = 180
15Q/15 = 180/15 (Divide both sides by 15.)
Q = 12
THERE ARE 12 QUARTERS TOTALING 12 x .25 = $3.00
The number of DIMES:
D + Q = 70
D + 12 = 70 (Since there are 12 QUARTERS.)
D = 58 (Subtract 12 from both sides.)
THERE ARE 58 DIMES TOTALING 58 x .10 = $5.80
ANSWER: 58 DIMES AND 12 QUARTERS
CHECKING:
Does D + Q = 70? 58 dimes + 12 quarters = 70 coins - YES.
Does .10D + .25Q = 8.80? (.10)(58) + (.25)(12) = 5.80 + 3.00 = 8.80 - YES.
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IF YOU USE $8.90 LIKE YOU ASKED.
D + Q = 70
.10D + .25Q = 8.90
10D + 25Q = 890 (Multiply second equation by 100.)
D = 70 - Q (Solve for D in top equation.)
10(70 - Q) + 25Q = 890 (Substitute 70 - Q for D into 10D + 25Q = 890.)
700 - 10Q + 25Q = 890 (Distribute 10 times 70 and Q.)
700 + 15Q = 890 (Combine like terms.)
15Q = 190 (Subtract 700 from both sides.)
Q = 190/15 (Divide both sides by 15.)
Q = 12.6666 YOU CANNOT HAVE 2/3 OF A QUARTER.
OK, so try 12 QUARTERS, which equals $3.00.
Since there are 70 coins, then there must be 58 DIMES, which is $5.80.
$3.00 + $5.80 = $8.80 and NOT $8.90.
OK, round up to 13 QUARTERS, which is $3.25.
Since there are 70 coins, there must be 57 DIMES, which is $5.70.
$3.25 + $5.70 = $8.95 and NOT $8.90
Hope this was helpful Rylan.
