Question 161472: Moolah has 27 coins that are all dimes and quarters. The value of th coins is $4.35. How many dimes and how many quarters does Moolah have? Found 3 solutions by checkley77, Alan3354, nerdybill:Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! Q+D=27 OR Q=27-D
.25Q+.10D=4.35
.25(27-D)+.10D=4.35
6.75-.25D+.10D=4.35
-.15D=4.35-6.75
-.15D=-2.40
D=-2.40/-.15
D=16 DIMES.
D+16=27
Q=27-16
Q=11 QUARTERS.PROOF:
.25*11+.10*16=4.35
2.75+1.60=4.35
4.35=4.35
You can put this solution on YOUR website! Moolah has 27 coins that are all dimes and quarters. The value of th coins is $4.35. How many dimes and how many quarters does Moolah have?
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D + Q = 27 (a total of 27 coins)
10D + 25Q = 435 (each dime is 10 cents, each quarter 25 cents. See that?)
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Now it's 2 eqns in 2 variables, no longer a "word problem"
Substitution is the easiest approach.
Since D + Q = 27, D = 27-Q.
Sub for D in the 2nd eqn
10*(27-Q) + 25Q = 435
270 - 10Q + 25Q = 435
270 + 15Q = 435
15Q = 165
Q = 11 (11 quarters)
D = 27-Q
D = 16
That's it.
You can put this solution on YOUR website! Let d = number of dimes
and q = number of quarters
.
From "Moolah has 27 coins" we get equation 1:
d+q = 27
.
From "The value of th coins is $4.35" we get equation 2:
.10d + .25q = 4.35
.
Solving equation 1 for d:
d+q = 27
d = 27-q
.
Substitute the above into equation 2 and solve for q:
.10d + .25q = 4.35
.10(27-q) + .25q = 4.35
2.7 - .10q + .25q = 4.35
.15q = 1.65
q = 11 (number of quarters)
.
Substitute the above back into equation 1 and solve fo d:
d+q = 27
d+11 = 27
d = 16 (number of dimes)
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