Question 1099859: A box contains $7.10 in nickels, dimes, and quarters. There are 42 coins in all, and the sum of the numbers of nickels and dimes is 2 less than the number of quarters. How many coins of each kind are there?
Found 2 solutions by ikleyn, richwmiller:
Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website! .
Let Q be the number of quarters.
Then the number of nickels and dimes is (42-Q), and the condition says that
this number (42-Q) is 2 less than Q.
It gives you an equation
42 - Q = Q - 2, which implies
2Q = 42 - 2 = 40 and then Q =  = 20.
So, we have 20 quarters that sum up to 20*25 = 5 dollars.
Taking these 20 quarters away, we have 42-20 = 22 coins that are nickels and dimes and worth 710 - 500 = 210 cents.
If you think one minute on it, you will find the solution MENTALLY: 20 dimes and 2 nickels.
Answer. 20 quarters, 20 dimes and 2 nickels.
The key to the solution of this problem is to choose the leading unknown correctly.
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! 42 - q = q - 2
add 2 to both sides
44-q=q
add q to both sides
44=2q
q=22 quarters
22 q=$5.50
42-22=20 more coins
7.10-5.50=$1.60
160=5n+10d
n+d=20
n=20-d
160=5(20-d)+10d
160=100-5d+10d
60=5d
d=12 dimes
n=8 nickels
q=22 quarters
22=12+8+2
22*.25=5.50
8*.05=.40
12*.10=1.20
5.50+.40+1.20=7.10
12+8+22=42
The sum of the numbers of nickels and dimes (20) is 2 LESS than the number of quarters (22)
20 is less than 22
ok
checking other tutors answers
20 quarters, 20 dimes and 2 nickels
20 *.25=5.00
20 *.10=2.00
2 *.05=.10
5.00+2.00+.10=$7.10 ok
The sum of the numbers of nickels and dimes (22) is 2 LESS (not more) than the number of quarters (20)
22 is more than 20
20 quarters, 20 dimes and 2 nickels does NOT satisfy the condition.
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