Question 1202628: A cash register contains only five dollar and ten dollar bills. It contains twice as many five's as ten's and the total amount of money in the cash register is 540 dollars. How many ten's are in the cash register?
Found 4 solutions by math_tutor2020, josgarithmetic, Edwin McCravy, ikleyn:
Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!
x = number of ten dollar bills
2x = number of five dollar bills
10x = value of the ten dollar bills only
5*2x = 10x = value of the five dollar bills only
tens+fives = 10x+10x = 20x = total value
20x = 540
x = 540/20
x = 27
Answer: 27 ten dollar bills
Answer by josgarithmetic(39613) (Show Source):
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
It takes half as many tens as it does fives to make any amount. So half of the
$540, or $270, would have to be in tens, which would take 27 tens.
Who needs algebra? J
Edwin
Answer by ikleyn(52750)  (Show Source):
You can put this solution on YOUR website! .
This problem is good to solve it MENTALLY without using equations.
Use the standard classic " grouping method ".
According to the problem, you can group the bills in sets,
each containing two 5-dollar bills and one 10-dollars bill.
The value of each such set is 5+5+10 = 20 dollars,
and the number of such groups is, OBVIOUSLY, 540/20 = 27.
Since each set contains precisely one 10-dollar bills,
the number of the 10-dollar bills is 27.
ANSWER. The number of the 10-dollar bills is 27.
Solved MENTALLY.
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