SOLUTION: Jarome has $880 in ten and twenty dollar bills. If he has 48 bills in total, how many of each bill does he have?
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Question 1207588: Jarome has $880 in ten and twenty dollar bills. If he has 48 bills in total, how many of each bill does he have?
Found 3 solutions by math_tutor2020, josgarithmetic, greenestamps:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
x = number of ten dollar bills
48-x = number of twenty dollar bills
10x = value of all the ten dollar bills
20(48-x) = value of all the twenty dollar bills
10x+20(48-x) = total value
10x+20(48-x) = 880
10x+960-20x = 880
-10x+960 = 880
-10x = 880-960
-10x = -80
x = -80/(-10)
x = 8
48-x = 48-8 = 40
Answer:
8 ten dollar bills
40 twenty dollar bills
Check:
8+40 = 48 bills total
8*10+40*20 = 880 dollars in total value
Both conditions are met to verify the answers above.
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
y of the twenty-dollar bills
48-y of the ten-dollar bills
----account for the money
------------so then 8 of the ten-dollar bills;
40 of the twenty-dollar bills.
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
If the speed of getting the answer is important and formal algebra is not required -- e.g., in a timed competitive math contest -- here is a quick and easy way to solve this kind of problem using logical reasoning and simple mental arithmetic.
If all 48 bills were $10 bills, the total would be $480.
The actual total is $880, which is $400 more than $480.
Each $20 bill is worth $10 more than each $10 bill. The number of $20 bills needed to make the additional $400 is 400/10 = 40.
ANSWER: 40 $20 bills; 48-40 = 8 $10 bills
NOTE: Even if the problem required a formal algebraic solution, solving problems using logical reasoning and mental arithmetic is excellent brain exercise.
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