Question 404677: I just can't seem to get how the word problems become equations.
John has $446. in ten-dollar, five-dollar, and one-dollar bills. There were 94 bills in all, and 10 more five-dollar bills than ten-dollar bills. How many one-dollar bills did John have?
Found 2 solutions by richard1234, nerdybill:
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! It takes time and practice to easily write a word problem into a set of equations...
Suppose the number of $10 bills, $5 bills, and $1 bills are denoted by a, b, c respectively. Since John has $446 in total, we can say that
446 = 10a + 5b + c
Also, there are 94 bills in all, so
94 = a + b + c
Since there are 10 more $5 bills than $10 bills,
b = c + 10
We wish to find the (numerical) value of c, in which I'll let you do, by using a system of equations and solving via substitution. Do you see where I got these equations from?
Answer by nerdybill(7384)  (Show Source):
You can put this solution on YOUR website! John has $446. in ten-dollar, five-dollar, and one-dollar bills. There were 94 bills in all, and 10 more five-dollar bills than ten-dollar bills. How many one-dollar bills did John have?
.
Let x = number ten dollar bills
then
x+10 = number of five dollar bills
94-x -(x+10) = number of one dollar bills
.
10x + 5(x+10) + 94-x -(x+10) = 446
10x + 5(x+10) + 94-x -x-10 = 446
10x + 5x+50 + 84-2x = 446
13x + 134 = 446
13x = 312
x = 24(tens)
.
Ones:
94-x -(x+10)
=94-x -x-10
=84-2x
=84-2(24)
=84-48
= 36 (number of one dollar bills)
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