SOLUTION: There was $2900 in the pot. If there were 293 more $1 bills than $10 bills, how many bills of each kind were there?

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Question 1202419: There was $2900 in the pot. If there were 293 more $1 bills than $10 bills, how many bills of each kind were there?
Found 4 solutions by math_tutor2020, mananth, greenestamps, josgarithmetic:
Answer by math_tutor2020(3817)   (Show Source):
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x = number of ten dollar bills
x+293 = number of one dollar bills

1*(x+293) = x+293 = total value of all the $1 bills only
10x = total value of all the $10 bills only
(x+293)+(10x) = 11x+293 = total value in dollars of both type of bills combined

total = $2900
11x+293 = 2900
11x = 2900-293
11x = 2607
x = 2607/11
x = 237 is the number of $10 bills
x+293 = 530 is the number of $1 bills

Check:
A = 530 one dollar bills = 530*1 = $530
B = 237 ten dollar bills = 237*10 = $2370
A+B = $530+$2370 = $2900
Solution is verified.

Answer:
530 one dollar bills
237 ten dollar bills

Answer by mananth(16946)   (Show Source):
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There was $2900 in the pot. If there were 293 more $1 bills than $10 bills, how many bills of each kind were there?
Let number of $ 10 be x
Number of $ 1 bills = x+293
Total amout = $2900
Equation of amount
10x+1$(x+293) = 2900
10x+x+293 =2900
11x = 2607
x = 237
complete it

Answer by greenestamps(13206)   (Show Source):
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Here is an informal solution using logical reasoning and simple arithmetic. Compare it to the solution from tutor @math_tutor2020 to see that the formal algebra does exactly what the informal solution does.

Subtract the "extra" 293 $1 bills from the total, leaving equal numbers of $1 and $10 bills with a total value of $2900-$293 = $2607.

The total value of one $1 bill and one $10 bill is $11.

So the number of bills of each kind remaining is $2607/$11 = 237.

ANSWERS:
$10 bills: 237
$1 bills: 237+293 = 530

CHECK: 237(10)+530(1) = 2370+530 = 2900

Answer by josgarithmetic(39627)   (Show Source):
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BILLS QTY. DOLLARS 1$ y+239 y+239 10$ y 10y Total 2900

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