Question

A student has some 1 bills and $5 bills in his wallet. he has a total of 15 bills that are worth $47. how many of each type of bill does he have?

Answer 1

B. seven $1 bills and eight $5 bills.

Explanation:

n = number of $1 bills

m = number of $5 bills

Given:

n + m = 15

n($1) + m($5) = $47

Just solve 2 equations in 2 unknowns n and m:

(i) n + m = 15

(ii) n + 5m = 47

Subtract (i) from (ii) and get

4m = 32

m = 32/4 = 8

Then from (i) get

n = 15 - m = 15 - 8 = 7

There are 7 $1 bills and 8 $5 bills.

Answer 2

*Given
Total number of bills - 15
Total value of the bills - $47
Bills - $1 and $5

*Solution
Let:
x - number of $1 bills
y - number of $5 bills

1. The total number of bills comprising of $1 and $5 bills is 15. Thus,

x + y = 15 (EQUATION 1)

2. The total value of the bills is $47. Thus,

$1 (x) + $5 (y) = $47 (EQUATION 2)

3. There are 2 ways to solve this (system of linear equations) mathematically. You can use the elimination method or the substitution method. Using the elimination method to eliminate the variable x... (subtract Equation 1 from Equation 2)

1x + 5y = 47
_
x + y = 15
0 + 4y = 32

4. Thus, the value of y, which is the number of $5 bills is,

4y = 32
4 4

y = 8

5. The number of $1 bills (x) can then be solved using either Equations 1 or 2. Using Equation 1 to determine x,

x + y = 15
x = 15 - y
x = 15 - 8
x = 7

From the solution, we have determined that there are 8 pieces of $5 bills and 7 pieces of $1 bills.