Question

Question 13 options:

Answer 1

Answer:

The jar has 32 dimes and 18 quarters

Step-by-step explanation:

To solve this problem we can create a system of linear equations in terms of two-variables (say $x$ and $y$) and solve it. To begin let us analyze the problem further. We know that the values of each coin type are:

Dimes ( $x$ )     = $0.10

Quarters( $y$ )  = $0.25

Total Value   = $7.70

Total Coins   = 50

Now let us set up our system of equations as:

$0.10x+0.25y=7.70$      Eqn.(1)

$x+y=50$                     Eqn.(2)

Lets take Eqn.(2) and rerrange it to solve for $x$ as:

$x=50-y$                     Eqn.(3)

Now lets plug this, in Eqn.(1) so we get the value of $y$ as:

$0.10(50-y)+0.25y=7.70\\\\5-0.10y+0.25y=7.70\\\\-0.10y+0.25y=7.70-5\\\\0.15y=2.7$

$y=\frac{2.70}{0.15} \\\\y=18$

Plugging in $y=18$ back in Eqn.(3) we finally have:

$x=50-18\\x=32$

Thus we conclude that in the jar the coins are:

Dimes ( $x$ ) $=32$

Quarters( $y$ ) $=18$