If evan has 10 dimes and quarters in his pocket, and they have combined value of 190 cents, how many of each coin does he have?

Question

Levi Adams · Accepted Answer

6 quarters and 4 dimes.

To solve this problem in two variables (dimes and quarters) it is necessary to write two equations. The first equation is dimes + quarters = #10# The second equation is #0.10# x dimes + #0.25# x quarters =$ 1.90 Now it is clear then dimes = #10# - quarters so this value can be put into the second equation giving #0.10 xx# ( 10 - quarters) + #0.25 xx# quarters = #$ 1.90# #$1.90 = 190# cents. This gives #100# cent - #10# quarters + #25# quarters = #190# cents Now #25# quarters - #10# quarters = #15# quarters. so #15 quarters + #100# cents = #190# cents now subtract #100# cents from both sides #15# quarters + #100# cents - #100# cents = #190# cents - #100# cents giving #15# quarters = #90# cents. It takes 6 quarters If there are #6# quarters there must be only #4# dimes #6 + 4 = 10# #90# cents = #0.90# # (0.90/.15 )xx (100/100 ) = 90/15 = 6 # #6# quarters x #$0.25# /quarter = #$ 1.50# #4# dimes x #$0.10# /dime = #$0.40#

Aurora Arias · Answer

undefined Let $ x $ be the number of dimes Evan has and $ y $ be the number of quarters he has.

Given:
1. $ x + y = 10 $ (since Evan has a total of 10 dimes and quarters)
2. $ 10x + 25y = 190 $ (since the total value of the coins is 190 cents)

Now, you can solve this system of equations to find the values of $ x $ and $ y $.

From equation 1:
$$ x = 10 - y $$

Substitute $ x $ in equation 2:
$$ 10(10 - y) + 25y = 190 $$

$$ 100 - 10y + 25y = 190 $$

$$ 100 + 15y = 190 $$

$$ 15y = 90 $$

$$ y = \frac{90}{15} = 6 $$

Now that you know $ y = 6 $, substitute this value back into equation 1 to find $ x $:
$$ x = 10 - 6 = 4 $$

So, Evan has 4 dimes and 6 quarters.