Answer:
B. 13 quarters
Step-by-step explanation:
Let:
quarters = x = $0.25
dimes = y = $0.10
Get the set of equation:
x + y = 20
0.10y + 0.25x = 3.95
First, solve for x. Use the first equation. Isolate the x by subtracting y from both sides:
x + y (-y) = 20 (-y)
x = 20 - y
Plug in 20 - y for x in the second equation:
0.10y + 0.25(2020 - y) = 3.95
Simplify. Remember to follow PEMDAS. Isolate the variable, y.
First, multiply 0.25 to all terms within the parenthesis:
0.10y + (0.25 * 20) + (0.25 * -y) = 3.95
0.10y + 5 - 0.25y = 3.95
Combine like terms:
(-0.25y + 0.10y) + 5 = 3.95
(-0.15y) + 5 = 3.95
Isolate the variable, y. Do the opposite of PEMDAS. First, subtract 5 from both sides:
(-0.15y) + 5 (-5) = 3.95 (-5)
(-0.15y) = 3.95 - 5
-0.15y = -1.05
Isolate the variable, y. Divide -0.15 from both sides:
(-0.15y)/-0.15 = (-1.05)/-0.15
y = 1.05/0.15 = 7
y = 7
There are 7 dimes.
Plug in 7 to one of the equations in the operation for y (dimes):
x + y = 20
x + (7) = 20
Isolate the variable, x. Note the equal sign, what you do to one side, you do to the other. Subtract 7 from both sides:
x + 7 (-7) = 20 (-7)
x = 20 - 7
x = 13
There are 13 quarters.
Check: Plug in 13 for x, and 7 for y in the second equation:
0.10y + 0.25x = 3.95
0.10(7) + 0.25(13) = 3.95
0.70 + 3.25 = 3.95
3.95 = 3.95
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Note that you are solving only for quarters. In this case, your answer will be B. 13.
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Answer:
B. 13 Quarters
Step-by-step explanation:
So since we don't have an amt. specified like e.g there are 10 more quarters than dimes, we can only rely on guess and check.
First we can try 10 quarters --> 10 x 25 = 250 --> 395 - 250 = 145 = 14.5 dimes
=> 10 is not equal to 14.5 so we know that the amt. of quarters must be more than 10 and must be an odd amt. => so now we can try 13
13 x 25 = 325 --> 395 - 325 = 70 => 70/10= 7 --> 13 + 7 = 20
Thus, we have our answer of 13
Hope this helps!
