Question

Marfugge had $72.50$ in quarters and half dollars. If he had 190 coins in all how many of each type did he have?

Answer 1

Hint: Here we are given a total number of coins equal to 190. From them we will consider x coins of half dollar and 190-x coins of quarter dollars. Sum of this value will be equal to $72.50$. On finding the value of x we will get to know the number of coins of quarter and half dollars both.

Complete step-by-step solution:
Marfugge had 190 coins in all. Let x be the number of coins of half dollar and 190-x be the number of coins of quarter dollars.
The sum of coins values $72.50$
So we can form an equation as
$\Rightarrow \dfrac{1}{2}x + \left( {190 - x} \right)\dfrac{1}{4} = 72.50$
Now using the distributive property,
$\Rightarrow \dfrac{1}{2}x + 190 \times \dfrac{1}{4} - \dfrac{1}{4}x = 72.50$
Taking like terms on one side,
$\Rightarrow \dfrac{1}{2}x - \dfrac{1}{4}x + 190 \times \dfrac{1}{4} = 72.50$
On performing the mathematical operations we get,
$\Rightarrow \dfrac{1}{4}x + 47.5 = 72.50$
Taking the constants on one side,
$\Rightarrow \dfrac{1}{4}x = 72.50 - 47.5$
Subtracting we get,
$\Rightarrow \dfrac{1}{4}x = 25$
Thus value of x will be,
$\Rightarrow x = 25 \times 4$
$\Rightarrow x = 100$

Thus number of coins of half quarter is 100 and that if quarter value will be $190 - 100 = 90$

Note: Note that don’t consider x as the number of coins of quarter dollar because that will lead the answer to a negative value and money is never negative. This can be judged by the fact that the value of half a dollar is always greater than that of a quarter dollar.