Question 1098807
<br>First I'll show a solution using the standard algebraic method found in most textbooks, at least in the US.  Then I'll show you an alternative method which I think is much faster and easier.<br>
By algebra...:
let q = number of quarters
let n = number of nickels
then (1) {{{q+n = 20}}}  (the total number of coins is 20)
and (2) {{{25q + 5n = 400}}}  (the total value of the coins, in cents, is 400)<br>
Now solve the pair of equations
(3) {{{5q+5n = 100}}}  (equation (1), multiplied by 5)
(4) {{{20q = 300}}}  (equation (2) minus equation (3))
{{{q = 15}}}
There are 15 quarters and 5 nickels.<br>
Using the method of alligation....
(1) If all 20 coins were quarters, the total value would be $5; if all were nickels, the total value would be $1.
(2) The actual total value, $4, is three-fourths of the way from $1 to $5.
(3) Therefore, 3/4 of the coins must be quarters; 3/4 of 20 is 15.
There are 15 quarters and 5 nickels.