Question 1169800

Hi
John has 36 coins  comprising 20c 50c and $1 coins. He has twice as many $1 coins as 20c coins and the value of the 50c coins $4.40 more than the total value of the 20c coins.

How many $1 coins does he have.

Thanks 
<pre>Theo is CONFUSED, and the fact that he makes these problems so COMPLEX when they don't have to be, has more than likely led him down the "WRONG-ANSWER" path. Hence, just IGNORE his response!
Also IGNORE the other response that brings in UNNECESSARY fractions, which most people dread dealing with!

Let number of 20c coins be T
Then he has 2T, $1 coins, and 36 - (T + 2T), or 36 - 3T, 50c coins
We then get: .5(36 - 3T) = .2T + 4.4
18 - 1.5T = .2T + 4.4
- 1.5T - .2T = 4.4 - 18
- 1.7T = - 13.6
Number of 20c coins, or {{{highlight_green(matrix(1,5, T, "=", (- 13.6)/(- 1.7), "=", 8))}}}
Number of $1 coins: {{{highlight_green(matrix(1,3, 2(8), "=", 16))}}}.
That's ALL!!