SOLUTION: Samy counts his change. He has twice as many 10 ¢ coins as he has 5 ¢ and 6 25 ¢ coins more than 10 ¢. He also has 4 $ 1 coins and 7 $ 2 coins. In all, he has $ 29.25. How

Question 1173505: Samy counts his change. He has twice as many 10 ¢ coins as he has 5 ¢ and 6 25 ¢ coins more than 10 ¢. He also has 4 $ 1 coins and 7 $ 2 coins. In all, he has $ 29.25.
How many of each of Samy's pieces does he have?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

let x be the number of 5 cent coins
then 2x is the number of 10 cent coins (twice as many as 5 cent)
then 2x+6 is the number of 25 cent coins (6 more than the number of 10 cent)

The total value of the coins, including 4 $1 coins and 7 $2 coins, is $29.25:

ANSWERS:
5 cent coins: x = 13
10 cent coins: 2x = 26
25 cent coins: 2x+6 = 32

CHECK: 13(5)+26(10)+32(25)+4(100)+7(200) = 65+260+800+400+1400 = 2925

A solution using logical reasoning can follow nearly the same path of calculations.

(1) Count the 4 $1 coins and the 7 $2 coins first. That is $18, leaving $11.25.
(2) Count the "extra" 6 quarters next; that is $1.50, leaving $9.75.
(3) What is left is x 5 cent coins, 2x 10 cent coins, and 2x quarters. Group those coins into groups of 1 5 cent coin, 2 10 cent coins, and 2 25 cent coins.
(4) The value of each of those groups is 5+20+50 = 75 cents. The number of groups, at 75 cents each, needed to make the remaining $9.75 is 975/75 = 13.

So there are 13 5 cent coins, 2*13=26 10 cent coins, and 2*13+6=32 25 cent coins.

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