Ismail & Rashad have $3.85 each person has only one type of coin. Rashad had 7 more.how many coins,and which kind could each person have?Have to...

As each of them has coins of only one type let it be worth
x dollars. Let the total number of coins be N.

Now the
total value of the coins is $3.85 or N*x= 3.85

Rashad has 7
coins more than Ismail. So if Rashad has n coins Ismail has n-7. The sum of n and n-7 is
N

So now we get 2n-7=N

Also
(2n-7)*x = 3.85

Now in the expression lets substitute
different values for n and x. We can only take those solutions as valid that yield a
whole number for n.

If x= .01 , 2n-7 = 385 => n=
(385 - 7)/2 = 189.

So the Rashad has 196
coins and Ismail has 189 coins of 0.01
each.

If x=0.05 , 2n-7 = 77 => n= (77
- 7)/2 = 35.

So Rashad has 42 coins and
Ismail has 35 coins of 0.05 each.

If x= 0.35
, 2n-7 = 11 => n= (11 - 7)/2 = 2.

So
Rashad has 9 coins and Ismail has 2 coins of 0.35
each.

If x= 0.07 , 2n-7 = 55 => n=
(55 - 7)/2 = 24.

So Rashad has 31 coins and
Ismail has 24 coins of 0.07 each.