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 find four different answers

william1941 | Student

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.

 

neela | Student

Let the value of each coin that Ismail and Rashad have  be    in dollars. As given we assumed that both have the same kind of coin.

Rsahad has 7 more coin than Ismail. So if Iamil has x coins, then Rshad has x+7.

So the total coins with the two people is x+x+7 = 2x+7.

So the value of the 2x+7 coins = $3.85

So the required equation to be solved is (2x+7)coins. = $3.85.

Let us put which value y of the coin  among values {$0.01 or $0.05 , $0.01, $0.25 , $0.50,  $1} dollar satisfies the equation with a solution  of a positive integer for x.

(2x +7)y= 3.85 Or

2x +7= 3.85/y

2x = 3.85/y - 7

x = (3.85-7y)/2y

x = 7(0.55 -y)/2y

x =189 if y = $0.01

x = 32 if y = $0.05

Therefore Ismail has 189 and rashad has189+97=196 coins of value $0.01 . So the total coins = 189+196 = 385 coins each of which has $0.01value (or cents).

Another soltion is Ismail has 32 coins each of ehich has value $0.05. And Rashad has  32+7 = 39. So the total coins together they possess = 32+39 = 71. The value of the 71 coins = 71*$0.05 = $3.85