A man has two more nickels than dimes. He has $1.15 in all. How many coins of each kind does he have?

Accepted Solution

Answer:Man has 9 Nickels and 7 dimes.Step-by-step explanation:Let Number of dimes be xand Number of Nickels be yAs Per given data he has two more nickels than dimes.[tex]y= 2+x\\[/tex]As we know that 1[tex]\$[/tex] = 100 cent1 nickel is worth = 5 cents = [tex]\frac{5}{100}[/tex]= 0.05[tex]\$[/tex]1 dime is worth = 10 cents = [tex]\frac{10}{100}[/tex] 0 .1[tex]\ $[/tex]According to given data he has [tex]\$\ 1.15[/tex] total nickel and dimesSo the equation becomes,[tex]0.05y+0.1x=1.15[/tex]but [tex]y= 2+x[/tex][tex]0.05(2+x)+0.1x=1.15\\0.1+0.05x+0.1x=1.15\\0.1+0.15x= 1.15\\0.15x= 1.15-0.1\\0.15x=1.05\\x=\frac{1.05}{0.15}= 7[/tex]Substituting value of x in equation [tex]y = 2+x[/tex] we get[tex]y= 2+7 =9[/tex]Man has 9 Nickels and 7 dimes.