Question 330520
let the no. of dimes be x
and no. of quarters be y

There were only 93 coins in the jar

so x+y=93 ----- eq. 1 

also one dime = 10 cents
and one quarter = 25 cents

so 10x+25y=1095

   10x+10y=930 (from equation 1)

 => 15y=165 (subtracting one eq. from another)

 => y=11

and x=82


so no. of dimes in the jar=82