(+/-) <span>1 x (+/-) 232 = 232
</span>(+/-) <span>2 x (+/-)116 = 232
</span>(+/-) 4 x (+/-) <span>58 = 232
</span>(+/-) 8 x (+/-) <span>29 = 232
</span>(+/-) 29 x (+/-) <span>8 = 232
</span>(+/-) 58 x (+/-) <span>4 = 232
</span>(+/-) 116 x (+/-) <span>2 = 232
</span>(+/-) 232 x (+/-) 1 = 232
If they set aside the 8 extra quarters, the remaining coins will be pairs consisting of one quarter and one dime--each pair worth $0.35. The value of those pairs is $5.15 -2.00 = $3.15, so there must be $3.15/$0.35 = 9 of them.
There are 9 dimes and 17 quarters.