Question 132011
Let x=total number of coins in the jar

(1/5)x are dimes valued at 0.10*(1/5)x dollars
(1/6)x are  nickels valued at 0.05(1/6)x dollars
(1/10)x are quarters valued at 0.25*(1/10)x
Now lets find out how many pennies before we go further.  We can do that two ways:
1.  We can find out the value of the dimes, nickels and quarters and subtract that from $3.52, or:
2.  We can just find out what fractional part of (x) is left.  Lets do #2:

1/5+1/6 + 1/10  LCM is 30, so:
(6/30)x+(5/30)x+(3/30)x=(14/30)x=dimes, nickels & quarters
So 30/30-14/30=16/30=8/15x number of pennies
Now we know that (8/15)x=number of pennies and 0.01*(8/15)x= value of pennies
We are told that:

0.10(1/5)x + 0.05(1/6)x+0.25(1/10x)+0.01(8/15)x=$3.52

30 is still LCM so lets multiply each term by 30 to get rid of fractions:

0.6x +0.25x+0.75x+0.16x=105.6 ($ are understood)  collect like terms
1.76x=105.6  divide both sides by 1.76

x=60 ------------------------total number of coins in the jar


(1/5)x=(1/5)*60=12--------------------number of dimes
(1/6)x=(1/6)*60=10---------------------number of nickels
(1/10)x=(1/10)*60=6---------------------number of quarters

(8/15)x=(8/15)*60=32----------------------number of pennies

CK
12+10+6+32=60
60=60
also
0.10*12+0.05*10+0.25*6+0.01*32=$3.52

$1.20+$0.50+$1.50+$0.32=$3.52
$3.52=$3.52

Extra: I went to the bank and got coin wrappers. It takes $10 worth of quarters to fill a quarter wrapper, $5 worth of dimes to fill a dime wrapper, $2 worth of nickels to fill a nickel wrapper, and $0.50 worth of pennies to fill a penny wrapper. 
How many more of each coin do I need before I can fill one wrapper of each?

Quarters  
Have 6 for $1.50  Need 34 for 34*0.25 or $8.50
Dimes
Have 12 for $1.20  Need 38 for 38*0.10 or $3.80  
Nickels
Have 10 for $0.50  Need 30 for 30*0.05 or $1.50 
Pennies
Have 32 for $0.32 Need 68 for 68*0.01 or $0.68  


Hope this helps---ptaylor