Omari (3 1/2)___________(0)school____________(3 1/4)daisy
3 1/2 + 3 1/4 = 6 + (1/2 + 1/4) = 6 + (2/4 + 1/4) = 6 3/4 blocks apart <==
Let the number of nickels found be n, the number of dimes be d and the number of quarters be q.
i) "you have twice as many quarters as dimes and 42 coins in all."
means that 2d=q, and n+d+q=42
we can reduce the number of unknowns by substituting q with 2d:
n+d+q=42
n+d+2d=42
n+3d=42
We can write all n, d and q in terms of d as follows:
there are n=42-3d nickels, d dimes and q=2d quarters.
ii) In total there are $6.60 dollars,
1 nickel = 5 cent = $0.05
1 dime = 10 cent = $0.1
1 quarter = 25 cent = $0.25
thus
(42-3d)*0.05 + d*0.1 +2d*0.25= $6.60
2.1 - 0.15d+0.1d+0.5d=6.60
2.1+0.45d=6.6
0.45d=6.6-2.1=4.5
d=4.5/0.45=10
iii)
so, there are 10 dimes, 2d=2*10=20 quarters and 42-3d=42-3*10=12 nickels.
Answer: 10 dimes, 20 quarters, 12 nickels
Let the price of 1 shirt be represented by = s
A pair of pants costs twice as much as a shirt, then the cost of pants is = 2s
The total cost of 1 pair of pants and 1 shirt is $18 , means : 
Solving this we get,
So, the price of 1 shirt is $6 and price of 1 pair of pants is = 2s = 2*6 = $12.
<span>88 stamps -------------------- > $15.56
A------------------------------>number of </span>stamps ------------ ><span>25 cent
B----------------------------- > </span>number of stamps ------------ >2 cent
A+B=88------------------ > A=88-B
0.25A+.02B=15.56
resolving
0.25(88-B)+0.02B=15.56
22-0.25B+0.02B=15.56
-0.23B=-6.44----------------------- > B=28 stamps of 2 cent
A=88-B-- >88-28=60------------- > A=60 stamps of 25 cent
