Warren has 40 coins (all nickels, dimes, and quarters worth $4.05. He has 7 more nickels than dimes. How many QUARTERS does Warren have? Use only one variable
2006-11-13 12:12:45 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(nickels, dimes, quarters) = 3x
Equation 1:
3x = 40
Equation 2:
.05x + .10x + .25x = 4.05
solve each for x and set them equal
2006-11-13 12:18:03 · answer #1 · answered by Anonymous · 0⤊ 1⤋
this question doesn't make sence if you have to use only one variable. There has to be three variable if there three different coins in different proportions. Here is how i did it: Let x,y,z be number of nickles, dimes and quarters respectively:
1. Total number of coins = 40
x+y+z=40
2. He has 7more nickles than dimes
y=z+7
3. Total amount = $4.05
0.05x+0.1y+0.25z=4.05
Using y=z+7, solve equation 1 and Eqn 3, and we get
x+2z=33---------------Eqn 4
0.05x+0.35z=3.35-----Eqn 5
Now solve Eqn 4 & 5 and substitue to ge the answer.
2006-11-13 20:33:40 · answer #2 · answered by Anonymous · 0⤊ 0⤋
4.05 = dimes+nickles+quarters
4.05=x*.1 + x+7*.05 + (40-2x-7)*.25
multiply both sides by 20 (keeps the equation, loses the decimals)
81=2x+x+7+(33-2x)*5
81=3x+7+165-10x
81=172 - 7x
.........add 7x to both sides
81+7x = 172
..... subtract 81 from both sides
7x=91
...... divide by 7
x=13
x.........x+7......40-2x-7
13 dimes, 20 nickles, 7 quarters
where do we get pizza ?
2006-11-13 20:37:49 · answer #3 · answered by tom4bucs 7 · 0⤊ 0⤋
Let n = number of nickles
n-7 = number of dimes
40 -n -(n-7) = 47 -2n number of quarters
.05n+.10(n-7)+.25(47-2n) = 4.05
-.35n=4.05+.70-11.75=-7.00
n=20
Number of quarters = 47 -2n = 47 -2(20) = 7
2006-11-13 20:46:27 · answer #4 · answered by The Quiet Cool 2 · 0⤊ 0⤋