If she has a total of 91 coins, how many dimes and how many quarters does she have?
2007-05-17 05:08:52 · 5 answers · asked by misscarter 1 in Science & Mathematics ➔ Mathematics
d is the number of dimes, 10d the value of them; q is the number of quarters, 25q the value of them. 2 equations, 1 for number, 1 for value:
d + q = 91
10d + 25q = 1360
subtract 10 times 1st equation from 2nd:
10d + 25q = 1360
10d + 10q = ..910
15q = 450
q = 30
d = 61
2007-05-17 05:15:46 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
An alternative method:
A dime is worth 10 cents
A quarter is worth 25 cents
We have d dimes and q quarters.
total amount in dimes: 10d
total amount in quarters: 25q
total in coins: 10d + 25q = 1360
d and q must be whole numbers
d = (1360 - 25q)/10 = 136 - 25(q/10)
q must be divisible by 10 and since d must be positive, the only valid values for q are 10, 20, 30, 40, 50
The values of d for these values of q are
111, 86, 61, 36, 11 respectively.
There are 91 coins altogether so we need to find the pair that adds to 91.
30 + 61 = 91
Therefore there are 30 quarters and 61 dimes
2007-05-17 05:26:36 · answer #2 · answered by peateargryfin 5 · 0⤊ 0⤋
If you want to solve the problem in a easier process,here it is
If all the 91 coins were dimes.the value would have been $9.10
But it is $4.50 less than what she actually had.This happened because she had some quarters value of each of which is $0.15 more than a dime
The no. of quarters=4.50/0.15=30
The no of dimes=91-30=61
2007-05-17 06:28:21 · answer #3 · answered by alpha 7 · 1⤊ 0⤋
D = 91 - Q
25Q + 10(91-Q) = 1360
Solve for Q.
Q=30
D = 91 - Q = 61
Simple (and fun) Algebra
2007-05-17 05:18:11 · answer #4 · answered by Jeff 3 · 0⤊ 0⤋
let x = number of dimes
let y = number of quarters
x+y = 91
10x + 25y = 1360
-10x - 10y = -910
10x + 25y = 1360
15y = 450
y = 30 quarters
x = 61 dimes
2007-05-17 05:24:22 · answer #5 · answered by Anonymous · 0⤊ 0⤋