A jar contains 42 10c and 25c coins. Total money equals $7.50. How many 25c coins are in the jar?

Question

Logic and methodology would be appreciated.

Answer 1

Let d = number of dimes, n = number of nickels
10d + 25n = 750
d + n = 42
n = 42 - d
10d + 42(25) - 25d = 750
-15d = 750 - 1050 = -300
d = 300/15 = 20
n = 42 - 20 = 22

Answer 2

Let x = number of 10c coins
Let y = number of 25c coins

(1) x + y = 42
(2) 10x + 25y = 750

multiply (1) by 10
(3) 10x + 10y = 420
subtract (3) from (2)
10x + 25y - 10x - 10y = 750 - 420
15y = 330
y = 22

Substitute for y in (1)
x + 22 = 42
x = 20

Check
20*10 + 22*25 + 750
200 + 550 = 750

Answer 3

Guess and check...

Start with 20 quarters, that equals $5.00
(42-20=12)...12 dimes that equal $1.20
$5.00 + $1.20 = $6.20 which is too low

Next guess: Try a larger amount for the quarters
23 quarters, equal $5.75
19 dimes, equal $1.90
$5.75 + $1.90 = $7.65 which is too high

Next guess: Try a number between 20 and 23
22 quarters, equal $5.50
20 dimes, equal $2.00
$5.50 + $2.00 = $7.50 which is right

Soo... if there are 42 coins including quarters and dimes, there are 22 quarters and 20 dimes which equal $7.50.

Answer 4

x + y = 42
.1x + .25y = 7.50

x + y = 42
10x + 25y = 750

Multiply top by -10

-10x - 10y = -420
10x + 25y = 750

15y = 330
y = 22

x + y = 42
x + 22 = 42
x = 20

20 dimes
22 quarters

Answer 5

Assume there are 42 dimes and zero quarters to start.

42d ($4.20), 0q($0.00) is $4.20

$7.50-$4.20=$3.30
The difference between $0.25 and $0.10 is $0.15
$3.30/$0.15=22

So, you need 22 fewer dimes and 22 more quarters.

(42-22)=20d ($2.00), (0+22)=22q ($5.50) is $7.50