Can someone solve this problem?

Question

A coin purse contains dimes and quarters. The number of dimes is five less than twice the number of quarters. The total value of the coins is $4.00. Find the number of dimes in the coin purse.

Answer 1

You have two simultaneous equations:

d = number of dimes
q = number of quarters

First equation (number of dimes is five less than twice number of quarters):
d = 2q - 5

Second equation (total value is $4.00, knowing dimes are worth $0.10 and quarters worth $0.25):
0.10d + 0.25q = 4.00

Substituting the first equation into the second:

0.10d + 0.25q = 4.00
0.10(2q - 5) + 0.25q = 4.00
0.20q - 0.50 + 0.25q = 4.00
0.45q = 4.50
q = 10

You have 10 quarters. Going back to the original equation:

d = 2q - 5
d = 2*10 - 5
d = 15

You have 15 dimes.

Checking the answer:

10 quarters = $2,50
15 dimes = $1.50
$1.50 + $2.50 = $4.00

15 is five less than twice 10.

Answer 2

This is a 2-equation/2-unknown case.

Let d = number of dimes
Let q = number of quarters

d=2q-5
.1d+.25q=4 (note that a dime is worth 10 cents or $0.1 and a quarter is worth 25 cents or $0.25 hence the coefficients)

Start by substituting the first equation into the second so,

.1(2q-5)+.25q=4
.2q-.5+.25q=4

Add .5 to both sides and add up the q variables, and you get:

.45q=4.5 or q=10

Now substitute 10 into the 1st equation, and you get:

d=2*10 - 5 or d=15

Answer 3

Let d = dimes, q = quarters

d = 2q - 5
10d + 25q = 400

20q - 50 + 25q = 400
45q = 450
q = 10

d = 2(10) - 5
d = 15

There are 15 dimes and 10 quarters.

Answer 4

Solve the following 2-equation linear system:

10d + 25q = 400
d = 2q - 5

where d = number of dimes and q = number of quarters.

q = 10, d = 15

Answer 5

Start with these two formulas. D is the number of dimes and Q is the number of quarters.

D = 2Q-5

10D + 25Q = 400

Answer 6

15 dimes, 10 quarters

Answer 7

D=# of dimes
Q=# of quarters

D = 2q-5
.1d+.25q=4

Multiply 2nd equation by 100 to get rid of decimals

10d+25q = 400
Substitute 1st equation into 2nd
10(2q-5) + 25q = 400
20q - 50 + 25q = 400
45 q = 450
q=10

d=2q-5
d=2(10)-5
d=20-5
d=15

Lets see if it works 10(.25) + 15(.1) = 2.50 + 1.5 = 4

Answer 8

d = 2q - 5
.10d + .25q = $4.00

solve for q:
.10(2q - 5) + .25q = 4.00
.20q - .5 + .25q = 4.00
.45q - .5 = 4.00
.45q = 4.50
q = 4.50/.45
q = 10

plug in your q value to solve for d:
d = 2q - 5
d = 2(10) - 5
d = 20 - 5
d = 15


10 quarters, 15 dimes
$2.50 + $1.50 = $4.00

Answer 9

10D+25Q=400
D=2Q-5
10(2Q-5)+25Q=400
20Q-50+25Q=400
45Q=450
Q=10
D=15