A person has $13.60 in dimes and quarters, 34 coins.?

Question

how many dimes could they possibly have? The problem could be solved using a linear system of equations. indicate the number of coins and another equation with their value.

Answer 1

Not possible. Even if all 34 coins were quarters, the total would be only $8.50.

Answer 2

Let D and Q be the number of dimes and quarters, respectively.
Since there are 34 coins, we know that the combined total of dimes and quarters should be 34. That is,

D + Q = 34

However, we're also given that this person has $13.60 in dimes and quarters. Note that

1 dime is represented by (0.10)(1)
2 dimes is represented by (0.10)(2)
3 dimes is represented by (0.10)(3).

Therefore, D dimes (which is what we have) is represented by (0.10)D. Similarly, Q quarters is represented by (0.25)Q. This means that the cost of D dimes and Q quarters should be 13.60. That is,

(0.10)D + (0.25)Q = 13.60.

So now, we have two equations and two unknowns.

D + Q = 34
(0.10)D + (0.25)Q = 13.60

One way to solve this is to solve one variable in terms of the other in the first equation, and then plug this into the second equation.

D + Q = 34, therefore D = 34 - Q

(0.10)D + (0.25)Q = 13.60, therefore
(0.10)[34 - Q] + (0.25)Q = 13.60

Then we solve as normal. Expanding the left hand side, we get

3.4 - (0.1)Q + (0.25)Q = 13.60

Moving the 3.4 over to the right hand side, we get
-(0.1)Q + (0.25)Q = 10.20

We can group like terms and add -0.1 and 0.25, but that might seem initially confusing. I'll just factor the Q out.

Q (-0.1 + 0.25) = 10.20
Q (0.15) = 10.20

And now, divide both sides by 0.15 gives us

Q = 68

So we have 68 quarters

Which makes absolutely no sense since it exceeds the number of coins.

I'll let someone else solve it as I can't find an error in my steps.

Answer 3

Let

d = number of dimes

q - Number of quarters

34 = total combines dimes and quarters

.1d = dime value

.25 = quarter value

13.60 = total value of dimes and quarters

- - - - - - - - - - - - - - - -

The equation

d + q = 34- - - - - - - - -Equation 1

.1d + .25q = 13.60- - - Equation 2
- - - - - - - - - - - --

Substitute method for equation 1

d + q = 34

d + q - d = 34 - d

q = 34 - d

The answer is q = 34 - d

Insert the q value into equation 2

- - - - - - - - - - - - - - - - - - - - - -

.1d + .25q = 13.60

.1d + .25(34 - d) = 13.60

.1d + 8.5 - .25d = 13.60

- .15d + 8.5 = 13.60

-.15d + 8.5 - 8.5 = 13.60 - 8.5

- .15d = 5.1

- .15d/.15 = 5.1 / .15

d = - 34

The answer is d = - 34

Insert the d value into equation 1

- - - - - - - - - - - - - - - - - - - - - -

d + q = 34

- 34 + q = 34

- 34 + q + 34 = 34 + 34

q = 68

The amount of quarters exceed the amount of total coins

by the equation d + q = 34

- 34 + 68 = 34

34 = 34

This implies there are no dimes. There is not a negative number of dimes

- - - - - - -s-

Answer 4

8 dimes, 26 quarters. That's the most she can have, if she had 9 dimes she'd have to have 25 quarters which would give a total of $7.15 which is wrong, so it's 8.

Answer 5

34 quarters = $8.5

So you can't get $13.60 with only a maximum of 24 dimes and quarters

Answer 6

25x + 10y = 1360
x + y = 34
y = 34 - x
25x + 10(34 - x) = 1360
25x + 340 - 10x = 1360
15x = 1020
x = 80.1333333

Answer 7

quarters: x
dime: x-34

.25x+.10(x-34)=13.60
.25x+.10x-3.4=13.60
.34x-3.4=13.60
.34x= 17
x=50

that is impossible because there are only 34 coins but there are 50 quarters.

Answer 8

.1d + .25q = 13.60
d + q = 34

10d + 25q = 1360
2d + 5q = 272

2d + 5q = 272
d + q = 34

Multiply bottom by -2

2d + 5q = 272
-2d - 2q = -68

3q = 204
q = 68

d + q = 34
d + 68 = 34
d = -34

q = 68
d = -34

So as you can see, something was written down wrong.

Answer 9

I get a circular problem when I try these