Math Problem Help

[AllSmiles]

I have a word problem that has me stumped. I'm having trouble even understanding the question. I'm sure I'm just missing something easy, but I was hoping someone could help. This problem needs to be solved using only one variable.

 

Leslie has 20 quarters and dimes. If she had as many dimes as she has quarters and as many quarters as dimes, she would have 30 cents less. How many of each coin does she have.

 

The answer: 11 quarters and 9 dimes

 

Thanks for you help!

AllSmiles

[regentrude]

Q= current number of quarters

D= current number of dimes

 

Q+D=20

 

Amount of money she has: 0.25*Q+0.10*D

 

Problem states: if she had Q dimes and D quarters instead, she'd have $0.30 less.

So:

0.10*Q +0.25*D= (0.25*Q+0.10*D) - 0.30

 

Solve the system of equations for Q and D.

Edited February 20, 2017 by regentrude

[AllSmiles]

Regentrude,

 

Thanks for the help. I can follow the reasoning of the solution you gave. Unfortunely, the problem can only be solved using one variable.

[regentrude]

Regentrude,

 

Thanks for the help. I can follow the reasoning of the solution you gave. Unfortunely, the problem can only be solved using one variable.

 

The bolded makes no sense since there are TWO variables: Q and D.

The two variables are related by the statement that you have 20 coins in total. Thus, Q+D=20.

Which lets you express one variable through the other, for example Q=20-D, and thus express the remaining equation in terms of only one variable. As is ALWAYS the case when you have a system of two linear equations with two variables.

[Kathy in Richmond]

Try thinking this way: 

 

Basically what Leslie did was to swap each quarter for a dime, and each dime for a quarter.

 

She gains 15 cents every time she swaps a dime for a quarter, and she loses 15 cents every time she swaps a quarter for a dime.

 

Since she lost a total of 30 cents, she must have done 2 more "quarter->dime swaps" than she die "dime-> quarter" swaps.

 

In other words she started with 2 more quarters than dimes. We can model it like this: (x= number of dimes at beginning)

 

|----------x------------|-2-|      quarters

|----------x------------|            dimes

 

So

x + 2 + x = 20 total coins

 

2x + 2 = 20

2x = 18

x= 9 dimes

x+2 = 11 quarters

 

 

 

[Noreen Claire]

now:

no. of quarters: Q

no. of dimes: 20-Q

amount of money she has: .25(Q) + .10(20-Q)

 

switch Qs and Ds:

no. of quarters: (20-Q)

no. of dimes: Q

amount of money she has: .25(20-Q) + .10(Q) - .30

 

set both scenarios equal and solve:

.25(Q) + .10(20-Q) = .25(20-Q) + .10(Q) - .30

.

.

if Q=9 and D=11, then total=$3.35

if Q=11 and D=9, then total=$3.65

 

(Last line reads "if she switched...sje would have less", so answer is 11 quarters and 9 dimes.)

 

 

[Noreen Claire]

Try thinking this way: 

 

Basically what Leslie did was to swap each quarter for a dime, and each dime for a quarter.

 

She gains 15 cents every time she swaps a dime for a quarter, and she loses 15 cents every time she swaps a quarter for a dime.

 

Since she lost a total of 30 cents, she must have done 2 more "quarter->dime swaps" than she die "dime-> quarter" swaps.

 

In other words she started with 2 more quarters than dimes. We can model it like this: (x= number of dimes at beginning)

 

|----------x------------|-2-|      quarters

|----------x------------|            dimes

 

So

x + 2 + x = 20 total coins

 

2x + 2 = 20

2x = 18

x= 9 dimes

x+2 = 11 quarters

 

Very elegant! Nice!

[AllSmiles]

Thanks for all the help! It was nice to see the explations. I think I have a handle on the problem now.

Thanks,

AllSmiles

Edited February 20, 2017 by AllSmiles