General Applied Problems - Coins

anastasios1984 wrote:I need help with the below.I am a bit confused...

Sue and Nancy wish to buy a snack. They combine their money and they have $4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have?

A - 3
B - 5
C - 6
D - 10
E - 20

Any ideas, i would like to see the approach to that kind of problems...

thanks in advance!

We can plug in the answers, which represent the number of quarters.
The number of nickels is double the number of quarters.
The remaining coins are dimes.

Answer choice C: 6 quarters, implying 12 nickels and 17 dimes
6(25) + 12(5) + 17(10) = 380 cents.
The sum must increase just a bit to 400 cents.
More quarters are needed.
Eliminate A, B, and C.
E more than triples the number of quarters; the result will be too great a sum.
Eliminate E.

The correct answer is D.

anastasios1984 wrote:I need help with the below.I am a bit confused...

Sue and Nancy wish to buy a snack. They combine their money and they have $4.00, consisting of quarters, dimes, and nickels. If they have 35 coins and the number of quarters is half the number of nickels, how many quarters do they have?

A - 3
B - 5
C - 6
D - 10
E - 20

Any ideas, i would like to see the approach to that kind of problems...

thanks in advance!

Mitch gives a really good way of using "reasoning".

You could also do it using traditional algebra.
Let number of nickels(5 cents) be n, dimes(10 cents) be d and quarters(25 cents) be q.
We need to find q.

We are told that
1. n=2q (quarters are half the number of nickels)
2. n+d+q=35. (total number of coins is 35)
3. 5n+10d+25q=400 (Total sum is $4.00 or 400 cents).

Now we substitute for n and d in final equation to get the answer.
5n+10d+25q = 400
=> n+2d+5q=80
=> 2q+2*(35-q-2q)+5q=80 (substituting n=2q and d=35-q-n)
=> 7q-6q+70=80
=> q=10

Let me know if this helps