A collection of 30 coins worth $5.50 consists of nickels,dimes and quarters. There are twice as many dimes as nickels. Howmany quarters are there?

sittesf

sittesf

Answered question

2022-08-08

A collection of 30 coins worth $5.50 consists of nickels,dimes and quarters. There are twice as many dimes as nickels. Howmany quarters are there?

Answer & Explanation

vibrerentb

vibrerentb

Beginner2022-08-09Added 21 answers

Let n=number of nickels, d=number of dimes, q=number ofquarters.
2n=d by the second statement for.
So n+d+q=n+2n+q=
3n+q=30. [Quantity function]
5n+10d+25q=5n+20n+25q=
25n+25q=550 [Value function]
n+q=22.
Subtract the quantity function from the rewritten value function toget:
2n=8
n=4.
Then q=18.
This mean there are 18 quarters.
darcybabe98ub

darcybabe98ub

Beginner2022-08-10Added 6 answers

To solve this problem, you need to make three different equationswith three different variables, and then use the combination methodto solve.
x=nickels
y=dimes
z=quarters
Here are the equations:
x+y+z=30 (all thecoins added together equal 30 coins in all)
0.05x+0.10y+0.25z=5.50(this is the dollar amount of all the coins added together equals$5.50)
2x=y (there are twice asmany dimes as nickels)
Rewrite the second equation as 5x+10y+25z=550 (multiply everythingby 100) and rewrite the third equation as 2x-y=0.
Now your equations are:
x+y+z=30
5x+10y+25z=550
2x-y=0
Now use the combination method to get rid of the variable z inorder to isolate the variables x and y.
5x+10y+25z=550
-25x-25y-25z=-750 (multiply everything by -25)
Add both of these equations together to get:
-20x-15y=-200
Then, add 2x-y=0 to this equation in order to isolate the variablex.
-20x-15y=-200
-30x+15y=0 (multiply everything by -15)
Add these two equations together and get:
x=4
This is the key that will allow you to solve theproblem. Plug x=4 into the equation 2x=y.
2(4)=y
y=8.
Then, plug x=4 and y=8 into x+y+z=30.
4+8+z=30
12+z=30
z=18.
The answer to your question: There are 18 quarters.

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