 # Help! Math problem!

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• #601178
supergirl613
Member

K, so I have this math problem that I have absolutely have NO clue how to do it. It’s an inequalities problem I think.

Jo has 37 coins(nickels, dimes and quarters) equaling up to \$5.50. She has four more quarters than nickels.

How many dimes does she have?

#835743

# P T

nickels x 5 5x

dimes 37-(x+x+4) 10 10(37-x+x+4)

quarters x+4 25 25(x+4)

5x+10(37-x+x+4)+25(x+4)=5.50

I hope that helps…that’s about all I remember 🙂

I don’t know why its not staying in chart form, sorry!

#835744
Lifeisgreat
Member

Let x=number of nickels

Then x+4=number of quarters

So the number of dimes would be:

37-x-(x+4)=

37-2x-4=

33-2x —- is the number of dimes

(lets deal in pennies)

5x+25(x+4)+10(33-2x)=550

so 5x+25x+100+330-20x=550

collect like terms

10x+430=550

subtract 430 from each side

10x+430-430=550-430=

10x=120

divide each side by 10

x=12

number of nickels

x+4=12+4=16

number of quarters

33-2x=33-24=9

number of dimes

Check

5*12=60

25*16=400

9*10=90

60+400+90=550

550=550

#835745
supergirl613
Member

#835746
Razzle
Member

Just google the question and a whole list of explanations come up. the answer is 9 dimes.

#835747

Here you go:

Set the number of nickels as “n”, the number of dimes as “d”, and the number of quarters as “q”.

There are a total of 37 coins. Therefore:

n+d+q=37

You also know that the total in dollar amounts is equal to \$5.50. Therefore:

.05n+.10d+.25q=5.50

In the first equation, we can set q = n+4 (as there are 4 more quarters than nickels). Resetting the first equation:

n+(n+4)+d=37

Solve for “d” in terms of “n”:

d=33-2n

Now we set the second equation (using q=n+4):

.05n+.10d+.25(n+4)=5.50

Now you can plug the “d=33-2n” value to the above equation to solve for “n”:

.05n+.10(33-2n)+.25(n+4)=5.50

Do all the annoying math, and the final answer is n=12. That gives you the number of nickels. Finally:

#n=12

#q=16

#d=37-28=9