SOLUTION: A jar contains 40 coins consisting of dimes and quarters and having a total value of $4.90. How many of each kind of coin are there? Having problems figuring out the word p

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Question 988424: A jar contains 40 coins consisting of dimes and quarters and having a total value of $4.90. How many of each kind of coin are there?

Having problems figuring out the word problem.
I have. d + q = 40
And
.1d + .25q = 4.90
Can you show me how to do it?
Found 5 solutions by CubeyThePenguin, ikleyn, greenestamps, MathTherapy, josgarithmetic:
Answer by CubeyThePenguin(3113)   (Show Source):
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d + q = 40
0.1d + 0.25q = 4.90

Use elimination by multiplying the second equation by 4 and subtracting it from the first equation.

d + q = 40
-(0.4d + q = 19.6)
----------------------
0.6d = 20.4
d = 34

There are 34 dimes and 6 quarters.

Answer by ikleyn(52787)   (Show Source):
You can put this solution on YOUR website!
.
On coin problems, see the lessons
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - Three methods for solving standard (typical) coin word problems
    - More complicated coin problems
    - Advanced coin problems
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem
    - OVERVIEW of lessons on coin word problems
in this site.

You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.

A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.

Read them and become an expert in solution of coin problems.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Coin problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

For one thing, I would definitely recommend using cents instead of dollars to write your second equation. Then you have

d+q=40
10d+25q=490

Multiply the first equation by 10 and then compare the two equations, giving you the number of quarters, q:

10d+10q=400
10d+25q=490

15q=90
q = 6

Then d+q=40 gives you d=34.

ANSWER: 6 quarters and 34 dimes

CHECK: 6(25)+34(10) = 150+340 = 490

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!

A jar contains 40 coins consisting of dimes and quarters and having a total value of $4.90. How many of each kind of coin are there?

Having problems figuring out the word problem.
I have. d + q = 40
And
.1d + .25q = 4.90
Can you show me how to do it?

There's no NEED to multiply the 2nd equation by 100 as 1 person suggested. Don't let anyone tell you RUBBISH!
d + q = 40____d = 40 - q ------ eq(i)
.1d + .25q = 4.9 ------ eq (ii)
The EASIEST "route" is the one that involves substituting 40 - q for d in eq (ii).
This gives: .1(40 - q) + .25q = 4.9
4 - .1q + .25q = 4.9
.15q = .9
Number of quarters, or
Now, substitute 6 for q in eq (i) to get "d," the number of dimes.
There's ABSOLUTELY nothing wrong with working with decimals!

Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website!

---------------six quarters
-
-----------thirty-four dimes