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Question
Hey I need help on this one word problem...

A girl has a bag of coins.  4/9 of the bag are nickles and 1/3 of the bag are dimes.  How many more nickles than dimes does she have? Keep answer in fraction form.

Answer
Hi Melanie,

The clue we get from "how many more" is that we need to subtract the smaller fraction from the larger fraction to find out the difference.

With practice, as you become familiar with fractions, you should be able to sense what ratio is represented by each fraction. In terms of proportion, 1/3 is much less than a half; whereas 4/9 is closer to (only slightly less than) a half. So, 4/9 is the larger fraction of the two.

To answer "how much more nickle than dimes", we have to evaluate (4/9) - (1/3).

We cannot subtract two fractions directly if their denominator (bottom part under the fraction bar) are different. Our goal is to rewrite 1/3 as something over 9, so that it shares a common denominator with the first fraction; and we must do this without changing its proportion.

Think, for a moment, what is (1/3) x 1?
It is still (1/3), isn't it.
In fact, we can write 1 as 3/3.
So, (1/3) x (3/3) still represents the same proportion.
Multiplying the top parts and bottom parts separately, we get
(1/3) = (1/3) x (3/3)
     = (1x3)/(3x3)
     = 3/9.

This is the form that we want. Now, 1/3 equals 3/9. So, we can say
(4/9) - (1/3) = (4/9) - (3/9)  ...since they share a common denominator, we can gather
                              ...the numerator terms (the "4" and the "-3")
             = (4-3)/9
             = 1/9

That is, in terms of proportion, there is 1/9 more nickels than dimes.

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When I work through problems, I like to emphasize concepts which I believe are worth noting. I will try to answer questions in the following areas, but not at the advanced level. Algebra. Sequences & Series. Trigonometry. Functions & Graphs. Coordinate Geometry. Quadratic Polynomials. Exponential & Logarithms. Basic Calculus. Probability, Permutation and Combination. Mathematical Induction. Complex numbers. Physics problems.

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