GRE Math : How to add fractions

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Example Question #1 : How To Add Fractions

What is the result of adding $20\%$ $\displaystyle 20\%$ of $\frac{2}{7}$ $\displaystyle \frac{2}{7}$ to $\frac{1}{4}$ $\displaystyle \frac{1}{4}$?

Possible Answers:

$\frac{23}{11}$ $\displaystyle \frac{23}{11}$

$\frac{47}{140}$ $\displaystyle \frac{47}{140}$

$\frac{3}{39}$ $\displaystyle \frac{3}{39}$

$\frac{43}{140}$ $\displaystyle \frac{43}{140}$

$\frac{3}{28}$ $\displaystyle \frac{3}{28}$

Correct answer:

$\frac{43}{140}$ $\displaystyle \frac{43}{140}$

Explanation:

Let us first get our value for the percentage of the first fraction. 20% of 2/7 is found by multiplying 2/7 by 2/10 (or, simplified, 1/5): (2/7) * (1/5) = (2/35)

Our addition is therefore (2/35) + (1/4). There are no common factors, so the least common denominator will be 35 * 4 or 140. Multiply the numerator and denominator of 2/35 by 4/4 and the numerator of 1/4 by 35/35.

This yields:

(8/140) + (35/140) = 43/140, which cannot be reduced.

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Example Question #1 : Operations

Reduce to simplest form: $\frac{1}{4}+(\frac{4}{3}\times \frac{3}{8})-(\frac{1}{4}\div \frac{3}{8})$ $\displaystyle \frac{1}{4}+(\frac{4}{3}\times \frac{3}{8})-(\frac{1}{4}\div \frac{3}{8})$

Possible Answers:

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\frac{3}{8}$ $\displaystyle \frac{3}{8}$

$\frac{3}{4}$ $\displaystyle \frac{3}{4}$

Correct answer:

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

Explanation:

Simplify expressions inside parentheses first: $\dpi{100} \small \left (\frac{4}{3} \times \frac{3}{8} \right ) = \frac{12}{24} = \frac{1}{2}$ $\displaystyle \dpi{100} \small \left (\frac{4}{3} \times \frac{3}{8} \right ) = \frac{12}{24} = \frac{1}{2}$ and $\dpi{100} \small \left (\frac{1}{4} \div \frac{3}{8} \right ) = \left (\frac{1}{4} \times \frac{8}{3} \right ) = \frac{8}{12} = \frac{2}{3}$ $\displaystyle \dpi{100} \small \left (\frac{1}{4} \div \frac{3}{8} \right ) = \left (\frac{1}{4} \times \frac{8}{3} \right ) = \frac{8}{12} = \frac{2}{3}$

Now we have: $\frac{1}{4} + \frac{1}{2} - \frac{2}{3}$ $\displaystyle \frac{1}{4} + \frac{1}{2} - \frac{2}{3}$

Add them by finding the common denominator (LCM of 4, 2, and 3 = 12) and then multiplying the top and bottom of each fraction by whichever factors are missing from this common denominator:

$\dpi{100} \small \frac{1\times 3}{4\times 3} + \frac{1\times 6}{2\times 6} - \frac{2\times 4}{3\times 4} =\frac{3}{12} + \frac{6}{12} - \frac{8}{12} = \frac{1}{12}$ $\displaystyle \dpi{100} \small \frac{1\times 3}{4\times 3} + \frac{1\times 6}{2\times 6} - \frac{2\times 4}{3\times 4} =\frac{3}{12} + \frac{6}{12} - \frac{8}{12} = \frac{1}{12}$

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Example Question #2 : Operations

Quantity A: $\frac{15}{x}+\frac{12}{y}$ $\displaystyle \frac{15}{x}+\frac{12}{y}$

Quantity B: $\frac{15y+12x}{xy}$ $\displaystyle \frac{15y+12x}{xy}$

Which of the following is true?

Possible Answers:

Quantity B is larger.

The two quantities are equal.

The relationship between the two quantities cannot be determined.

Quantity A is larger.

Correct answer:

The two quantities are equal.

Explanation:

Start by looking at Quantity A. The common denominator for this expression is $\displaystyle xy$. To calculate this, you perform the following multiplications:

$\frac{15}{x}*\frac{y}{y}+\frac{12}{y}*\frac{x}{x}$ $\displaystyle \frac{15}{x}*\frac{y}{y}+\frac{12}{y}*\frac{x}{x}$

This is the same as:

$\frac{15y}{xy}+\frac{12x}{xy}$ $\displaystyle \frac{15y}{xy}+\frac{12x}{xy}$, or $\frac{15y+12x}{xy}$ $\displaystyle \frac{15y+12x}{xy}$

This is the same as Quantity B. They are equal!

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GRE Math : How to add fractions

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Example Question #1 : How To Add Fractions

Example Question #1 : Operations

Example Question #2 : Operations

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