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SAT Mathematics : Simplifying Fractions

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Example Questions

Example Question #1 : Simplifying Fractions

Simplify $\frac{x}{2}-\frac{x}{5}$

Possible Answers:

$\frac{2x}{7}$

$\frac{3x}{7}$

$\frac{5x}{3}$

$\frac{3x}{10}$

Correct answer:

$\frac{3x}{10}$

Explanation:

Simplifying this expression is similar to $\frac{1}{2}-\frac{1}{5}$ . The denominators are relatively prime (have no common factors) so the least common denominator (LCD) is 2*5=10 . So the problem becomes $\frac{1}{2}-\frac{1}{5}=\frac{5}{10}-\frac{2}{10}=\frac{3}{10}$ .

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Example Question #1 : Simplifying Fractions

If $\frac{p}{6}$ is an integer, which of the following is a possible value of ?

Possible Answers:

Correct answer:

Explanation:

$\frac{0}{6}=0$ , which is an integer (a number with no fraction or decimal part). All the other choices reduce to non-integers.

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Example Question #21 : Divisibility & Number Fluency

Simplify: $\frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}$

Possible Answers:

$\frac{1}{3x^{2}y^{3}z}$

$\frac{x^{2}}{8y^{3}z}$

$\frac{3x^{2}y^{3}}{z}$

$\frac{x^{2}}{3y^{3}z}$

Correct answer:

$\frac{x^{2}}{3y^{3}z}$

Explanation:

$\frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}=\frac{x^{2}}{3y^{3}z}$

First, let's simplify $\frac{4}{12}$ . The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore $\frac{4}{12}=\frac{1}{3}$ .

To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with $\frac{1}{3}x^{2}y^{-3}z^{-1}$ or $\frac{x^{2}}{3y^{3}z}$

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Example Question #22 : Divisibility & Number Fluency

Which of the following is not equal to $\frac{32}{24}$ ?

Possible Answers:

$\frac{4}{3}$

$\frac{16}{12}$

$\frac{160}{96}$

$\frac{224}{168}$

Correct answer:

$\frac{160}{96}$

Explanation:

$\frac{32}{24}=1.33$ , so the answer choice needs to be something that is not equal to 1.33 in order to be correct.

$\frac{16}{12}=1.33$

$\frac{224}{168}=1.33$

$\frac{4}{3}=1.33$

$\frac{160}{96}=1.67$

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Example Question #23 : Divisibility & Number Fluency

Find the root of:

$y=\frac{x^2-\sqrt3x+\sqrt5x-\sqrt{15}}{x-\sqrt3}$

Possible Answers:

$-\sqrt3$

$\sqrt3$

$\sqrt5$

$-\sqrt{5 }$

Correct answer:

$-\sqrt{5 }$

Explanation:

$y=\frac{x^2-\sqrt3x+\sqrt5x-\sqrt{15}}{x-\sqrt3}=\frac{(x-\sqrt3)(x+\sqrt5)}{x-\sqrt3}=x+\sqrt5$

The root occurs where y=0 . So we substitute 0 for .

$y=x+\sqrt{5}$

$0=x+\sqrt{5}$

This means that the root is at $x=-\sqrt5$ .

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Example Question #24 : Divisibility & Number Fluency

Simplify $\frac{32}{88}$

Possible Answers:

$\frac{4}{11}$

$\frac{1}{4}$

$\frac{8}{11}$

$\frac{2}{11}$

Correct answer:

$\frac{4}{11}$

Explanation:

Find the common factors of the numerator and denominator. They both share factors of 2,4, and 8. For simplicity, factor out an 8 from both terms and simplify.

$\frac{32}{88}=\frac{8\times 4}{8\times 11}=\frac{4}{11}$

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Example Question #2 : Simplifying Fractions

Simply the following fraction:

$\frac{\frac{a^2}{b}}{\frac{a}{b}}$

Possible Answers:

$\infty$

$\frac{a^2}{b}$

$\frac{a^3}{b^2}$

Correct answer:

Explanation:

Remember that when you divide a fraction by a fraction, that is the same as multiplying the fraction in the numerator by the reciprocal of the fraction in the denominator.

In other words,

$\frac{\frac{a^2}{b}}{\frac{a}{b}} = \frac{a^2}{b}\cdot\frac{b}{a}=\frac{a^2b}{ab}$

Simplifying this final fraction gives us our correct answer, .

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Example Question #8 : Simplifying Fractions

Solve for .

$\frac{\frac{x-2}{4^2}}{\frac{x^2-4}{2}}=1$

Possible Answers:

$x=-\frac{15}{8}$

$x=\frac{15}{8}$

$x=\frac{8}{15}$

$x=-\frac{8}{15}$

Correct answer:

$x=-\frac{15}{8}$

Explanation:

To solve for , simplify the fraction. In order to do this, recall that dividing by a fraction is the same as multiplying by its reciprocal. Therefore, rewrite the equation as follows.

$\frac{\frac{x-2}{4^2}}{\frac{x^2-4}{2}}=1\Rightarrow \frac{x-2}{4^2}\times \frac{2}{x^2-4}=1$

Now, simplify the first fraction by calculating four squared.

$\frac{x-2}{16}\times \frac{2}{x^2-4}=1$

From here, factor the denominator of the second fraction.

$\frac{x-2}{16}\times \frac{2}{(x-2)(x+2)}=1$

Next, factor the 16.

$\frac{x-2}{2(8)}\times \frac{2}{(x-2)(x+2)}=1$

From here, cancel out like terms that are in both the numerator and denominator. In this particular case that includes (x-2) and 2.

$\frac{1}{8(x+2)}=1$

Now, distribute the eight.

$\frac{1}{8x+16}=1$

Next, multiply both sides by the denominator.

$(8x+16)\times \frac{1}{8x+16}=1\times (8x+16)$

The (8x+16) cancels out and leaves the following equation.

1=8x+16

Now to solve for perform opposite operations to move all numerical values to one side of the equation leaving by itself on the other side of the equation.

$\\1=8x+16 \\-15=8x \\\\\frac{-15}{8}=x$

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Example Question #25 : Divisibility & Number Fluency

Which of the following fractions is not equivalent to $\frac{6}{45}$ ?

Possible Answers:

$\frac{4}{30}$

$\frac{12}{89}$

$\frac{3}{22.5}$

$\frac{2}{15}$

Correct answer:

$\frac{12}{89}$

Explanation:

Let us simplify $\frac{6}{45}$ :

$\frac{6}{45}=\frac{3\times 2}{3\times 15}=\frac{2}{15}$

We can get alternate forms of the same fraction by multiplying the denominator and the numerator by the same number:

$\frac{2\times 2}{15\times 2}=\frac{4}{30}$

$\frac{2\times 1.5}{15\times 1.5}=\frac{3}{22.5}$

Now let's look at $\frac{12}{89}$ :

$6 \times 2 = 12$ , but $45 \times 2 = 90$ .

Therefore, $\frac{12}{89}$ is the correct answer, as it is not equivalent to $\frac{6}{45}$ .

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Example Question #10 : Simplifying Fractions

Which of the following fractions has the greatest value?

Possible Answers:

$\frac{77}{65}$

$\frac{7007}{6006}$

$\frac{152}{132}$

$\frac{7}{6}$

Correct answer:

$\frac{77}{65}$

Explanation:

To best compare these fractions, it is helpful to reduce them to their simplest form. $\frac{7}{6}$ is already as reduced as it can be.

For $\frac{152}{132}$ , you can factor out a 2 from the numerator and denominator to get to $\frac{76}{66}$ . Note that you could again divide by 2 to get $\frac{38}{33}$ , but here you can also be creative: if you were comparing to $\frac{7}{6}$ and $\frac{77}{65}$ as two options, you should see that $\frac{76}{66}$ is a hair less than $\frac{77}{66}$ , which would reduce directly to $\frac{7}{6}$ . So you can eliminate $\frac{152}{132}$ by that comparison.

7007 and 6006 are each divisible by 1001 (something that should be somewhat evident by the similar way the numbers look). If you then factor out the 1001 you have $\frac{7(1001)}{6(1001)}$ and the 1001s cancel, leaving $\frac{7}{6}$ . Since you cannot have two correct answers, then $\frac{7007}{6006}$ and $\frac{7}{6}$ are a "tie" and both are incorrect. And when you see that $\frac{77}{65}$ is just a bit larger than $\frac{77}{66}$ , a number that would also equal $\frac{7}{6}$ , you can see that $\frac{77}{65}$ is correct.

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SAT Mathematics : Simplifying Fractions

Study concepts, example questions & explanations for SAT Mathematics

All SAT Mathematics Resources

Example Questions

Example Question #1 : Simplifying Fractions

Example Question #1 : Simplifying Fractions

Example Question #21 : Divisibility & Number Fluency

Example Question #22 : Divisibility & Number Fluency

Example Question #23 : Divisibility & Number Fluency

Example Question #24 : Divisibility & Number Fluency

Example Question #2 : Simplifying Fractions

Example Question #8 : Simplifying Fractions

Example Question #25 : Divisibility & Number Fluency

Example Question #10 : Simplifying Fractions

All SAT Mathematics Resources

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