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SAT Mathematics : Divisibility & Number Fluency

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Example Questions

Example Question #1 : Divisibility & Number Fluency

Jesse has a large movie collection containing movies. $\frac{1}{3}$ of his movies are action movies, $\frac{3}{5}$ of the remainder are comedies, and the rest are historical movies. How many historical movies does Jesse own?

Possible Answers:

$\left ( \frac{11}{15} \right )*x$

$\left ( \frac{4}{15} \right )*x$

$\left ( \frac{2}{5} \right )*x$

$\left ( \frac{7}{12} \right )*x$

Correct answer:

$\left ( \frac{4}{15} \right )*x$

Explanation:

$\frac{1}{3}$ of the movies are action movies. $\frac{3}{5}$ of $\frac{2}{3}$ of the movies are comedies, or $\frac{3}{5}*\frac{2}{3}$ , or $\frac{6}{15}$ . Combining the comedies and the action movies ( $\frac{1}{3}$ or $\frac{5}{15}$ ), we get $\frac{11}{15}$ of the movies being either action or comedy. Thus, $\frac{4}{15}$ of the movies remain and all of them have to be historical.

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Example Question #2 : Divisibility & Number Fluency

If $x=\frac{1}{3}$ and $y=\frac{1}{2}$ , find the value of 2x+3y .

Possible Answers:

$\frac{5}{6}$

$\frac{13}{6}$

$\frac{6}{5}$

Correct answer:

$\frac{13}{6}$

Explanation:

Substitute the values of and into the given expression:

$2(\frac{1}{3})+3\left ( \frac{1}{2} \right )$

$=\frac{2}{3}+\frac{3}{2}$

$=\frac{4}{6}+\frac{9}{6}$

$=\frac{13}{6}$

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Example Question #3 : Divisibility & Number Fluency

Let the nth term of a sequence be denoted as $a_{n}$ and given by the following equation:

$a_{n}=\frac{\left ( -1 \right )^{n-1}}{n}$

For example, the tenth term of the sequence is $a_{10}=\frac{\left ( -1 \right )^{10-1}}{10}$ .

What is the sum of the first five terms of the sequence?

Possible Answers:

$\frac{43}{60}$

$-\frac{47}{60}$

$\frac{47}{60}$

$\frac{17}{60}$

Correct answer:

$\frac{47}{60}$

Explanation:

In order to find the sum of the first five terms, we will need to find the values of each of the first five terms using the equation given above. Essentially, we will let range from to to determine each term.

$a_{1}=\frac{\left ( -1 \right )^{1-1}}{1}=\frac{\left ( -1 \right )^{0}}{1}=\frac{1}{1}=1$

Remember that anything to the power of zero is equal to 1. Therefore, raised to the zero power is also .

$a_{2}=\frac{\left ( -1 \right )^{2-1}}{2}=\frac{\left ( -1 \right )^{1}}{2}=-\frac{1}{2}$

$a_{3}=\frac{\left ( -1 \right )^{3-1}}{3}=\frac{\left ( -1 \right )^{2}}{3}=\frac{\left (-1 \right )\left ( -1 \right )}{3}=\frac{1}{3}$

In general, when is raised to an even power, the result is . Conversely, if is raised to an odd power, the result is .

$a_{4}=\frac{\left ( -1 \right )^{4-1}}{4}=\frac{\left ( -1 \right )^{3}}{4}=\frac{\left (-1 \right )\left ( -1 \right )\left ( -1 \right )}{4}=-\frac{1}{4}$

$a_{5}=\frac{\left ( -1 \right )^{5-1}}{5}=\frac{\left ( -1 \right )^{4}}{5}=\frac{1}{5}$

Thus, the first five terms of the sequence are $1, -\frac{1}{2},\frac{1}{3},-\frac{1}{4}, and\, \frac{1}{5}$ . We must now add these all together. Because we are adding fractions with unlike denominators, we need to find the smallest multiple that $2, 3, 4, and\, 5$ have in common. Because is a multiple of , we really only need to worry about $3, 4, and \, 5$ . If we were to list out the multiple of $3, 4, and \, 5$ , e would see that the smallest one they have in common is . Sometimes, the easiest way to find the least common multiple of several numbers is o multiply them together. The products of $3, 4, and \, 5$ is .

We will now convert each fraction to an equivalent form with a denominator of . For example, if we were to convert $\frac{1}{2}$ to a fraction with a denominator of , we would multiple both the numerator and denominator by , as shown below:

$\frac{1}{2}=\frac{\left ( 1 \right )\left ( 30 \right )}{\left ( 2 \right )\left ( 30 \right )}=\frac{30}{60}$

$1+\frac{1}{2}+\frac{1}{3}+-\frac{1}{4}+\frac{1}{5}=\frac{60}{60}-\frac{30}{60}+\frac{20}{60}-\frac{15}{60}+\frac{12}{60}=\frac{60-30+20-15+12}{60}=\frac{47}{60}$

The answer is $\frac{47}{60}$ .

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Example Question #1 : Divisibility & Number Fluency

What is the solution, reduced to its simplest form, for $x = \frac{7}{9}+\frac{3}{5}+\frac{2}{15}+\frac{7}{45}}$ ?

Possible Answers:

$x=\frac{7}{15}$

x=2

$x=\frac{75}{45}$

$x=\frac{5}{3}$

Correct answer:

$x=\frac{5}{3}$

Explanation:

$x=\frac{7}{9}+\frac{3}{5}+\frac{2}{15}+\frac{7}{45}=\frac{35}{45}+\frac{27}{45}+\frac{6}{45}+\frac{7}{45}=\frac{75}{45}=\frac{5}{3}$

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Example Question #5 : Divisibility & Number Fluency

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

Possible Answers:

$\frac{3}{28}$

$\frac{43}{140}$

$\frac{3}{39}$

$\frac{47}{140}$

Correct answer:

$\frac{43}{140}$

Explanation:

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

Our addition is therefore $\frac{2}{35}+\frac{1}{4}$ . There are no common factors, so the least common denominator will be 35*4 or 140 . Multiply the numerator and denominator of $\frac{2}{35}$ by $\frac{4}{4}$ and the numerator of $\frac{1}{4}$ by $\frac{35}{35}$

This yields:

$\frac{8}{140}+\frac{35}{140}=\frac{43}{140}$ , which cannot be reduced.

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Example Question #1 : Working With Fractions

Add:

$\frac{2}{9x}+\frac{4}{27}$

Possible Answers:

$\frac{6}{27x}$

$\frac{4x+6}{27x}$

$\frac{4x+6}{27}$

$\frac{10}{27x}$

Correct answer:

$\frac{4x+6}{27x}$

Explanation:

Find the least common denominator to solve this problem

Multiply 27 with , and multiply with 3 to obtain common denominators.

Convert the fractions.

$\frac{2}{9x}+\frac{4}{27}=\frac{6}{27x}+\frac{4x}{27x}$

Combine the terms as one fraction.

The answer is: $\frac{4x+6}{27x}$

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Example Question #6 : Divisibility & Number Fluency

A record store has stocked up on new supplies so that their current inventory consists of pop, rock, and hip-hop. $\frac{1}{4}$ of their records are pop records and $\frac{3}{7}$ of their records are rock records. If the store has a total of 364 records, how many hip-hop records are stocked in the store?

Possible Answers:

156

232

117

Correct answer:

117

Explanation:

First we must find what fraction of hip-hop records are there in the store. We find a common denominator for the faction of pop records and rock records by multiplying $\frac{1}{4}$ by $\frac{7}{7}$ and $\frac{3}{7}$ by $\frac{4}{4}$ to get $\frac{7}{28}$ records being pop records and $\frac{12}{28}$ records being rock records.

In order to get the faction of hip-hop records remaining in the store, we must write out:

$\frac{7}{28}+\frac{12}{28}+\frac{x}{28}=\frac{28}{28}$ since the sum total of the fractions must add up to . We solve to get x=9 .

This tells us that fraction of hip-hop records must be $\frac{9}{28}$ .

The question is asking how many records are stocked in the store, therefore we need a denominator of 364 . We can write out $\frac{9}{28}=\frac{h}{364}$ and solve for h=117 .

Similarly, this question can be solved using deduction. If $\frac{1}{4}$ of 364 is and $\frac{3}{7}$ of 364 is 156 , knowing that we need the amount remaining, we can subtract 91+156 from 364 and get 117 .

Answer choice is wrong because it is finding the total amount of Pop records.

Answer choice 156 is wrong because it is finding the total amount of rock records.

Answer choice 156 is wrong because the factions were added incorrectly $\frac{1}{4}+\frac{3}{7}\neq \frac{4}{11}$ .

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Example Question #7 : Divisibility & Number Fluency

Jackson wrote a song with equal parts pre-chorus, chorus, and bridge. The song ends up being minutes and seconds with a beat occurring every seconds. How many beats are there in the chorus?

Possible Answers:

Correct answer:

Explanation:

First, we must convert the song length from minutes and seconds to simply seconds.
$\frac{2\, minutes}{x\, seconds}=\frac{1\, minute}{60\, seconds}$ . We add seconds to 120 seconds and get 146 seconds total song length.

If the song is equal parts pre-chorus, chorus, and bridge that means $\frac{1}{3}$ of the song is composed of pre-chorus, $\frac{1}{3}$ of the song is composed of chorus, and $\frac{1}{3}$ of the song is composed of bridges. Therefore, we divide 146 by and get 48.67 seconds of chorus.
Since a beat occurs every five seconds, we divide 48.67 seconds by seconds to get the amount of beats in that time frame. $\frac{48.67\, seconds}{5\, seconds}=9.73\, beats$ . Since a beat can not be sectioned, the answer is rounded down to .

Answer choice is incorrect because it is finding the total number of beats in the song.

Answer choice is incorrect because it rounds up to the nearest beat.

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Example Question #8 : Divisibility & Number Fluency

Please simplify the following:

$\frac{(\frac{a}{b})}{(\frac{c}{d})}$

Possible Answers:

$\frac{cb}{ad}$

$\frac{a}{bdc}$

$\frac{ac}{db}$

$\frac{ad}{bc}$

Correct answer:

$\frac{ad}{bc}$

Explanation:

The rule for dividing fractions is invert the second fraction and multiply. We keep the $\frac{a}{b}$ as it is then change the division sign to a multiplication sign and invert $\frac{c}{d}$ into $\frac{d}{c}$ .
$\frac{a}{b}\div\frac{c}{d}$ becomes $\frac{a}{b}\times \frac{d}{c}=\frac{ad}{bc}$ .

Similarly, you can memorize, “top, bottom, bottom, top” meaning is on top and it stays on top, is labeled bottom and it stays on the bottom, is labeled bottom and it moved to the bottom, then is labeled top so it moves to the top.

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Example Question #1 : Working With Fractions

Find the value of if $\frac{x}{x+4}=\frac{9}{15}$

Possible Answers:

Correct answer:

Explanation:

To find the answer we must cross multiply. When we cross multiply we get:

$x\times 15=9\times (x+4)$ or 15x=9x+36

We subtract from 15x

6x=36 Then we solve for x by dividing by 6,

x=6

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SAT Mathematics : Divisibility & Number Fluency

Study concepts, example questions & explanations for SAT Mathematics

All SAT Mathematics Resources

Example Questions

Example Question #1 : Divisibility & Number Fluency

Example Question #2 : Divisibility & Number Fluency

Example Question #3 : Divisibility & Number Fluency

Example Question #1 : Divisibility & Number Fluency

Example Question #5 : Divisibility & Number Fluency

Example Question #1 : Working With Fractions

Example Question #6 : Divisibility & Number Fluency

Example Question #7 : Divisibility & Number Fluency

Example Question #8 : Divisibility & Number Fluency

Example Question #1 : Working With Fractions

All SAT Mathematics Resources

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