Working with Fractions - SAT Math

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Question

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?

Answer

$\frac{1}{3}$ 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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Question

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

Answer

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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Question

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?

Answer

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 have in common. Because is a multiple of , we really only need to worry about . If we were to list out the multiple of , 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 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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Question

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

Answer

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

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Question

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

Answer

Let us first get our value for the percentage of the first fraction. 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 or . 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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Question

Add:

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

Answer

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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Question

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 records, how many hip-hop records are stocked in the store?

Answer

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 .

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 . We can write out $\frac{9}{28}=\frac{h}{364}$ and solve for .

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

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

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

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

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Question

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?

Answer

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 seconds and get 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 by and get seconds of chorus.
Since a beat occurs every five seconds, we divide 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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Question

Please simplify the following:

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

Answer

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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Question

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

Answer

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

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

We subtract from

Then we solve for x by dividing by 6,

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Question

$x+\frac{7}{15}=\frac{2}{3}$

For what value of is this equation true?

Answer

To solve this problem we must first set a common denominator. To set a common denominator we multiply $\frac{2}{3}$ by , by doing this we get $x+\frac{7}{15}=\frac{10}{15}$ and see that the common denominator is . Then, since the denominators are now the same, we subtract $\frac{8}{15}$ from $\frac{10}{15}$ and move $\frac{7}{15}$ to the other side to get the alone and simultaneously solve for .

$x=\frac{10}{15}-\frac{7}{15}=\frac{3}{15}$

We simplify $\frac{3}{15}$ by dividing the numerator and denominator by , and get the answer $\frac{1}{5}$ .

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Question

Add:

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

Answer

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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Question

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?

Answer

$\frac{1}{3}$ 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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Question

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

Answer

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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Question

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?

Answer

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 have in common. Because is a multiple of , we really only need to worry about . If we were to list out the multiple of , 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 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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Question

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

Answer

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

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Question

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

Answer

Let us first get our value for the percentage of the first fraction. 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 or . 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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Question

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 records, how many hip-hop records are stocked in the store?

Answer

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 .

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 . We can write out $\frac{9}{28}=\frac{h}{364}$ and solve for .

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

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

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

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

Compare your answer with the correct one above

Question

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?

Answer

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 seconds and get 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 by and get seconds of chorus.
Since a beat occurs every five seconds, we divide 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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Question

Please simplify the following:

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

Answer

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.

Compare your answer with the correct one above

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