Operations with Fractions - PSAT Math

Card 0 of 231

Question

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

Answer

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

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Question

If x = 1/3 and y = 1/2, find the value of 2_x_ + 3_y_.

Answer

Substitute the values of x and y into the given expression:

2(1/3) + 3(1/2)

= 2/3 + 3/2

= 4/6 + 9/6

= 13/6

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Question

Answer

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Question

Solve $\frac{3}{7}+\frac{5}{8}-\frac{1}{2}$ .

Answer

Finding the common denominator yields $\frac{24}{56}+\frac{35}{56}-\frac{28}{56}$ . We can then evaluate leaving $\frac{31}{56}$ .

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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 sum of $\small \frac{4}{5}$ and $\small 2 \frac{5}{6}$ ?

Answer

We can begin by eliminating the obviously wrong answers. We know that the sum of the two fractions will be more than 1, so the answer choices $\tiny \tiny \frac{49}{109}$ and $\small \frac{30}{109}$ are out. Now, let's add the two fractions:

Begin by converting $\small 2 \frac{5}{6}$ to $\tiny \frac{17}{6}$ .

Now find the common denominator of $\small \frac{4}{5}$ and $\tiny \frac{17}{6}$ . The least common multiple of 5 and 6 is 30, so 30 is the common denominator. Now alter both fractions so that they use the common denominator:

$\tiny \small \frac{4}{5}= \frac{46}{56}=\frac{24}{30}$

$\tiny \small \frac{17}{6}=\frac{175}{65}=\frac{85}{30}$

Now we can easily add the two fractions together:

$\small \frac{24}{30}+\frac{85}{30}=\frac{109}{30}$

The answer is $\small \frac{109}{30}$ .

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Question

Simplify:

Answer

Division is the same as multiplying by the reciprocal. Thus, a/b ÷ c/d = a/b x d/c = ad/bc

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Question

If p is a positive integer, and 4 is the remainder when p-8 is divided by 5, which of the following could be the value of p?

Answer

Remember that if x has a remainder of 4 when divided by 5, xminus 4 must be divisible by 5. We are therefore looking for a number p such that p - 8 - 4 is divisible by 5. The only answer choice that fits this description is 17.

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Question

If $\dpi{100} \small x=\frac{2}{3}$ and $\dpi{100} \small y= \frac {3}{4}$ , then what is the value of $\dpi{100} \small \frac {x}{y}$ ?

Answer

Dividing by a number (in this case $\dpi{100} \small \frac {3}{4}$ ) is equivalent to multiplying by its reciprocal (in this case $\dpi{100} \small \frac {4}{3}$ ). Therefore:

$\dpi{100} \small \frac {2}{3}\div \frac{3}{4} = \frac{2}{3}\times \frac{4}{3} = \frac{8}{9}$

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Question

Evaluate the following:

$\left ( \frac{5}{6}\times \frac{12}{13} \right )^{2}\div \frac{3}{4}$

Answer

$\left ( \frac{5}{6}\times \frac{12}{13} \right )^{2}\div \frac{3}{4}$

First we will evaluate the terms in the parentheses:

$\left ( \frac{5}{6}\times \frac{2\times 6}{13} \right )^{2}\div \frac{3}{4}$

$\left ( \frac{5}{1}\times \frac{2}{13} \right )^{2}\div \frac{3}{4}$

$\left ( \frac{10}{13} \right )^{2}\div \frac{3}{4}$

Next, we will square the first fraction:

$\left ( \frac{10^{2}}{13^{2}} \right )\div \frac{3}{4}$

$\frac{100}{169}\div \frac{3}{4}$

We can evaluate the division as such:

$\frac{100}{169}\times\frac{4}{3}=\frac{400}{507}$

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Question

Simplify: $\frac{\frac{3x}{2}}{\frac{5x}{4}}$

Answer

Start by rewriting this fraction as a division problem:

$\frac{3x}{2}\div \frac{5x}{4}$

When dividing fractions, you multiply by the reciprocal of the second fraction, so you can rewrite your problem like this:

$\frac{3x}{2}\times \frac{4}{5x}$

Multiply across the numerators and then across the denominators to get $\frac{12x}{10x}$ . The x's cancel, and you can reduce the fraction to be $\frac{6}{5}$ .

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Question

Define an operation $\blacklozenge$ as follows:

For all real numbers ,

$a ; \blacklozenge ; b = a \div b^{2}$ .

Evaluate $25 ; \blacklozenge ; 6\frac{1}{4}$ .

Answer

$a ; \blacklozenge ; b = a \div b^{2}$

$25 ; \blacklozenge ; 6\frac{1}{4}= 25 \div \left ( 6\frac{1}{4} \right ) ^{2}$

$= 25 \div \left ( \frac{25}{4} \right ) ^{2}$

$= 25 \div \left ( \frac{625}{16} \right )$

$= \frac{25 }{1}\times \left ( \frac{16} {625}\right )$

$= \frac{1}{1}\times \left ( \frac{16} {25}\right )$

$= \frac{16} {25}$

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Question

Define an operation $\odot$ as follows:

For all real numbers ,

$a \odot b = \frac{a}{b+2}$ .

Evaluate $\frac{3}{5} \odot \frac{1}{5}$ .

Answer

$a \odot b = \frac{a}{b+2}$ ,

or, equivalently,

$a \odot b = a \div\left ( b+2 \right )$

$\frac{3}{5} \odot \frac{1}{5}= \frac{3}{5} \div \left (\frac{1}{5}+2 \right )$

$= \frac{3}{5} \div \frac{11}{5}$

$= \frac{3}{5} \cdot \frac{5}{11}$

$= \frac{3}{1} \cdot \frac{1}{11}$

$= \frac{3} {11}$

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Question

Define an operation $\ominus$ as follows:

For all real numbers ,

$a ; \ominus ; b = a^{2} \div b$ .

Evaluate $16 ; \ominus ; 3\frac{1}{5}$ .

Answer

$a ; \ominus ; b = a^{2} \div b$

$16 ; \ominus ; 3\frac{1}{5} = 16^{2} \div 3\frac{1}{5}$

$= 256 \div \frac{16}{5}$

$= \frac{256 }{1}\times \frac{5}{16}$

$= \frac{16}{1}\times \frac{5}{1}$

$= 16 \times 5 = 80$

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Question

Define an operation $\otimes ;$ as follows:

For all real numbers ,

$a \otimes b = \frac{4-a}{b}$ .

Evaluate $1 \frac{1}{2} \otimes 2 \frac{1}{3}$ .

Answer

$a \otimes b = \frac{4-a}{b}$ , or, equivalently,

$a \otimes b =( 4-a ) \div b$

$1 \frac{1}{2} \otimes 2 \frac{1}{3} =\left ( 4-1 \frac{1}{2} \right ) \div 2 \frac{1}{3}$

$= 2 \frac{1}{2} \div 2 \frac{1}{3}$

$= \frac{5}{2} \div \frac{7}{3}$

$= \frac{5}{2} \times \frac{3}{7}$

$= \frac{15}{14}$

$= 1 \frac{1}{14}$

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Question

Define an operation $\ddagger$ as follows:

For all real numbers ,

$a \ddagger b = \frac{a}{b} - \frac{b}{a}$ .

Evaluate $\frac{3}{4} \ddagger\frac{4}{3}$ .

Answer

$a \ddagger b = \frac{a}{b} - \frac{b}{a}$ ,

or, equivalently,

$a \ddagger b = (a \div b) - (b \div a)$

$\frac{3}{4} \ddagger\frac{4}{3} =\left (\frac{3}{4} \div \frac{4}{3} \right ) - \left (\frac{4}{3} \div \frac{3}{4} \right )$

$=\left (\frac{3}{4} \times \frac{3}{4} \right ) - \left (\frac{4}{3} \times \frac{4} {3}\right )$

$= \frac{9}{16} -\frac{16}{9}$

From here we need to find a common denominator.

$=\frac{9}{16}\bigg(\frac{9}{9}\bigg)- \frac{16}{9}\bigg(\frac{16}{16}\bigg)$

$= \frac{81}{144} -\frac{256}{144}$

$= -\frac{175}{144}$

$= -1 \frac{31}{144}$

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Question

Define an operation $\amalg$ as follows:

For all real numbers ,

$a \amalg b = 5 - \frac{a}{b}$ .

Evaluate $1 \frac{1}{2}; \amalg ; 2 \frac{1}{3}$ .

Answer

$a \amalg b = 5 - \frac{a}{b}$

or, equivalently,

$a \amalg b = 5 - a \div b$

$1 \frac{1}{2}; \amalg ; 2 \frac{1}{3}= 5 - 1 \frac{1}{2} \div 2 \frac{1}{3}$

$= 5 - \frac{3}{2} \div \frac{7}{3}$

$= 5 - \frac{3}{2} \times \frac{3}{7}$

$= 5 - \frac{9}{14}$

$= 4 \frac{5}{14}$

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Question

Simplify:

$\frac{1+5}{\frac{6}{3+1}}$

Answer

Begin by simplifying any additions that need to be done:

$\frac{1+5}{\frac{6}{3+1}}$

becomes

$\frac{6}{\frac{6}{4}}$

Now, remember that the numerator can be rewritten $\frac{6}{1}$ :

$\frac{\frac{6}{1}}{\frac{6}{4}}$

Now, when you divide fractions, you multiply the numerator by the reciprocal of the denominator:

$\frac{6}{1} * \frac{4}{6}$

Cancel the s and you get:

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Question

If xy = 1 and 0 < x < 1, then which of the following must be true?

Answer

If x is between 0 and 1, it must be a proper fraction (e.g., ½ or ¼). Solving the first equation for y, y = 1/x. When you divide 1 by a proper fraction between 0 and 1, the result is the reciprocal of that fraction, which will always be greater than 1.

To test this out, pick any fraction. Say x = ½. This makes y = 2.

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Question

Before going to school, Joey ran 1/3 of his daily total miles. In gym class, Joey did 2/3 of the remainder. What part of his daily total miles was left for after school?

Answer

Before school, Joey did 1/3 of the total miles. In school, Joey did 2/3 of the remaining 2/3, or 4/9 of the running. When added to his in school run, his before school run of 3/9 brings his completed miles to 7/9 of his dialy total. Thus, only 2/9 of the total miles are left for after school.

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