Operations with Fractions - PSAT Math

Card 0 of 231

In a group of 20 children, 25% are girls. How many boys are there?

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Since $\frac{1}{4}$ of the children are girls, this totals to 20 times $\frac{1}{4}$ = 5 girls in the group.

20-5=15 boys.

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?

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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.

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

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

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

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

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

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

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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}$$ .

Simplify:

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

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?

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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.

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

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

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

Evaluate the following:

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

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$\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}$

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

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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 $\frac{6}{5}$ .

Define an operation $\blacklozenge$ as follows:

For all real numbers ,

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

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

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$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}$

Define an operation $\odot$ as follows:

For all real numbers ,

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

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

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$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}$

Define an operation $\ominus$ as follows:

For all real numbers ,

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

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

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$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}$$

$= 16 \times 5 = 80$

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}$$ .

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$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}$$

Define an operation $\ddagger$ as follows:

For all real numbers ,

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

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

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$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}$$

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}$$ .

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$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}$$

Simplify:

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

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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:

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

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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.