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SAT Math : Algebraic Fractions

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Example Question #21 : Algebraic Fractions

If pizzas cost dollars and sodas cost dollars, what is the cost of pizzas and sodas in terms of and ?

Possible Answers:

3x+5y

15x+5y

$\frac{3x+5y}{15}$

15xy

$5x+\frac{3y}{15}$

Correct answer:

$\frac{3x+5y}{15}$

Explanation:

If 10 pizzas cost x dollars, then each pizza costs x/10. Similarly, each soda costs y/6. We can add the pizzas and sodas together by finding a common denominator:

$2*\frac{x}{10}+2*\frac{y}{6}=\frac{x}{5}+\frac{y}{3}=\frac{3x}{3*5}+\frac{5y}{5*3}=\frac{3x+5y}{15}$

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Example Question #22 : Algebraic Fractions

Gre9

According the pie chart, the degree measure of the sector representing the number of workers spending 5 to 9 years in the same role is how much greater in the construction industry chart than in the financial industry chart?

Possible Answers:

$7^{\circ}$

$18^{\circ}$

$35^{\circ}$

$70^{\circ}$

$25^{\circ}$

Correct answer:

$25^{\circ}$

Explanation:

Since the values in the pie charts are currently in terms of percentages (/100), we must convert them to degrees (/360, since within a circle) to solve the question. The "5 to 9 years" portion for the financial and construction industries are 18 and 25 percent, respectively. As such, we can cross-multiply both:

18/100 = x/360

x = 65 degrees

25/100 = y/360

y = 90 degrees

Subtract: 90 – 65 = 25 degrees

Alternatively, we could first subtract the percentages (25 – 18 = 7), then convert the 7% to degree form via the same method of cross-multiplication.

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Example Question #23 : Algebraic Fractions

6 contestants have an equal chance of winning a game. One contestant is disqualified, so now the 5 remaining contestants again have an equal chance of winning. How much more likely is a contestant to win after the disqualification?

Possible Answers:

$\frac{2}{5}$

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{5}$

$\frac{1}{30}$

Correct answer:

$\frac{1}{30}$

Explanation:

When there are 6 people playing, each contestant has a 1/6 chance of winning. After the disqualification, the remaining contestants have a 1/5 chance of winning.

1/5 – 1/6 = 6/30 – 5/30 = 1/30.

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Example Question #33 : Algebraic Fractions

Simplify:

$\frac{\frac{1}{4}+\frac{5}{7}}{\frac{2}{5}}$

Possible Answers:

$\frac{81}{7}$

$\frac{27}{70}$

$\frac{135}{56}$

$\frac{47}{5}$

$\frac{98}{71}$

Correct answer:

$\frac{135}{56}$

Explanation:

Begin by simplifying the numerator.

$\frac{1}{4}+\frac{5}{7}$ has a common denominator of . Therefore, we can rewrite it as:

$\frac{7}{28}+\frac{20}{28}=\frac{27}{28}$

Now, in our original problem this is really is:

$\frac{\frac{27}{28}}{\frac{2}{5}}$

When you divide by a fraction, you really multiply by the reciprocal:

$\frac{\frac{27}{28}}{\frac{2}{5}}=\frac{27}{28} \cdot\frac{5}{2}=\frac{135}{56}$

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Example Question #34 : Algebraic Fractions

Simplify:

$\frac{\frac{81}{5}+\frac{7}{25}}{\frac{1}{9}-\frac{2}{7}}$

Possible Answers:

$\frac{8414}{41}$

$-\frac{25956}{275}$

$-\frac{4532}{1575}$

$-\frac{551}{12}$

$\frac{8417}{14}$

Correct answer:

$-\frac{25956}{275}$

Explanation:

Begin by simplifying the numerator and the denominator.

Numerator

$\frac{81}{5}+\frac{7}{25}$ has a common denominator of . Therefore, we have:

$\frac{405}{25}+\frac{7}{25} = \frac{412}{25}$

Denominator

$\frac{1}{9}-\frac{2}{7}$ has a common denominator of . Therefore, we have:

$\frac{7}{63}-\frac{18}{63}=-\frac{11}{63}$

Now, reconstructing our fraction, we have:

$\frac{\frac{412}{25}}{-\frac{11}{63}}$

To make this division work, you multiply the numerator by the reciprocal of the denominator:

$\frac{\frac{412}{25}}{-\frac{11}{63}} = \frac{412}{25} \cdot -\frac{63}{11} =-\frac{25956}{275}$

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Example Question #31 : Algebraic Fractions

Simplify: $(\frac{x-4}{0.5})\div (\frac{1}{x+4})$

Possible Answers:

$\frac{3}{2x^2}-16$

$\frac{1}{2}$

2x^2-32

None of the other answer choices are correct.

Correct answer:

2x^2-32

Explanation:

Recall that dividing is equivalent multiplying by the reciprocal. Therefore, ((x - 4) / (1 / 2)) / (1 / (x + 4)) = ((x - 4) * 2) * (x + 4) / 1.

Let's simplify this further:

(2x – 8) * (x + 4) = 2x² – 8x + 8x – 32 = 2x² – 32

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Example Question #32 : Algebraic Fractions

Solve for :

$\frac{3x}{8}+12=\frac{1}{3}-2x$

Possible Answers:

$\frac{100}{21}$

$-\frac{81}{13}$

$-\frac{280}{57}$

$\frac{43}{2}$

$-\frac{5}{7}$

Correct answer:

$-\frac{280}{57}$

Explanation:

Begin by isolating the variables:

$\frac{3x}{8}+2x=\frac{1}{3}-12$

Now, the common denominator of the variable terms is . The common denominator of the constant values is . Thus, you can rewrite your equation:

$\frac{3x}{8}+\frac{16x}{8}=\frac{1}{3}-\frac{36}{3}$

Simplify:

$\frac{19x}{8}=-\frac{35}{3}$

Cross-multiply:

19x * 3 = -35*8

Simplify:

57x = -280

Finally, solve for :

$x = -\frac{280}{57}$

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Example Question #36 : Algebraic Fractions

Solve: $\frac{8}{3} + \frac{x}{2}=4$

Possible Answers:

$\frac{6}{5}$

$\frac{3}{16}$

$\frac{8}{3}$

$\frac{4}{3}$

Correct answer:

$\frac{8}{3}$

Explanation:

In order to solve $\frac{8}{3} + \frac{x}{2}=4$ , identify the least common denominator, or LCD. Multiply the uncommon denominators, and the LCD is 6.

Rewrite the equation.

$\frac{16}{6} + \frac{3x}{6}=\frac{24}{6}$

Multiply by six on both sides of the equation to cancel the denominators.

16+3x=24

3x=8

$x=\frac{8}{3}$

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Example Question #37 : Algebraic Fractions

Evaluate: $\frac{x}{3}+\frac{11}{6}=6$

Possible Answers:

$\frac{6}{11}$

$\frac{11}{25}$

$\frac{2}{9}$

$\frac{25}{2}$

$\frac{51}{6}$

Correct answer:

$\frac{25}{2}$

Explanation:

Find the least common denominator, or LCD of $\frac{x}{3}+\frac{11}{6}=6$ is six.

Rewrite the equation with the correct denominator.

$\frac{2x}{6}+\frac{11}{6}=\frac{36}{6}$

Multiply by six on both sides of the equation and solve for .

2x+11=36

2x=25

$x=\frac{25}{2}$

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Example Question #31 : Algebraic Fractions

Solve the following: $\frac{1}{3}-\frac{2}{9}$

Possible Answers:

$\frac{1}{9}$

$-\frac{1}{3}$

$-\frac{1}{9}$

$-\frac{5}{9}$

$\frac{7}{27}$

Correct answer:

$\frac{1}{9}$

Explanation:

In order to subtract the fractions, the denominator must be the same. The common denominator is 9. Rewrite the fractions.

$\frac{1}{3}-\frac{2}{9}=\frac{3}{9}-\frac{2}{9}= \frac{1}{9}$

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SAT Math : Algebraic Fractions

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #21 : Algebraic Fractions

Example Question #22 : Algebraic Fractions

Example Question #23 : Algebraic Fractions

Example Question #33 : Algebraic Fractions

Example Question #34 : Algebraic Fractions

Example Question #31 : Algebraic Fractions

Example Question #32 : Algebraic Fractions

Example Question #36 : Algebraic Fractions

Example Question #37 : Algebraic Fractions

Example Question #31 : Algebraic Fractions

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