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SAT Math : SAT Mathematics

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Example Questions

Example Question #2601 : Sat Mathematics

Simplify the expression:

$3x^{2}+4x-2+x^{2}-2x+6$

Possible Answers:

$3x^{2}-2x+4$

$4x^{2}+2x+4$

$2x^{2}-x+2$

$4x^{2}-2x+5$

Correct answer:

$4x^{2}+2x+4$

Explanation:

In order to simplify an expression, we rearrange it to put terms with the same base or type of variable together, then add or subtract accordingly. That would look as follows:

$3x^{2}+4x-2+x^{2}-2x+6$

$3x^{2}+x^{2}+4x-2x-2+6$

$4x^{2}+2x+4$

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Example Question #2602 : Sat Mathematics

Given $a\neq b$ , simplify the following expression.

$\frac{\frac{a^0}{b}}{\frac{a^2}{b^3}}$

Possible Answers:

$\frac{a^2}{b^2}$

$\frac{b^2}{a^2}$

$\frac{b}{a}$

Correct answer:

$\frac{b^2}{a^2}$

Explanation:

Taking a look at the given expression, we can see that we have two fractions divided by one another. The first fraction in the numerator is $\frac{a^0}{b}$ , and the second fraction in the denominator is $\frac{a^2}{b^3}$ .

Remember that when we have a fraction divided by a fraction, that is the same thing as multiplying the numerator by the reciprocal of the denominator. To simplify, we will do just that.

$\frac{a^0}{b}\cdot \frac{b^3}{a^2} = \frac{1}{b}\cdot \frac{b^3}{a^2}= \frac{b^2}{a^2}$ .

To double check your answer, you can choose a numerical value for a and b and plug them into the expression.

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Example Question #2602 : Sat Mathematics

Simplify:

$\frac{\frac{a}{b}}{\frac{a}{b}}$

Possible Answers:

$\frac{a}{b}$

$\frac{b^2}{a}$

$\frac{b}{a}$

Correct answer:

Explanation:

Remember that anything divided by itself equals .

To double check, let's turn this into a multiplication problem. Take the numerator and multiply it by the denominator's reciprocal:

$\frac{a}{b} \cdot \frac{b}{a} = 1$

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Example Question #2603 : Sat Mathematics

Simplify:

$\frac{\frac{a^2}{b^3}}{\frac{a}{b}}$

Possible Answers:

$\frac{a}{b}$

$\frac{a}{b^2}$

None of the given answers.

$\frac{b^2}{a}$

Correct answer:

$\frac{a}{b^2}$

Explanation:

Remember that dividing two fractions is the same as multiplying the numerator by the reciprocal of the denominator, like so:

$\frac{\frac{a^2}{b^3}}{\frac{a}{b}} = \frac{a^2}{b^3} \cdot \frac{b}{a} = \frac{a^2}{ab^3} = \frac{a}{b^2}$

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Example Question #2604 : Sat Mathematics

Simplify:

$abc \cdot a^2 b^2 c^2$

Possible Answers:

$\frac{1}{abc}$

a^3b^3c^3

None of the given answers.

a^2b^2c^2

Correct answer:

a^3b^3c^3

Explanation:

When we multiply two polynomials together, we add their exponents.

$abc \cdot a^2 b^2 c^2 = a^{1+2}b^{1+2}c^{1+2} = a^3b^3c^3$

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Example Question #2605 : Sat Mathematics

Simplify:

(a^0)^7

Possible Answers:

a^7

$\frac{1}{a^7}$

None of the given answers.

Correct answer:

Explanation:

When we raise a monomial with an exponent to the power of another exponent, we multiply their exponents, like so:

(a^0)^7 = a^0

Remember that anything raised to the power of zero is equal to .

Therefore, (a^0)^7 = a^0 = 1 .

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Example Question #831 : Algebra

Simplify:

$\frac{\frac{a^2b^3c}{a^5b^7c^4}}{abc}$

Possible Answers:

$\frac{1}{a^4b^5c^4}$

$\frac{a}{b}$

$\frac{a^5b^7c^4}{ab^2}$

None of the given answers

a^2b^3c

Correct answer:

$\frac{1}{a^4b^5c^4}$

Explanation:

We can rewrite this problem so that it's a simpler multiplication problem. Take the bottom of the express ( abc ) and write its reciprocal, then multiply it by the fraction in the numerator, like so:

$\frac{\frac{a^2b^3c}{a^5b^7c^4}}{abc} = \frac{a^2b^3c}{a^5b^7c^4} \cdot \frac{1}{abc}$

Now, we can simplify using cross-division.

$\frac{a^2b^3c}{a^5b^7c^4} \cdot \frac{1}{abc} = \frac{ab^2}{a^5b^7c^4}$

Now, we can simplify this expression. When we divide these terms, we subtract their exponents.

$\frac{ab^2}{a^5b^7c^4} = \frac{1}{a^4b^5c^4}$

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Example Question #69 : How To Simplify An Expression

Simplify.

$\frac{a^{4}b^{-2}c^{5}d^{14}e^{10}}{a^{6}b^{3}c^{7}d^{-7}e^{8}}$

Possible Answers:

$\frac{d^{7}e^{2}}{a^{2}b^{1}c^{2}}$

$a^{10}b^{1}c^{12}d^{7}e^{18}$

$\frac{d^{21}e^{2}}{a^{2}b^{5}c^{2}}$

$a^{\frac{2}3{}}b^{\frac{-2}{3}}c^{\frac{5}{7}}d^{-2}e^{\frac{5}{4}}$

Correct answer:

$\frac{d^{21}e^{2}}{a^{2}b^{5}c^{2}}$

Explanation:

Separate each group of variables to simplify.

$\frac{a^{4}}{a^{6}} =$ $\frac{a\times a\times a\times a}{a \times a \times a \times a \times a \times a}$ $= \frac{1}{a^{2}}$

$\frac{b^{-2}}{b^{3}}=$ $\frac{1}{b^{2}}\times\frac{1}{b^{3}}=$ $\frac{1}{b\times b\times b\times b\times b}=\frac{1}{b^{5}}$

$\frac{c^{5}}{c^{7}} =$ $\frac{c\times c\times c\times c\times c}{c\times c\times c\times c\times c\times c\times c}$ $= \frac{1}{c^{2}}$

$\frac{d^{14}}{d^{-7}}=\frac{d^{14}}{1}\times \frac{d^{7}}{1}$ $= \frac{d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d}{1}$

$= \frac{d^{21}}{1}$

$\frac{e^{10}}{e^{8}}=\frac{e\times e\times e\times e\times e\times e\times e\times e\times e\times e}{e\times e\times e\times e\times e\times e\times e\times e}$ $= \frac{e^{2}}{1}$

$\frac{1}{a^{2}}\times \frac{1}{b^{5}}\times \frac{1}{c^{2}}\times \frac{d^{21}}{1}\times \frac{e^{2}}{1}= \frac{d^{21}e^{2}}{a^{2}b^{5}c^{2}}$

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Example Question #141 : Expressions

$\frac{1}{u+1} = t$ . $u \ne -1, u \ne 4$ .

Which of the following is equal to $\frac{1}{u-4}$ ?

Possible Answers:

$\frac{5t}{1-t}$

t+5

$\frac{1-5t}{t}$

$\frac{t}{1-5t}$

$\frac{t}{t-5}$

Correct answer:

$\frac{t}{1-5t}$

Explanation:

Take the reciprocal of both expressions:

$\frac{1}{u+1} = t$

$u+1 = \frac{1}{t}$

Subtract 5 from both sides:

$u+1-5 = \frac{1}{t}-5$

$u-4 = \frac{1}{t}-5$

Rewrite the expression at left and simplify it:

$u-4 = \frac{1}{t}-\frac{5t}{t}$

$u-4 = \frac{1-5t}{t}$

Take the reciprocal of both expressions:

$\frac{1}{u-4} = \frac{t}{1-5t}$

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Example Question #141 : Expressions

$\sqrt{x+ 1} = t$ ; $x \ge 7$

Which of the following is equal to $\sqrt{x- 7}$ ?

Possible Answers:

$\sqrt{t^{2} -64 }$

$\sqrt{t^{2} -8}$

$t - 2 \sqrt{2}$

t -8

$t -1 - \sqrt{7}$

Correct answer:

$\sqrt{t^{2} -8}$

Explanation:

Square both expressions

$(\sqrt{x+ 1} )^{2}= t^{2}$

$x+1 = t^{2}$

Subtract 8 from both sides:

$x+1-8 = t^{2} -8$

$x-7= t^{2} - 8$

Take the positive square root of both sides:

$\sqrt{x-7}= \sqrt{t^{2} -8}$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2601 : Sat Mathematics

Example Question #2602 : Sat Mathematics

Example Question #2602 : Sat Mathematics

Example Question #2603 : Sat Mathematics

Example Question #2604 : Sat Mathematics

Example Question #2605 : Sat Mathematics

Example Question #831 : Algebra

Example Question #69 : How To Simplify An Expression

Example Question #141 : Expressions

Example Question #141 : Expressions

All SAT Math Resources

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