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SAT Math : SAT Mathematics

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Example Questions

Example Question #1 : Rational Expressions

Function_part1

Possible Answers:

9/5

37/15

–37/15

–11/5

–9/5

Correct answer:

–11/5

Explanation:

Fraction_part2

Fraction_part3

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Example Question #1 : Expressions

If Jill walks $100\:\text{ft}$ in $20\:\sec$ , how long will it take Jill to walk $1500\:\text{ft}$ ?

Possible Answers:

$200\sec$

$400\sec$

$100\sec$

$30\sec$

$300\sec$

Correct answer:

$300\sec$

Explanation:

To solve this, we need to set a proportion.

$\frac{100\:\text{ft}}{20\sec}=\frac{1500\:\text{ft}}{x}$

Cross Multiply

$100\cdot x=1500\cdot 20$

$x=\frac{30000}{100}=300$

So it will take Jill $300 \sec$ to walk $1500\: \text{ft}$

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Example Question #2471 : Sat Mathematics

If $\frac{b+c}{c}=\frac{14}{15}$ , then which of the following must be also true?

Possible Answers:

$\frac{b}{c}=10$

$\frac{b}{c}=\frac{7}{10}$

$\frac{b}{c}=1$

$\frac{b}{c}=-\frac{1}{15}$

Correct answer:

$\frac{b}{c}=-\frac{1}{15}$

Explanation:

$\frac{b+c}{c}=\frac{14}{15}$

$\frac{b}{c}+\frac{c}{c}=\frac{14}{15}$

$\frac{b}{c}+1=\frac{14}{15}$

$\frac{b}{c}+1\ {\color{Red} -1}=\frac{14}{15}\ {\color{Red} -1}$

$\frac{b}{c}=\frac{14}{15}-\frac{15}{15}$

$\frac{b}{c}=-\frac{1}{15}$

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Example Question #2472 : Sat Mathematics

Which of the following is equivalent to $\dpi{100} \frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ ? Assume that denominators are always nonzero.

Possible Answers:

$t-x$

$x^{2}-t^{2}$

$x-t$

$(xt)^{-1}$

$\frac{x}{t}$

Correct answer:

$(xt)^{-1}$

Explanation:

We will need to simplify the expression $\frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ . We can think of this as a large fraction with a numerator of $\frac{1}{t}-\frac{1}{x}$ and a denominator of $\dpi{100} x-t$ .

In order to simplify the numerator, we will need to combine the two fractions. When adding or subtracting fractions, we must have a common denominator. $\frac{1}{t}$ has a denominator of $\dpi{100} t$ , and $\dpi{100} -\frac{1}{x}$ has a denominator of $\dpi{100} x$ . The least common denominator that these two fractions have in common is $\dpi{100} xt$ . Thus, we are going to write equivalent fractions with denominators of $\dpi{100} xt$ .

In order to convert the fraction $\dpi{100} \frac{1}{t}$ to a denominator with $\dpi{100} xt$ , we will need to multiply the top and bottom by $\dpi{100} x$ .

$\frac{1}{t}=\frac{1\cdot x}{t\cdot x}=\frac{x}{xt}$

Similarly, we will multiply the top and bottom of $\dpi{100} -\frac{1}{x}$ by $\dpi{100} t$ .

$\frac{1}{x}=\frac{1\cdot t}{x\cdot t}=\frac{t}{xt}$

We can now rewrite $\frac{1}{t}-\frac{1}{x}$ as follows:

$\frac{1}{t}-\frac{1}{x}$ = $\frac{x}{xt}-\frac{t}{xt}=\frac{x-t}{xt}$

Let's go back to the original fraction $\frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ . We will now rewrite the numerator:

$\frac{(\frac{1}{t}-\frac{1}{x})}{x-t}$ = $\frac{\frac{x-t}{xt}}{x-t}$

To simplify this further, we can think of $\frac{\frac{x-t}{xt}}{x-t}$ as the same as $\frac{x-t}{xt}\div (x-t)$ . When we divide a fraction by another quantity, this is the same as multiplying the fraction by the reciprocal of that quantity. In other words, $a\div b=a\cdot \frac{1}{b}$ .

$\frac{x-t}{xt}\div (x-t)$ = $\frac{x-t}{xt}\cdot \frac{1}{x-t}$ $=\frac{x-t}{xt(x-t)}$ $= \frac{1}{xt}$

Lastly, we will use the property of exponents which states that, in general, $\frac{1}{a}=a^{-1}$ .

$\frac{1}{xt}=(xt)^{-1}$

The answer is $(xt)^{-1}$ .

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Example Question #2473 : Sat Mathematics

Simplify (4x)/(x²– 4) * (x + 2)/(x²– 2x)

Possible Answers:

(4x²+ 8x)/(x⁴+ 8x)

4/(x + 2)²

x/(x + 2)

x/(x – 2)²

4/(x – 2)²

Correct answer:

4/(x – 2)²

Explanation:

Factor first. The numerators will not factor, but the first denominator factors to (x – 2)(x + 2) and the second denomintaor factors to x(x – 2). Multiplying fractions does not require common denominators, so now look for common factors to divide out. There is a factor of x and a factor of (x + 2) that both divide out, leaving 4 in the numerator and two factors of (x – 2) in the denominator.

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Example Question #4 : Expressions

what is 6/8 X 20/3

Possible Answers:

18/160

120/11

3/20

9/40

Correct answer: 5

Explanation:

6/8 X 20/3 first step is to reduce 6/8 -> 3/4 (Divide top and bottom by 2)

3/4 X 20/3 (cross-cancel the threes and the 20 reduces to 5 and the 4 reduces to 1)

1/1 X 5/1 = 5

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Example Question #8 : Rational Expressions

Evaluate and simplify the following product:

$\frac{15x^2+4xy^3-3y}{x^2y^7}\times\frac{5y^4}{4x^3}$

Possible Answers:

$\frac{20x^2+9xy^3-2y}{4x^5y^3}$

$\frac{75x^2+20xy^3-15y}{4x^5y^3}$

$\frac{75x^6+20xy^1^2-15y^4}{4x^6y^3}$

$\frac{75x^2y^-^3+20x-15y}{4x^5}$

Correct answer:

$\frac{75x^2+20xy^3-15y}{4x^5y^3}$

Explanation:

The procedure for multplying together two rational expressions is the same as multiplying together any two fractions: find the product of the numerators and the product of the denominators separately, and then simplify the resulting quotient as far as possible, as shown:

$\frac{15x^2+4xy^3-3y}{x^2y^7}\times\frac{5y^4}{4x^3}$

$=\frac{(15x^2+4xy^3-3y)(5y^4)}{(x^2y^7)(4x^3)}$

$=\frac{75x^2y^4+20xy^7-15y^5}{4x^5y^7}$

$=\frac{75x^2+20xy^3-15y}{4x^5y^3}$

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Example Question #1 : How To Add Rational Expressions With A Common Denominator

Simplify the following rational expression: (9x - 2)/(x²) MINUS (6x - 8)/(x²)

Possible Answers:

$\frac{3x+6}{x^{2}}$

$\frac{15x+6}{x}$

$\frac{12x-6}{x}$

3x-10

Correct answer:

$\frac{3x+6}{x^{2}}$

Explanation:

Since both expressions have a common denominator, x², we can just recopy the denominator and focus on the numerators. We get (9x - 2) - (6x - 8). We must distribute the negative sign over the 6x - 8 expression which gives us 9x - 2 - 6x + 8 ( -2 minus a -8 gives a +6 since a negative and negative make a positive). The numerator is therefore 3x + 6.

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Example Question #2474 : Sat Mathematics

Simplify the following rational expression:

$\frac{7x-18}{x^{2}}+\frac{6x-14}{x^{2}}$

Possible Answers:

$\frac{13x-4}{x^{2}}$

$\frac{13x-32}{x^{2}}$

$\frac{x-32}{x^{2}}$

$\frac{13x-28}{x^{2}}$

$\frac{x-4}{x^{2}}$

Correct answer:

$\frac{13x-32}{x^{2}}$

Explanation:

Since both fractions in the expression have a common denominator of $x^{2}$ , we can combine like terms into a single numerator over the denominator:

$\frac{7x-18}{x^{2}}+\frac{6x-14}{x^{2}}$

$=\frac{(7x-18)+(6x-14)}{x^{2}}$

$=\frac{13x-32}{x^{2}}$

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Example Question #3 : How To Add Rational Expressions With A Common Denominator

Simplify the following rational expression:

$\frac{5x-5}{2x^{2}} + \frac{7x+9}{2x^{2}}$

Possible Answers:

$\frac{6x+4}{2x^{2}}$

$\frac{12x+8}{2x^{2}}$

$\frac{8x+4}{2x^{2}}$

$\frac{6x+8}{2x^{2}}$

$\frac{6x+2}{x^{2}}$

Correct answer:

$\frac{6x+2}{x^{2}}$

Explanation:

Since both rational terms in the expression have the common denominator $2x^{2}$ , combine the numerators and simplify like terms:

$\frac{5x-5}{2x^{2}} + \frac{7x+9}{2x^{2}}$

$=\frac{(5x-5)+(7x+9)}{2x^{2}}$

$=\frac{12x+4}{2x^{2}}$

$=\frac{6x+2}{x^2}$

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #1 : Rational Expressions

Example Question #1 : Expressions

Example Question #2471 : Sat Mathematics

Example Question #2472 : Sat Mathematics

Example Question #2473 : Sat Mathematics

Example Question #4 : Expressions

Example Question #8 : Rational Expressions

Example Question #1 : How To Add Rational Expressions With A Common Denominator

Example Question #2474 : Sat Mathematics

Example Question #3 : How To Add Rational Expressions With A Common Denominator

All SAT Math Resources

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