Expressions - GRE Quantitative Reasoning

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Question

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

Answer

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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Question

Simplify the following rational expression:

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

Answer

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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Question

Simplify the following rational expression:

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

Answer

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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Question

Simplify the following expression:

$\frac{10x-9}{x^{3}}+\frac{11x+12}{x^{3}}$

Answer

Since both terms in the expression have the common denominator $x^{3}$ , combine the fractions and simplify the numerators:

$\frac{10x-9}{x^{3}}+\frac{11x+12}{x^{3}}$

$=\frac{(10x-9)+(11x+12)}{x^{3}}$

$=\frac{21x+3}{x^{3}}$

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Question

Add and simplify:

$\frac{3x^2 + 3x + 3}{2y} + \frac{10x^2 + 3x}{2y}$

Answer

When adding rational expressions with common denominators, you simply need to add the like terms in the numerator.

Therefore, $\frac{13x^2 + 6x + 3}{2y}$ is the best answer.

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Question

Simplify the expression.

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

Answer

To add rational expressions, first find the least common denominator. Because the denominator of the first fraction factors to 2(x+2), it is clear that this is the common denominator. Therefore, multiply the numerator and denominator of the second fraction by 2.

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

$\frac{x}{2x+4}+\frac{(2)x}{(2)(x+2)}$

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

$\frac{3x}{2x+4}$

This is the most simplified version of the rational expression.

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Question

Simplify the following:

$\frac{x+3}{x-2}+\frac{x-2}{x+3}$

Answer

To simplify the following, a common denominator must be achieved. In this case, the first term must be multiplied by (x+2) in both the numerator and denominator and likewise with the second term with (x-3).

$\frac{(x+3)(x+3)}{(x-2)(x+3)}+\frac{(x-2)(x-2)}{(x+3)(x-2)}$

$\frac{(x+3)^2+(x-2)^2}{(x+3)(x-2)}$

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Question

Choose the answer which best simplifies the following expression:

$\frac{3p}{2} + \frac{2p}{p^3}$

Answer

To simplify this expression, you have to get both numerators over a common denominator. The best way to go about doing so is to multiply both expressions by the others denominator over itself:

$\frac{3p}{2} \cdot \frac{p^3}{p^3 } + \frac{2p}{p^3} \cdot \frac{2}{2}$

Then you are left with:

$\frac{3p^4}{2p^3} + \frac{4p}{2p^3}$

Which you can simplify into:

$\frac{3p^4+4p}{2p^3}$

From there, you can take out a :

$\frac{p(3p^3 + 4)}{p(2p^2)}$

Which gives you your final answer:

$\frac{3p^3 +4}{2p^2}$

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Question

Choose the answer which best simplifies the following expression:

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

Answer

To solve this problem, first multiply both terms of the expression by the denominator of the other over itself:

$\frac{2a + 2}{3} \cdot \frac{a^2}{a^2} + \frac{3}{a^2} \cdot \frac{3}{3}$

$\frac{2a^3 +2a^2}{3a^2} + \frac{9}{3a^2}$

Now that both terms have a common denominator, you can add them together:

$\frac{2a^3 +2a^2 + 9}{3a^2}$

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Question

Choose the answer which best simplifies the following expression:

$\frac{3a^2 + a}{2} + \frac{10}{2a^2-a}$

Answer

To simplify, first multiply both terms by the denominator of the other term over itself:

$\frac{3a^2 + a}{2} \left(\frac{2a^2 - a}{2a^2 - a}\right) + \frac{10}{2a^2-a}\left(\frac{2}{2}\right)$

$\frac{6a^4 + 2a^3 -3a^3 - a^2}{4a^2 -2a} + \frac{20}{4a^2 - 2a}$

Then, you can combine the terms, now that they share a denominator:

$\frac{6a^4 - a^3 -a^2 +20}{4a^2 - 2a}$

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Question

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

Answer

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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Question

Simplify:

$\frac{x^{2}+x-2}{3x^{2} + 9x + 6} \div (x-1)$

Answer

Multiply by the reciprocal of $\frac{x-1}1{}$ .

$\frac{x^{2}+x-2}{3x^{2}+9x+6} \cdot \frac{1}{x-1}$

Factor $\frac{(x-1)(x+2)}{3(x+2)(x+1)} \cdot \frac{1}{x-1}$

Divide by common factors.

$= \frac{1}{3(x+1)}$

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Question

Solve for :

$\frac{\frac{2m^2 + 5}{4}}{ \frac{m}{2}} = 20$

Answer

To tackle this problem, you need to invert and multiply:

$\frac{\frac{2m^2 + 5}{4}} { \frac{m}{2}} = 20$

$\frac{2m^2 + 5}{4} \cdot \frac{2}{m} = 20$

$\frac{4m^2 + 10}{4m} = 20$

Here we see that we have created a quadratic equation. Therefore, we get all terms to one side, set it equal to zero and use the quadratic formula to solve.

The quadratic formula is:

$m=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where

Plugging these values in we get the following:

$\frac{80\pm\sqrt{80^2-4(4)(10)}}{2(4)}=\frac{80\pm\sqrt{6240}}{8}=\frac{80\pm78.99}{8}$

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Question

Express the following as a single rational expression:

$\frac{\frac{2x^2 + 7}{12b} }{ \frac{y^2 + 2y}{3a^2 + 2}}$

Answer

To divide one rational expression by another, invert and multiply:

$\frac{\frac{2x^2 + 7}{12b}}{ \frac{y^2 + 2y}{3a^2 + 2}}$

$\frac{2x^2 + 7 }{12b} \cdot \frac{3a^2 + 2}{y^2 + 2y}$

Remember to foil the numerator meaning, multiply the first components of each binomial. Then multiply the outer components of each binomial. After that, multiply the inner components together, and lastly, multiply the components in the last position of the binomials together.

This arrives at the following:

$\frac{6x^2a^2 +21a^2 +4x^2 +14}{12by^2 + 24by}$

You can't factor anything out, so that's your final answer.

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Question

Quantitative Comparison

$x<0\ and\ y<0$

A

---

$\frac{x^{-1}}{y^{-1}}$

B

---

$\left | \frac{y}{x} \right |$

Answer

Since both values are negative, in both situations the final result of the fraction will be positive regardless of the absolute value sign. Additionally, the definition of the exponent "–1" means the value is becoming its reciprocal. Thus, "x" → "1/x" and "1/y" → "y", equating to y/x.

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Question

Quantitative Comparison

0 < x < 1

A

---

(2x + 5)/(x2)

B

---

5x

Answer

Since A is a fraction with an exponential term in the denominator, its maximum value is when x is at a minimum. In B, the maximum value is when x approaches its maximum. Therefore, we can check whether there is overlap between the two quantities: No matter how close to either 0 or 1 x reaches, A will always be greater than B. (In fact, the minimum value for A is ~7, while the maximum value of B is ~5)

Be sure to keep your value of x consistent when plugging between the two fractions! The question asks for when they have the same x-value, not for when they are solved independently.

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Question

If x = -4 and y = 7, what is the value of 3x-5y?

Answer

Substitute the values into equation: 3(-4) - 5(7) = -12 - 35 = -47.

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Question

If the average (arithmetic mean) of the above terms is , what is the median?

Answer

Since we know the mean is 7, and the number of terms is 5, we can first find the value of x:

((x – 2)+ (x) + (x + 1) + (x + 4) + (x + 7))/5 = (5x + 10)/5 = 7

x = 5

Since the terms are already listed in order, we can simpy plug in x = 5 to the middle value to get 5 + 1 = 6 as the median term.

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Question

Quantitative Comparison

Quantity A: x

Quantity B: 2_x_

Answer

For a quantitative comparison question such as this one, it is best to first plug in the numbers 0, 1, and –1. Plugging in 0 gets the same answer for both columns. Plugging in 1 makes Quantity B bigger. Plugging in –1 makes Quantity A bigger. Therefore the answer cannot be determined.

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Question

A store sells 17 coffee mugs for $169. Some of the mugs are $12 each and some are $7 each. How many $7 coffee mugs were sold?

Answer

The answer is 7.

Write two independent equations that represent the problem.

x + y = 17 and 12_x_ + 7_y_ = 169

If we solve the first equation for x, we get x = 17 – y and we can plug this into the second equation.

12(17 – y) + 7_y_ = 169

204 – 12_y_ + 7_y_ =169

–5_y_ = –35

y = 7

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