GMAT Math : Real Numbers

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Example Question #1 : Real Numbers

Solve $\dpi{100} \small 2x-3> 0$ for x.

Possible Answers:

$\dpi{100} \small x\leq \frac{3}{2}$

$\dpi{100} \small x> \frac{-3}{2}$

$\dpi{100} \small x> \frac{3}{2}$

$\dpi{100} \small x< \frac{3}{2}$

$\dpi{100} \small x\geq \frac{3}{2}$

Correct answer:

$\dpi{100} \small x> \frac{3}{2}$

Explanation:

$\dpi{100} \small 2x-3> 0$

Add 3 to both sides: $\dpi{100} \small 2x> 3$

Divide both sides by 2: $\dpi{100} \small x> \frac{3}{2}$

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Example Question #1 : Real Numbers

Of 200 students, 80 take biology, 40 take chemistry, 60 take physics, 13 take two science courses, and no one takes three science courses. How many students are not taking a science course?

Possible Answers:

180

Correct answer:

Explanation:

To calculate the number of students taking at least 1 science course, add the number of students who are taking each course and subtract the number of students who are taking 2 (to ensure they're not counting twice).

$\small 80\ +\ 40\ +\ 60\ -\ 13\ =167$

To calculate the number of students NOT taking a class, subtract this number by the total number of students.

$\small 200\ -\ 167\ =\ 33$

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Example Question #3 : Real Numbers

Which of the following expressions is equal to $\sqrt[3]{4} \cdot \sqrt{2}$

Possible Answers:

$4 \sqrt[6]{4}$

$4 \sqrt[6]{2}$

$2 \sqrt[6]{2}$

$\sqrt[6]{8}$

$\sqrt[8]{8}$

Correct answer:

$2 \sqrt[6]{2}$

Explanation:

$\small \sqrt[3]{4} \cdot \sqrt{2} = 4^{\frac{1}{3}} \cdot 2^{\frac{1}{2}} = (2^{2})^{\frac{1}{3}} \cdot 2^{\frac{1}{2}} = 2^{2\; \cdot \frac{1}{3}} \cdot 2^{\frac{1}{2}}= 2^{ \frac{2}{3}} \cdot 2^{\frac{1}{2}}= 2^{ \frac{2}{3} + \frac{1}{2}}$

$\small = 2^{ \frac{7}{6} }} = \sqrt[6]{2^7} = 2\sqrt[6]{2}$

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Example Question #4 : Real Numbers

Given that $\small \small x + y = 9$ , $\small xy = 16$ , and $\small \small x < y$ , evaluate $\small \frac{1}{x}-\frac{1}{y}$ .

Possible Answers:

$\small \frac{\sqrt{17}}{16}$

$\small \small \small \frac{{9}}{16}$

Not enough information is given to answer the question

$\small \small \small \small \frac{{3}}{16}$

$\small \small \frac{{17}}{16}$

Correct answer:

$\small \frac{\sqrt{17}}{16}$

Explanation:

$\small \frac{1}{x}-\frac{1}{y} = \frac{y-x}{xy}$

To find $\small y - x$ :

$\small \small \small y + x = 9$ ,

so $\small \small \left (y + x \right )^{2} =y^{2} + 2xy +x^{2} = 81$

$\small \small y^{2} + 2xy +x^{2} - 4xy= 81 - 4 \cdot 16$

$\small \small y^{2} - 2xy +x^{2} = (y - x)^{2}= 17$

$\small \small y - x = \pm \sqrt{17}$

Since $\small \small x < y$ , $\small \small \small \small y - x > 0$

and we choose the positive square root

$\small \small \small y - x = \sqrt{17}$

$\small \small \small \frac{1}{x}-\frac{1}{y} = \frac{y-x}{xy}= \frac{\sqrt{17}}{16}$

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Example Question #5 : Real Numbers

Today, Becky's age (B) is 3 times Charlie's age. In 3 years, what will Charlie's age be in terms of B?

Possible Answers:

$\small \frac{B}{3}+3$

$\small \frac{B+3}{3}$

$\small 3(B+3)$

$\small 3B+3$

$\small 3(B+3)$

$\small 3B+3$

Correct answer:

$\small \frac{B}{3}+3$

Explanation:

Today, $\small C=\frac{B}{3}$ . In 3 years, $\small C=\frac{B}{3}+3$ .

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Example Question #6 : Real Numbers

Given that $6 \leq A \leq 12$ and $4 \leq B \leq 6$ , what is the range of possible values for $\frac{A}{B}$ ?

Possible Answers:

$1 \leq \frac{A}{B} \leq 3$

$\frac{3}{2} \leq \frac{A}{B} \leq 2$

$24 \leq \frac{A}{B} \leq 72$

$0 \leq \frac{A}{B} \leq 3$

$\frac{1}{2} \leq \frac{A}{B} \leq \frac{2}{3}$

Correct answer:

$1 \leq \frac{A}{B} \leq 3$

Explanation:

The lowest possible value of $\frac{A}{B}$ is the lowest possible value of divided by the highest possible value of :

$\frac{A}{B} \geq 6 \div 6 = 1$

The highest possible value of $\frac{A}{B}$ is the highest possible value of divided by the lowest possible value of :

$\frac{A}{B} \leq 12 \div 4 = 3$

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Example Question #7 : Real Numbers

If and are composite integers, which of the following can be prime?

Possible Answers:

x^2-y^2

None of these can be prime.

x^2y+2y

x^2+y^2

$x\cdot y$

Correct answer:

x^2+y^2

Explanation:

x^2-y^2 = (x-y)(x+y) so this is a composite number for all and .

$x\cdot y$ is by definition a composite number.

$x^2y+2y = y\cdot (x^2+2)$ the product of 2 numbers.

This leaves just x^2+y^2 . For a number to be prime, it must be odd (except for 2) so we need to have either or be odd (but not both). The first composite odd number is 9. 9^2 = 81 . The smallest composite number is 4. 4^2 = 16 .

81+16 = 97 is a prime number.

So the answer is x^2+y^2

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Example Question #1 : Real Numbers

If $\frac{122}{x-3}$ is a real number, which of the following CANNOT be a value for x?

Possible Answers:

122

-122

125

-3

Correct answer:

Explanation:

The definition of the set of real numbers is the set of all numbers that can fit into a/b where a and b are both integers and b does not equal 0.

So, since we see a fraction here, we know a non-real number occurs if the denominator is 0. Therefore we can find where the denominator is 0 by setting x-3 =0 and solving for x. In this case, x=3 would create a non-real number. Thus our answer is that x CANNOT be 3 for our expression to be a real number.

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Example Question #8 : Real Numbers

Let be the product of integers from 18 to 33, inclusive. If g=2x , how many more unique prime factors does have than ?

Possible Answers:

Greater than

Not enough information given.

Correct answer:

Explanation:

This question does not require any calculation. Given that 32 (an even number) is a factor of , then 2 must be a prime factor. If is then multiplied by 2 (to get ) then has no additional unique prime factors (its only additional prime factor, 2, is NOT unique).

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Example Question #1 : Real Numbers

If $\dpi{100} \small x\ and\ y$ are both negative, then $\dpi{100} \small \frac{x+y}{-xy}$ could NOT be equal to....

Possible Answers:

$\dpi{100} \small \frac{3}{4}$

$\dpi{100} \small \frac{\sqrt{8}}{4}$

$\dpi{100} \small -5$

$\dpi{100} \small 5$

Correct answer:

$\dpi{100} \small -5$

Explanation:

$\dpi{100} \small x+y$ is negative and $\dpi{100} \small xy$ is positive

$\dpi{100} \small \frac{Negative}{-Positive} = \frac{Negative}{Negative} = Positive$

Therefore, the solution cannot be negative.

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GMAT Math : Real Numbers

Study concepts, example questions & explanations for GMAT Math

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