Real Numbers - GMAT Quantitative Reasoning

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Question

Each of $\bigcirc, \square, \bigtriangleup$ stands for a real number; if one appears more than once in a choice, it stands for the same number each time.

Which of the following diagrams demonstrates an associative property?

Answer

Addition and multiplication are both associative, which means that a sum or product of three numbers will result in the same value regardless of which two are added or multiplied first. This is demonstrated in multiplication by the diagram

$\bigcirc \times (\square \times \bigtriangleup) = ( \bigcirc \times \square) \times \bigtriangleup)$ .

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Question

Each of $\bigcirc, \square, \bigtriangleup$ stands for a real number; if one appears more than once in a choice, it stands for the same number each time.

Which of the following diagrams demonstrates the distributive property?

Answer

The distributive property of multiplication over addition is the idea that, if a number is multiplied by a sum, the result will be the same as if the products of that number and each individual addend are added. The diagram that demonstrates this is

$\bigcirc \times (\square + \bigtriangleup) = \bigcirc \times \square + \bigcirc \times \bigtriangleup$ .

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Question

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

Answer

$\dpi{100} \small x+y$ 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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Question

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

Answer

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

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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Question

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?

Answer

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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Question

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

Answer

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

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Question

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

Answer

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

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

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Question

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

Answer

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

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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 . 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. . The smallest composite number is 4. .

is a prime number.

So the answer is

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Question

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

Answer

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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Question

If $\frac{873}{x^3-8}$ is a real number, which one of these cannot be a value of ?

Answer

For the expression to be defined, the denominator needs to be different from 0. Therefore:

$x^{3}-8=0$

$x^3 = 8 \Leftrightarrow x = \sqrt[3]{8} \Leftrightarrow x = (2^{3})^{\frac{1}{3}} = 2$

So the correct answer is 2.

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Question

If , and , and , what can we say for sure about ?

Answer

To show the other answers aren't always true, we need to choose 2 numbers that satisfy the given inequalities but also contradict each answer one-by-one.

Let this choice shows that sometimes , which rules out the answers , is negative, and

Now Let , this will still satisfy the given inequalities, but now . This means that the answer " is positive" isn't always true either.

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Question

Add one fourth to one seventh. Subtract this sum from one. What is the result?

Answer

The expression is equal to $1 - \left ( \frac{1}{4} + \frac{1}{7}\right )$ , which is calculated as follows:

$1 - \left ( \frac{1}{4} + \frac{1}{7}\right )$

$= 1 - \left ( \frac{1\cdot 7}{4 \cdot 7} + \frac{4 \cdot 1}{4 \cdot 7}\right )$

$= 1 - \left ( \frac{7}{28} + \frac{4 }{28}\right )$

$= 1 - \left ( \frac{7+4}{28} \right )$

$= 1 - \frac{11}{28}$

$= \frac{28}{28} - \frac{11}{28}$

$= \frac{28-11}{28}$

$= \frac{17}{28}$ ,

or seventeen twenty-eighths.

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Question

Which of the following is equal to 0.0407?

Answer

The last nonzero digit is located in the fourth place right of the decimal point - the ten-thousandths place. Put the number, without decimal point or leading zeroes, over 10,000. This number is $\frac{407}{10,000}$ , or four hundred seven ten-thousandths.

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Question

Which of the following is equal to the sum of thirty-three one-thousandths and three hundred three ten-thousandths?

Answer

The one-thousandths and ten-thousandths places are the third places and the fourth places, respectively, to the right of the decimal point. Therefore:

Thirty-three one-thousandths =

Three hundred three ten-thousandths =

Add:

$\begin{matrix} : : : 0.0330\ +\underline{0.0303}\ : : : 0.0633 \end{matrix}$

The last nonzero digit ends at the ten-thousandths place, so this is $\frac{633}{10,000}$ , or

six hundred thirty-three ten-thousandths.

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Question

Which of the following is equal to three one-hundredths subtracted from one fourth?

Answer

The one-hundredths place is the second place to the right of the decimal point; therefore, three one-hundredths is equal to 0.03. One fourth can be converted to decimal form as follows:

$\frac{1}{4} = \frac{1 \cdot 25}{4\cdot 25} = \frac{25}{100}$ ,

which is twenty-five one-hundredths, or 0.25. Subtract:

,

or twenty-two one-hundredths.

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Question

$\small \frac{a}{b}$ and $\small \frac{b}{c}$ are integers. Which of the following need not be an integer?

Answer

$\frac{a}{b}\cdot \frac{b}{c} = \frac{a} {c}$ . As the product of integers, $\frac{a}{c}$ must be an integer.

$\frac{a+b}{b} = \frac{a }{b} + \frac{ b}{b} = \frac{a }{b} + 1$ , making $\frac{a+b}{b}$ the sum of integers, and, consequently, an integer.

$\frac{a+b}{c} = \frac{a }{c} + \frac{ b}{c}$ , making $\frac{a+b}{b}$ the sum of integers, and, consequently, an integer.

$\frac{b-c}{c} = \frac{b }{c} -\frac{ c}{c} = \frac{b}{c} - 1$ , making $\frac{b-c}{c}$ the difference of integers, and, consequently, an integer.

We demonstrate that $\small \frac{a+c}{b}$ need not be an integer through a counterexample. Let .

$\frac{a}{b} = \frac{20}{10} = 2$ and $\frac{b}{c} = \frac{10}{5} = 2$ , so the conditions of the problem are met. However,

$\small \frac{a+c}{b} =\small \frac{20+5}{10} = \frac{25}{10} = 2.5$ , which is not an integer. This makes $\small \frac{a+c}{b}$ the correct response.

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Question

Order $\small a,b, c,d$ from least to greatest.

Answer

We can find each in terms of .

In ascending order, the numbers are:

The correct choice is $\small b,a,d,c$ .

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