Arithmetic Practice Questions — GRE General Test — Page 3

Question 41Question →

Let $m$ and $n$ be negative integers such that $m < n$ and $m - n$ is an odd integer. Which of the following expressions must be a positive even integer?

(m - n)^2 + 1

m + n + 1

(-1)^n (m - n)

m^2 + n^2

(m - n)^3

Show answer & explanation

Answer: $(m - n)^2 + 1$

Answer

(m - n)^2 + 1

must be a positive even integer.

Because

m - n

is given as an odd integer and

m < n

,

m - n

is a negative odd integer. Squaring any negative odd integer gives a positive odd integer. Adding

1

to a positive odd integer yields a positive even integer, which guarantees the result is always positive and even.

Step-by-Step Solution

1

Analyze the parity and sign of

m - n

m - n

is a negative odd integer.

Since

m < n

,

m - n < 0

. The problem specifies that

m - n

is odd.

2

Evaluate the expression

(m - n)^2

(m - n)^2

is a positive odd integer.

Squaring any non-zero real number yields a positive result. Squaring an odd integer always yields an odd integer.

3

Add 1 to

(m - n)^2

(m - n)^2 + 1

is an even integer greater than or equal to 2.

Adding 1 to a positive odd integer converts it into a positive even integer.

Key Concept

Parity rules under arithmetic operations and exponents with signed integers

Estimated Time:1m 15s

Question 42Question →

Let $x$ and $y$ be non-zero integers that satisfy all of the following conditions:
I. $(-x)^{y + 1} < 0$
II. $x^2 y + x$ is odd
III. $x < y$

Which of the following statements MUST be true? Select all that apply.

Select all that apply

y^x

is an even integer

x^y - y

is an odd integer

-x < -y

(x + y)^2

is an odd integer

(-x)^y < 0

Show answer & explanation

Answer: $y^x$ is an even integer; $x^y - y$ is an odd integer; $(x + y)^2$ is an odd integer

Answer

The statements that must be true are '

y^x

is an even integer', '

x^y - y

is an odd integer', and '

(x + y)^2

is an odd integer'.

From the given conditions, we deduce that

x

is a positive odd integer and

y

is a positive even integer with

x < y

. Therefore:
- Raising the positive even integer

y

to the positive power

x

yields an even integer (

y^x

is even).
- Raising the odd integer

x

to the positive power

y

yields an odd integer, and subtracting the even integer

y

leaves an odd integer (

x^y - y

is odd).
- The sum of odd

x

and even

y

is odd, and squaring an odd integer yields an odd integer (

(x + y)^2

is odd).

Step-by-Step Solution

1

Analyze Condition I:

(-x)^{y + 1} < 0

x > 0

(a positive integer) and

y

is an even integer.

For a number raised to an integer power to be strictly negative, the base must be negative and the exponent must be odd. Hence,

-x < 0 \implies x > 0

, and

y + 1

is odd

\implies y

is even.

2

Analyze Condition II:

x^2 y + x

is odd

x

is an odd integer.

Factor the expression as

x(xy + 1)

. A product of two integers is odd if and only if both factors are odd. Thus,

x

must be odd. (Additionally,

xy + 1

must be odd

\implies xy

is even, which holds since

y

is even).

3

Analyze Condition III:

x < y

y

is a positive even integer.

Since

x

is a positive odd integer (

x \ge 1

) and

x < y

,

y

must also be a positive integer (

y \ge 2

).

4

Evaluate the statements based on the derived properties (

x

is positive odd,

y

is positive even,

x < y

)

The statements '

y^x

is an even integer', '

x^y - y

is an odd integer', and '

(x + y)^2

is an odd integer' are guaranteed to be true.

1)

\text{even}^{\text{positive odd}} = \text{even}

. 2)

\text{odd}^{\text{positive even}} - \text{even} = \text{odd} - \text{even} = \text{odd}

. 3)

(\text{odd} + \text{even})^2 = \text{odd}^2 = \text{odd}

.

Key Concept

Deducing parity and signs of variables using exponent rules and arithmetic properties of even and odd numbers.

Estimated Time:2m 0s

Question 43Question →

If $x$ is a real number such that $|2x - 5| = |x + 4|$ , which of the following could be the value of $x^2$ ? Select all such values.

Select all that apply

\frac{1}{9}

\frac{1}{3}

1

9

81

Show answer & explanation

Answer: $\frac{1}{9}$ ; $81$

Answer

The possible values of

x^2

are

\frac{1}{9}

and

81

.

To solve

|2x - 5| = |x + 4|

, set

2x - 5

equal to both

x + 4

and

-(x + 4)

. Solving

2x - 5 = x + 4

gives

x = 9

, which leads to

x^2 = 81

. Solving

2x - 5 = -x - 4

gives

3x = 1

, so

x = \frac{1}{3}

, which leads to

x^2 = \frac{1}{9}

. Therefore, the options representing

\frac{1}{9}

and

81

are correct.

Step-by-Step Solution

1

Set up equations to remove the absolute value bars

Two linear equations:

2x - 5 = x + 4

or

2x - 5 = -(x + 4)

.

For real numbers

A

and

B

,

|A| = |B|

implies

A = B

or

A = -B

.

2

Solve the first equation

2x - 5 = x + 4

x = 9

.

Subtract

x

and add

5

to both sides.

3

Solve the second equation

2x - 5 = -x - 4

3x = 1 \implies x = \frac{1}{3}

.

Add

x

and add

5

to both sides.

4

Compute

x^2

for each possible value of

x

If

x = 9

, then

x^2 = 81

. If

x = \frac{1}{3}

, then

x^2 = \frac{1}{9}

.

The question asks for the values of

x^2

, not

x

.

Key Concept

Solving equations with absolute values on both sides requires considering positive and negative case equivalences.

Question 44Question →

If $x$ and $y$ are real numbers such that $|x - 4| \le 3$ and $|y + 2| \le 5$ , what is the maximum possible value of $|x - y|$ ?

Show answer & explanation

Answer: 14

Answer

The maximum possible value of

|x - y|

is 14.

Solving

|x - 4| \le 3

gives the closed interval

[1, 7]

for

x

. Solving

|y + 2| \le 5

gives the closed interval

[-7, 3]

for

y

. The maximum possible value of

|x - y|

is the maximum distance between a point in

[1, 7]

and a point in

[-7, 3]

, which is

7 - (-7) = 14

.

Step-by-Step Solution

1

Determine the range of possible values for

x

.

1 \le x \le 7

The inequality

|x - 4| \le 3

represents all numbers within distance 3 of 4 on the number line.

2

Determine the range of possible values for

y

.

-7 \le y \le 3

The inequality

|y + 2| \le 5

represents all numbers within distance 5 of -2 on the number line.

3

Find the maximum distance between any point

x

in

[1, 7]

and any point

y

in

[-7, 3]

.

14

The maximum absolute difference

|x - y|

occurs between the upper endpoint of the

x

-interval (

7

) and the lower endpoint of the

y

-interval (

-7

).

Key Concept

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables

Question 45Question →

If $n$ is a positive integer such that $2^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480$ , what is the value of $n$ ?

Show answer & explanation

Answer: 7

Answer

The value of

n

is 7.

Factoring out

2^{n-2}

converts the sum into

2^{n-2}(1 + 2 + 4 + 8) = 15 \cdot 2^{n-2} = 480

. Dividing 480 by 15 gives

2^{n-2} = 32

. Since

32 = 2^5

, setting

n - 2 = 5

yields

n = 7

.

Step-by-Step Solution

1

Factor out the lowest power of 2, which is

2^{n-2}

, from all terms on the left side of the equation.

2^{n-2}(1 + 2 + 4 + 8) = 480

Factoring out a common exponential term simplifies the addition of powers into a product of a single exponential term and a constant.

2

Evaluate the constant factor inside the parentheses and solve for the exponential expression

2^{n-2}

.

15 \cdot 2^{n-2} = 480 \implies 2^{n-2} = 32

Summing

1 + 2 + 4 + 8

yields 15. Dividing both sides by 15 isolates the base-2 term.

3

Express 32 as a power with base 2 and equate the exponents.

2^{n-2} = 2^5 \implies n - 2 = 5 \implies n = 7

Since

32 = 2^5

and the bases are identical, the exponents must be equal.

Key Concept

Factoring sum of exponential terms with common bases

Question 46Question →

On the real number line, $x$ is a negative real number such that $|5 - 2x| = 11$ , and $y$ is a positive real number such that $|3y + 1| = 13$ . What is the distance on the real number line between $x$ and $y$ ?

Show answer & explanation

Answer: 7

Answer

The distance between

x

and

y

on the real number line is 7.

Solving

|5 - 2x| = 11

with

x < 0

yields

x = -3

. Solving

|3y + 1| = 13

with

y > 0

yields

y = 4

. The distance between

-3

and

4

on the number line is

|4 - (-3)| = 7

.

Step-by-Step Solution

1

Solve for the negative real number

x

using the equation

|5 - 2x| = 11

.

x = -3

The equation splits into

5 - 2x = 11

(yielding

x = -3

) and

5 - 2x = -11

(yielding

x = 8

). Since

x

must be negative,

x = -3

is selected.

2

Solve for the positive real number

y

using the equation

|3y + 1| = 13

.

y = 4

The equation splits into

3y + 1 = 13

(yielding

y = 4

) and

3y + 1 = -13

(yielding

y = -\frac{14}{3}

). Since

y

must be positive,

y = 4

is selected.

3

Compute the distance between

x

and

y

on the number line.

7

The distance between two points on the number line is given by

|y - x| = |4 - (-3)| = 7

.

Key Concept

Distance on the real number line between two points

a

and

b

is given by

|a - b|

, solved by evaluating absolute value equations under given sign constraints.

Alternative Method

Plot the candidate solutions for

x

(

x = -3

and

x = 8

) and

y

(

y = 4

and

y = -14/3

) on a number line, then directly count units between the valid points

x = -3

and

y = 4

.

Estimated Time:1m 15s

Question 47Question →

In a school club of $40$ students, the ratio of the number of boys to the number of girls is $3 : 5$ . Which of the following statements must be true? Select all such statements.

Select all that apply

The number of boys in the club is

15

.

The ratio of the number of boys to the total number of students is

3 : 8

.

The number of girls in the club exceeds the number of boys by

10

.

The ratio of the number of boys to the total number of students is

3 : 5

.

The number of girls in the club is

20

.

Show answer & explanation

Answer: The number of boys in the club is $15$ .; The ratio of the number of boys to the total number of students is $3 : 8$ .; The number of girls in the club exceeds the number of boys by $10$ .

Answer

The correct statements are those indicating that there are 15 boys in the club, that the ratio of boys to total students is 3:8, and that there are 10 more girls than boys in the club.

The total number of ratio parts is

3 + 5 = 8

, representing the entire group of

40

students. Dividing

40

by

8

gives

5

students per part. Therefore, the number of boys is

3 \times 5 = 15

, the number of girls is

5 \times 5 = 25

, and the ratio of boys to total students is

15:40 = 3:8

. Subtracting the number of boys from girls (

25 - 15 = 10

) shows there are

10

more girls than boys.

Step-by-Step Solution

1

Calculate total ratio parts and determine the value of one part.

Total ratio parts =

3 + 5 = 8

. Each part corresponds to

\frac{40}{8} = 5

students.

Converting ratio terms into parts allows calculation of actual quantities from the total group size.

2

Compute the total number of boys and girls.

Number of boys =

3 \times 5 = 15

. Number of girls =

5 \times 5 = 25

.

Multiply each component's ratio share by the value of one part.

3

Evaluate the given statements against calculated quantities.

There are

15

boys; the ratio of boys to total students is

15:40 = 3:8

; and the difference between girls and boys is

25 - 15 = 10

.

Determines which options represent true statements about the group.

Key Concept

Part-to-Part vs. Part-to-Whole Ratios

Question 48Question →

Let $x$ and $y$ be real numbers such that $|x^2 - 4| = 2y$ and $|y - 3| \le 1$ . Which of the following could be the value of $x$ ? Select all such values.

Select all that apply

-3

0

2

3

4

Show answer & explanation

Answer: $-3$ ; $0$ ; $3$

Answer

The values that

x

could take are

-3

,

0

, and

3

.

The inequality

|y - 3| \le 1

restricts

y

to the closed interval

[2, 4]

. Consequently,

2y

lies in

[4, 8]

, which means

4 \le |x^2 - 4| \le 8

. Breaking this compound inequality into cases yields

x = 0

(from

x^2 = 0

) or

x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]

. Since

\sqrt{8} \approx 2.83

and

\sqrt{12} \approx 3.46

, the integer values

-3

and

3

fall in these intervals, alongside

0

. Thus,

-3

,

0

, and

3

are all valid choices.

Step-by-Step Solution

1

Determine the acceptable range for

y

from the given absolute value inequality.

Solving

|y - 3| \le 1

gives

-1 \le y - 3 \le 1

, which simplifies to

2 \le y \le 4

.

The inequality

|y - c| \le r

represents all values of

y

within distance

r

from

c

.

2

Relate the range of

y

to the absolute value expression

|x^2 - 4|

.

Since

2 \le y \le 4

, multiplying by

2

yields

4 \le 2y \le 8

. Therefore,

4 \le |x^2 - 4| \le 8

.

Substitute

2y = |x^2 - 4|

into the inequality derived for

y

.

3

Analyze the two cases for the absolute value equation

|x^2 - 4|

.

Case 1:

4 \le x^2 - 4 \le 8 \implies 8 \le x^2 \le 12

, which gives

x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]

. Case 2:

4 \le -(x^2 - 4) \le 8 \implies 4 \le 4 - x^2 \le 8 \implies -4 \le -x^2 \le 4 \implies -4 \le x^2 \le 0

. Since

x^2 \ge 0

for all real numbers,

x^2 = 0 \implies x = 0

.

Absolute value

|A|

splits into

A

when

A \ge 0

and

-A

when

A < 0

.

4

Evaluate the test values against the valid domains for

x

.

For

x = -3

,

x^2 = 9

, which satisfies

8 \le 9 \le 12

. For

x = 0

,

x^2 = 0

, which satisfies

x^2 = 0

. For

x = 3

,

x^2 = 9

, which satisfies

8 \le 9 \le 12

. The values

x = 2

and

x = 4

give

x^2 = 4

and

x^2 = 16

, neither of which fall into the valid intervals.

Checking test options confirms which values fall within the solution intervals.

Key Concept

Absolute Value Equations and System Constraints on Real Numbers

Question 49Question →

Let $x$ and $y$ be positive integers such that $\gcd(x, y) = 14$ and $\text{lcm}(x, y) = 420$ . Which of the following values could be the sum $x + y$ ? Select all such values.

Select all that apply

154

182

210

238

435

Show answer & explanation

Answer: 154; 182; 238

Answer

The possible values for the sum

x + y

are 154, 182, and 238.

By writing

x = 14a

and

y = 14b

with

\gcd(a, b) = 1

, the relation

\text{lcm}(x, y) = 14ab = 420

requires

ab = 30

. The positive coprime factor pairs of 30 are

(1, 30)

,

(2, 15)

,

(3, 10)

, and

(5, 6)

. Multiplying these pairs by 14 gives the possible sums 434, 238, 182, and 154. Therefore, the options equal to 154, 182, and 238 are all correct.

Step-by-Step Solution

1

Express

x

and

y

in terms of their greatest common divisor.

Let

x = 14a

and

y = 14b

, where

a

and

b

are positive integers such that

\gcd(a, b) = 1

.

Factoring out the greatest common divisor leaves coprime quotient factors

a

and

b

.

2

Relate the least common multiple to

a

and

b

.

\text{lcm}(x, y) = 14ab = 420 \implies ab = \frac{420}{14} = 30$.

The least common multiple of two numbers sharing a GCD of

g

is given by

g \cdot a \cdot b

.

3

Find all coprime pairs

(a, b)

with

a \le b

whose product is 30.

The prime factorization of 30 is

2 \times 3 \times 5

. The valid coprime pairs

(a, b)

are

(1, 30)

,

(2, 15)

,

(3, 10)

, and

(5, 6)

.

Since

\gcd(a, b) = 1

, all factors of 30 split into pairs of coprime integers.

4

Calculate the corresponding values of

x

,

y

, and their sum

x + y

for each pair.

Pair (1, 30):

x = 14, y = 420 \implies x + y = 434

.
Pair (2, 15):

x = 28, y = 210 \implies x + y = 238

.
Pair (3, 10):

x = 42, y = 140 \implies x + y = 182

.
Pair (5, 6):

x = 70, y = 84 \implies x + y = 154

.

Multiplying each pair by the GCD of 14 yields the original integers

x

and

y

.

Key Concept

Relationship between GCD, LCM, and prime factorization of quotient factors

Question 50Question →

A logistics company received a shipment of identical freight containers. On Monday, the crew unloaded $\frac{2}{7}$ of the total shipment. On Tuesday, they unloaded $\frac{3}{5}$ of the remaining containers. On Wednesday, they unloaded $\frac{1}{2}$ of the containers that remained after Tuesday's work. If 30 containers remained unloaded at the end of Wednesday, what was the total number of containers in the original shipment?

80

105

140

210

525

Show answer & explanation

Answer: 210

Answer

The total number of containers in the original shipment was 210.

The correct answer is 210. Working forward, after Monday

\frac{5}{7}

of the initial shipment

N

remains. On Tuesday,

\frac{2}{5}

of that remainder stays unloaded, which equals

\frac{2}{5} \times \frac{5}{7}N = \frac{2}{7}N

. On Wednesday,

\frac{1}{2}

of that remainder stays unloaded, yielding

\frac{1}{2} \times \frac{2}{7}N = \frac{1}{7}N

. Setting

\frac{1}{7}N = 30

gives

N = 210

.

Step-by-Step Solution

1

Determine the fraction of containers remaining after Monday.

Since

\frac{2}{7}

of the total shipment

N

was unloaded on Monday, the fraction remaining is

1 - \frac{2}{7} = \frac{5}{7}

of

N

.

Subtracting the fraction unloaded on Monday from 1 gives the remaining fraction.

2

Calculate the fraction of containers remaining after Tuesday.

On Tuesday,

\frac{3}{5}

of the remaining

\frac{5}{7}N

was unloaded, which is

\frac{3}{5} \times \frac{5}{7}N = \frac{3}{7}N

. The remaining fraction after Tuesday is

\frac{5}{7}N - \frac{3}{7}N = \frac{2}{7}N

.

Alternatively, if

\frac{3}{5}

of the remainder was unloaded, then

1 - \frac{3}{5} = \frac{2}{5}

of the remainder was left:

\frac{2}{5} \times \frac{5}{7}N = \frac{2}{7}N

.

3

Calculate the fraction of containers remaining after Wednesday.

On Wednesday,

\frac{1}{2}

of the remaining

\frac{2}{7}N

was unloaded, leaving

1 - \frac{1}{2} = \frac{1}{2}

of that remainder. Thus, the final fraction remaining is

\frac{1}{2} \times \frac{2}{7}N = \frac{1}{7}N

.

Multiplying the remaining fraction after Tuesday by the fraction left unhandled on Wednesday yields the overall fraction of the original shipment remaining.

4

Solve for the total initial number of containers

N

.

Set

\frac{1}{7}N = 30

, which gives

N = 30 \times 7 = 210

.

Equating the calculated final remaining fraction to the given numerical count allows solving for the total original quantity.

Key Concept

Sequential Fraction of Remaining Quantities

Question 51Question →

If $k$ and $m$ are integers such that $k < 0 < m$ , $(-1)^k = -1$ , and $(-1)^m = 1$ , which of the following expressions MUST be a negative odd integer?

k - m

k + m

k m

m - k

k^2 - m

Show answer & explanation

Answer: $k - m$

Answer

The expression

k - m

MUST be a negative odd integer.

Given

k < 0 < m

,

k

is negative and

m

is positive. The relation

(-1)^k = -1

shows

k

is odd, while

(-1)^m = 1

shows

m

is even. Subtracting a positive even integer

m

from a negative odd integer

k

yields

k - m

, which must be less than 0 (negative) and odd (odd minus even).

Step-by-Step Solution

1

Determine the parity and sign of

k

k

is a negative odd integer.

k < 0

specifies that

k

is negative, and

(-1)^k = -1

implies that the exponent

k

must be odd.

2

Determine the parity and sign of

m

m

is a positive even integer.

m > 0

specifies that

m

is positive, and

(-1)^m = 1

implies that the exponent

m

must be even.

3

Evaluate the sign and parity of

k - m

k - m

is strictly negative and odd.

Since

k < 0

and

m > 0

,

k - m = k + (-m) < 0

. By parity rules,

\text{odd} - \text{even} = \text{odd}

.

Key Concept

Even-Odd Properties and Sign Rules

Question 52Question →

On the real number line, the distance between two real numbers $x$ and $y$ is $d$ . If the midpoint of $x$ and $y$ is $7$ and $|x - 3| = 2|y - 3|$ , what is the maximum possible value of $d$ ?

12

16

24

28

32

Show answer & explanation

Answer: 24

Answer

The maximum possible value of

d

is 24.

The midpoint condition dictates that

x + y = 14

, so

y = 14 - x

and the distance between them is

d = |x - y| = 2|x - 7|

. Substituting

y = 14 - x

into

|x - 3| = 2|y - 3|

gives

|x - 3| = 2|11 - x|

. Solving the positive case

x - 3 = 2(x - 11)

gives

x = 19

and

y = -5

, resulting in distance

d = |19 - (-5)| = 24

. Solving the negative case

x - 3 = -2(x - 11)

gives

x = 25/3

and

y = 17/3

, resulting in distance

d = 8/3

. Thus, 24 is the maximum possible value of

d

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the midpoint formula

Since the midpoint of

x

and

y

is

7

,

\frac{x + y}{2} = 7

, which gives

y = 14 - x

. The distance

d = |x - y| = |x - (14 - x)| = |2x - 14| = 2|x - 7|

.

Relating

y

to

x

reduces the problem to a single variable.

2

Substitute

y = 14 - x

into the absolute value equation

|x - 3| = 2|(14 - x) - 3| \implies |x - 3| = 2|11 - x| = 2|x - 11|

.

Setting up the single-variable absolute value equation allows finding all possible values for

x

.

3

Solve the absolute value equation for all possible cases

Case 1:

x - 3 = 2(x - 11) \implies x - 3 = 2x - 22 \implies x = 19

. Then

y = 14 - 19 = -5

.
Case 2:

x - 3 = -2(x - 11) \implies x - 3 = -2x + 22 \implies 3x = 25 \implies x = \frac{25}{3}

. Then

y = 14 - \frac{25}{3} = \frac{17}{3}

.

Absolute value equations

|A| = B

split into

A = B

and

A = -B

.

4

Calculate the distance

d

for each case and select the maximum

For Case 1 (

x = 19, y = -5

):

d = |19 - (-5)| = 24

.
For Case 2 (

x = 25/3, y = 17/3

):

d = |25/3 - 17/3| = 8/3

.
The maximum possible value of

d

is

24

.

Comparing the distance values determined in each case identifies the maximum distance.

Key Concept

Absolute Value as Distance and Multi-Case Equations on the Real Number Line

Estimated Time:2m 30s

Question 53Question →

For how many integer values of $n$ in the interval $-15 \le n \le 15$ is the value of the expression $(-1)^{n^2 + n} - (-1)^{3n}$ equal to $2$ ?

Show answer & explanation

Answer: 16

Answer

The correct answer is 16.

Because

n^2 + n = n(n + 1)

is the product of two consecutive integers, it is guaranteed to be even for every integer

n

. Consequently,

(-1)^{n^2 + n} = 1

. Substituting this into the given equation yields

1 - (-1)^{3n} = 2

, which reduces to

(-1)^{3n} = -1

. A power of

-1

equals

-1

if and only if the exponent is odd, so

3n

must be odd, which requires

n

itself to be odd. In the interval

[-15, 15]

, there are 16 odd integers: 8 negative odd integers and 8 positive odd integers.

Step-by-Step Solution

1

Determine the parity of

n^2 + n

The expression

n^2 + n = n(n + 1)

represents the product of two consecutive integers. Because one of any two consecutive integers is even, their product is always even. Therefore,

(-1)^{n^2 + n} = 1

for all integers

n

.

Simplifying the exponent with a known parity rule reduces the expression to a constant.

2

Isolate

(-1)^{3n}

in the equation

Substituting

1

into the original equation gives

1 - (-1)^{3n} = 2

, which simplifies to

(-1)^{3n} = -1

.

Isolating the exponential term reveals the sign condition required for the equality to hold.

3

Find the parity condition for

n

For

(-1)^{3n}

to equal

-1

, the exponent

3n

must be an odd integer. Since

3

is odd, the product

3n

is odd if and only if

n

is odd.

Applying the product parity rule (

\text{odd} \times \text{odd} = \text{odd}

) relates the condition on

3n

back to

n

.

4

Count the odd integers in the interval

[-15, 15]

The odd integers in the interval are

-15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15

. There are 16 such integers.

Counting all qualifying values within the specified range yields the final numeric answer.

Key Concept

Parity rules for consecutive integers and exponents of negative numbers

Question 54Question →

At the beginning of Year 1, a commercial logistics company allocated its total storage space between two facilities, Facility X and Facility Y, such that Facility X contained $60\%$ of the total space and Facility Y contained the remaining $40\%$ . During Year 1, the storage space of Facility X was increased by $x\%$ , while the storage space of Facility Y was increased by $25\%$ . During Year 2, the storage space of Facility X was decreased by $20\%$ from its Year 1 level, while the storage space of Facility Y was increased by $x\%$ from its Year 1 level. If the total combined storage space of both facilities at the end of Year 2 was $17.6\%$ greater than the total combined storage space at the beginning of Year 1, what is the value of $x$ ?

15

18.5

19.6

20

25

Show answer & explanation

Answer: 20

Answer

The value of

x

is 20.

Let the initial total space be

A

. Facility X initially has

0.60A

space and Facility Y has

0.40A

space. After Year 1, Facility X space becomes

0.60A(1 + x/100)

and Facility Y space becomes

0.40A(1.25) = 0.50A

. At the end of Year 2, Facility X space decreases by

20\%

to

0.48A(1 + x/100)

, while Facility Y space increases by

x\%

to

0.50A(1 + x/100)

. Factoring out

(1 + x/100)

, the total final combined space is

(0.48A + 0.50A)(1 + x/100) = 0.98A(1 + x/100)

. Setting this equal to

1.176A

(a

17.6\%

increase over

A

) yields

1 + x/100 = 1.176 / 0.98 = 1.20

, which gives

x = 20

.

Step-by-Step Solution

1

Define initial storage spaces for Facility X and Facility Y in terms of total initial area

A

.

Facility X initial space =

0.60A

, Facility Y initial space =

0.40A

.

Establishing explicit algebraic representations based on the given

60\% / 40\%

ratio.

2

Calculate the storage space of each facility after Year 1 changes.

Facility X after Year 1 =

0.60A \left(1 + \frac{x}{100}\right)

; Facility Y after Year 1 =

0.40A (1 + 0.25) = 0.50A

.

Facility X increased by

x\%

and Facility Y increased by

25\%

of its original base.

3

Calculate the storage space of each facility at the end of Year 2.

Facility X final =

0.60A \left(1 + \frac{x}{100}\right) (1 - 0.20) = 0.48A \left(1 + \frac{x}{100}\right)

; Facility Y final =

0.50A \left(1 + \frac{x}{100}\right)

.

Facility X decreased by

20\%

from its Year 1 amount, and Facility Y increased by

x\%

from its Year 1 amount.

4

Sum the final spaces of both facilities and set equal to the total space given (

17.6\%

overall increase over

A

).

Total final space =

0.48A \left(1 + \frac{x}{100}\right) + 0.50A \left(1 + \frac{x}{100}\right) = 0.98A \left(1 + \frac{x}{100}\right) = 1.176A

.

The total space at the end of Year 2 is

100\% + 17.6\% = 117.6\%

of initial total space

A

.

5

Solve the linear equation for

x

.

1 + \frac{x}{100} = \frac{1.176}{0.98} = 1.20 \implies \frac{x}{100} = 0.20 \implies x = 20

.

Dividing both sides by

0.98A

isolates the percentage factor.

Key Concept

Successive Percent Change and Multi-Base Percentage Problems

Estimated Time:2m 30s

Question 55Question →

A positive integer $n$ has the prime factorization $n = 2^a \cdot 3^b \cdot 5^c$ , where $a$ , $b$ , and $c$ are positive integers. If $n$ is divisible by $36$ and is a divisor of $5{}400$ , which of the following values could be the total number of positive divisors of $n$ ? Select all such values.

Select all that apply

18

20

27

32

40

Show answer & explanation

Answer: 18; 27; 32

Answer

18, 27, and 32 are all possible total numbers of positive divisors for

n

.

Prime factorization gives

36 = 2^2 \cdot 3^2

and

5{}400 = 2^3 \cdot 3^3 \cdot 5^2

. For

n = 2^a \cdot 3^b \cdot 5^c

to be a multiple of 36 and a divisor of 5,400 with positive integer exponents,

a \in \{2,3\}

,

b \in \{2,3\}

, and

c \in \{1,2\}

. The total number of divisors is

(a+1)(b+1)(c+1)

. The possible values for this product are 18, 24, 27, 32, 36, and 48. Among the choices, 18, 27, and 32 are valid values.

Step-by-Step Solution

1

Find the prime factorizations of the boundary numbers 36 and 5,400.

36 = 2^2 \cdot 3^2

and

5{}400 = 2^3 \cdot 3^3 \cdot 5^2

.

Establishing the prime factor bounds determines the range of possible values for exponents

a

,

b

, and

c

.

2

Determine the constraints on exponents

a

,

b

, and

c

.

Since

36 \mid n

,

a \ge 2

,

b \ge 2

, and

c \ge 1

(given

c

is a positive integer). Since

n \mid 5{}400

,

a \le 3

,

b \le 3

, and

c \le 2

. Thus,

a \in \{2, 3\}

,

b \in \{2, 3\}

, and

c \in \{1, 2\}

.

Divisibility rules require prime factor exponents of a multiple to be greater than or equal to those of the divisor, and exponents of a divisor to be less than or equal to those of the multiple.

3

Calculate all possible total divisor counts using the formula

d(n) = (a+1)(b+1)(c+1)

.

Possible factor values are

(a+1) \in \{3, 4\}

,

(b+1) \in \{3, 4\}

, and

(c+1) \in \{2, 3\}

. Evaluating all combinations yields:

3 \cdot 3 \cdot 2 = 18

,

3 \cdot 3 \cdot 3 = 27

,

3 \cdot 4 \cdot 2 = 24

,

3 \cdot 4 \cdot 3 = 36

,

4 \cdot 4 \cdot 2 = 32

, and

4 \cdot 4 \cdot 3 = 48

.

The total number of positive integer divisors is found by adding 1 to each exponent in the prime factorization and multiplying the results.

4

Compare the calculated divisor counts with the options provided.

The values 18, 27, and 32 appear in the calculated set of possible total divisors.

Direct matching identifies all valid options.

Key Concept

Prime factor exponent bounds and total number of positive divisors formula

Question 56Question →

If $x$ is a real number such that $|x^2 - 9| = 5|x - 3|$ , which of the following could be the value of $x$ ? Select all such values.

Select all that apply

-8

-2

2

3

8

Show answer & explanation

Answer: $-8$ ; $2$ ; $3$

Answer

The values of

x

that satisfy the equation are

-8

,

2

, and

3

.

Factoring the left side of

|x^2 - 9| = 5|x - 3|

yields

|x - 3||x + 3| = 5|x - 3|

. Setting the common factor

|x - 3| = 0

gives

x = 3

. Dividing both sides by the non-zero quantity

|x - 3|

leaves

|x + 3| = 5

, which splits into

x + 3 = 5 \implies x = 2

and

x + 3 = -5 \implies x = -8

. Therefore,

-8

,

2

, and

3

are all valid solutions.

Step-by-Step Solution

1

Apply the product rule for absolute values to factor the left-hand side.

|x^2 - 9| = |(x - 3)(x + 3)| = |x - 3| \cdot |x + 3|

, rewriting the equation as

|x - 3| \cdot |x + 3| = 5|x - 3|

.

The absolute value of a product is equal to the product of the individual absolute values.

2

Evaluate the case where the shared factor is zero:

|x - 3| = 0

.

x - 3 = 0 \implies x = 3

. Both sides equal

0

, making

x = 3

a valid solution.

Dividing by

|x - 3|

without checking if it can be zero would cause the loss of the root

x = 3

.

3

Evaluate the case where

|x - 3| \neq 0

by dividing both sides of the equation by

|x - 3|

.

|x + 3| = 5

.

Since

|x - 3| > 0

, we can safely divide both sides by this non-zero quantity.

4

Solve the remaining absolute value equation

|x + 3| = 5

.

x + 3 = 5 \implies x = 2

, and

x + 3 = -5 \implies x = -8

.

An expression inside an absolute value equal to a positive number

k

can equal either

k

or

-k

.

Key Concept

Factoring absolute value expressions using

|ab| = |a||b|

and systematically considering all cases to avoid dropping zero-roots or negative solutions.

Question 57Question →

Two positive integers $m$ and $n$ satisfy $\gcd(m, n) = 20$ and $\text{lcm}(m, n) = 4{}200$ . It is given that $m$ is divisible by $7$ and has exactly $24$ positive integer divisors. If $n$ is a multiple of $3$ , what is the value of $n$ ?

60

200

300

420

600

Show answer & explanation

Answer: $60$

Answer

The value of

n

is

60

.

Prime factorization reveals that

\gcd(m,n) = 2^2 \cdot 3^0 \cdot 5^1 \cdot 7^0

and

\text{lcm}(m,n) = 2^3 \cdot 3^1 \cdot 5^2 \cdot 7^1

. The condition that

n

is a multiple of

3

fixes the exponent of

3

in

n

to

1

(so

m

has exponent

0

). The condition that

m

is a multiple of

7

fixes the exponent of

7

in

m

to

1

(so

n

has exponent

0

). For

m = 2^a \cdot 3^0 \cdot 5^c \cdot 7^1

, its divisor count equation

2(a+1)(c+1) = 24

simplifies to

(a+1)(c+1) = 12

. Since

a \in \{2, 3\}

and

c \in \{1, 2\}

, the only valid solution is

a=3

and

c=2

. This leaves

n

with exponents

2

for prime

2

,

1

for prime

3

,

1

for prime

5

, and

0

for prime

7

, giving

n = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 60

.

Step-by-Step Solution

1

Express the GCD and LCM in prime factorized form.

\gcd(m, n) = 20 = 2^2 \cdot 3^0 \cdot 5^1 \cdot 7^0

and

\text{lcm}(m, n) = 4{}200 = 2^3 \cdot 3^1 \cdot 5^2 \cdot 7^1

.

The prime factor exponents of

m

and

n

must have minimums equal to the GCD exponents and maximums equal to the LCM exponents.

2

Determine the exponents of prime factors

3

and

7

for

m

and

n

.

Since

n

is divisible by

3

, exponent of

3

in

n

is

1

, so exponent of

3

in

m

is

0

. Since

m

is divisible by

7

, exponent of

7

in

m

is

1

, so exponent of

7

in

n

is

0

.

Each prime exponent in

\text{lcm}(m, n)

must belong to at least one of the numbers.

3

Use the divisor count of

m

to find its remaining exponents for primes

2

and

5

.

Let

m = 2^a \cdot 3^0 \cdot 5^c \cdot 7^1

, where

a \in \{2, 3\}

and

c \in \{1, 2\}

. The number of positive divisors is

(a+1)(0+1)(c+1)(1+1) = 2(a+1)(c+1) = 24

, giving

(a+1)(c+1) = 12

. Testing

a=2 \implies c=3

(invalid range). Testing

a=3 \implies c=2

(valid). Thus

m = 2^3 \cdot 3^0 \cdot 5^2 \cdot 7^1 = 1{}400

.

The total number of positive integer divisors of

p_1^{e_1} p_2^{e_2} \cdots

is given by

(e_1+1)(e_2+1)\cdots

.

4

Determine the prime exponents for

n

and calculate its value.

Since

a=3

,

n

gets

2^2

. Since

c=2

,

n

gets

5^1

. Together with

3^1

and

7^0

,

n = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 60

.

For each prime

p

, the exponent in

n

must match the bound opposite to

m

to satisfy both

\gcd

and

\text{lcm}

.

Key Concept

Prime exponent analysis of GCD and LCM alongside the divisor count formula

Question 58Question →

Consider three non-zero integers $x$ , $y$ , and $z$ that satisfy all of the following conditions:

I. $(-1)^{x^2 y + z} = -1$
II. $x y z^2 < 0$
III. $x + y$ is an even integer

Which of the following expressions MUST be an odd integer?

(x + z)(y + z)

x y z

x + y + z

x(y + z)

x z + y

Show answer & explanation

Answer: $(x + z)(y + z)$

Answer

The expression

(x + z)(y + z)

MUST be an odd integer.

Condition I dictates that

x^2 y + z

is odd. Condition III states that

x + y

is even, meaning

x

and

y

share the same parity. If

x

and

y

are both even,

x^2 y

is even, forcing

z

to be odd. If

x

and

y

are both odd,

x^2 y

is odd, forcing

z

to be even. Consequently,

z

always has the opposite parity of both

x

and

y

. Therefore,

(x + z)

is always odd and

(y + z)

is always odd. The product of two odd integers,

(x + z)(y + z)

, is guaranteed to be odd.

Step-by-Step Solution

1

Analyze Condition I for exponent parity

x^2 y + z

must be an odd integer

For

(-1)^k = -1

, the exponent

k

must be odd.

2

Analyze Condition III for shared parity of

x

and

y

x

and

y

are either both even or both odd

The sum of two integers is even if and only if they share the same parity.

3

Deduce parity relationship for

z

across cases

In Case 1 (

x, y

even),

x^2 y

is even, so

z

must be odd. In Case 2 (

x, y

odd),

x^2 y

is odd, so

z

must be even.

To satisfy

x^2 y + z = \text{odd}

,

x^2 y

and

z

must have opposite parities.

4

Evaluate the parity of

(x + z)(y + z)

(x + z)

is odd and

(y + z)

is odd, so their product is odd

In both cases,

z

has opposite parity to both

x

and

y

. Adding two integers of opposite parity always yields an odd integer, and the product of two odd integers is always odd.

Key Concept

Parity rules under exponentiation and algebraic combination

Estimated Time:2m 0s

Question 59Question →

If $x$ is a negative real number, which of the following expressions are equivalent to $\sqrt{-x^3}$ ? Select all such expressions.

Select all that apply

-x\sqrt{-x}

x\sqrt{-x}

|x|\sqrt{-x}

(-x)^{3/2}

-x^{3/2}

Show answer & explanation

Answer: $-x\sqrt{-x}$ ; $|x|\sqrt{-x}$ ; $(-x)^{3/2}$

Answer

The expressions equivalent to

\sqrt{-x^3}

are

-x\sqrt{-x}

,

|x|\sqrt{-x}

, and

(-x)^{3/2}

.

Because

x

is negative,

-x

is a positive quantity. We can express

-x^3

as

(-x)^3 = (-x)^2 \cdot (-x)

. Taking the principal square root yields

\sqrt{(-x)^2 \cdot (-x)} = \sqrt{(-x)^2} \cdot \sqrt{-x} = -x\sqrt{-x}

. Because

|x| = -x

for negative numbers, the expression

|x|\sqrt{-x}

is identical to

-x\sqrt{-x}

. Furthermore, converting to rational exponents gives

(-x)^{3/2} = \sqrt{(-x)^3} = \sqrt{-x^3}

. Thus, all three of these expressions are mathematically equivalent to the original radical expression.

Step-by-Step Solution

1

Analyze the sign of the base and inside of the radical

Since

x < 0

, the quantity

-x

is strictly positive (

-x > 0

). Consequently,

-x^3 = (-x)^3 > 0

, ensuring

\sqrt{-x^3}

is a real, non-negative number.

Principal square roots require a non-negative radicand and yield a non-negative result in real arithmetic.

2

Simplify the radical expression using perfect squares

\sqrt{-x^3} = \sqrt{(-x)^2 \cdot (-x)} = \sqrt{(-x)^2} \cdot \sqrt{-x} = (-x)\sqrt{-x} = -x\sqrt{-x}

.

Because

-x > 0

,

\sqrt{(-x)^2} = -x

.

3

Evaluate equivalent representations using absolute value and rational exponents

Since

x < 0

,

|x| = -x

, so

|x|\sqrt{-x} = -x\sqrt{-x}

. Also,

(-x)^{3/2} = \sqrt{(-x)^3} = \sqrt{-x^3}

.

Both rewrite rules preserve both magnitude and non-negative sign for all

x < 0

.

Key Concept

Simplifying radicals and fractional exponents with negative variable bases

Question 60Question →

A pharmaceutical facility uses two liquid compounding lines, Line $A$ and Line $B$ , to produce a standardized saline solution. Line $A$ operates at a constant flow rate of $120$ liters per hour, producing a solution with active compound and purified water in a ratio of $1 : 3$ by volume. Line $B$ operates at a constant flow rate of $180$ liters per hour, producing a solution with active compound and purified water in a ratio of $2 : 3$ by volume. If Line $A$ runs for $4$ hours and Line $B$ runs for $5$ hours, and all output is collected into a single storage tank, what is the ratio of active compound to purified water in the storage tank?

8 : 15

8 : 23

13 : 27

1 : 2

38 : 31

Show answer & explanation

Answer: $8 : 15$

Answer

The ratio of active compound to purified water in the storage tank is

8 : 15

.

The correct answer is derived by first converting the flow rates and operating hours into total volumes (

480

L for Line A and

900

L for Line B). Using the part-to-whole fraction conversion, Line A contributes

120

L of active compound and

360

L of water, while Line B contributes

360

L of active compound and

540

L of water. Adding these quantities yields

480

L of active compound and

900

L of water, which simplifies to the ratio

8 : 15

.

Step-by-Step Solution

1

Calculate total output volume and component quantities from Line A

Total Line A Volume =

120 \text{ L/hr} \times 4 \text{ hrs} = 480 \text{ liters}

. Active Compound =

480 \times \frac{1}{1+3} = 120 \text{ liters}

. Purified Water =

480 - 120 = 360 \text{ liters}

.

Line A produces a total volume of 480 liters, and a

1 : 3

ratio means active compound constitutes

\frac{1}{4}

of the total volume.

2

Calculate total output volume and component quantities from Line B

Total Line B Volume =

180 \text{ L/hr} \times 5 \text{ hrs} = 900 \text{ liters}

. Active Compound =

900 \times \frac{2}{2+3} = 360 \text{ liters}

. Purified Water =

900 - 360 = 540 \text{ liters}

.

Line B produces a total volume of 900 liters, and a

2 : 3

ratio means active compound constitutes

\frac{2}{5}

of the total volume.

3

Sum the components from both lines in the storage tank

Total Active Compound =

120 + 360 = 480 \text{ liters}

. Total Purified Water =

360 + 540 = 900 \text{ liters}

.

Combining the output of both lines adds the respective volumes of active compound and purified water.

4

Compute and simplify the final ratio of active compound to purified water

Ratio =

\frac{480}{900} = \frac{8}{15}

, expressed as

8 : 15

.

Dividing both numerator and denominator by their greatest common divisor, 60, yields the simplest integer ratio.

Key Concept

Weighted mixture ratios combining multi-stage rates and part-to-whole fraction conversions

Arithmetic

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution