Algebra Practice Questions — GRE General Test — Page 3

If $|2x - 4| \le 6$ , which of the following values could be a solution for $x$ ? Select all that apply.

Select all that apply

-3

-1

2

5

6

Show answer & explanation

Answer: $-1$ ; $2$ ; $5$

Answer

The values

-1

,

2

, and

5

are solutions to the inequality.

Solving

|2x - 4| \le 6

yields

-6 \le 2x - 4 \le 6

. Adding

4

gives

-2 \le 2x \le 10

, and dividing by

2

gives

-1 \le x \le 5

. The options stating

-1

,

2

, and

5

are all within this interval

[-1, 5]

.

Step-by-Step Solution

1

Rewrite the absolute value inequality as a compound inequality.

-6 \le 2x - 4 \le 6

An inequality of the form

|u| \le c

(where

c \ge 0

) is equivalent to

-c \le u \le c

.

2

Add

4

to all three parts of the compound inequality.

-2 \le 2x \le 10

Isolate the variable term

2x

in the middle.

3

Divide all three parts by

2

.

-1 \le x \le 5

Isolate

x

. Since

2

is positive, the inequality signs remain unchanged.

4

Test which of the given choices fall within the interval

[-1, 5]

.

-1

,

2

, and

5

lie within the range

[-1, 5]

, while

-3

and

6

lie outside.

Values equal to or between

-1

and

5

inclusive are valid solutions.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:1m 0s

Question 42Question →

If $x > 1$ , which of the following expressions are equivalent to $\frac{\sqrt{x^5 \cdot x^3}}{(x^{-2})^{-1}}$ ? Select all such expressions.

Select all that apply

(x^{\frac{2}{3}})^3

\sqrt{x^4 + x^4}

\frac{x^{-1}}{x^{-3}}

(x^{-1})^{-2} \cdot x^{-4}

\sqrt[4]{x^4}

Show answer & explanation

Answer: $(x^{\frac{2}{3}})^3$ ; $\frac{x^{-1}}{x^{-3}}$

Answer

The expressions equivalent to the target expression are

(x^{\frac{2}{3}})^3

and

\frac{x^{-1}}{x^{-3}}

.

The original expression simplifies step by step to

x^2

. The expression

(x^{2/3})^3

equals

x^{(2/3)\cdot 3} = x^2

, and the expression

\frac{x^{-1}}{x^{-3}}

equals

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

. Both are equal to

x^2

.

Step-by-Step Solution

1

Simplify the numerator of the given expression

\sqrt{x^5 \cdot x^3} = \sqrt{x^{5+3}} = \sqrt{x^8} = x^{8/2} = x^4

Combine bases using product rule for exponents, then convert radical to fractional exponent.

2

Simplify the denominator of the given expression

(x^{-2})^{-1} = x^{(-2) \cdot (-1)} = x^2

Multiply exponents when raising a power to a power.

3

Divide the simplified numerator by the simplified denominator

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

Subtract denominator exponent from numerator exponent using the quotient rule.

4

Evaluate each choice against

x^2

Only

(x^{2/3})^3 = x^2

and

\frac{x^{-1}}{x^{-3}} = x^2

are equal to

x^2

.

Apply basic exponent rules to each proposed expression.

Key Concept

Algebraic Exponents and Radicals Rules

Question 43Question →

If $x$ satisfies the inequality $|3 - 2x| > 9$ , which of the following could be the value of $x$ ? Select all that apply.

Select all that apply

-5

-3

0

4

8

Show answer & explanation

Answer: $-5$ ; $8$

Answer

The values

-5

and

8

satisfy the inequality.

The absolute value inequality

|3 - 2x| > 9

splits into

3 - 2x > 9

or

3 - 2x < -9

. Solving these yields

x < -3

or

x > 6

. The value

-5

is less than

-3

and the value

8

is greater than

6

, so both are valid solutions.

Step-by-Step Solution

1

Set up the two separate linear inequalities based on the definition of absolute value.

3 - 2x > 9

or

3 - 2x < -9

An absolute value expression

|u| > c

(where

c > 0

) breaks into two disjoint cases:

u > c

or

u < -c

.

2

Solve the first case:

3 - 2x > 9

.

-2x > 6 \implies x < -3

Subtract 3 from both sides to get

-2x > 6

, then divide by

-2

and reverse the inequality sign.

3

Solve the second case:

3 - 2x < -9

.

-2x < -12 \implies x > 6

Subtract 3 from both sides to get

-2x < -12

, then divide by

-2

and reverse the inequality sign.

4

Combine the solution sets and evaluate the given options.

The valid solution set is

x < -3

or

x > 6

. Among the choices,

-5

(since

-5 < -3

) and

8

(since

8 > 6

) fall into the solution set.

Values between

-3

and

6

, inclusive, do not satisfy the original inequality.

Key Concept

Solving absolute value inequalities of the form

|u| > c

by splitting into compound inequalities and reversing the direction of inequality signs when dividing by negative numbers.

Estimated Time:1m 15s

Question 44Question →

In the linear equation below,

k

is a constant:

\frac{x - k}{2} + \frac{2x + 1}{5} = \frac{3x - 4}{10}

If

x

must be a positive integer, which of the following values could be

k

? Select all such values.

Select all that apply

-6

6

10

12

15

Show answer & explanation

Answer: $6$ ; $12$

Answer

The values of

k

that satisfy the condition are

6

and

12

.

Clearing denominators gives

6x = 5k - 6

, or

x = \frac{5k - 6}{6}

. For

x

to be a positive integer,

5k - 6

must be positive and divisible by

6

. This requires

k

to be a positive multiple of

6

. The options

6

and

12

satisfy both criteria, yielding

x = 4

and

x = 9

respectively.

Step-by-Step Solution

1

Clear denominators by multiplying the entire equation by the least common denominator (LCD), which is 10.

5(x - k) + 2(2x + 1) = 3x - 4

Eliminating fractions simplifies further algebraic manipulation.

2

Expand terms and combine like terms on the left side of the equation.

5x - 5k + 4x + 2 = 3x - 4 \implies 9x - 5k + 2 = 3x - 4

Distribute constants across parentheses to isolate variable expressions.

3

Isolate

x

in terms of

k

.

6x = 5k - 6 \implies x = \frac{5k - 6}{6}

Express

x

as an explicit linear function of

k

to test divisibility and positivity conditions.

4

Determine constraints on

k

for

x

to be a positive integer.

5k

must be divisible by

6

(so

k

must be a multiple of

6

) and

5k - 6 > 0 \implies k > 1.2

.

Since

5

and

6

share no common factors,

k

must be a positive multiple of

6

.

5

Evaluate the choices against the constraints.

Among the options, only

6

and

12

are positive multiples of

6

, producing

x = 4

and

x = 9

respectively.

Both values produce integer solutions greater than zero.

Key Concept

Solving single-variable linear equations with rational coefficients and applying divisibility and inequality constraints.

Question 45Question →

If $x$ and $y$ are real numbers satisfying the inequalities $|2x - 7| \le 5$ and $|3y + 2| < 8$ , which of the following inequalities must be true? Select all that apply.

Select all that apply

x - y > -1

xy < 12

x + y > 0

|y| < 3

y - x > -3

Show answer & explanation

Answer: $x - y > -1$ ; $xy < 12$

Answer

The inequalities that must be true are

x - y > -1

and

xy < 12

.

Solving

|2x - 7| \le 5

yields

1 \le x \le 6

, and solving

|3y + 2| < 8

yields

-\frac{10}{3} < y < 2

. Combining

x \ge 1

with

-y > -2

gives

x - y > -1

, which is always true. Furthermore, since

x

is positive and bounded above by

6

while

y

is bounded above by

2

, the product

xy

must be strictly less than

12

.

Step-by-Step Solution

1

Solve the absolute value inequality for

x

.

-5 \le 2x - 7 \le 5 \implies 2 \le 2x \le 12 \implies 1 \le x \le 6

.

Unfolding the absolute value

|2x - 7| \le 5

gives a compound linear inequality.

2

Solve the absolute value inequality for

y

.

-8 < 3y + 2 < 8 \implies -10 < 3y < 6 \implies -\frac{10}{3} < y < 2

.

Unfolding the absolute value

|3y + 2| < 8

gives a strict compound linear inequality.

3

Evaluate the inequality

x - y > -1

.

Since

x \ge 1

and

y < 2 \implies -y > -2

, adding the inequalities gives

x + (-y) > 1 + (-2) = -1

.

This establishes that

x - y > -1

is always true.

4

Evaluate the inequality

xy < 12

.

Since

1 \le x \le 6

(all positive) and

y < 2

, if

y > 0

,

xy < 6 \cdot 2 = 12

. If

y \le 0

,

xy \le 0 < 12

.

In all cases within the domain,

xy < 12

holds strictly.

5

Test counterexamples for the remaining statements.

For

x + y > 0

,

x=1, y=-3 \implies x+y=-2 \ngtr 0

. For

|y| < 3

,

y=-3.2 \implies |-3.2|=3.2 \nless 3

. For

y - x > -3

,

x=6, y=0 \implies y-x=-6 \ngtr -3

.

Counterexamples disprove that these remaining statements must be true.

Key Concept

Linear Inequalities and Absolute Value Bounds

Question 46Question →

Which of the following values of $x$ are solutions to the equation $(x - 2)^{\frac{4}{3}} - 5(x - 2)^{\frac{2}{3}} + 4 = 0$ ? Select all that apply.

Select all that apply

-6

1

10

-2

6

Show answer & explanation

Answer: -6; 1; 10

Answer

The correct values of

x

that satisfy the equation are

-6

,

1

, and

10

.

Substituting

u = (x - 2)^{\frac{2}{3}}

yields

u^2 - 5u + 4 = 0

, which factors as

(u - 1)(u - 4) = 0

, giving

u = 1

and

u = 4

. Solving

(x - 2)^{\frac{2}{3}} = 1

gives

(x - 2)^2 = 1 \implies x - 2 = \pm 1

, yielding

x = 3

and

x = 1

. Solving

(x - 2)^{\frac{2}{3}} = 4

gives

(x - 2)^2 = 64 \implies x - 2 = \pm 8

, yielding

x = 10

and

x = -6

. Thus, the values

-6

,

1

, and

10

are all valid solutions.

Step-by-Step Solution

1

Perform a substitution to rewrite the equation in quadratic form.

Let

u = (x - 2)^{\frac{2}{3}}

. Then

u^2 = (x - 2)^{\frac{4}{3}}

, giving

u^2 - 5u + 4 = 0

.

Recognizing quadratic structure simplifies equations with rational exponents.

2

Solve the quadratic equation for

u

.

(u - 1)(u - 4) = 0 \implies u = 1

or

u = 4

.

Factoring determines the values of the substituted variable

u

.

3

Solve for

x

when

u = 1

.

(x - 2)^{\frac{2}{3}} = 1 \implies (x - 2)^2 = 1^3 = 1 \implies x - 2 = \pm 1 \implies x = 3

or

x = 1

.

Raising both sides to the power of

\frac{3}{2}

requires taking both positive and negative roots because the numerator of the power is even.

4

Solve for

x

when

u = 4

.

(x - 2)^{\frac{2}{3}} = 4 \implies (x - 2)^2 = 4^3 = 64 \implies x - 2 = \pm 8 \implies x = 10

or

x = -6

.

Squaring and taking square roots yields two solutions,

10

and

-6

.

5

Match calculated solutions with the given choices.

The solutions present among the options are

-6

,

1

, and

10

.

Comparing all valid algebraic solutions to the available choices identifies all correct options.

Key Concept

Solving quadratic-form equations with fractional exponents and accounting for negative base branches when taking even roots.

Question 47Question →

If $a > 0$ , which of the following is equivalent to the expression $\sqrt{\frac{a^8 \cdot a^4}{a^{-2}}}$ ?

a^7

a^5

a^6

a^{14}

a^{10}

Show answer & explanation

Answer: $a^7$

Answer

a^7

Multiplying the terms in the numerator gives

a^{12}

. Dividing by

a^{-2}

gives

a^{12 - (-2)} = a^{14}

. Taking the square root of

a^{14}

gives

(a^{14})^{1/2} = a^7

.

Step-by-Step Solution

1

Simplify the numerator inside the square root using the product rule

a^m \cdot a^n = a^{m+n}

.

a^8 \cdot a^4 = a^{8+4} = a^{12}

Powers with the same base are multiplied by adding their exponents.

2

Divide by the denominator using the quotient rule

\frac{a^m}{a^n} = a^{m-n}

.

\frac{a^{12}}{a^{-2}} = a^{12 - (-2)} = a^{14}

Dividing powers with the same base requires subtracting the lower exponent from the upper exponent.

3

Apply the fractional exponent rule for radicals

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

.

\sqrt{a^{14}} = (a^{14})^{\frac{1}{2}} = a^{14 \cdot \frac{1}{2}} = a^7

Taking the square root of a power is equivalent to multiplying the exponent by

\frac{1}{2}

.

Key Concept

Simplifying expressions using exponent laws and fractional radical powers

Question 48Question →

If $x$ is a real number that satisfies the inequality $|2x - 7| < 5$ , which of the following represents all possible values of the expression $1 - 3x$ ?

-17 < 1 - 3x < -2

-2 < 1 - 3x < 17

-17 < 1 - 3x < 1

-17 < 1 - 3x < 17

-5 < 1 - 3x < 5

Show answer & explanation

Answer: $-17 < 1 - 3x < -2$

Answer

-17 < 1 - 3x < -2

Solving

|2x - 7| < 5

gives

-5 < 2x - 7 < 5

. Adding

7

to all three parts yields

2 < 2x < 12

, which simplifies to

1 < x < 6

. Multiplying this inequality by

-3

reverses the direction of the inequalities, resulting in

-18 < -3x < -3

. Adding

1

to each part gives

-17 < 1 - 3x < -2

. Therefore, the range of possible values for the expression is strictly between

-17

and

-2

.

Step-by-Step Solution

1

Express the absolute value inequality as a compound inequality.

-5 < 2x - 7 < 5

An inequality of the form

|u| < k

for

k > 0

is equivalent to

-k < u < k

.

2

Isolate

x

in the compound inequality.

1 < x < 6

Add

7

to all parts to get

2 < 2x < 12

, then divide all parts by

2

to obtain

1 < x < 6

.

3

Multiply the compound inequality by

-3

.

-18 < -3x < -3

Multiplying an inequality by a negative number reverses the direction of the inequality signs:

-3(6) < -3(x) < -3(1)

.

4

Add

1

to all parts of the compound inequality.

-17 < 1 - 3x < -2

Adding a constant to an inequality preserves the inequality direction.

Key Concept

Linear Inequalities and Absolute Value Transformations

Estimated Time:1m 30s

Question 49Question →

If $x$ is a real number that satisfies the inequality $||2x - 5| - 7| \le 4$ , and $y = |3 - x|$ , what is the difference between the maximum possible value and the minimum possible value of $y$ ?

3

4

5

6

7

Show answer & explanation

Answer: 5

Answer

The difference between the maximum possible value and the minimum possible value of

y

is 5.

Solving

||2x - 5| - 7| \le 4

gives

3 \le |2x - 5| \le 11

, which restricts

x

to the disconnected domain

[-3, 1] \cup [4, 8]

. Evaluating

y = |3 - x|

across these intervals gives a maximum value of 6 (at

x = -3

) and a minimum value of 1 (at

x = 4

). The difference between the maximum and minimum values is

6 - 1 = 5

.

Step-by-Step Solution

1

Unpack the outer absolute value inequality

||2x - 5| - 7| \le 4

.

-4 \le |2x - 5| - 7 \le 4 \implies 3 \le |2x - 5| \le 11

.

An absolute value inequality

|u| \le a

(for

a \ge 0

) is equivalent to

-a \le u \le a

.

2

Solve the double inequality

3 \le |2x - 5| \le 11

by splitting it into two conditions.

Condition 1:

|2x - 5| \le 11 \implies -11 \le 2x - 5 \le 11 \implies -3 \le x \le 8

.
Condition 2:

|2x - 5| \ge 3 \implies 2x - 5 \ge 3

or

2x - 5 \le -3 \implies x \ge 4

or

x \le 1

.

The quantity

|2x - 5|

must simultaneously satisfy upper and lower absolute value bounds.

3

Intersect Condition 1 and Condition 2 to determine the complete domain of

x

.

x \in [-3, 1] \cup [4, 8]

.

Values in the open interval

(1, 4)

make

|2x - 5| < 3

and must be excluded from the domain.

4

Evaluate the range of

y = |3 - x|

over the valid domain of

x

.

On

[-3, 1]

,

y = 3 - x

decreases from

3 - (-3) = 6

to

3 - 1 = 2

, giving

y \in [2, 6]

.
On

[4, 8]

,

y = x - 3

increases from

4 - 3 = 1

to

8 - 3 = 5

, giving

y \in [1, 5]

.
The complete range of

y

is

[1, 6]

.

Combining the output ranges of both disjoint intervals yields all possible values for

y

.

5

Calculate the difference between the maximum and minimum values of

y

.

Maximum

y = 6

, Minimum

y = 1

, Difference =

6 - 1 = 5

.

Subtracting the minimum value 1 from the maximum value 6 gives the required difference.

Key Concept

Solving compound nested absolute value inequalities and finding the extreme values of a transformed function over disconnected solution intervals.

Estimated Time:2m 0s

Question 50Question →

If $2^{x + 3} \cdot 4^{x - 1} = 16^x$ , what is the value of $x$ ?

Show answer & explanation

Answer: 1

Answer

The value of

x

is 1.

Rewriting

4^{x-1}

as

2^{2x-2}

and

16^x

as

2^{4x}

transforms the left side into

2^{x+3} \cdot 2^{2x-2} = 2^{3x+1}

. Setting exponents equal gives

3x + 1 = 4x

, which simplifies to

x = 1

.

Step-by-Step Solution

1

Express all terms with a common base of 2

2^{x+3} \cdot (2^2)^{x-1} = (2^4)^x

Converting

4

to

2^2

and

16

to

2^4

allows all terms to share the base 2.

2

Apply the power of a power rule

(a^m)^n = a^{m \cdot n}

and the product rule

a^m \cdot a^n = a^{m+n}

2^{3x+1} = 2^{4x}

Multiplying exponents gives

(2^2)^{x-1} = 2^{2x-2}

and

(2^4)^x = 2^{4x}

. Adding exponents on the left gives

(x+3) + (2x-2) = 3x+1

.

3

Equate the exponents and solve for

x

x = 1

Since

2^A = 2^B

implies

A = B

, setting

3x + 1 = 4x

directly yields

x = 1

.

Key Concept

Solving exponential equations using common bases and exponent properties

Question 51Question →

If $x$ is an integer that satisfies both $|5 - 2x| \le 9$ and $|x + 1| > 3$ , what is the least possible value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The least possible value of

x

is 3.

Solving

|5 - 2x| \le 9

gives

-2 \le x \le 7

. Solving

|x + 1| > 3

gives

x > 2

or

x < -4

. Taking the intersection of both regions yields

2 < x \le 7

. The integer values satisfying this combined inequality are

3, 4, 5, 6,

and

7

. The smallest among these integer values is

3

.

Step-by-Step Solution

1

Solve the first absolute value inequality

|5 - 2x| \le 9

-2 \le x \le 7

Remove the absolute value bars to set up the compound inequality

-9 \le 5 - 2x \le 9

. Subtracting

5

from all parts gives

-14 \le -2x \le 4

. Dividing all parts by

-2

and reversing the inequality signs yields

7 \ge x \ge -2

, or

-2 \le x \le 7

.

2

Solve the second absolute value inequality

|x + 1| > 3

x > 2

or

x < -4

Remove the absolute value bars to create two separate cases:

x + 1 > 3 \implies x > 2

, or

x + 1 < -3 \implies x < -4

.

3

Determine the set of values that satisfy both inequalities simultaneously

2 < x \le 7

The intersection of the interval

[-2, 7]

and

(-\infty, -4) \cup (2, \infty)

is

(2, 7]

, because

x < -4

does not overlap with

[-2, 7]

.

4

Find the smallest integer within the interval

(2, 7]

3

The integers included in the interval

(2, 7]

are

3, 4, 5, 6,

and

7

. Note that

2

is excluded due to the strict inequality

x > 2

. Therefore, the least possible integer value is

3

.

Key Concept

Linear Inequalities and Absolute Value

Question 52Question →

Consider the linear equation in one variable

x

:

\frac{2x - 3}{3} - \frac{x - 5}{4} = \frac{x + 9}{6}

If

x

is the solution to this equation, what is the value of

4x - 1

?

5

7

11

19

59

Show answer & explanation

Answer: 19

Answer

19

To solve the linear equation, first multiply all terms by the least common denominator, 12, yielding 4(2x - 3) - 3(x - 5) = 2(x + 9). Expanding both sides yields 8x - 12 - 3x + 15 = 2x + 18. Combining like terms on the left side yields 5x + 3 = 2x + 18. Subtracting 2x and 3 from both sides gives 3x = 15, so x = 5. Substituting x = 5 into the target expression 4x - 1 gives 4(5) - 1 = 19.

Step-by-Step Solution

1

Clear the fractional denominators by multiplying every term by the least common multiple (LCM) of 3, 4, and 6, which is 12.

12 \cdot \left(\frac{2x - 3}{3}\right) - 12 \cdot \left(\frac{x - 5}{4}\right) = 12 \cdot \left(\frac{x + 9}{6}\right) \implies 4(2x - 3) - 3(x - 5) = 2(x + 9)

Multiplying every term by 12 eliminates all denominators without changing the equation's solution set.

2

Distribute the factors across the binomial terms inside parentheses.

8x - 12 - 3x + 15 = 2x + 18

Distributing -3 across (x - 5) yields -3x + 15 because the product of two negative numbers is positive.

3

Combine like terms on the left side of the equation.

5x + 3 = 2x + 18

Combining 8x - 3x gives 5x, and -12 + 15 gives +3.

4

Isolate x by subtracting 2x and 3 from both sides of the equation.

3x = 15 \implies x = 5

Subtracting 2x from both sides yields 3x + 3 = 18, then subtracting 3 yields 3x = 15, giving x = 5.

5

Substitute x = 5 into the expression 4x - 1 to find the final value.

4(5) - 1 = 20 - 1 = 19

The question asks for the evaluation of 4x - 1 rather than the value of x alone.

Key Concept

Solving linear equations in one variable involving fractional expressions and evaluating algebraic expressions.

Estimated Time:2m 0s

Question 53Question →

If $x$ is a real number that satisfies both $|3x - 12| \le 9$ and $|2 - x| \ge 4$ , the maximum possible value of the expression $5 - 2x$ is $M$ and the minimum possible value is $m$ . What is the value of $M - m$ ?

Show answer & explanation

Answer: 2

Answer

The value of

M - m

is 2.

Solving

|3x - 12| \le 9

yields

1 \le x \le 7

. Solving

|2 - x| \ge 4

yields

x \le -2

or

x \ge 6

. The set of

x

-values satisfying both conditions is the intersection

[6, 7]

. Since

5 - 2x

is a linear expression with a negative coefficient, its maximum value

M

occurs at the smallest value of

x

(

x = 6

), giving

M = 5 - 2(6) = -7

. Its minimum value

m

occurs at the largest value of

x

(

x = 7

), giving

m = 5 - 2(7) = -9

. Thus,

M - m = -7 - (-9) = 2

.

Step-by-Step Solution

1

Solve the inequality

|3x - 12| \le 9

1 \le x \le 7

Expanding absolute value yields

-9 \le 3x - 12 \le 9

. Adding 12 gives

3 \le 3x \le 21

, then dividing by 3 yields

1 \le x \le 7

.

2

Solve the inequality

|2 - x| \ge 4

x \le -2

or

x \ge 6

Absolute value inequality

|u| \ge a

splits into

u \ge a

or

u \le -a

. Here

2 - x \ge 4 \implies x \le -2

, and

2 - x \le -4 \implies x \ge 6

.

3

Determine the overlapping domain for

x

6 \le x \le 7

Combining

1 \le x \le 7

with

x \le -2

or

x \ge 6

leaves only the interval

6 \le x \le 7

.

4

Evaluate maximum

M

and minimum

m

of

5 - 2x

on

6 \le x \le 7

M = -7

and

m = -9

Since

-2x

decreases as

x

increases, the maximum occurs at

x = 6

(

M = 5 - 12 = -7

) and the minimum occurs at

x = 7

(

m = 5 - 14 = -9

).

5

Calculate the difference

M - m

2

Subtracting

m

from

M

gives

-7 - (-9) = 2

.

Key Concept

Linear Inequalities and Absolute Value

Estimated Time:2m 0s

Question 54Question →

Two water pumps, Pump A and Pump B, were used to drain a reservoir containing 12,000 gallons of water. Pump A operates at a constant rate that is 50 gallons per hour greater than the rate of Pump B. Pump A worked alone for 4 hours, after which both pumps worked together for another 8 hours to completely empty the reservoir. What is the pumping rate of Pump B, in gallons per hour?

Show answer & explanation

Answer: 570

Answer

570

Let

r

represent the rate of Pump B in gallons per hour. Pump A's rate is

(r + 50)

gallons per hour. In the first 4 hours, Pump A drains

4(r + 50) = 4r + 200

gallons. In the next 8 hours, both pumps operate together at a combined rate of

(r + 50) + r = 2r + 50

gallons per hour, draining

8(2r + 50) = 16r + 400

gallons. Adding both quantities gives total volume drained:

(4r + 200) + (16r + 400) = 12,000

. Simplifying gives

20r + 600 = 12,000

, so

20r = 11,400

, which yields

r = 570

gallons per hour.

Step-by-Step Solution

1

Define variables for the rate of each pump

Rate of Pump B =

r

gal/hr; Rate of Pump A =

r + 50

gal/hr

Establishing a single unknown variable allows setting up a one-variable linear equation.

2

Write expressions for water drained during each time period

Period 1 (Pump A alone for 4 hrs):

4(r + 50) = 4r + 200

; Period 2 (Both pumps for 8 hrs):

8(2r + 50) = 16r + 400

Work done equals rate multiplied by time for each phase of operation.

3

Sum the work done in both periods to equal total volume and solve for

r

20r + 600 = 12,000 \Rightarrow 20r = 11,400 \Rightarrow r = 570

Solving the linear equation yields the exact rate of Pump B.

Key Concept

Linear Equations in One Variable

Estimated Time:2m 0s

Question 55Question →

For all $x > 1$ , which of the following expressions is equivalent to $\frac{\sqrt{x^3 \cdot \sqrt[3]{x^2}}}{(x^{-1/3})^2}$ ?

x^{5/2}

x^{7/6}

x^{5/3}

x^{13/3}

x^{35/12}

Show answer & explanation

Answer: $x^{5/2}$

Answer

x^{5/2}

Converting all radical expressions into rational exponent form simplifies the numerator to

(x^{11/3})^{1/2} = x^{11/6}

and the denominator to

x^{-2/3}

. Applying the quotient rule

x^{11/6} / x^{-2/3} = x^{11/6 - (-2/3)}

yields

x^{15/6} = x^{5/2}

.

Step-by-Step Solution

1

Convert radicals in the numerator to fractional exponents and combine terms inside the square root

x^3 \cdot \sqrt[3]{x^2} = x^3 \cdot x^{2/3} = x^{3 + 2/3} = x^{11/3}

When multiplying exponential expressions with the same base, add the exponents.

2

Apply the outer square root to the simplified expression in the numerator

\sqrt{x^{11/3}} = (x^{11/3})^{1/2} = x^{11/6}

Taking the square root is equivalent to raising an expression to the power of

1/2

.

3

Simplify the denominator using the power of a power rule

(x^{-1/3})^2 = x^{(-1/3) \cdot 2} = x^{-2/3}

When raising a power to another power, multiply the exponents.

4

Divide the simplified numerator by the simplified denominator

\frac{x^{11/6}}{x^{-2/3}} = x^{11/6 - (-2/3)} = x^{11/6 + 4/6} = x^{15/6} = x^{5/2}

When dividing exponential expressions with the same base, subtract the denominator exponent from the numerator exponent.

Key Concept

Simplifying nested algebraic radicals using rational exponent laws

Estimated Time:1m 30s

Question 56Question →

Two commercial printing presses, Press A and Press B, operate at constant rates to print a total order of $N$ pages. Press A prints at a constant rate of $x$ pages per minute, while Press B prints at a constant rate that is $25$ pages per minute faster than Press A. Press A begins printing alone. After $20$ minutes, Press B is turned on, and both presses work simultaneously for an additional $30$ minutes. At that point, Press A stops, and Press B works alone for $10$ final minutes to complete the order. If Press B printed exactly $\frac{6}{11}$ of the total number of pages in the order, what is the value of $x$ ?

Show answer & explanation

Answer: 50

Answer

The value of

x

is 50.

To find

x

, calculate the total time each press operated. Press A ran for 20 minutes alone plus 30 minutes with Press B, giving 50 minutes total at

x

pages per minute (

50x

pages). Press B ran for 30 minutes with Press A plus 10 minutes alone, giving 40 minutes total at

(x + 25)

pages per minute (

40x + 1000

pages). The total order size is

N = 90x + 1000

. Setting Press B's output equal to

\frac{6}{11}N

yields

40x + 1000 = \frac{6}{11}(90x + 1000)

. Multiplying both sides by 11 gives

440x + 11000 = 540x + 6000

, which simplifies to

100x = 5000

, giving

x = 50

.

Step-by-Step Solution

1

Determine total operating times and express pages printed by each press in terms of

x

Press A printed

50x

pages; Press B printed

40(x + 25) = 40x + 1000

pages.

Press A operated for 20 minutes alone plus 30 minutes together (50 minutes total). Press B operated for 30 minutes together plus 10 minutes alone (40 minutes total).

2

Write the expression for total pages

N

N = 50x + (40x + 1000) = 90x + 1000

The total pages in the order is the sum of the pages printed by Press A and Press B.

3

Formulate the linear equation in one variable using the given ratio

40x + 1000 = \frac{6}{11}(90x + 1000)

Press B printed exactly

\frac{6}{11}

of the total pages

N

.

4

Solve the linear equation for

x

x = 50

Multiplying both sides by 11 clears the fraction to give

440x + 11000 = 540x + 6000

, which simplifies to

100x = 5000

.

Key Concept

Formulating and solving linear equations in one variable from multi-step rate and work scenarios.

Question 57Question →

If $x$ is an integer that satisfies the inequality $|6 - 3x| - 4 \le 5$ , which of the following could be the value of $x$ ? Select all such values.

Select all that apply

-3

-1

2

4

7

Show answer & explanation

Answer: $-1$ ; $2$ ; $4$

Answer

The correct values of

x

are

-1

,

2

, and

4

.

Isolating the absolute value yields

|6 - 3x| \le 9

, which expands to

-9 \le 6 - 3x \le 9

. Subtracting

6

gives

-15 \le -3x \le 3

. Dividing by

-3

and reversing the inequality signs results in

-1 \le x \le 5

. Among the given choices,

-1

,

2

, and

4

lie within this solution interval

[-1, 5]

.

Step-by-Step Solution

1

Isolate the absolute value expression

|6 - 3x| \le 9

Add

4

to both sides of the inequality to isolate the absolute value term.

2

Rewrite as a compound inequality

-9 \le 6 - 3x \le 9

The property

|u| \le a

(where

a \ge 0

) expands to

-a \le u \le a

.

3

Subtract 6 from all parts

-15 \le -3x \le 3

Isolate the variable term

-3x

by subtracting

6

across the compound inequality.

4

Divide by -3 and reverse inequality signs

5 \ge x \ge -1

, which is equivalent to

-1 \le x \le 5

Dividing an inequality by a negative quantity requires flipping the inequality direction.

Key Concept

Solving absolute value inequalities of the form

|ax + b| \le c

by expanding into a compound inequality

-c \le ax + b \le c

and correctly reversing inequality signs when multiplying or dividing by negative numbers.

Question 58Question →

For all positive real numbers $x$ and $y$ , which of the following expressions are equivalent to $\left(\frac{x^{-2} y^{3/2}}{\sqrt{x^4 y^{-1}}}\right)^{-1/2}$ ? Select all that apply.

Select all that apply

x^2 y^{-1}

\sqrt{\frac{x^4}{y^2}}

\frac{y}{x^2}

\frac{x^3 y}{\sqrt{x^2 y^4}}

\frac{x^2}{\sqrt{y}}

Show answer & explanation

Answer: $x^2 y^{-1}$ ; $\sqrt{\frac{x^4}{y^2}}$ ; $\frac{x^3 y}{\sqrt{x^2 y^4}}$

Answer

The expressions equivalent to the given expression are

x^2 y^{-1}

,

\sqrt{\frac{x^4}{y^2}}

, and

\frac{x^3 y}{\sqrt{x^2 y^4}}

.

Simplifying the original expression step-by-step yields

\frac{x^2}{y}

, which is equal to

x^2 y^{-1}

. Taking the square root of

\frac{x^4}{y^2}

gives

\frac{x^2}{y}

, and simplifying

\frac{x^3 y}{\sqrt{x^2 y^4}} = \frac{x^3 y}{x y^2}

also gives

\frac{x^2}{y}

. Thus, these three choices are equivalent to the target expression.

Step-by-Step Solution

1

Simplify the denominator inside the parentheses

\sqrt{x^4 y^{-1}} = (x^4 y^{-1})^{1/2} = x^2 y^{-1/2}

Apply the power of a product rule and principal square root properties for positive variables.

2

Simplify the fraction inside the parentheses

\frac{x^{-2} y^{3/2}}{x^2 y^{-1/2}} = x^{-2 - 2} y^{3/2 - (-1/2)} = x^{-4} y^2

Subtract exponents of like bases when dividing.

3

Apply the outer exponent

-1/2

(x^{-4} y^2)^{-1/2} = x^{(-4)(-1/2)} y^{(2)(-1/2)} = x^2 y^{-1} = \frac{x^2}{y}

Multiply exponents when raising a power to a power.

4

Evaluate each option against

\frac{x^2}{y}

x^2 y^{-1}

,

\sqrt{\frac{x^4}{y^2}}

, and

\frac{x^3 y}{\sqrt{x^2 y^4}}

all simplify to

\frac{x^2}{y}

.

Check algebraic equivalence for each provided choice.

Key Concept

Algebraic Exponents and Radicals

Question 59Question →

If

x

and

y

are positive real numbers satisfying the system of exponential radical equations

\sqrt{x\sqrt{y}} = 3

\sqrt{y\sqrt{x}} = 9

which of the following statements must be true? Select all that apply.

Select all that apply

xy = 81

x^{1/2} + y^{1/2} = 10

\sqrt{xy} = 27

x

has two valid real solutions,

x = 1

and

x = -1

y^x - x^y = 80

Show answer & explanation

Answer: $xy = 81$ ; $x^{1/2} + y^{1/2} = 10$ ; $y^x - x^y = 80$

Answer

The statements asserting that

xy = 81

,

x^{1/2} + y^{1/2} = 10

, and

y^x - x^y = 80

are true.

Solving the system of radical equations gives the unique positive solution pair

x = 1

and

y = 81

. Substituting these values into the statements shows that

xy = 1 \cdot 81 = 81

,

x^{1/2} + y^{1/2} = \sqrt{1} + \sqrt{81} = 10

, and

y^x - x^y = 81^1 - 1^{81} = 80

are all mathematically valid.

Step-by-Step Solution

1

Square both sides of each equation to remove the outer radicals

x\sqrt{y} = 9 \implies x^2 y = 81

and

y\sqrt{x} = 81 \implies y^2 x = 6561

Squaring

\sqrt{A} = B

yields

A = B^2

, and rewriting radicals as rational powers gives

x y^{1/2} = 9

and

y x^{1/2} = 81

.

2

Express

y

in terms of

x

from the first equation and substitute into the second

y = \frac{81}{x^2} \implies \left(\frac{81}{x^2}\right)^2 x = 6561 \implies \frac{6561}{x^3} = 6561

Substituting

y

eliminates the variable

y

, leaving a single equation in terms of

x

.

3

Solve for

x

and

y

x^3 = 1 \implies x = 1

, which gives

y = \frac{81}{1^2} = 81

Since

x > 0

is a positive real number,

x = 1

is the unique real solution, yielding

y = 81

.

4

Evaluate each given statement using

x = 1

and

y = 81

xy = 81

(True),

x^{1/2} + y^{1/2} = 1 + 9 = 10

(True),

\sqrt{xy} = 9 \neq 27

(False),

x = -1

is invalid (False),

81^1 - 1^{81} = 80

(True)

Direct calculation confirms which individual statements hold.

Key Concept

Simplifying nested radical equations using fractional exponent rules

Estimated Time:2m 0s

Question 60Question →

If $x$ is a real number such that $2^{x+1} + 2^{x-1} = 40$ , what is the value of the expression $(x+1)^{x-1}$ ?

65

64

125

243

15

Show answer & explanation

Answer: 125

Answer

125

Factoring out

2^x

from the given equation yields

2^x(2 + 0.5) = 40

, which gives

2.5 \cdot 2^x = 40

and

2^x = 16

. This determines that

x = 4

. Substituting

x = 4

into

(x+1)^{x-1}

gives

(4+1)^{3} = 5^3 = 125

.

Step-by-Step Solution

1

Rewrite the given exponential terms with a common power of 2.

2^{x+1} = 2^x \cdot 2^1

and

2^{x-1} = 2^x \cdot 2^{-1} = \frac{2^x}{2}

.

Applying product rule for exponents allows factoring out

2^x

.

2

Factor out

2^x

and solve for

x

.

2^x \left(2 + \frac{1}{2}\right) = 40 \implies 2^x \cdot \frac{5}{2} = 40 \implies 2^x = 16 \implies x = 4

.

Combining fractional coefficients isolates the exponential term

2^x

.

3

Substitute

x = 4

into the target expression

(x+1)^{x-1}

.

(4+1)^{4-1} = 5^3 = 125

.

Simplifying the base and exponent yields the final numerical value.

Key Concept

Solving exponential equations using power distribution rules and factoring.

Algebra

Answer

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Key Concept

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Answer

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