Algebra and Functions Practice Questions — GMAT — Page 3

Question 41Question →

An investor allocates a sum of money across three accounts— $A$ , $B$ , and $C$ —which earn simple annual interest rates of $5\%$ , $4\%$ , and $6\%$ , respectively. The total interest earned from all three accounts in one year is $\$1,060$ . The amount invested in Account $C$ equals the total amount invested in Accounts $A$ and $B$ combined. If the amount in Account $A$ were increased by $50\%$ and the amount in Account $B$ were decreased by $25\%$ while Account $C$ remained unchanged, the total annual interest earned would increase by $\$110$ . What is the total amount, in dollars, invested across all three accounts?

\$10,000

\$15,000

\$18,000

\$20,000

\$24,000

Show answer & explanation

Answer: $\$20,000$

Answer

The total amount invested across all three accounts is

\$20,000

.

The system of linear equations derived from the problem constraints uniquely solves to

a = \$6,000

,

b = \$4,000

, and

c = \$10,000

. Summing these three amounts gives a total investment of

\$20,000

.

Step-by-Step Solution

1

Define variables and translate the problem statements into algebraic equations.

Let

a

,

b

, and

c

be the amounts invested in Accounts

A

,

B

, and

C

.
Equation 1 (Total Interest):

0.05a + 0.04b + 0.06c = 1,060 \implies 5a + 4b + 6c = 106,000

.
Equation 2 (Relationship among principal amounts):

c = a + b \implies a + b - c = 0

.
Equation 3 (Change in interest): Increasing

a

by

50\%

adds

0.05(0.50a) = 0.025a

interest. Decreasing

b

by

25\%

reduces interest by

0.04(0.25b) = 0.01b

. The net change is

0.025a - 0.01b = 110 \implies 25a - 10b = 110,000 \implies 5a - 2b = 22,000

.

Establishing a complete 3-variable system of linear equations is necessary to determine the unknowns.

2

Substitute

c = a + b

into Equation 1 to reduce the system to two variables.

5a + 4b + 6(a + b) = 106,000 \implies 11a + 10b = 106,000

.

Using substitution eliminates variable

c

, leaving a linear system in

a

and

b

.

3

Solve the two-variable system using elimination.

We have:
(1)

11a + 10b = 106,000

(2)

5a - 2b = 22,000

Multiply (2) by 5:

25a - 10b = 110,000

.
Add this to (1):

(11a + 10b) + (25a - 10b) = 106,000 + 110,000 \implies 36a = 216,000 \implies a = 6,000

.
Substitute

a = 6,000

back into (2):

5(6,000) - 2b = 22,000 \implies 30,000 - 2b = 22,000 \implies 2b = 8,000 \implies b = 4,000

.

Elimination allows straightforward calculation of individual values for

a

and

b

.

4

Calculate

c

and find the total sum invested across all three accounts.

c = a + b = 6,000 + 4,000 = 10,000

.
Total investment

= a + b + c = 6,000 + 4,000 + 10,000 = 20,000

.

The question asks for the total amount invested in all three accounts combined.

Key Concept

Systems of Linear Equations in Three Variables

Estimated Time:2m 0s

Question 42Question →

A commercial bakery operates two automated production lines, Line A and Line B, to produce specialized pastry boxes. Line A requires $4$ minutes of mixing and $2$ minutes of baking per box. Line B requires $3$ minutes of mixing and $5$ minutes of baking per box. On a given shift, the bakery utilizes a total of $180$ minutes of mixing time and $160$ minutes of baking time, with both lines running continuously without downtime.

If $x$ represents the number of pastry boxes produced by Line A and $y$ represents the number of pastry boxes produced by Line B during the shift, which of the following statements regarding the production parameters must be true? Select all that apply.

Select all that apply

Line A produced

10

more pastry boxes than Line B during the shift.

The total number of pastry boxes produced by both lines combined is equal to

50

.

The linear equation modeling total mixing time across both lines is

4x + 2y = 180

.

Line B accounted for

60\%

of the total pastry boxes produced during the shift.

The system of linear equations modeling the resource constraints yields a unique solution of

x = 30

and

y = 20

.

Show answer & explanation

Answer: Line A produced $10$ more pastry boxes than Line B during the shift.; The total number of pastry boxes produced by both lines combined is equal to $50$ .; The system of linear equations modeling the resource constraints yields a unique solution of $x = 30$ and $y = 20$ .

Answer

The statements that must be true are: Line A produced 10 more pastry boxes than Line B during the shift; the total number of pastry boxes produced by both lines combined is equal to 50; and the system of linear equations modeling the resource constraints yields a unique solution of

x = 30

and

y = 20

.

Solving the system of equations representing mixing time (

4x + 3y = 180

) and baking time (

2x + 5y = 160

) gives

x = 30

boxes for Line A and

y = 20

boxes for Line B. Consequently, Line A produced

30 - 20 = 10

more boxes than Line B, the total production is

30 + 20 = 50

boxes, and the system indeed yields the unique solution

(30, 20)

.

Step-by-Step Solution

1

Set up the linear equations based on mixing time and baking time constraints.

Mixing constraint:

4x + 3y = 180

; Baking constraint:

2x + 5y = 160

.

Each unit of

x

requires

4

min mixing and

2

min baking; each unit of

y

requires

3

min mixing and

5

min baking.

2

Solve the system of equations using elimination.

Multiply the baking equation by

2

:

4x + 10y = 320

. Subtracting the mixing equation (

4x + 3y = 180

) yields

7y = 140 \implies y = 20

. Substituting

y = 20

into

4x + 3(20) = 180

gives

4x = 120 \implies x = 30

.

Eliminating

x

yields the exact value for

y

, which then gives

x

.

3

Evaluate each candidate statement against the solution

(x, y) = (30, 20)

.

Difference:

x - y = 30 - 20 = 10

(True). Total:

x + y = 30 + 20 = 50

(True). Mixing equation:

4x + 3y = 180

, not

4x + 2y = 180

(False). Line B share:

\frac{20}{50} = 40\%

, not

60\%

(False). Unique solution

(30, 20)

(True).

Checking each statement determines which options are valid.

Key Concept

Algebraic modeling of multi-variable resource constraint word problems using linear systems of equations

Estimated Time:2m 0s

Question 43Question →

If $x$ is a real number satisfying the inequality $\frac{|2x - 5|}{x + 1} \le 1$ , which of the following represents the complete set of all possible values of $x$ ?

x < -1

or

\frac{4}{3} \le x \le 6

\frac{4}{3} \le x \le 6

-1 < x \le 6

x < -1

or

x \ge \frac{4}{3}

x \le -1

or

\frac{4}{3} \le x \le 6

Show answer & explanation

Answer: $x < -1$ or $\frac{4}{3} \le x \le 6$

Answer

x < -1

or

\frac{4}{3} \le x \le 6

The correct answer accounts for both possible sign states of the denominator

x + 1

. When

x > -1

, multiplying gives

|2x - 5| \le x + 1

, yielding

\frac{4}{3} \le x \le 6

. When

x < -1

, multiplying flips the inequality to

|2x - 5| \ge x + 1

, which is trivially satisfied by all

x < -1

because an absolute value is non-negative and

x + 1

is negative. Combining both cases yields

x < -1

or

\frac{4}{3} \le x \le 6

.

Step-by-Step Solution

1

Determine domain restrictions and break the inequality into cases based on the denominator's sign.

The expression is undefined at

x = -1

, so

x \neq -1

. We evaluate Case 1 (

x > -1

) and Case 2 (

x < -1

).

Multiplying an inequality by an algebraic expression requires knowing whether that expression is positive or negative to maintain or flip the inequality sign.

2

Solve Case 1 where

x + 1 > 0

(

x > -1

).

|2x - 5| \le x + 1 \implies -(x + 1) \le 2x - 5 \le x + 1

. Solving

-x - 1 \le 2x - 5

yields

x \ge \frac{4}{3}

. Solving

2x - 5 \le x + 1

yields

x \le 6

. Combining gives

\frac{4}{3} \le x \le 6

.

For positive denominators, multiplying both sides by

x + 1

preserves the inequality direction.

3

Solve Case 2 where

x + 1 < 0

(

x < -1

).

|2x - 5| \ge x + 1

. Since

|2x - 5| \ge 0

for all real

x

and

x + 1 < 0

, the left-hand side is non-negative while the right-hand side is strictly negative, which is always true for all

x < -1

.

Multiplying by a negative expression flips the inequality sign, and any non-negative quantity is strictly greater than any negative quantity.

4

Combine solutions from both cases.

The final solution set is

x < -1

or

\frac{4}{3} \le x \le 6

.

Taking the union of valid solution regions from both disjoint cases gives the complete solution.

Key Concept

Solving Rational Inequalities with Absolute Values by Case Analysis

Question 44Question →

If $a$ and $b$ are positive integers such that $2^{a+3} \cdot 5^b - 2^a \cdot 5^{b+1} = 6,000$ , what is the value of $a \cdot b$ ?

7

10

12

15

16

Show answer & explanation

Answer: 12

Answer

12

Factoring out

2^a \cdot 5^b

yields

2^a \cdot 5^b (2^3 - 5^1) = 3 \cdot 2^a \cdot 5^b = 6,000

. Dividing by 3 yields

2^a \cdot 5^b = 2,000

. Expressing

2,000

in prime factored form gives

2^4 \cdot 5^3

. Matching exponents for the prime bases gives

a = 4

and

b = 3

. The product

a \cdot b

is

4 \times 3 = 12

.

Step-by-Step Solution

1

Rewrite the terms in the expression using exponent rules to isolate common bases.

2^{a+3} \cdot 5^b - 2^a \cdot 5^{b+1} = (2^a \cdot 2^3) \cdot 5^b - 2^a \cdot (5^b \cdot 5^1)

Applying

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

allows us to extract common powers of

2^a

and

5^b

.

2

Factor out the common term

2^a \cdot 5^b

from the left side of the equation.

2^a \cdot 5^b (2^3 - 5^1) = 2^a \cdot 5^b (8 - 5) = 3 \cdot 2^a \cdot 5^b

Simplifying the constant factor in parentheses simplifies the equation.

3

Divide both sides of the equation by 3 and perform prime factorization on the resulting integer.

3 \cdot 2^a \cdot 5^b = 6,000 \implies 2^a \cdot 5^b = 2,000 = 2^4 \cdot 5^3

Prime factorization of

2,000

determines the unique integer exponents for bases 2 and 5.

4

Equate the corresponding exponents and calculate the requested product

a \cdot b

.

a = 4

and

b = 3 \implies a \cdot b = 4 \times 3 = 12

Since 2 and 5 are prime numbers, the prime factorization representation is unique.

Key Concept

Factoring Exponents and Prime Factorization

Estimated Time:2m 0s

Question 45Question →

If $x = \sqrt[3]{5 + 2\sqrt{13}} + \sqrt[3]{5 - 2\sqrt{13}}$ , what is the value of $x^3 + 9x$ ?

Show answer & explanation

Answer: 10

Answer

10

By defining

x = u + v

with

u = \sqrt[3]{5 + 2\sqrt{13}}

and

v = \sqrt[3]{5 - 2\sqrt{13}}

, cubing both sides gives

x^3 = u^3 + v^3 + 3uv(u + v)

. Evaluating the components yields

u^3 + v^3 = 10

and

uv = \sqrt[3]{25 - 52} = -3

. Substituting these back gives

x^3 = 10 - 9(x)

, which rearranges to

x^3 + 9x = 10

.

Step-by-Step Solution

1

Express

x

as a sum of two variables

u

and

v

x = u + v

, where

u = \sqrt[3]{5 + 2\sqrt{13}}

and

v = \sqrt[3]{5 - 2\sqrt{13}}

Grouping the binomial terms simplifies algebraic expansion using standard polynomial identities.

2

Calculate the sum of the cubes

u^3 + v^3

u^3 + v^3 = (5 + 2\sqrt{13}) + (5 - 2\sqrt{13}) = 10

Eliminating the cube roots allows for simple additive cancellation of the radical terms.

3

Calculate the product

uv

uv = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{5^2 - (2\sqrt{13})^2} = \sqrt[3]{25 - 52} = \sqrt[3]{-27} = -3

Applying the difference of squares property inside the cube root simplifies the product of conjugate radicals to a single integer.

4

Cube both sides of

x = u + v

and substitute evaluated terms

x^3 = u^3 + v^3 + 3uv(u + v) = 10 + 3(-3)x = 10 - 9x

Using

(u+v)^3 = u^3 + v^3 + 3uv(u+v)

connects

x^3

directly to

x

without expanding long radical terms.

5

Rearrange the equation to isolate

x^3 + 9x

x^3 + 9x = 10

Adding

9x

to both sides gives the exact numerical value of the requested expression.

Key Concept

Algebraic Identities with Polynomials and Radicals

Question 46Question →

A logistics company offers two freight pricing models for oversized cargo. Model X charges a fixed monthly account fee of $\$8$ plus $\$0.60$ per kilometer traveled. Model Y charges a fixed monthly account fee of $\$54$ plus $\$0.20$ per kilometer traveled. For how many kilometers in a month will the total monthly charge under Model X be exactly $20\%$ less than the total monthly charge under Model Y?

Show answer & explanation

Answer: 80

Answer

80 kilometers

The total monthly cost under Model X is

C_X = 8 + 0.60k

and under Model Y is

C_Y = 54 + 0.20k

. The condition that Model X is 20% less than Model Y means

C_X = 0.80 C_Y

. Substituting the expressions gives

8 + 0.60k = 0.80(54 + 0.20k) = 43.2 + 0.16k

. Subtracting

0.16k

and

8

from both sides yields

0.44k = 35.2

, which simplifies to

k = 80

.

Step-by-Step Solution

1

Define variables and establish linear cost equations for both models.

Model X cost:

C_X = 8 + 0.60k

; Model Y cost:

C_Y = 54 + 0.20k

, where

k

is kilometers traveled.

Linear modeling translates flat fees and variable rates into algebraic expressions.

2

Formulate the linear equation based on the condition that Model X is 20% less than Model Y.

C_X = 0.80 C_Y \implies 8 + 0.60k = 0.80(54 + 0.20k)

.

Being 20% less than a base value means taking 80% (or 0.80) of that value.

3

Expand and simplify the algebraic equation.

8 + 0.60k = 43.2 + 0.16k \implies 0.44k = 35.2

.

Distributing 0.80 across

(54 + 0.20k)

yields

43.2 + 0.16k

, and subtracting

0.16k

and

8

isolates

k

on one side.

4

Calculate the value of

k

.

k = \frac{35.2}{0.44} = 80

.

Dividing

35.2

by

0.44

gives the exact number of kilometers required.

Key Concept

Linear Equations in One and Two Variables

Question 47Question →

If $x = \sqrt{7 - 4\sqrt{3}}$ and $y = \sqrt{7 + 4\sqrt{3}}$ , what is the value of $\frac{x^{-2} + y^{-2}}{(x+y)^{-1}}$ ?

\frac{1}{4}

\frac{7}{2}

14

56

64

Show answer & explanation

Answer: $56$

Answer

56

The expression

x = \sqrt{7 - 4\sqrt{3}}

can be unnested by recognizing

7 - 4\sqrt{3}

as the perfect square

(2 - \sqrt{3})^2

, so

x = 2 - \sqrt{3}

and

y = 2 + \sqrt{3}

. This gives

x+y = 4

and

xy = 1

. The numerator

x^{-2} + y^{-2} = \frac{x^2+y^2}{(xy)^2} = \frac{(x+y)^2 - 2xy}{1} = 14

. The denominator

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

. Dividing

14

by

\frac{1}{4}

yields

56

.

Step-by-Step Solution

1

Simplify the radical expressions for

x

and

y

.

x = 2 - \sqrt{3}

and

y = 2 + \sqrt{3}

.

Note that

7 - 4\sqrt{3} = 4 - 4\sqrt{3} + 3 = (2 - \sqrt{3})^2

, so

\sqrt{7 - 4\sqrt{3}} = 2 - \sqrt{3}

. Similarly,

7 + 4\sqrt{3} = (2 + \sqrt{3})^2

.

2

Find the sum

x+y

and product

xy

.

x+y = 4

and

xy = 1

.

(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4

, and

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

.

3

Simplify the numerator

x^{-2} + y^{-2}

.

x^{-2} + y^{-2} = 14

.

x^{-2} + y^{-2} = \frac{1}{x^2} + \frac{1}{y^2} = \frac{x^2 + y^2}{(xy)^2} = \frac{(x+y)^2 - 2xy}{(xy)^2} = \frac{4^2 - 2(1)}{1^2} = 14

.

4

Evaluate the full expression

\frac{x^{-2} + y^{-2}}{(x+y)^{-1}}

.

\frac{14}{1/4} = 56

.

The denominator is

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

. Dividing

14

by

\frac{1}{4}

equals

14 \cdot 4 = 56

.

Key Concept

Nested Radical Simplification & Algebraic Exponent Identities

Alternative Method

Instead of unnesting the radicals first, observe that

x^2 = 7 - 4\sqrt{3}

and

y^2 = 7 + 4\sqrt{3}

. Then

x^2 y^2 = (7-4\sqrt{3})(7+4\sqrt{3}) = 49 - 48 = 1

, and

x^2 + y^2 = 14

. Thus

x^{-2} + y^{-2} = \frac{x^2+y^2}{x^2 y^2} = 14

. Next, find

(x+y)^2 = x^2 + y^2 + 2xy = 14 + 2(1) = 16

, so

x+y = 4

. Then

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

, leading directly to

\frac{14}{1/4} = 56

.

Estimated Time:2m 0s

Question 48Question →

An investor holds $x$ shares of Stock X and $y$ shares of Stock Y in a financial portfolio. The total monetary value of these holdings is $\$6,300$ , modeled by the linear equation $40x + 20y = 6,300$ . If the ratio of the number of shares of Stock X to the total number of shares held is $2$ to $5$ , what is the total number of shares of Stock X and Stock Y in the portfolio?

90

135

210

225

245

Show answer & explanation

Answer: 225

Answer

225

The correct answer is 225. Expressing the part-to-whole ratio

\frac{x}{x+y} = \frac{2}{5}

algebraically gives

3x = 2y

, or

y = 1.5x

. Substituting

1.5x

for

y

in the value equation

40x + 20y = 6,300

yields

70x = 6,300

, so

x = 90

. Then

y = 135

, and the total number of shares is

90 + 135 = 225

.

Step-by-Step Solution

1

Set up the linear relationship between

x

and

y

using the given ratio.

\frac{x}{x + y} = \frac{2}{5} \implies 5x = 2x + 2y \implies 3x = 2y \implies y = 1.5x

The problem specifies that the ratio of Stock X shares (

x

) to total shares (

x + y

) is

2:5

.

2

Substitute

y = 1.5x

into the portfolio value equation

40x + 20y = 6,300

.

40x + 20(1.5x) = 6,300 \implies 40x + 30x = 6,300 \implies 70x = 6,300 \implies x = 90

Replacing

y

with an equivalent expression in terms of

x

reduces the system to a single linear equation in one variable.

3

Calculate

y

and determine the total number of shares

x + y

.

y = 1.5(90) = 135 \implies x + y = 90 + 135 = 225

The question asks for the total combined number of shares of both stocks held in the portfolio.

Key Concept

Solving systems of linear equations formed by combining a linear value equation with a ratio relationship.

Question 49Question →

Machine X, working alone at a constant rate, can complete a manufacturing order in $4$ hours. Machine Y, working alone at a constant rate, can complete the same manufacturing order in $6$ hours. If Machine X and Machine Y work simultaneously at their respective constant rates, how many hours will it take them to complete $\frac{5}{6}$ of the manufacturing order?

2

hours

2.4

hours

4.17

hours

5

hours

8.33

hours

Show answer & explanation

Answer: $2$ hours

Answer

2

hours

Machine X completes

\frac{1}{4}

of the order per hour and Machine Y completes

\frac{1}{6}

of the order per hour. Working together, their combined rate is

\frac{1}{4} + \frac{1}{6} = \frac{5}{12}

of the order per hour. To determine the time required to finish

\frac{5}{6}

of the order, divide the targeted work amount by the combined rate:

\frac{5/6}{5/12} = 2

hours.

Step-by-Step Solution

1

Determine the individual work rates for Machine X and Machine Y

Rate of Machine X =

\frac{1}{4}

order per hour; Rate of Machine Y =

\frac{1}{6}

order per hour.

Work rate is defined as

\text{Rate} = \frac{\text{Work}}{\text{Time}}

.

2

Calculate the combined work rate when both machines work together

Combined Rate =

\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}

order per hour.

When machines work simultaneously, their individual rates add up.

3

Calculate the time required to complete

\frac{5}{6}

of the order using the combined rate

\text{Time} = \frac{\text{Work}}{\text{Combined Rate}} = \frac{5/6}{5/12} = \frac{5}{6} \times \frac{12}{5} = 2

hours.

Rearranging

\text{Work} = \text{Rate} \times \text{Time}

gives

\text{Time} = \frac{\text{Work}}{\text{Rate}}

.

Key Concept

Combined Work Rates

Estimated Time:2m 0s

Question 50Question →

If $x$ is a real number that satisfies the equation $|2x - 1| = 3x + 11$ , what is the value of $x^2 + 2x$ ?

0

3

8

35

120

Show answer & explanation

Answer: $0$

Answer

The value of

x^2 + 2x

is

0

.

Solving the equation

|2x - 1| = 3x + 11

produces two algebraic candidates:

x = -12

and

x = -2

. Substituting

x = -12

into the right side yields

3(-12) + 11 = -25

. Since absolute value expressions cannot be negative,

x = -12

is an extraneous solution. Substituting

x = -2

yields

|-5| = 5

, which is valid. Evaluating

x^2 + 2x

at

x = -2

gives

(-2)^2 + 2(-2) = 0

.

Step-by-Step Solution

1

Set up the two linear cases for the absolute value equation

|2x - 1| = 3x + 11

.

Case 1:

2x - 1 = 3x + 11

; Case 2:

2x - 1 = -(3x + 11)

.

By definition,

|u| = c

implies

u = c

or

u = -c

(provided

c \geq 0

).

2

Solve each linear equation for candidate values of

x

.

From Case 1:

x = -12

. From Case 2:

2x - 1 = -3x - 11 \implies 5x = -10 \implies x = -2

.

Isolate

x

algebraically in both equations.

3

Check candidate solutions in the original equation to eliminate extraneous roots.

For

x = -12

:

|2(-12) - 1| = |-25| = 25

, but

3(-12) + 11 = -25 \neq 25

(extraneous). For

x = -2

:

|2(-2) - 1| = |-5| = 5

, and

3(-2) + 11 = 5

(valid solution).

An absolute value cannot equal a negative number; substituting back is mandatory.

4

Evaluate the target expression

x^2 + 2x

using the valid root

x = -2

.

(-2)^2 + 2(-2) = 4 - 4 = 0

.

Substitute the verified real solution into the given expression.

Key Concept

Solving Absolute Value Linear Equations and Checking for Extraneous Solutions

Estimated Time:1m 30s

Question 51Question →

If $n$ is an integer such that $|2n + 1| \le |n - 4|$ , how many distinct integer values of $n$ satisfy the inequality?

5

6

7

8

9

Show answer & explanation

Answer: 7

Answer

There are 7 distinct integer values of n that satisfy the inequality.

Squaring both non-negative sides of the absolute value inequality

|2n + 1| \le |n - 4|

eliminates the absolute value bars without requiring multiple case splits. Expanding

(2n + 1)^2 \le (n - 4)^2

yields

4n^2 + 4n + 1 \le n^2 - 8n + 16

, which simplifies to

3n^2 + 12n - 15 \le 0

. Dividing the inequality by 3 gives

n^2 + 4n - 5 \le 0

, which factors as

(n + 5)(n - 1) \le 0

. The inequality is satisfied for all values in the closed interval

[-5, 1]

. Listing the integers in this range yields

-5, -4, -3, -2, -1, 0, 1

, which comprises exactly 7 integers.

Step-by-Step Solution

1

Square both sides of the inequality since absolute values are non-negative.

(2n + 1)^2 \le (n - 4)^2

Since

|A| \le |B|

is equivalent to

A^2 \le B^2

for all real numbers.

2

Expand both algebraic expressions and move all terms to one side.

4n^2 + 4n + 1 \le n^2 - 8n + 16 \implies 3n^2 + 12n - 15 \le 0

Standard quadratic inequality form requires comparing the quadratic polynomial to zero.

3

Simplify by dividing by 3 and factor the quadratic expression.

n^2 + 4n - 5 \le 0 \implies (n + 5)(n - 1) \le 0

Dividing by a positive constant preserves the inequality sign and allows factoring into linear binomials.

4

Determine the solution set for the inequality and count the integer solutions.

The inequality holds for

-5 \le n \le 1

. The integer solutions are

-5, -4, -3, -2, -1, 0, 1

, giving a total of

1 - (-5) + 1 = 7

integers.

The quadratic expression is non-positive between its two real roots, inclusive of the endpoints.

Key Concept

Solving Absolute Value Inequalities via Squaring and Quadratic Factoring

Estimated Time:1m 30s

Question 52Question →

A boutique bakery sells custom gift baskets containing two types of pastries: almond tarts and chocolate croissants. Basket A contains 4 almond tarts and 3 chocolate croissants and costs $62. Basket B contains 3 almond tarts and 4 chocolate croissants and costs$ 57. What is the combined cost of 1 almond tart and 1 chocolate croissant?

$15

$17

$19

$30

$34

Show answer & explanation

Answer: $17

Answer

The combined cost of 1 almond tart and 1 chocolate croissant is $17.

By representing the prices of an almond tart and a chocolate croissant as

t

and

c

, we form the system

4t + 3c = 62

and

3t + 4c = 57

. Adding both equations yields

7t + 7c = 119

. Dividing both sides by 7 gives

t + c = 17

, which directly provides the combined price of 1 almond tart and 1 chocolate croissant.

Step-by-Step Solution

1

Set up a system of two linear equations using variables for the prices of the pastries.

Let

t

be the price of one almond tart and

c

be the price of one chocolate croissant.
Equation 1:

4t + 3c = 62

Equation 2:

3t + 4c = 57

Translating the problem statement into standard linear algebraic equations.

2

Add the two linear equations together.

(4t + 3c) + (3t + 4c) = 62 + 57 \implies 7t + 7c = 119

Symmetric coefficients allow finding the sum of

t + c

without needing to solve for individual variables first.

3

Divide the combined equation by 7 to solve for

(t + c)

.

t + c = \frac{119}{7} = 17

Factoring out 7 gives

7(t + c) = 119

, which simplifies directly to the requested sum.

Key Concept

Solving Systems of Linear Equations via Symmetric Coefficient Addition

Alternative Method

Alternatively, solve for one variable first: multiply Equation 1 by 3 (

12t + 9c = 186

) and Equation 2 by 4 (

12t + 16c = 228

). Subtracting the equations gives

7c = 42 \implies c = 6

. Substituting

c = 6

into Equation 1 gives

4t + 18 = 62 \implies 4t = 44 \implies t = 11

. Thus,

t + c = 11 + 6 = 17

.

Estimated Time:1m 15s

Question 53Question →

A municipal water treatment facility operates three types of filtration units: Model X, Model Y, and Model Z.

\bullet \text{ 2 Model X units, 3 Model Y units, and 1 Model Z unit together process 134,000 gallons per hour.}

\bullet \text{ 1 Model X unit, 4 Model Y units, and 2 Model Z units together process 156,000 gallons per hour.}

\bullet \text{ 3 Model X units, 1 Model Y unit, and 4 Model Z units together process 196,000 gallons per hour.}

What is the processing capacity, in thousands of gallons per hour, of a single Model Y filtration unit?

Show answer & explanation

Answer: 20

Answer

The processing capacity of a single Model Y filtration unit is 20 thousand gallons per hour.

By setting up the system of three linear equations

2x + 3y + z = 134

,

x + 4y + 2z = 156

, and

3x + y + 4z = 196

, and eliminating variables systematically through substitution and combination, we find

y = 20

. Thus, a single Model Y filtration unit processes 20 thousand gallons per hour.

Step-by-Step Solution

1

Set up a system of linear equations in three variables.

Let

x

,

y

, and

z

represent the hourly capacities (in thousands of gallons) of Model X, Model Y, and Model Z respectively.

\begin{aligned} (1) \quad 2x + 3y + z &= 134 \\ (2) \quad x + 4y + 2z &= 156 \\ (3) \quad 3x + y + 4z &= 196 \end{aligned}

Expressing the given conditions algebraically translates the word problem into a solvable linear system.

2

Express

x

in terms of

y

and

z

using Equation (2).

x = 156 - 4y - 2z

Equation (2) has a coefficient of 1 for

x

, making it ideal for algebraic substitution.

3

Substitute

x

into Equations (1) and (3) to eliminate

x

.

Substituting into Equation (1):

2(156 - 4y - 2z) + 3y + z = 134 \implies 312 - 8y - 4z + 3y + z = 134

312 - 5y - 3z = 134 \implies (4) \quad 5y + 3z = 178

Substituting into Equation (3):

3(156 - 4y - 2z) + y + 4z = 196 \implies 468 - 12y - 6z + y + 4z = 196

468 - 11y - 2z = 196 \implies (5) \quad 11y + 2z = 272

Reducing a 3-variable system to a 2-variable system simplifies the calculation.

4

Eliminate variable

z

from Equations (4) and (5) to solve for

y

.

Multiply Equation (4) by 2:

10y + 6z = 356

Multiply Equation (5) by 3:

33y + 6z = 816

Subtract the first result from the second:

(33y + 6z) - (10y + 6z) = 816 - 356

23y = 460 \implies y = 20

Eliminating

z

directly yields the required value of

y

, which is the capacity of Model Y.

Key Concept

Solving systems of three linear equations in three variables via substitution and elimination.

Question 54Question →

If $x = \sqrt{7 + 4\sqrt{3}}$ , what is the value of $x - \frac{1}{x}$ ?

2

4

\sqrt{3}

2\sqrt{3}

4\sqrt{3}

Show answer & explanation

Answer: $2\sqrt{3}$

Answer

2\sqrt{3}

The expression under the outer radical,

7 + 4\sqrt{3}

, can be rewritten as

(2 + \sqrt{3})^2

because

2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

. Taking the square root gives

x = 2 + \sqrt{3}

. The reciprocal

\frac{1}{x}

is

\frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}

. Thus,

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

.

Step-by-Step Solution

1

Un-nest the radical

\sqrt{7 + 4\sqrt{3}}

by expressing

7 + 4\sqrt{3}

as a perfect square

(a + b\sqrt{3})^2

.

(a + b\sqrt{3})^2 = a^2 + 3b^2 + 2ab\sqrt{3} = 7 + 4\sqrt{3}

, which yields

a = 2

and

b = 1

, so

x = 2 + \sqrt{3}

.

Recognizing nested radicals in the form

\sqrt{A + B\sqrt{C}}

allows simplification into a binomial radical.

2

Find the reciprocal

\frac{1}{x}

by rationalizing the denominator.

\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3})(2 - \sqrt{3})} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.

Multiplying the numerator and denominator by the conjugate clears the radical from the denominator.

3

Calculate the difference

x - \frac{1}{x}

.

(2 + \sqrt{3}) - (2 - \sqrt{3}) = 2 + \sqrt{3} - 2 + \sqrt{3} = 2\sqrt{3}.

Subtracting the reciprocal isolates the irrational component.

Key Concept

Simplification of nested radicals and rationalizing denominators using algebraic conjugates.

Estimated Time:1m 30s

Question 55Question →

If real numbers

x

and

y

satisfy the system of equations:

3x - 5y = 28

x(x - 2y) = 3y(x - 2y)

where

x \neq 2y

, what is the value of

x + y

?

7

21

28

35

84

Show answer & explanation

Answer: 28

Answer

The value of

x + y

is 28.

Because

x \neq 2y

, the expression

(x - 2y)

is non-zero. Dividing both sides of the second equation

x(x - 2y) = 3y(x - 2y)

by

(x - 2y)

gives

x = 3y

. Substituting

x = 3y

into the first equation

3x - 5y = 28

yields

3(3y) - 5y = 28

, which simplifies to

4y = 28

, so

y = 7

. Then

x = 3(7) = 21

. The required sum

x + y

is

21 + 7 = 28

.

Step-by-Step Solution

1

Simplify the second equation using the given constraint.

Since

x \neq 2y

, the factor

(x - 2y) \neq 0

. Dividing both sides of

x(x - 2y) = 3y(x - 2y)

by

(x - 2y)

yields

x = 3y

.

Dividing by a non-zero expression is algebraically valid and simplifies the second equation into a linear relationship.

2

Substitute

x = 3y

into the first linear equation.

3(3y) - 5y = 28 \implies 9y - 5y = 28 \implies 4y = 28 \implies y = 7

.

Substitution reduces the system to a single linear equation in one variable.

3

Calculate

x

and the required sum

x + y

.

x = 3(7) = 21

, so

x + y = 21 + 7 = 28

.

Demonstrates full solution of the target expression.

Key Concept

Solving Systems of Linear Equations by Algebraic Reduction

Question 56Question →

A logistics company models its total daily delivery cost $C$ , in dollars, as a linear function of the total distance driven $d$ , in miles, given by $C = m \cdot d + b$ , where $m$ and $b$ are constants. The total daily cost increases by $\$180$ when the distance driven increases by $120$ miles. If the total daily delivery cost is $\$450$ when $250$ miles are driven, what is the total daily delivery cost when $400$ miles are driven?

$600

$630

$675

$720

$810

Show answer & explanation

Answer: $675

Answer

The total daily delivery cost when 400 miles are driven is $675.

The cost function follows the linear form

C = m \cdot d + b

. The marginal rate

m = \frac{\Delta C}{\Delta d} = \frac{180}{120} = 1.5

dollars per mile. Substituting

d = 250

and

C = 450

gives

450 = 1.5(250) + b

, which yields

b = 75

. Substituting

d = 400

into

C = 1.5d + 75

produces

C = 600 + 75 = 675

dollars.

Step-by-Step Solution

1

Determine the slope (variable cost per mile)

m

from the rate of change.

m = \frac{\Delta C}{\Delta d} = \frac{180}{120} = 1.5

dollars per mile.

The slope of a linear equation represents the constant rate of change between the dependent and independent variables.

2

Calculate the fixed base cost

b

using the given data point

(d = 250, C = 450)

.

450 = 1.5(250) + b \implies 450 = 375 + b \implies b = 75

.

Substituting a known point into the linear model determines the vertical intercept (fixed cost).

3

Evaluate the linear function for

d = 400

miles.

C = 1.5(400) + 75 = 600 + 75 = 675

dollars.

Applying the fully determined linear equation

C = 1.5d + 75

yields the total cost for the required distance.

Key Concept

Linear Modeling in Two Variables (

y = mx + b

)

Question 57Question →

If $3^x + 3^{x+1} + 3^{x+2} = 351$ , what is the value of $(x + 1)^x$ ?

12

27

64

81

243

Show answer & explanation

Answer: $64$

Answer

The value of

(x + 1)^x

is

64

.

Factoring

3^x

out of the sum gives

3^x(1 + 3 + 9) = 13 \cdot 3^x = 351

. Dividing

351

by

13

gives

3^x = 27

, so

x = 3

. Substituting

x = 3

into

(x + 1)^x

yields

(3 + 1)^3 = 4^3 = 64

.

Step-by-Step Solution

1

Factor out the common exponential term

3^x

from the left side of the equation.

3^x(1 + 3^1 + 3^2) = 351

, which simplifies to

3^x(1 + 3 + 9) = 351

or

13 \cdot 3^x = 351

.

When adding terms with identical bases and variable exponents, factor out the term with the smallest exponent.

2

Solve for

3^x

and determine the value of

x

.

3^x = \frac{351}{13} = 27

. Since

27 = 3^3

,

x = 3

.

Divide both sides by

13

and set the powers of equal bases equal to find

x

.

3

Substitute

x = 3

into the target expression

(x + 1)^x

.

(3 + 1)^3 = 4^3 = 64

.

Evaluate the expression as requested by the question stem.

Key Concept

Factoring exponential expressions with common bases and applying exponent rules

Estimated Time:1m 30s

Question 58Question →

For all real numbers $x$ , the identity $\sqrt{x^2} = x$ holds true.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false because the principal square root

\sqrt{x^2}

is defined as the absolute value of

x

,

|x|

, which equals

-x

when

x < 0

.

The statement is false because the expression

\sqrt{x^2}

represents the principal (non-negative) square root of

x^2

, which simplifies to

|x|

. When

x

is negative,

|x| = -x

, which is not equal to

x

.

Step-by-Step Solution

1

Recall the definition of the principal square root radical symbol

\sqrt{\vphantom{x}}

By definition,

\sqrt{a} \ge 0

for any real number

a \ge 0

.

The principal square root function always yields a non-negative output.

2

Test a negative value for the variable

x

Let

x = -3

. Then

\sqrt{(-3)^2} = \sqrt{9} = 3

.

Evaluating a specific negative number tests if the equality holds universally for all real numbers.

3

Compare the evaluated result with the original value of

x

Since

3 \neq -3

,

\sqrt{x^2} \neq x

when

x < 0

.

A single counterexample disproves a universal mathematical identity statement.

Key Concept

Principal Square Root and Absolute Value Property (\sqrt{x^2} = |x|)

Estimated Time:45s

Question 59Question →

How many integer values of $x$ satisfy the inequality $|x^2 - 4x| \leq 3x - 6$ ?

3

4

5

6

8

Show answer & explanation

Answer: 4

Answer

4 integer values satisfy the inequality.

The solution requires breaking the inequality into cases based on the sign of

x^2 - 4x

while enforcing that the right-hand side

3x - 6

must be non-negative (meaning

x \geq 2

). Combining the valid sub-intervals

3 \leq x < 4

and

4 \leq x \leq 6

yields

3 \leq x \leq 6

. The integers satisfying this inequality are 3, 4, 5, and 6, giving a total of 4 integers.

Step-by-Step Solution

1

Establish the domain restriction based on non-negativity.

Since the left side

|x^2 - 4x| \geq 0

for all real

x

, the right side must also be non-negative:

3x - 6 \geq 0 \implies x \geq 2

.

An absolute value quantity can never be less than a negative number.

2

Analyze Case 1 where the expression inside the absolute value is non-negative (

x^2 - 4x \geq 0

).

For

x \geq 2

,

x(x - 4) \geq 0

implies

x \geq 4

. The inequality becomes

x^2 - 4x \leq 3x - 6 \implies x^2 - 7x + 6 \leq 0 \implies (x - 1)(x - 6) \leq 0

, so

1 \leq x \leq 6

. Intersecting with

x \geq 4

yields

4 \leq x \leq 6

.

When

x \geq 4

,

|x^2 - 4x| = x^2 - 4x

.

3

Analyze Case 2 where the expression inside the absolute value is negative (

x^2 - 4x < 0

).

For

x \geq 2

,

x(x - 4) < 0

implies

2 \leq x < 4

. The inequality becomes

-(x^2 - 4x) \leq 3x - 6 \implies -x^2 + 4x \leq 3x - 6 \implies x^2 - x - 6 \geq 0 \implies (x - 3)(x + 2) \geq 0

, so

x \leq -2

or

x \geq 3

. Intersecting with

2 \leq x < 4

yields

3 \leq x < 4

.

When

2 \leq x < 4

,

|x^2 - 4x| = -(x^2 - 4x)

.

4

Combine the valid intervals from both cases and count the integer solutions.

Combining

[3, 4)

and

[4, 6]

gives the interval

[3, 6]

. The integer values in this range are

x = 3, 4, 5, 6

. Total count = 4.

The union of the solution intervals yields all real numbers between 3 and 6 inclusive.

Key Concept

Solving Absolute Value Inequalities with Variable RHS

Estimated Time:2m 30s

Question 60Question →

If $|x - 5| = 2x - 4$ , what is the value of $x$ ?

3

-1

-1 and 3

1

5

Show answer & explanation

Answer: 3

Answer

The value of

x

is 3.

Solving the absolute value equation yields two candidate values,

x = 3

and

x = -1

. Substituting

x = 3

into the original equation gives

|3 - 5| = 2

and

2(3) - 4 = 2

, which makes the statement true. Therefore,

x = 3

is the only valid solution.

Step-by-Step Solution

1

Set up the two cases for the absolute value equation

|x - 5| = 2x - 4

.

Case 1:

x - 5 = 2x - 4

; Case 2:

x - 5 = -(2x - 4)

.

By definition,

|a| = b

implies

a = b

or

a = -b

when

b \ge 0

.

2

Solve Case 1:

x - 5 = 2x - 4

.

x = -1

.

Subtract

x

and add

4

to both sides.

3

Solve Case 2:

x - 5 = -2x + 4

.

3x = 9 \implies x = 3

.

Add

2x

and add

5

to both sides.

4

Test potential solutions in the original equation to eliminate extraneous roots.

For

x = -1

:

|-1 - 5| = 6 \neq 2(-1) - 4 = -6

(Invalid). For

x = 3

:

|3 - 5| = 2 = 2(3) - 4 = 2

(Valid).

The absolute value of a real number cannot be negative, so the right-hand side

2x - 4

must be non-negative.

Key Concept

Absolute Value Equations and Extraneous Solutions

Algebra and Functions

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

Alternative Method

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Common Mistakes

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Common Mistakes

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Common Mistakes

Alternative Method

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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