Algebra and Functions Practice Questions — GMAT

Question 1Question →

If $r$ and $s$ are the two distinct real roots of the quadratic equation $x^2 - 6x + 4 = 0$ , what is the value of $\frac{r^3 + s^3}{r + s}$ ?

24

28

32

36

44

Show answer & explanation

Answer: $24$

Answer

The value of

\frac{r^3 + s^3}{r + s}

is

24

.

Using Vieta's formulas for

x^2 - 6x + 4 = 0

, the sum of the roots is

r + s = 6

and the product is

rs = 4

. Factoring

r^3 + s^3

gives

(r + s)(r^2 - rs + s^2)

. Dividing by

(r + s)

leaves

r^2 - rs + s^2

, which can be rewritten as

(r + s)^2 - 3rs

. Substituting the Vieta values yields

6^2 - 3(4) = 36 - 12 = 24

.

Step-by-Step Solution

1

Apply Vieta's formulas to the given quadratic equation

x^2 - 6x + 4 = 0

.

The sum of the roots is

r + s = 6

, and the product of the roots is

rs = 4

.

For a quadratic equation

x^2 + bx + c = 0

, Vieta's formulas state that the sum of roots is

-b

and the product of roots is

c

.

2

Factor the sum of cubes expression

r^3 + s^3

.

\frac{r^3 + s^3}{r + s} = \frac{(r + s)(r^2 - rs + s^2)}{r + s} = r^2 - rs + s^2

.

The sum of cubes factors algebraically into

(r + s)(r^2 - rs + s^2)

, and

r + s = 6 \neq 0

allows cancellation.

3

Express

r^2 - rs + s^2

in terms of

(r + s)

and

rs

.

r^2 - rs + s^2 = (r + s)^2 - 3rs

.

Since

(r + s)^2 = r^2 + 2rs + s^2

, subtracting

3rs

yields

r^2 - rs + s^2

.

4

Substitute the known values

r + s = 6

and

rs = 4

into the expression.

6^2 - 3(4) = 36 - 12 = 24

.

Evaluating the algebraic expression yields the final value.

Key Concept

Polynomial Factoring and Vieta's Formulas for Quadratic Equations

Estimated Time:1m 30s

Question 2Question →

Determine the sum of all real solutions to the equation $|x^2 - 4x| = 3x - 6$ .

7

8

9

10

12

Show answer & explanation

Answer: $9$

Answer

The sum of all valid real solutions is

9

.

The correct answer is

9

. Setting up the two cases

x^2 - 4x = 3x - 6

and

x^2 - 4x = -(3x - 6)

yields candidate roots

x = 1, 6, 3,

and

-2

. Because the absolute value expression

|x^2 - 4x|

cannot be negative,

3x - 6

must be non-negative, requiring

x \ge 2

. Evaluating each candidate shows that

x = 1

and

x = -2

produce negative right-hand sides and are extraneous. The only valid solutions are

x = 3

and

x = 6

, whose sum is

3 + 6 = 9

.

Step-by-Step Solution

1

Establish the domain condition for the right-hand side of the absolute value equation.

Since absolute values are non-negative,

|x^2 - 4x| \geq 0

requires

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

.

An absolute value expression cannot equal a negative number.

2

Solve Case 1 where

x^2 - 4x = 3x - 6

.

x^2 - 7x + 6 = 0 \implies (x - 1)(x - 6) = 0 \implies x = 1

or

x = 6

.

This corresponds to the positive branch of the absolute value.

3

Solve Case 2 where

x^2 - 4x = -(3x - 6)

.

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

or

x = -2

.

This corresponds to the negative branch of the absolute value.

4

Test all candidate solutions (

x = -2, 1, 3, 6

) against the domain constraint

x \geq 2

.

x = -2

yields

3(-2)-6 = -12 < 0

(extraneous).

x = 1

yields

3(1)-6 = -3 < 0

(extraneous).

x = 3

yields

|9-12| = 3 = 3(3)-6

(valid).

x = 6

yields

|36-24| = 12 = 3(6)-6

(valid).

Extraneous roots introduced by unconstrained case splitting must be eliminated.

5

Sum the valid real solutions.

3 + 6 = 9

.

The question asks specifically for the sum of all valid real solutions.

Key Concept

Absolute Value Equations and Extraneous Solution Verification

Question 3Question →

How many integer values of $y$ satisfy the inequality $|2y - 7| \le |y + 2|$ ?

Show answer & explanation

Answer: 8

Answer

The total number of integer values of

y

satisfying the inequality is 8.

Squaring both sides of

|2y - 7| \le |y + 2|

gives

3y^2 - 32y + 45 \le 0

, which factors into

(3y - 5)(y - 9) \le 0

. The solution range for

y

is

\frac{5}{3} \le y \le 9

. Since

\frac{5}{3} \approx 1.67

, the integer values of

y

satisfying this range are

2, 3, 4, 5, 6, 7, 8, 9

, giving a total of 8 integers.

Step-by-Step Solution

1

Square both sides of the inequality

|2y - 7| \le |y + 2|

(2y - 7)^2 \le (y + 2)^2

Since both sides of an absolute value expression are non-negative, squaring both sides maintains the inequality direction.

2

Expand terms and move all terms to the left side

3y^2 - 32y + 45 \le 0

Expanding gives

4y^2 - 28y + 49 \le y^2 + 4y + 4

. Subtracting

(y^2 + 4y + 4)

from both sides produces the standard quadratic inequality.

3

Factor the quadratic expression to find critical points

(3y - 5)(y - 9) \le 0

, yielding

\frac{5}{3} \le y \le 9

The roots are

y = \frac{5}{3}

and

y = 9

. A quadratic with a positive leading coefficient is non-positive between its roots.

4

Determine all integers within the range

\left[\frac{5}{3}, 9\right]

2, 3, 4, 5, 6, 7, 8, 9

(8 integers total)

Because

\frac{5}{3} \approx 1.67

, the smallest integer within the range is 2 and the largest is 9.

Key Concept

Solving absolute value inequalities of the form

|A| \le |B|

by squaring both sides and determining integer solutions.

Question 4Question →

How many integer values of $x$ satisfy the inequality $||2x - 5| - 7| \leq 4$ ?

Show answer & explanation

Answer: 10

Answer

The total number of integer values of

x

satisfying the inequality is 10.

To solve

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

, rewrite the inequality without the outer absolute value as

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

. Adding 7 across all parts yields

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

. The inequality

|2x - 5| \leq 11

simplifies to

-3 \leq x \leq 8

. The inequality

|2x - 5| \geq 3

simplifies to

x \leq 1

or

x \geq 4

. Intersecting these two regions gives the set of real numbers

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

. The integer solutions within

[-3, 1]

are

-3, -2, -1, 0, 1

(5 integers), and within

[4, 8]

are

4, 5, 6, 7, 8

(5 integers). The total number of valid integer solutions is

5 + 5 = 10

.

Step-by-Step Solution

1

Unfold the outer absolute value expression.

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

An inequality of the form

|U| \leq C

with

C > 0

is equivalent to

-C \leq U \leq C

.

2

Isolate the inner absolute value expression by adding 7 throughout.

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

Adding a constant to all parts preserves the direction of the inequality.

3

Solve the upper bound

|2x - 5| \leq 11

.

-3 \leq x \leq 8

-11 \leq 2x - 5 \leq 11

adds 5 to give

-6 \leq 2x \leq 16

, which divides by 2 to yield

-3 \leq x \leq 8

.

4

Solve the lower bound

|2x - 5| \geq 3

.

x \leq 1

or

x \geq 4

An inequality

|U| \geq C

splits into

U \geq C

(

2x - 5 \geq 3 \implies x \geq 4

) or

U \leq -C

(

2x - 5 \leq -3 \implies x \leq 1

).

5

Find the intersection of the upper and lower bound conditions and count the integers.

10 integer solutions:

\{-3, -2, -1, 0, 1, 4, 5, 6, 7, 8\}

.

Combining

-3 \leq x \leq 8

with (

x \leq 1

or

x \geq 4

) produces two disjoint intervals

[-3, 1]

and

[4, 8]

, containing 5 integers each.

Key Concept

Solving nested absolute value inequalities using double inequalities and boundary region intersections.

Question 5Question →

What is the sum of all integer values of $x$ that satisfy the inequality $|x^2 - 6x + 5| < 2x - 2$ ?

Show answer & explanation

Answer: 15

Answer

The sum of all integer values of

x

satisfying the inequality is

15

.

Factoring both sides gives

|(x - 1)(x - 5)| < 2(x - 1)

. Since the absolute value is non-negative, the right-hand side requires

2x - 2 > 0 \implies x > 1

. Under

x > 1

, the term

x - 1

is strictly positive, allowing us to simplify to

|x - 5| < 2

, which yields

3 < x < 7

. The integer solutions are

4, 5,

and

6

, and their sum is

15

.

Step-by-Step Solution

1

Determine the necessary condition for the right-hand side of the inequality.

Because the left-hand side

|x^2 - 6x + 5|

is non-negative for all real

x

, the right-hand side must be strictly positive. Thus,

2x - 2 > 0 \implies x > 1

.

An absolute value expression cannot be strictly less than a zero or negative quantity.

2

Factor the quadratic expression inside the absolute value and the linear expression on the right.

|(x - 1)(x - 5)| < 2(x - 1)

.

Factoring reveals a common linear factor

(x - 1)

on both sides.

3

Simplify the inequality using the condition

x > 1

.

Since

x > 1

, we know

x - 1 > 0

, so

|x - 1| = x - 1

. Splitting the product inside the absolute value gives

(x - 1)|x - 5| < 2(x - 1)

. Dividing both sides by

(x - 1)

yields

|x - 5| < 2

.

Dividing an inequality by a strictly positive number preserves the direction of the inequality sign.

4

Solve the simplified absolute value inequality and sum the integer solutions.

|x - 5| < 2 \implies -2 < x - 5 < 2 \implies 3 < x < 7

. The integer solutions strictly within this range are

x = 4, 5,

and

6

. Their sum is

4 + 5 + 6 = 15

.

The integers strictly between

3

and

7

are

4

,

5

, and

6

.

Key Concept

Solving Quadratic Absolute Value Inequalities via Domain Constraints and Factoring

Estimated Time:2m 0s

Question 6Question →

A manufacturing shop produces two models of custom bicycle frames: Standard and Deluxe. Producing each Standard frame requires $2$ hours of welding and $1$ hour of painting. Producing each Deluxe frame requires $3$ hours of welding and $2$ hours of painting. During a single week, the shop logged a total of $130$ hours of welding and $75$ hours of painting for these two models. How many Deluxe bicycle frames were produced during that week?

Show answer & explanation

Answer: 20

Answer

The number of Deluxe bicycle frames produced during that week is 20.

Translating the resource limitations into a system of two linear equations yields

2x + 3y = 130

for welding hours and

x + 2y = 75

for painting hours, where

x

and

y

represent the number of Standard and Deluxe frames respectively. Expressing

x

in terms of

y

from the painting equation gives

x = 75 - 2y

. Substituting this expression into the welding equation gives

2(75 - 2y) + 3y = 130

, which simplifies to

150 - y = 130

, giving

y = 20

Deluxe frames.

Step-by-Step Solution

1

Define variables for the unknowns.

Let

x

be the number of Standard bicycle frames produced and

y

be the number of Deluxe bicycle frames produced.

Assigning variables allows us to translate the problem into algebraic expressions.

2

Set up a system of linear equations.

Welding constraint:

2x + 3y = 130

Painting constraint:

x + 2y = 75

Each constraint represents the sum of hours spent on Standard and Deluxe frames for that process.

3

Solve the system using elimination or substitution.

Multiply the painting equation by

2

:

2x + 4y = 150

.
Subtract the welding equation (

2x + 3y = 130

) from this result:

(2x + 4y) - (2x + 3y) = 150 - 130 \Rightarrow y = 20

.

Eliminating

x

directly solves for

y

, which is the requested quantity (Deluxe frames).

Key Concept

Solving systems of two linear equations with two variables by substitution or elimination.

Estimated Time:1m 30s

Question 7Question →

What is the sum of all real solutions to the equation $\sqrt{2x^2 - 5x + 3} = 1 - x$ ?

1

2

3

5

0

Show answer & explanation

Answer: 1

Answer

1

Squaring both sides of

\sqrt{2x^2 - 5x + 3} = 1 - x

yields

2x^2 - 5x + 3 = (1 - x)^2 = x^2 - 2x + 1

. Moving all terms to one side yields the quadratic equation

x^2 - 3x + 2 = 0

, which factors as

(x - 1)(x - 2) = 0

. The algebraic solutions are

x = 1

and

x = 2

. Testing

x = 1

in the original equation gives

\sqrt{0} = 0

, which is valid. Testing

x = 2

gives

\sqrt{1} = -1

, which is invalid because the principal square root cannot be negative. Thus,

x = 1

is the sole valid real solution, and its sum is 1.

Step-by-Step Solution

1

Square both sides of the equation to eliminate the radical

2x^2 - 5x + 3 = (1 - x)^2 = 1 - 2x + x^2

Squaring both sides converts the radical equation into a standard algebraic polynomial equation.

2

Rearrange into standard quadratic form

ax^2 + bx + c = 0

x^2 - 3x + 2 = 0

Subtracting

(x^2 - 2x + 1)

from both sides collects all terms on one side.

3

Factor the quadratic expression to find candidate roots

(x - 1)(x - 2) = 0 \implies x = 1 \text{ or } x = 2

Factoring determines the values of

x

that satisfy the squared equation.

4

Substitute candidate roots back into the original equation to check for extraneous solutions

For

x = 1

:

\sqrt{2(1)^2 - 5(1) + 3} = \sqrt{0} = 0

, and

1 - 1 = 0

(Valid).
For

x = 2

:

\sqrt{2(2)^2 - 5(2) + 3} = \sqrt{1} = 1

, but

1 - 2 = -1

(Invalid, since

1 \neq -1

).

Squaring an equation can introduce extraneous roots where the principal square root would equal a negative quantity.

5

Sum the valid real solutions

The only valid solution is

x = 1

, so the sum is 1.

Only valid roots that satisfy the original equation may be summed.

Key Concept

Quadratic Factoring and Extraneous Solutions in Radical Equations

Estimated Time:1m 30s

Question 8Question →

What is the sum of all real solutions to the equation $(x - 3)^2 = 4(x - 3)$ ?

4

7

10

12

3

Show answer & explanation

Answer: 10

Answer

The sum of all real solutions to the equation is 10.

To find all solutions to

(x - 3)^2 = 4(x - 3)

, bring all terms to the left side:

(x - 3)^2 - 4(x - 3) = 0

. Factoring out

(x - 3)

yields

(x - 3)(x - 7) = 0

. This gives two distinct real roots,

x = 3

and

x = 7

. Adding these together yields a sum of 10.

Step-by-Step Solution

1

Rearrange the given equation so that all terms are on one side.

(x - 3)^2 - 4(x - 3) = 0

Moving all terms to one side allows for factoring without losing variable solutions.

2

Factor out the common algebraic factor

(x - 3)

.

(x - 3)\,[(x - 3) - 4] = 0 \implies (x - 3)(x - 7) = 0

Factoring avoids the mistake of dividing by a variable expression.

3

Set each factor equal to zero to find the real roots.

x - 3 = 0 \implies x = 3

and

x - 7 = 0 \implies x = 7

A product of factors equals zero if and only if at least one factor is zero.

4

Calculate the sum of the real solutions.

3 + 7 = 10

The question asks for the sum of all distinct real solutions.

Key Concept

Factoring Quadratic Expressions and Avoiding Variable Cancellation

Estimated Time:1m 0s

Question 9Question →

If $m$ and $n$ are non-zero real numbers such that $m$ and $n$ are the roots of the quadratic equation $x^2 + mx + n = 0$ , what is the value of $m - n$ ?

-3

-1

0

1

3

Show answer & explanation

Answer: 3

Answer

3

The correct answer is 3. By Vieta's formulas, for

x^2 + mx + n = 0

, the sum of roots

m + n = -m

gives

2m + n = 0

, and the product of roots

m \cdot n = n

gives

m = 1

since

n \neq 0

. Substituting

m = 1

into

2m + n = 0

yields

n = -2

. Thus,

m - n = 1 - (-2) = 3

.

Step-by-Step Solution

1

Apply Vieta's formulas to express the sum and product of roots in terms of coefficients.

For the quadratic equation

x^2 + mx + n = 0

, the sum of roots is

m + n = -m

, and the product of roots is

m \cdot n = n

.

By Vieta's relations for a standard quadratic equation

ax^2 + bx + c = 0

, the sum of roots is

-\frac{b}{a}

and the product is

\frac{c}{a}

.

2

Solve the product relation for

m

.

Since

n \neq 0

, divide both sides of

m \cdot n = n

by

n

to get

m = 1

.

Because

n

is specified to be non-zero, division by

n

is valid and yields a unique non-zero value for

m

.

3

Substitute

m = 1

into the sum relation to determine

n

.

1 + n = -1 \implies n = -2

.

Rearranging

m + n = -m

gives

2m + n = 0

, so

n = -2(1) = -2

.

4

Calculate the target value

m - n

.

m - n = 1 - (-2) = 3

.

Subtracting

-2

from

1

produces

1 + 2 = 3

.

Key Concept

Vieta's Formulas and Root-Coefficient Relationships in Quadratic Equations

Estimated Time:2m 0s

Question 10Question →

If $|2x - 5| = 9$ , what is the sum of all possible values of $x$ ?

5

7

-2

9

14

Show answer & explanation

Answer: $5$

Answer

The sum of all possible values of

x

is

5

.

To solve

|2x - 5| = 9

, split the equation into two linear cases:

2x - 5 = 9

and

2x - 5 = -9

. Solving

2x - 5 = 9

gives

2x = 14

, so

x = 7

. Solving

2x - 5 = -9

gives

2x = -4

, so

x = -2

. Summing these two solutions gives

7 + (-2) = 5

.

Step-by-Step Solution

1

Set up the two equations based on the definition of absolute value.

2x - 5 = 9

and

2x - 5 = -9

For any expression

A

and real number

B \ge 0

,

|A| = B

implies

A = B

or

A = -B

.

2

Solve the first equation for

x

.

2x = 14 \implies x = 7

Add

5

to both sides and divide by

2

.

3

Solve the second equation for

x

.

2x = -4 \implies x = -2

Add

5

to both sides and divide by

2

.

4

Calculate the sum of all valid solutions.

7 + (-2) = 5

The question asks for the sum of all possible solutions for

x

.

Key Concept

Solving Absolute Value Equations

Question 11Question →

What is the product of all real solutions to the equation

(x^2 - 6x + 10)^2 - 2(x^2 - 6x + 10) - 35 = 0?

Show answer & explanation

Answer: 3

Answer

The product of all real solutions to the equation is 3.

Substituting

u = x^2 - 6x + 10

converts the given equation into

u^2 - 2u - 35 = 0

, which factors as

(u - 7)(u + 5) = 0

. Substituting back yields two quadratic equations:

x^2 - 6x + 3 = 0

and

x^2 - 6x + 15 = 0

. Checking the discriminants reveals that

x^2 - 6x + 3 = 0

has

D = 24 > 0

, producing two real roots with product

\frac{c}{a} = 3

, whereas

x^2 - 6x + 15 = 0

has

D = -24 < 0

, producing no real roots. Therefore, the product of all real solutions is 3.

Step-by-Step Solution

1

Substitute a dummy variable to simplify the equation structure.

Let

u = x^2 - 6x + 10

. The equation becomes

u^2 - 2u - 35 = 0

.

Recognizing the repeated quadratic expression allows transforming a fourth-degree equation into a standard quadratic form.

2

Factor the quadratic equation in terms of

u

.

(u - 7)(u + 5) = 0 \implies u = 7 \text{ or } u = -5

.

Finding the values of

u

establishes the intermediate equations for

x

.

3

Analyze the first case

u = 7

for real solutions.

x^2 - 6x + 3 = 0

. Discriminant

D_1 = 36 - 12 = 24 > 0

. Product of real roots is

\frac{3}{1} = 3

.

A positive discriminant guarantees two real roots, and Vieta's formulas give their product directly.

4

Analyze the second case

u = -5

for real solutions.

x^2 - 6x + 15 = 0

. Discriminant

D_2 = 36 - 60 = -24 < 0

. No real roots.

A negative discriminant indicates complex conjugate roots, which must be excluded when finding the product of real solutions.

5

Combine results to find the final product of all real solutions.

Product = 3.

Only the two roots from the first case are real, so their product is the product of all real solutions.

Key Concept

Solving disguised quadratics via algebraic substitution and testing discriminants to filter out non-real roots before applying Vieta's formulas.

Question 12Question →

A manufacturing plant has a fixed daily setup cost of $\$1,500$ . For a specific product, the production cost for each of the first $50$ units is $\$60$ per unit, while each additional unit produced beyond the first $50$ costs $\$30$ per unit. If the average total cost per unit produced on a given day was $\$45$ , how many total units were produced on that day?

Show answer & explanation

Answer: 200

Answer

The total number of units produced on that day was 200.

Setting up the piecewise linear total cost expression

C(n) = 1,500 + 50(60) + (n-50)30 = 3,000 + 30n

and equating average cost

\frac{3,000 + 30n}{n}

to

45

yields

15n = 3,000

, which solves to

n = 200

.

Step-by-Step Solution

1

Model total daily production cost as a piecewise linear algebraic equation.

C(n) = 1,500 + (50 \times 60) + (n - 50) \times 30 = 3,000 + 30n

for

n > 50

.

Total cost combines fixed setup fees, cost of the initial 50 units, and tier-2 cost for units exceeding 50.

2

Formulate the equation for average cost per unit.

\frac{C(n)}{n} = \frac{3,000 + 30n}{n} = 45

Average cost is total daily cost divided by total quantity produced, given as $45 per unit.

3

Solve the algebraic equation for n.

3,000 + 30n = 45n \implies 15n = 3,000 \implies n = 200

Isolating n yields the exact volume of units needed to satisfy the average cost target.

Key Concept

Algebraic Equation Modeling with Piecewise Cost Functions

Estimated Time:2m 0s

Question 13Question →

An arithmetic sequence $a_1, a_2, a_3, \dots$ with a positive common difference $d$ and a geometric sequence $b_1, b_2, b_3, \dots$ with a positive common ratio $r$ both have the same positive first term ( $a_1 = b_1 > 0$ ). If the 3rd term of the arithmetic sequence equals the 3rd term of the geometric sequence ( $a_3 = b_3$ ), and the 7th term of the arithmetic sequence equals the 5th term of the geometric sequence ( $a_7 = b_5$ ), what is the value of $r$ ?

\sqrt{2}

\sqrt{3}

2

\frac{2\sqrt{3}}{3}

2\sqrt{2}

Show answer & explanation

Answer: $\sqrt{2}$

Answer

\sqrt{2}

Using the formulas for the

n

-th terms,

a_3 = a_1 + 2d

and

b_3 = a_1 r^2

, giving

2d = a_1(r^2 - 1)

. Substituting

6d = 3a_1(r^2 - 1)

into

a_7 = a_1 + 6d = a_1 r^4

yields

a_1 + 3a_1(r^2 - 1) = a_1 r^4

. Dividing by

a_1 > 0

gives

1 + 3r^2 - 3 = r^4

, which simplifies to

r^4 - 3r^2 + 2 = 0

. Factoring gives

(r^2 - 1)(r^2 - 2) = 0

. Since

d > 0

and

a_1 > 0

, we must have

r^2 > 1

, eliminating

r^2 = 1

. Therefore,

r^2 = 2

, and since

r > 0

,

r = \sqrt{2}

.

Step-by-Step Solution

1

Express sequence terms in terms of first term

a_1

, common difference

d

, and common ratio

r

a_3 = a_1 + 2d

,

a_7 = a_1 + 6d

,

b_3 = a_1 r^2

, and

b_5 = a_1 r^4

Apply the standard formulas for the

n

-th term of arithmetic (

a_n = a_1 + (n-1)d

) and geometric (

b_n = b_1 r^{n-1}

) sequences.

2

Set up equations based on given equality of terms

Equation 1:

a_1 + 2d = a_1 r^2 \implies 2d = a_1(r^2 - 1)

;
Equation 2:

a_1 + 6d = a_1 r^4

Translate the given conditions

a_3 = b_3

and

a_7 = b_5

into algebraic relations.

3

Substitute

2d

from Equation 1 into Equation 2

a_1 + 3(2d) = a_1 + 3a_1(r^2 - 1) = a_1 r^4

Express

6d

as

3(2d)

to eliminate

d

from the system.

4

Divide by

a_1

(since

a_1 > 0

) and simplify to solve for

r

1 + 3r^2 - 3 = r^4 \implies r^4 - 3r^2 + 2 = 0 \implies (r^2 - 1)(r^2 - 2) = 0

Reduce the equation to a quadratic in terms of

r^2

.

5

Determine the valid root for

r

r^2 = 2 \implies r = \sqrt{2}

(since

r > 0

and

d > 0

implies

r^2 > 1

)

Since

d > 0

and

a_1 > 0

,

2d = a_1(r^2 - 1) > 0

, which requires

r^2 > 1

. Thus

r^2 = 1

is rejected, leaving

r^2 = 2

.

Key Concept

Arithmetic and Geometric Sequences Alignment

Estimated Time:2m 0s

Question 14Question →

A manufacturing plant uses three assembly lines, $L_1$ , $L_2$ , and $L_3$ , to produce three custom components, $X$ , $Y$ , and $Z$ .

- Producing one unit of component $X$ requires 2 hours on $L_1$ , 1 hour on $L_2$ , and 3 hours on $L_3$ .
- Producing one unit of component $Y$ requires 3 hours on $L_1$ , 4 hours on $L_2$ , and 2 hours on $L_3$ .
- Producing one unit of component $Z$ requires 1 hour on $L_1$ , 2 hours on $L_2$ , and 4 hours on $L_3$ .

During a given week, assembly lines $L_1$ , $L_2$ , and $L_3$ were operated for a total of 140 hours, 165 hours, and 235 hours, respectively, with zero idle time. Assuming full capacity utilization, how many units of component $Z$ were produced during that week?

Show answer & explanation

Answer: 30

Answer

30 units of component Z were produced.

Translating the assembly line operational hours into a 3x3 system of linear equations yields

2X + 3Y + Z = 140

,

X + 4Y + 2Z = 165

, and

3X + 2Y + 4Z = 235

. Eliminating

X

results in two equations in

Y

and

Z

:

5Y + 3Z = 190

and

5Y + Z = 130

. Subtracting these equations gives

2Z = 60

, so

Z = 30

.

Step-by-Step Solution

1

Formulate a system of 3 linear equations representing total hours logged on each assembly line.

Line 1:

2X + 3Y + Z = 140

; Line 2:

X + 4Y + 2Z = 165

; Line 3:

3X + 2Y + 4Z = 235

.

Each component requires specific line processing time, and total time per line equals total available capacity.

2

Eliminate variable

X

by substituting

X = 165 - 4Y - 2Z

into the other two equations.

Equation A:

5Y + 3Z = 190

and Equation B:

5Y + Z = 130

.

Reducing a 3-variable system to a 2-variable system simplifies linear elimination.

3

Subtract Equation B from Equation A.

2Z = 60, so Z = 30.

Since the coefficients of

Y

in both reduced equations are identical (

5Y

), subtraction directly isolates

Z

.

Key Concept

Solving Systems of Three Linear Equations via Variable Substitution and Elimination

Question 15Question →

Four numerical quantities $K$ , $L$ , $M$ , and $N$ are defined based on arithmetic and geometric sequences as follows. Arrange these four quantities in ascending order (from smallest to largest value):

- **Quantity $K$ **: The sum of the first 5 terms of an arithmetic sequence with first term $a_1 = 2$ and common difference $d = 3$ .
- **Quantity $L$ **: The 4th term of a geometric sequence with first term $b_1 = 3$ and common ratio $r = 2$ .
- **Quantity $M$ **: The sum of an infinite geometric series with first term $c_1 = 18$ and common ratio $r = \frac{1}{2}$ .
- **Quantity $N$ **: The 7th term of an arithmetic sequence with first term $d_1 = 50$ and common difference $d = -4$ .

Which of the following represents the correct ascending order of the four quantities?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order of the quantities is Quantity L (24) < Quantity N (26) < Quantity M (36) < Quantity K (40).

Evaluating each quantity gives Quantity L = 24, Quantity N = 26, Quantity M = 36, and Quantity K = 40. Arranging these in ascending numerical order produces the sequence Quantity L, Quantity N, Quantity M, Quantity K.

Step-by-Step Solution

1

Calculate Quantity K

K = 40

The sum of an arithmetic sequence is given by

S_n = \frac{n}{2}[2a_1 + (n-1)d]

. For

n = 5

,

a_1 = 2

, and

d = 3

,

S_5 = \frac{5}{2}[2(2) + (5-1)(3)] = \frac{5}{2}[4 + 12] = \frac{5}{2}(16) = 40

.

2

Calculate Quantity L

L = 24

The

n

-th term of a geometric sequence is

b_n = b_1 \cdot r^{n-1}

. For

n = 4

,

b_1 = 3

, and

r = 2

,

b_4 = 3 \cdot 2^{4-1} = 3 \cdot 2^3 = 3 \cdot 8 = 24

.

3

Calculate Quantity M

M = 36

The sum of an infinite geometric series with

|r| < 1

is

S_\infty = \frac{c_1}{1 - r}

. For

c_1 = 18

and

r = \frac{1}{2}

,

S_\infty = \frac{18}{1 - 1/2} = \frac{18}{1/2} = 36

.

4

Calculate Quantity N

N = 26

The

n

-th term of an arithmetic sequence is

d_n = d_1 + (n-1)d

. For

n = 7

,

d_1 = 50

, and

d = -4

,

d_7 = 50 + (7-1)(-4) = 50 + 6(-4) = 50 - 24 = 26

.

5

Compare the evaluated quantities to arrange them in ascending order

24 < 26 < 36 < 40

, which corresponds to

L < N < M < K

Comparing the numeric values directly yields

24 \text{ (L)} < 26 \text{ (N)} < 36 \text{ (M)} < 40 \text{ (K)}

.

Key Concept

Arithmetic and Geometric Sequence Formulas

Question 16Question →

A software analytics company offers three annual subscription plans: Standard, Professional, and Enterprise.

• A client purchasing 3 Standard, 2 Professional, and 1 Enterprise plan pays a total of $1,110. �� A client purchasing 1 Standard, 4 Professional, and 2 Enterprise plans pays a total of$ 1,620.
• A client purchasing 4 Standard, 1 Professional, and 3 Enterprise plans pays a total of $1,730.

What is the cost, in dollars, of 1 Enterprise plan?

Show answer & explanation

Answer: 350

Answer

The cost of 1 Enterprise plan is 350 dollars.

Setting up equations for each purchase tier gives

3x + 2y + z = 1110

,

x + 4y + 2z = 1620

, and

4x + y + 3z = 1730

. Isolating

x

in the second equation gives

x = 1620 - 4y - 2z

. Substituting

x

into the first and third equations yields

2y + z = 750

and

3y + z = 950

, respectively. Subtracting these reduced equations gives

y = 200

, which upon back-substitution into

2y + z = 750

reveals

z = 350

. Thus, 1 Enterprise plan costs 350 dollars.

Step-by-Step Solution

1

Formulate the linear system of equations from the given conditions.

Let

x

be the price of a Standard plan,

y

the price of a Professional plan, and

z

the price of an Enterprise plan:
(1)

3x + 2y + z = 1110

(2)

x + 4y + 2z = 1620

(3)

4x + y + 3z = 1730

Translating the verbal conditions into algebraic equations creates a solvable system.

2

Isolate variable

x

in equation (2) and substitute it into equations (1) and (3).

From (2),

x = 1620 - 4y - 2z

.
Substituting into (1):

3(1620 - 4y - 2z) + 2y + z = 1110 \implies 4860 - 12y - 6z + 2y + z = 1110 \implies 10y + 5z = 3750 \implies 2y + z = 750

(Equation 4)

Substituting into (3):

4(1620 - 4y - 2z) + y + 3z = 1730 \implies 6480 - 16y - 8z + y + 3z = 1730 \implies 15y + 5z = 4750 \implies 3y + z = 950

(Equation 5)

Eliminating

x

reduces the system to two linear equations with two variables.

3

Solve the 2x2 system of equations for

y

and

z

.

Subtracting Equation (4) from Equation (5):

(3y + z) - (2y + z) = 950 - 750 \implies y = 200

.

Substitute

y = 200

back into Equation (4):

2(200) + z = 750 \implies 400 + z = 750 \implies z = 350

.

Solving the reduced system yields the exact values of

y

and

z

.

Key Concept

Solving Systems of Three Linear Equations via Gaussian Elimination / Variable Substitution

Question 17Question →

The sum of the first three terms of an increasing geometric sequence of positive numbers is $21$ , and the sum of the squares of these same three terms is $189$ . What is the first term of the sequence?

2

3

6

9

12

Show answer & explanation

Answer: 3

Answer

The first term of the sequence is 3.

By setting up the equations

a(1+r+r^2) = 21

and

a^2(1+r^2+r^4) = 189

, we use the identity

1+r^2+r^4 = (1+r+r^2)(1-r+r^2)

to find

a(1-r+r^2) = 9

. Subtracting this from the first equation gives

2ar = 12

, so

ar = 6

. Solving for

r

gives

r = 2

for an increasing sequence, which yields

a = 3

.

Step-by-Step Solution

1

Set up the algebraic equations for the sum of terms and sum of squared terms.

Let the first three terms be

a

,

ar

, and

ar^2

. The given conditions yield

a(1 + r + r^2) = 21

and

a^2(1 + r^2 + r^4) = 189

.

Standard representation of a geometric sequence with first term

a

and common ratio

r

.

2

Factor the sum of squares expression using algebraic identities.

Note that

1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2)

. Thus,

a^2(1 + r + r^2)(1 - r + r^2) = 189

.

Factoring allows substitution of the first equation into the second.

3

Substitute

a(1 + r + r^2) = 21

into the factored equation.

Substituting

21

gives

21 \cdot a(1 - r + r^2) = 189 \implies a(1 - r + r^2) = 9

.

Simplifies the second-degree term expression to a system of two linear equations in terms of

a

and

ar

.

4

Subtract the simplified equation from the initial sum equation to isolate

ar

.

a(1 + r + r^2) - a(1 - r + r^2) = 21 - 9 \implies 2ar = 12 \implies ar = 6

.

Eliminating terms isolates the product of the first term and ratio, which is the second term.

5

Solve for

r

and find

a

.

Substitute

a = \frac{6}{r}

into

a(1 + r + r^2) = 21

to get

6r^2 - 15r + 6 = 0

, which factors as

(2r - 1)(r - 2) = 0

. Since the sequence is increasing,

r = 2

, yielding

a = 3

.

Determines the specific parameters satisfying the increasing constraint.

Key Concept

Properties and algebraic manipulation of terms in geometric sequences and series.

Estimated Time:2m 0s

Question 18Question →

For all positive real numbers $x$ and $y$ , if $x^2 y^3 = 108$ and $x^3 y^2 = 72$ , then $x + y = 5$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

Multiplying the two equations yields

(xy)^5 = 7776 = 6^5

, so

xy = 6

. Dividing the second equation by the first yields

\frac{x}{y} = \frac{2}{3}

, meaning

x = \frac{2}{3}y

. Substituting this into

xy = 6

gives

y = 3

and

x = 2

, making

x + y = 5

.

Step-by-Step Solution

1

Multiply the two system equations to find the product

xy

.

(x^2 y^3)(x^3 y^2) = x^5 y^5 = (xy)^5 = 108 \times 72 = 7776 = 6^5

, so

xy = 6

.

Multiplying exponential expressions with common bases allows adding their exponents to form a unified power

(xy)^5

.

2

Divide the second equation by the first equation to find the relationship between

x

and

y

.

\frac{x^3 y^2}{x^2 y^3} = \frac{x}{y} = \frac{72}{108} = \frac{2}{3}, which simplifies to

x = \frac{2}{3}y

.

Dividing exponential expressions subtracts their exponents, yielding the simple ratio of the variables.

3

Solve for individual values of

x

and

y

and evaluate

x + y

.

Substituting

x = \frac{2}{3}y

into

xy = 6

gives

\frac{2}{3}y^2 = 6 \implies y = 3

and

x = 2

. Therefore,

x + y = 5

.

Solving the system confirms that

x = 2

and

y = 3

are the unique positive real solutions.

Key Concept

System of exponential equations solved by combining power products and ratios.

Estimated Time:1m 30s

Question 19Question →

If $4^x + 4^x + 4^x + 4^x = 2^{10}$ , what is the value of $x$ ?

\frac{5}{4}

2

\frac{5}{2}

4

5

Show answer & explanation

Answer: $4$

Answer

The value of

x

is

4

.

Combining the four identical terms on the left side gives

4 \cdot 4^x = 4^{1+x}

. Converting

4

to base

2

yields

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

. Setting this equal to the right side

2^{10}

gives the linear equation

2x + 2 = 10

, which solves to

x = 4

.

Step-by-Step Solution

1

Combine the repeated addition on the left side of the equation.

4^x + 4^x + 4^x + 4^x = 4 \cdot 4^x = 4^{x+1}

Adding four identical terms is equivalent to multiplying the term by 4.

2

Convert all bases to 2 so that both sides can be compared.

4^{x+1} = (2^2)^{x+1} = 2^{2(x+1)} = 2^{2x+2}

Since

4 = 2^2

, applying the power of a power rule

(a^m)^n = a^{mn}

expresses the left side with base 2.

3

Equate the exponents and solve for

x

.

2x + 2 = 10 \implies 2x = 8 \implies x = 4

If

b^u = b^v

for a positive base

b \neq 1

, then

u = v

.

Key Concept

Combining like exponential terms by factoring and applying base conversion rules

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

and

(a^m)^n = a^{mn}

.

Question 20Question →

A boutique bookstore sells hardcover books for $\$25$ each and paperback books for $\$12$ each. On a certain day, the store sold a total of $85$ books and generated a total revenue of $\$1,579$ . How many hardcover books were sold on that day?

Show answer & explanation

Answer: 43

Answer

The total number of hardcover books sold on that day is 43.

By setting up the two linear equations

h + p = 85

(quantity) and

25h + 12p = 1579

(revenue), eliminating

p

yields

13h = 559

, which gives

h = 43

.

Step-by-Step Solution

1

Define variables and construct the system of linear equations

Let

h

be the number of hardcover books and

p

be the number of paperback books. The equations are

h + p = 85

and

25h + 12p = 1579

.

The total quantity of items sold gives a sum equation, and the individual prices multiplied by their quantities yield the total revenue equation.

2

Use substitution or elimination to solve for

h

Multiplying the total books equation by 12 yields

12h + 12p = 1020

. Subtracting this from

25h + 12p = 1579

gives

13h = 559

.

Eliminating

p

directly isolates the variable

h

representing hardcover books.

3

Calculate the value of

h

h = \frac{559}{13} = 43

.

Dividing the remaining total revenue by the coefficient of

h

yields the precise quantity.

Key Concept

Solving two-variable systems of linear equations using elimination or substitution

Algebra and Functions

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