Algebra and Functions Practice Questions — GMAT — Page 2

Question 21Question →

A research laboratory placed two orders for high-precision sensors. The first order, consisting of $5$ acoustic sensors and $3$ thermal sensors, cost a total of $\$1{}160$ . The second order, consisting of $3$ acoustic sensors and $5$ thermal sensors, cost a total of $\$1{}080$ . If each acoustic sensor costs the same amount and each thermal sensor costs the same amount, what is the combined cost of $1$ acoustic sensor and $1$ thermal sensor?

\$140

\$160

\$280

\$320

\$560

Show answer & explanation

Answer: $\$280$

Answer

The combined cost of 1 acoustic sensor and 1 thermal sensor is

\$280

.

By representing the acoustic sensor price as

a

and thermal sensor price as

t

, we form two equations:

5a + 3t = 1160

and

3a + 5t = 1080

. Adding both equations gives

8a + 8t = 2240

. Dividing the entire equation by

8

isolates

a + t = 280

.

Step-by-Step Solution

1

Define variables and write the system of equations based on the problem statement.

Let

a

be the price of one acoustic sensor and

t

be the price of one thermal sensor.
Equation 1:

5a + 3t = 1160

Equation 2:

3a + 5t = 1080

Translate word problem conditions into algebraic equations.

2

Add the two equations together to find a direct linear combination.

(5a + 3t) + (3a + 5t) = 1160 + 1080 \implies 8a + 8t = 2240

Since the target is

a + t

, adding symmetric coefficients yields a common factor of

8

, avoiding the need to solve for

a

and

t

individually.

3

Factor out

8

and solve for

(a + t)

.

8(a + t) = 2240 \implies a + t = \frac{2240}{8} = 280

Dividing the combined equation by

8

directly isolates the required quantity

(a + t)

.

Key Concept

Solving Linear Systems via Algebraic Combinations

Question 22Question →

How many integer values of $n$ satisfy the inequality $|n^2 - 9| \leq 2n + 6$ ?

Show answer & explanation

Answer: 6

Answer

There are 6 integer values of

n

that satisfy the given inequality.

Because

|n^2 - 9| \geq 0

for all real numbers,

2n + 6

must be non-negative, requiring

n \geq -3

. At

n = -3

, the inequality reduces to

0 \leq 0

, which is valid. For

n > -3

, the term

n + 3

is positive, so factoring

|n^2 - 9|

into

|n - 3|(n + 3)

and dividing by

n + 3

simplifies the inequality to

|n - 3| \leq 2

. This yields the integer range

1 \leq n \leq 5

, containing 5 integers. Including

n = -3

gives a total of 6 integer solutions.

Step-by-Step Solution

1

Determine the domain constraint from the non-negativity of the absolute value expression.

2n + 6 \geq 0 \implies n \geq -3

The absolute value expression

|n^2 - 9|

is non-negative for all real

n

, so the right-hand side must also be non-negative.

2

Evaluate the boundary case

n = -3

.

|(-3)^2 - 9| = 0

and

2(-3) + 6 = 0 \implies 0 \leq 0

(True)

When

n = -3

, both sides equal 0, making

n = -3

a valid integer solution.

3

Factor the quadratic inside the absolute value for

n > -3

and simplify.

|n - 3||n + 3| \leq 2(n + 3) \implies |n - 3| \leq 2

Since

n > -3

, the term

n + 3

is strictly positive, allowing division of both sides by

n + 3

without reversing the inequality.

4

Solve the linear absolute value inequality

|n - 3| \leq 2

.

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

Removing the absolute value creates a compound inequality bounded between

-2

and

2

.

5

List all valid integer solutions and count them.

n \in \{-3, 1, 2, 3, 4, 5\} \implies 6

integer solutions

Combining the boundary root

n = -3

with the five consecutive integers from

1

to

5

gives 6 integer solutions.

Key Concept

Absolute Value Inequalities with Variable RHS and Boundary Factors

Question 23Question →

What is the numerical value of the expression $\sqrt[3]{7 + 5\sqrt{2}} - \sqrt[3]{5\sqrt{2} - 7}$ ?

Show answer & explanation

Answer: 2

Answer

The numerical value of the expression is 2.

The value of the expression is 2. This can be demonstrated either by setting the expression equal to

x

, cubing both sides to construct the cubic equation

x^3 + 3x - 14 = 0

, and factoring out the real root

x = 2

, or by recognizing that

(1 + \sqrt{2})^3 = 7 + 5\sqrt{2}

and

(\sqrt{2} - 1)^3 = 5\sqrt{2} - 7

, which simplifies the expression directly to

(1 + \sqrt{2}) - (\sqrt{2} - 1) = 2

.

Step-by-Step Solution

1

Define variables for the two cubic terms and write the target expression as a difference.

Let

u = \sqrt[3]{7 + 5\sqrt{2}}

and

v = \sqrt[3]{5\sqrt{2} - 7}

, so the target value is

x = u - v

.

Grouping nested radical terms simplifies the algebraic manipulation.

2

Cube both sides of

x = u - v

using the algebraic identity

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

.

x^3 = u^3 - v^3 - 3uv \cdot x

.

Cubing eliminates the outer radical signs on the cubed terms.

3

Evaluate

u^3 - v^3

and the product

uv

.

u^3 - v^3 = (7 + 5\sqrt{2}) - (5\sqrt{2} - 7) = 14

, and

uv = \sqrt[3]{(5\sqrt{2}+7)(5\sqrt{2}-7)} = \sqrt[3]{50 - 49} = 1

.

Using the difference of squares under the cube root simplifies the product term to 1.

4

Substitute the evaluated terms into the cubic equation and solve for the real root

x

.

x^3 = 14 - 3x \implies x^3 + 3x - 14 = 0 \implies (x - 2)(x^2 + 2x + 7) = 0 \implies x = 2

.

The quadratic factor

x^2 + 2x + 7

has negative discriminant (

4 - 28 = -24

), leaving

x = 2

as the unique real solution.

Key Concept

Simplifying nested radicals using cubic algebraic identities and binomial expansions

Question 24Question →

A craft brewery produces two specialty beverages, Batch A and Batch B, using two primary ingredients: hops and malt. To produce 1 barrel of Batch A, the facility requires 4 kilograms of hops and 10 kilograms of malt. To produce 1 barrel of Batch B, the facility requires 6 kilograms of hops and 15 kilograms of malt. Let $a$ represent the number of barrels of Batch A produced and $b$ represent the number of barrels of Batch B produced in a week, where $a > 0$ and $b > 0$ . The total mass of hops used is $H$ kilograms, and the total mass of malt used is $M$ kilograms. Which of the following statements must be true for any valid production quantities of $a$ and $b$ ? Select all that apply.

Select all that apply

2M = 5H

M > H

3H - 2M = 0

The ratio of total malt used to total hops used,

\frac{M}{H}

, depends on the specific values of

a

and

b

.

If

H = 120

, then the number of barrels of Batch A produced (

a

) must be equal to 15.

Show answer & explanation

Answer: $2M = 5H$ ; $M > H$

Answer

The statements

2M = 5H

and

M > H

must both be true.

Expressing

H = 4a + 6b = 2(2a + 3b)

and

M = 10a + 15b = 5(2a + 3b)

reveals that

M = 2.5H

, or equivalently

2M = 5H

. Because

H

is strictly positive for positive values of

a

and

b

,

M = 2.5H

also guarantees that

M > H

.

Step-by-Step Solution

1

Set up algebraic equations for total hops (

H

) and total malt (

M

) in terms of

a

and

b

.

H = 4a + 6b

and

M = 10a + 15b

.

Each barrel of Batch A uses 4 kg hops and 10 kg malt, while each barrel of Batch B uses 6 kg hops and 15 kg malt.

2

Factor common numerical terms from both algebraic expressions.

H = 2(2a + 3b)

and

M = 5(2a + 3b)

.

Factoring isolates the common linear factor

(2a + 3b)

present in both quantities.

3

Calculate the ratio

\frac{M}{H}

and clear fractions to find the invariant linear equation.

\frac{M}{H} = \frac{5(2a + 3b)}{2(2a + 3b)} = \frac{5}{2} \implies 2M = 5H.

Since

a > 0

and

b > 0

,

(2a + 3b) \neq 0

, so the variable terms cancel completely.

4

Evaluate the inequality relationship between

M

and

H

.

M = 2.5H \implies M - H = 1.5H > 0 \implies M > H

.

Because

a, b > 0

,

H > 0

, making

2.5H

strictly greater than

H

.

Key Concept

Linear Equation Modeling and Proportional Invariants

Question 25Question →

If $x$ is a real number such that $x^2 - 5x + 1 = 0$ , what is the value of $x^2 + \frac{1}{x^2}$ ?

Show answer & explanation

Answer: 23

Answer

The value of

x^2 + \frac{1}{x^2}

is 23.

Dividing the quadratic equation

x^2 - 5x + 1 = 0

by

x

yields

x + \frac{1}{x} = 5

. Squaring both sides of this identity gives

\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} = 25

. Subtracting 2 from both sides results in

x^2 + \frac{1}{x^2} = 23

.

Step-by-Step Solution

1

Divide the given quadratic equation

x^2 - 5x + 1 = 0

by

x

x - 5 + \frac{1}{x} = 0 \implies x + \frac{1}{x} = 5

Since

x \neq 0

, dividing by

x

isolates the sum of

x

and its reciprocal.

2

Square both sides of the expression

x + \frac{1}{x} = 5

\left(x + \frac{1}{x}\right)^2 = 25 \implies x^2 + 2 + \frac{1}{x^2} = 25

Expanding the binomial square produces the required quadratic sum along with a constant cross-term.

3

Isolate

x^2 + \frac{1}{x^2}

by subtracting 2 from 25

x^2 + \frac{1}{x^2} = 23

Subtracting the constant cross-term yields the exact requested numeric value.

Key Concept

Algebraic transformation of quadratic equations into reciprocal power sums

Question 26Question →

A bakery sells blueberry muffins for $3 each and chocolate chip muffins for$ 4 each. On Saturday, the bakery sold a total of 50 muffins for $170. How many chocolate chip muffins were sold on Saturday?

Show answer & explanation

Answer: 20

Answer

The bakery sold 20 chocolate chip muffins on Saturday.

Let

b

represent the number of blueberry muffins and

c

represent the number of chocolate chip muffins sold. Based on the problem text, we construct the system of equations:

b + c = 50

and

3b + 4c = 170

. From the first equation,

b = 50 - c

. Substituting this into the second equation yields

3(50 - c) + 4c = 170

, which expands to

150 - 3c + 4c = 170

. Simplifying gives

150 + c = 170

, so

c = 20

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations.

b + c = 50

and

3b + 4c = 170

The total number of muffins sold is 50, and the total revenue from selling blueberry muffins at

3 each and chocolate chip muffins at

4 each is $170.

2

Substitute

b = 50 - c

into the cost equation.

3(50 - c) + 4c = 170 \implies 150 + c = 170

Substituting

b

in terms of

c

creates a single-variable linear equation for

c

.

3

Solve for

c

.

c = 20

Subtracting 150 from both sides gives the exact number of chocolate chip muffins sold.

Key Concept

Solving Systems of Linear Equations by Substitution or Elimination

Question 27Question →

What is the sum of all real solutions to the absolute value equation $|2x - 3| + |x + 5| = 12 - x$ ?

-\frac{9}{2}

\frac{5}{2}

-\frac{1}{2}

\frac{9}{2}

\frac{19}{2}

Show answer & explanation

Answer: $-\frac{9}{2}$

Answer

The sum of all real solutions to the equation is

-\frac{9}{2}

.

By splitting the real number line into three intervals based on the critical points

x = -5

and

x = \frac{3}{2}

, we find two valid solutions:

x = \frac{5}{2}

(from

x \ge \frac{3}{2}

) and

x = -7

(from

x < -5

). Their sum is

\frac{5}{2} + (-7) = -\frac{9}{2}

.

Step-by-Step Solution

1

Identify critical points for the absolute value expressions

The critical points are

x = \frac{3}{2}

and

x = -5

.

The expressions inside the absolute values,

2x - 3

and

x + 5

, change signs at

x = \frac{3}{2}

and

x = -5

respectively.

2

Evaluate Region 1 (

x \ge \frac{3}{2}

)

(2x - 3) + (x + 5) = 12 - x \implies 3x + 2 = 12 - x \implies 4x = 10 \implies x = \frac{5}{2}

.

In this region, both

2x - 3 \ge 0

and

x + 5 > 0

, so absolute value bars can be removed directly. Since

\frac{5}{2} \ge \frac{3}{2}

,

x = \frac{5}{2}

is a valid solution.

3

Evaluate Region 2 (

-5 \le x < \frac{3}{2}

)

-(2x - 3) + (x + 5) = 12 - x \implies -x + 8 = 12 - x \implies 8 = 12

(No solution).

In this interval,

2x - 3 < 0

while

x + 5 \ge 0

. The resulting equation produces a contradiction, so there are no solutions in this interval.

4

Evaluate Region 3 (

x < -5

)

-(2x - 3) - (x + 5) = 12 - x \implies -3x - 2 = 12 - x \implies -2x = 14 \implies x = -7

.

In this region, both

2x - 3 < 0

and

x + 5 < 0

. Since

-7 < -5

,

x = -7

is a valid solution.

5

Calculate the sum of all valid solutions

\frac{5}{2} + (-7) = \frac{5}{2} - \frac{14}{2} = -\frac{9}{2}

.

Combining the valid roots from Region 1 and Region 3 gives the final requested sum.

Key Concept

Solving piecewise linear equations involving multiple absolute value terms.

Estimated Time:2m 30s

Question 28Question →

An event planner ordered a total of $30$ gift baskets for a corporate conference. Standard gift baskets cost $\$25$ each, and Deluxe gift baskets cost $\$40$ each. If the total cost of all $30$ gift baskets was $\$900$ , how many Deluxe gift baskets were ordered?

10

12

15

20

25

Show answer & explanation

Answer: $10$

Answer

The correct number of Deluxe gift baskets ordered is

10

.

By setting up the linear system

x + y = 30

and

25x + 40y = 900

, substituting

x = 30 - y

into the cost equation yields

750 + 15y = 900

, which solves directly to

y = 10

. Therefore,

10

Deluxe baskets were ordered.

Step-by-Step Solution

1

Define variables and set up the system of linear equations

Let

x

be the number of Standard baskets and

y

be the number of Deluxe baskets. Equation 1:

x + y = 30

. Equation 2:

25x + 40y = 900

.

The total number of baskets establishes a quantity relation, while the individual prices establish a cost relation.

2

Express

x

in terms of

y

using Equation 1

x = 30 - y

Isolating

x

allows direct substitution into the total cost equation to solve for

y

.

3

Substitute

x = 30 - y

into Equation 2 and solve for

y

25(30 - y) + 40y = 900 \implies 750 - 25y + 40y = 900 \implies 15y = 150 \implies y = 10

.

Simplifying the single-variable equation yields the exact number of Deluxe gift baskets.

Key Concept

Solving Systems of Linear Equations using Substitution or Elimination

Alternative Method

Use the elimination method: multiply the total quantity equation by

25

(

25x + 25y = 750

) and subtract it from the total cost equation (

25x + 40y = 900

) to get

15y = 150

, leading directly to

y = 10

.

Estimated Time:45s

Question 29Question →

A courier service calculates its total delivery charge using a fixed base fee plus a constant per-mile rate. On Monday, a delivery of 15 miles received a 20% discount on the fixed base fee and a 25% surcharge on the per-mile rate, resulting in a total charge of $30.50. On Tuesday, a delivery of 20 miles incurred a 40% increase on the fixed base fee and received a 15% discount on the per-mile rate, resulting in a total charge of$ 34.40. What is the standard total delivery charge, in dollars, for a 25-mile delivery with no fee adjustments or rate changes?

Show answer & explanation

Answer: 40

Answer

The standard total delivery charge for a 25-mile delivery is 40 dollars.

Translating the scenario into linear equations gives

0.80B + 18.75r = 30.50

and

1.40B + 17.00r = 34.40

. Solving this linear system yields a standard base fee of

B = 10.00

dollars and a standard per-mile rate of

r = 1.20

dollars. Substituting these into the standard 25-mile cost expression

B + 25r

yields

10.00 + 25(1.20) = 40.00

dollars.

Step-by-Step Solution

1

Formulate linear equations from the word problem context.

System of equations:

0.80B + 18.75r = 30.50

and

1.40B + 17.00r = 34.40

.

Applying the percentage adjustments to the fixed base fee

B

and the rate per mile

r

for the given distances yields exact linear expressions.

2

Eliminate variable

B

to solve for

r

.

Multiply equations to equate coefficients of

B

:

5.60B + 131.25r = 213.50

and

5.60B + 68.00r = 137.60

. Subtracting gives

63.25r = 75.90

, so

r = 1.20

.

Finding the per-mile rate

r

allows determination of the standard mileage component.

3

Solve for base fee

B

using

r = 1.20

.

1.40B + 17(1.20) = 34.40 \implies 1.40B = 14.00 \implies B = 10.00

.

Substituting

r

into either linear equation gives the fixed base fee.

4

Calculate the target standard cost for 25 miles.

B + 25r = 10.00 + 25(1.20) = 40.00

.

Evaluating the standard pricing expression

B + 25r

with

B = 10

and

r = 1.20

gives the total cost.

Key Concept

Solving Systems of Two-Variable Linear Equations from Word Problems

Question 30Question →

If $3x + 2y = 18$ and $x + 4y = 16$ , what is the value of $x + y$ ?

3

4

7

10

17

Show answer & explanation

Answer: 7

Answer

7

Solving the system of linear equations

3x + 2y = 18

and

x + 4y = 16

yields

x = 4

and

y = 3

. Adding these values together produces

4 + 3 = 7

.

Step-by-Step Solution

1

Express

x

in terms of

y

using the second equation

x = 16 - 4y

Isolating one variable allows straightforward substitution into the other equation.

2

Substitute

x = 16 - 4y

into the first equation

3(16 - 4y) + 2y = 18 \implies 48 - 12y + 2y = 18 \implies 48 - 10y = 18

This creates an equation containing only the variable

y

.

3

Solve for

y

-10y = 18 - 48 = -30 \implies y = 3

Isolating the numerical term isolates the value of

y

.

4

Substitute

y = 3

back to find

x

x = 16 - 4(3) = 16 - 12 = 4

Determines the specific value of

x

.

5

Calculate the requested sum

x + y

x + y = 4 + 3 = 7

Provides the final requested expression value.

Key Concept

Solving a system of linear equations by substitution or elimination to evaluate a combined linear expression.

Estimated Time:45s

Question 31Question →

If $x$ and $y$ are positive integers such that $2^x - 2^{x-2} = 3^y$ , what is the value of $2^{x+y} + 3^{x-y}$ ?

11

13

25

35

73

Show answer & explanation

Answer: 11

Answer

The value of

2^{x+y} + 3^{x-y}

is 11.

Factoring the left side of the equation yields

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

. Setting this equal to

3^y

and dividing both sides by 3 gives

2^{x-2} = 3^{y-1}

. Because 2 and 3 are coprime bases, equality is possible only when both exponents equal zero. This determines

x = 2

and

y = 1

, both of which are positive integers. Substituting these values into

2^{x+y} + 3^{x-y}

results in

2^3 + 3^1 = 8 + 3 = 11

.

Step-by-Step Solution

1

Factor out the common term on the left side of the given equation

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

Factoring out the smallest power of 2 simplifies the subtraction into a multiplication.

2

Equate the factored expression to the right side and simplify

3 \cdot 2^{x-2} = 3^y \implies 2^{x-2} = \frac{3^y}{3} = 3^{y-1}

Dividing both sides by 3 isolates the prime bases 2 and 3 on opposite sides of the equation.

3

Apply fundamental properties of prime factorizations

x - 2 = 0 \implies x = 2

and

y - 1 = 0 \implies y = 1

Since 2 and 3 are distinct prime numbers,

2^a = 3^b

can hold for integer exponents only if

a = 0

and

b = 0

.

4

Evaluate the target expression using

x = 2

and

y = 1

2^{2+1} + 3^{2-1} = 2^3 + 3^1 = 8 + 3 = 11

Substituting the unique positive integer solution pair

(2, 1)

into the expression yields the final value.

Key Concept

Uniqueness of Prime Base Exponent Equality

Question 32Question →

How many integer values of $x$ satisfy the inequality $\frac{|x - 2| - 3}{|x + 1| - 4} \leq 0$ ?

Show answer & explanation

Answer: 6

Answer

The correct answer is 6.

Analyzing the signs of the numerator

|x - 2| - 3

and denominator

|x + 1| - 4

shows that the quotient is non-positive when the numerator and denominator have opposite signs or when the numerator is zero. This yields two intervals:

(-5, -1]

and

(3, 5]

. The integers contained in these intervals are

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

, which totals 6 integer values.

Step-by-Step Solution

1

Analyze the sign of the numerator

|x - 2| - 3

The numerator is zero at

x = -1

and

x = 5

, negative for

-1 < x < 5

, and positive for

x < -1

or

x > 5

.

Solving

|x - 2| = 3

yields

x - 2 = 3 \implies x = 5

and

x - 2 = -3 \implies x = -1

.

2

Analyze the sign of the denominator

|x + 1| - 4

The denominator is zero at

x = -5

and

x = 3

, negative for

-5 < x < 3

, and positive for

x < -5

or

x > 3

. Exclude

x = -5

and

x = 3

.

Solving

|x + 1| = 4

yields

x + 1 = 4 \implies x = 3

and

x + 1 = -4 \implies x = -5

. Denominators cannot be zero.

3

Find intervals where numerator and denominator have opposite signs or numerator is zero

The solution set is the union of

(-5, -1]

and

(3, 5]

.

A fraction

\frac{N}{D} \leq 0

requires

N \ge 0, D < 0

or

N \le 0, D > 0

.

4

Count the total number of integer solutions

The valid integers are

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

, giving a total count of 6.

Listing integers in

(-5, -1]

yields

-4, -3, -2, -1

, and in

(3, 5]

yields

4, 5

.

Key Concept

Solving Rational Inequalities with Absolute Values

Question 33Question →

Consider the system of linear equations in

x

and

y

:

kx + 6y = 18

(k + 1)x + 9y = 27

where

k

is a constant. If this system has infinitely many solutions

(x, y)

, what is the sum of all real values of

x

that satisfy the equation

|kx - 10| = 3x - 4

?

\frac{14}{5}

-\frac{16}{5}

-6

10

\frac{6}{5}

Show answer & explanation

Answer: $\frac{14}{5}$

Answer

The sum of all real values of

x

satisfying the equation is

\frac{14}{5}

(or

2.8

).

The system of linear equations has infinitely many solutions when the ratio of coefficients is constant:

\frac{k}{k+1} = \frac{6}{9} = \frac{18}{27} = \frac{2}{3}

, which yields

k = 2

. Substituting

k = 2

gives

|2x - 10| = 3x - 4

. Breaking this into cases yields

2x - 10 = 3x - 4 \implies x = -6

and

-(2x - 10) = 3x - 4 \implies x = \frac{14}{5}

. Testing

x = -6

in the original equation gives

|-22| = -22

, which is false, making

x = -6

an extraneous solution. Testing

x = \frac{14}{5}

gives

|-4.4| = 4.4

, which is true. Thus, the only real solution is

\frac{14}{5}

, making the option equal to

\frac{14}{5}

correct.

Step-by-Step Solution

1

Determine the value of the constant

k

from the system of linear equations.

k = 2

For a system of two linear equations in two variables to have infinitely many solutions, the ratio of coefficients of

x

,

y

, and the constant terms must all be equal:

\frac{k}{k+1} = \frac{6}{9} = \frac{18}{27}

. Since

\frac{6}{9} = \frac{2}{3}

, setting

\frac{k}{k+1} = \frac{2}{3}

gives

3k = 2k + 2 \implies k = 2

.

2

Substitute

k = 2

into the absolute value equation.

|2x - 10| = 3x - 4

Replacing

k

with

2

sets up the single-variable linear absolute value equation to solve.

3

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

2x - 10 \ge 0

).

x = -6

(Extraneous)

Setting

2x - 10 = 3x - 4

yields

x = -6

. However, substituting

x = -6

back into the right side gives

3(-6) - 4 = -22

. Since an absolute value cannot equal a negative number,

x = -6

is an extraneous solution.

4

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

2x - 10 < 0

).

x = \frac{14}{5}

(Valid)

Setting

-(2x - 10) = 3x - 4

gives

-2x + 10 = 3x - 4 \implies 5x = 14 \implies x = \frac{14}{5}

. Substituting

x = 2.8

into the right side yields

3(2.8) - 4 = 4.4 \ge 0

, and

|2(2.8) - 10| = |-4.4| = 4.4

. Thus,

x = \frac{14}{5}

is valid.

5

Sum all valid real solutions.

\frac{14}{5}

Since

x = \frac{14}{5}

is the only valid solution, the sum of all solutions is

\frac{14}{5}

.

Key Concept

Systems of linear equations with infinitely many solutions require identical proportional coefficients. Absolute value equations of the form

|A| = B

require checking candidate solutions in

B \ge 0

to eliminate extraneous roots.

Estimated Time:3m 0s

Question 34Question →

A technology manufacturing company produces two server models: Model Alpha and Model Beta. The manufacturing cost of one Model Alpha server is $400 more than twice the manufacturing cost of one Model Beta server. Last month, the company produced 15 Model Alpha servers and 40 Model Beta servers, incurring a total manufacturing cost of$ 76,000. What is the manufacturing cost, in dollars, of one Model Alpha server?

Show answer & explanation

Answer: 2400

Answer

The manufacturing cost of one Model Alpha server is $2,400.

By defining the costs as

A = 2B + 400

and setting up the total cost equation

15A + 40B = 76,000

, substituting

A

yields

70B = 70,000

, giving

B = 1,000

. Substituting

B = 1,000

back into

A = 2B + 400

gives

A = 2,400

.

Step-by-Step Solution

1

Define variables and construct the linear relation equation.

A = 2B + 400

The cost of one Model Alpha server (

A

) is

400 more than twice the cost of Model Beta (

B$).

2

Formulate the total production cost equation.

15A + 40B = 76,000

15 units of Model Alpha and 40 units of Model Beta sum to a total cost of $76,000.

3

Substitute

A

into the total cost equation and solve for

B

.

15(2B + 400) + 40B = 76,000 \implies 70B + 6,000 = 76,000 \implies B = 1,000

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

4

Calculate the value of

A

.

A = 2(1,000) + 400 = 2,400

Substitute

B = 1,000

back into the equation for

A

.

Key Concept

Solving Systems of Linear Equations by Substitution

Question 35Question →

An express train travels a distance of $180$ miles from Station A to Station B at a constant speed of $v$ miles per hour. On the return trip from Station B to Station A, the train travels along the same route at a constant speed that is $15$ miles per hour slower. If the return trip takes $1$ hour longer than the outbound trip, what is the value of $v$ ?

45

50

60

75

90

Show answer & explanation

Answer: $60$

Answer

60

The correct answer is

60

. The outbound time is

\frac{180}{v}

hours and the return time is

\frac{180}{v - 15}

hours. Setting their difference equal to

1

gives

\frac{180}{v - 15} - \frac{180}{v} = 1

. Solving the resulting quadratic equation

v^2 - 15v - 2700 = 0

yields

v = 60

miles per hour for the positive root.

Step-by-Step Solution

1

Express the time taken for each leg of the journey in terms of

v

.

Outbound time

t_1 = \frac{180}{v}

hours, return time

t_2 = \frac{180}{v - 15}

hours.

Time is equal to distance divided by speed.

2

Set up the equation based on the given time difference of

1

hour.

\frac{180}{v - 15} - \frac{180}{v} = 1

The return trip takes

1

hour longer than the outbound trip.

3

Clear the denominators by multiplying the equation by

v(v - 15)

.

180v - 180(v - 15) = v(v - 15) \implies 2700 = v^2 - 15v

Simplifying rational expressions into standard quadratic form.

4

Rearrange into

v^2 - 15v - 2700 = 0

and factor to solve for

v

.

(v - 60)(v + 45) = 0 \implies v = 60 \text{ or } v = -45

Factoring the quadratic equation.

5

Select the valid positive speed value.

Since speed must be positive,

v = 60

miles per hour.

Negative values for speed are not physically valid in this context.

Key Concept

Algebraic Word Problems and Equation Modeling

Question 36Question →

A logistics company ships three types of packages: Small, Medium, and Large. A shipment containing $3$ Small, $5$ Medium, and $2$ Large packages has a total weight of $170$ kilograms. A second shipment containing $1$ Small, $2$ Medium, and $1$ Large package has a total weight of $65$ kilograms. What is the total weight, in kilograms, of a shipment containing $5$ Small, $9$ Medium, and $4$ Large packages?

235

275

300

335

405

Show answer & explanation

Answer: $300$

Answer

The total weight of the shipment is

300

kilograms.

The target quantity

5S + 9M + 4L

can be expressed directly as

1 \times (3S + 5M + 2L) + 2 \times (1S + 2M + 1L)

. Substituting the given total weights yields

1(170) + 2(65) = 300

kilograms.

Step-by-Step Solution

1

Formulate linear equations representing the package weights.

Let

S

,

M

, and

L

represent the weight of a Small, Medium, and Large package respectively. Equation 1:

3S + 5M + 2L = 170

. Equation 2:

1S + 2M + 1L = 65

.

Setting up algebraic representations for the system based on the problem statement.

2

Determine if the requested expression

5S + 9M + 4L

can be formed as a linear combination

c_1(3S + 5M + 2L) + c_2(1S + 2M + 1L)

.

Match coefficients:

3c_1 + c_2 = 5

,

5c_1 + 2c_2 = 9

, and

2c_1 + c_2 = 4

.

Since individual values of

S

,

M

, and

L

cannot be uniquely determined from two equations with three variables, we seek scalar constants

c_1

and

c_2

.

3

Solve for the multipliers

c_1

and

c_2

.

Subtracting

2c_1 + c_2 = 4

from

3c_1 + c_2 = 5

yields

c_1 = 1

. Substituting

c_1 = 1

into

2c_1 + c_2 = 4

gives

c_2 = 2

. Checking the middle equation:

5(1) + 2(2) = 9

, which holds true.

Finding the scalar weights that recreate the exact combination requested.

4

Compute the total weight using the linear combination of the known values.

Total Weight

= 1 \times 170 + 2 \times 65 = 170 + 130 = 300

kilograms.

Applying the solved linear combination to the total weights of the shipments.

Key Concept

Solving for a linear combination of variables in an underdetermined system without finding individual variable values.

Question 37Question →

If $x$ is a real number that satisfies the equation $|x + 7| = 2x + 3$ , what is the sum of all valid real solutions for $x$ ?

\frac{2}{3}

4

-\frac{10}{3}

-\frac{2}{3}

\frac{22}{3}

Show answer & explanation

Answer: $4$

Answer

The sum of all valid real solutions is 4.

To solve

|x + 7| = 2x + 3

, we break the absolute value into two linear cases:

x + 7 = 2x + 3

giving

x = 4

, and

x + 7 = -(2x + 3)

giving

x = -\frac{10}{3}

. We must test both candidate solutions in the original equation because absolute value outputs must be non-negative. Testing

x = 4

gives

|11| = 11

, which is true. Testing

x = -\frac{10}{3}

gives

|\frac{11}{3}| = -\frac{11}{3}

, which is false because an absolute value cannot equal a negative number. Thus,

x = 4

is the only valid solution, making the sum of all valid solutions equal to

4

.

Step-by-Step Solution

1

Set up the two linear cases for the absolute value equation

|x + 7| = 2x + 3

.

Case 1:

x + 7 = 2x + 3

; Case 2:

x + 7 = -(2x + 3)

.

By definition,

|A| = B

implies

A = B

or

A = -B

, provided

B \ge 0

.

2

Solve Case 1 for

x

.

x + 7 = 2x + 3 \implies 2x - x = 7 - 3 \implies x = 4

.

Isolating

x

gives the first candidate solution.

3

Solve Case 2 for

x

.

x + 7 = -2x - 3 \implies 3x = -10 \implies x = -\frac{10}{3}

.

Expanding the negative sign and isolating

x

gives the second candidate solution.

4

Check candidate solutions in the original equation

|x + 7| = 2x + 3

to eliminate extraneous solutions.

For

x = 4

:

|4 + 7| = 11

and

2(4) + 3 = 11

(Valid).
For

x = -\frac{10}{3}

:

|-\frac{10}{3} + 7| = \frac{11}{3}

, but

2(-\frac{10}{3}) + 3 = -\frac{11}{3}

(Extraneous, since

\frac{11}{3} \neq -\frac{11}{3}

).

An absolute value expression cannot equal a negative number, so candidate values resulting in a negative right side are invalid.

5

Calculate the sum of all valid real solutions.

Sum =

4

.

Since

x = 4

is the only valid solution, the sum is simply

4

.

Key Concept

Solving Absolute Value Linear Equations and Validating against Extraneous Solutions

Estimated Time:2m 0s

Question 38Question →

A hardware manufacturer produces two models of enterprise servers: Server X and Server Y. Each Server X requires $3$ processors of Type A and $2$ processors of Type B. Each Server Y requires $2$ processors of Type A and $5$ processors of Type B. In a given production run, the manufacturer assembled a total of $40$ servers using exactly $115$ processors of Type A. What is the total number of Type B processors used in this production run?

Show answer & explanation

Answer: 95

Answer

The total number of Type B processors used in the production run is 95.

By representing the count of Server X models as

x

and Server Y models as

y

, the given conditions yield two equations:

x + y = 40

and

3x + 2y = 115

. Solving this system gives

x = 35

and

y = 5

. Substituting these values into the expression for Type B processors (

2x + 5y

) produces

2(35) + 5(5) = 70 + 25 = 95

.

Step-by-Step Solution

1

Define variables and establish equations for server totals and Type A processor usage.

x + y = 40

and

3x + 2y = 115

, where

x

is Server X count and

y

is Server Y count.

Word problems require translating relationships into algebraic linear equations.

2

Solve the system of equations for

x

and

y

using substitution.

x = 35

Server X models and

y = 5

Server Y models.

Finding the specific quantities of each server model is required to compute total Type B processors.

3

Substitute

x = 35

and

y = 5

into the expression for Type B processors (

2x + 5y

).

2(35) + 5(5) = 70 + 25 = 95

.

Determines the target unknown value requested in the problem stem.

Key Concept

Modeling word problems using systems of linear equations in two variables.

Estimated Time:2m 0s

Question 39Question →

If $x$ is a real number satisfying the equation $|2x - 3| = 5x - 12$ , what is the sum of all valid real solutions for $x$ ?

3

\frac{15}{7}

\frac{36}{7}

-3

\frac{45}{7}

Show answer & explanation

Answer: $3$

Answer

The sum of all valid real solutions is

3

.

The option specifying

3

correctly identifies

x = 3

as the only valid solution after discarding the extraneous root

x = \frac{15}{7}

, which produces a negative right-hand side in the original equation.

Step-by-Step Solution

1

Set up the two algebraic cases for the absolute value equation

|2x - 3| = 5x - 12

.

Case 1:

2x - 3 = 5x - 12

Case 2:

2x - 3 = -(5x - 12) = -5x + 12

By definition,

|a| = b

implies

a = b

or

a = -b

, provided

b \ge 0

.

2

Solve Case 1 for

x

.

2x - 3 = 5x - 12 \implies 3x = 9 \implies x = 3

Isolate the variable

x

algebraically.

3

Solve Case 2 for

x

.

2x - 3 = -5x + 12 \implies 7x = 15 \implies x = \frac{15}{7}

Isolate the variable

x

algebraically.

4

Check candidate solutions for extraneous roots in the original equation.

For

x = 3

:

|2(3) - 3| = |3| = 3

and

5(3) - 12 = 3

. Valid.
For

x = \frac{15}{7}

:

|2(\frac{15}{7}) - 3| = |\frac{9}{7}| = \frac{9}{7}

, but

5(\frac{15}{7}) - 12 = -\frac{9}{7}

. Since

\frac{9}{7} \neq -\frac{9}{7}

,

x = \frac{15}{7}

is extraneous.

The output of an absolute value expression cannot be negative, so any candidate root making the right side negative must be discarded.

5

Calculate the sum of all valid real solutions.

Sum =

3

There is only one valid solution,

x = 3

.

Key Concept

Solving absolute value equations with a variable expression on the right-hand side requires checking candidate solutions to eliminate extraneous roots.

Question 40Question →

Sequence $A$ is an arithmetic sequence with first term $a_1 = 5$ and common difference $d = 3$ .
Sequence $B$ is a geometric sequence with first term $b_1 = 2$ and common ratio $r = \sqrt{3}$ .

Arrange the four quantities defined below in ascending order (from smallest to largest value).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct ascending order is Quantity K, followed by Quantity N, Quantity M, and finally Quantity L.

Evaluating each expression gives Quantity K = 47, Quantity N ≈ 142.07, Quantity M = 162, and Quantity L = 185. Comparing these values from smallest to largest yields the order K, N, M, L.

Step-by-Step Solution

1

Calculate Quantity K (

a_{15}

for Sequence A)

a_{15} = 5 + (15 - 1) \times 3 = 47

The

n

-th term of an arithmetic sequence is given by

a_n = a_1 + (n - 1)d

.

2

Calculate Quantity L (

S_{10}

for Sequence A)

S_{10} = \frac{10}{2} \times [2(5) + (10 - 1) \times 3] = 5 \times (10 + 27) = 185

The sum of the first

n

terms of an arithmetic sequence is given by

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

.

3

Calculate Quantity M (

b_9

for Sequence B)

b_9 = 2 \times (\sqrt{3})^{9 - 1} = 2 \times (\sqrt{3})^8 = 2 \times 3^4 = 162

The

n

-th term of a geometric sequence is given by

b_n = b_1 r^{n-1}

.

4

Calculate and approximate Quantity N (

S_6

for Sequence B)

S_6 = \frac{2((\sqrt{3})^6 - 1)}{\sqrt{3} - 1} = \frac{2(27 - 1)}{\sqrt{3} - 1} = \frac{52}{\sqrt{3} - 1} = 52(\sqrt{3} + 1) \approx 52 \times 2.732 = 142.07

The sum of a geometric series is

S_n = \frac{b_1(r^n - 1)}{r - 1}

. Rationalizing the denominator yields

52(\sqrt{3} + 1)

.

5

Compare all four values to establish ascending order

47 < 142.07 < 162 < 185

, which corresponds to

K < N < M < L

Ordering the numerical outputs from least to greatest gives the required sequence.

Key Concept

Calculating specific terms and sums of arithmetic and geometric sequences using explicit formulas and ordering calculated quantities.

Algebra and Functions

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution