Linear Equations in One and Two Variables Practice Questions — GMAT — Page 2

Question 21Question →

A corporate advisory firm executes two types of client projects: Type A and Type B. Each Type A project requires $x$ hours of financial modeling and $y$ hours of executive reporting. Each Type B project requires $x + 3$ hours of financial modeling and $2y - 2$ hours of executive reporting. During the first quarter, the firm completed 6 Type A projects and 4 Type B projects. If the total time spent on financial modeling across all 10 projects was 72 hours and the total time spent on executive reporting was 62 hours, what is the value of $3x + 4y$ ?

Show answer & explanation

Answer: 38

Answer

The value of

3x + 4y

is 38.

Setting up total financial modeling hours gives

6x + 4(x + 3) = 72

, which yields

x = 6

. Setting up total executive reporting hours gives

6y + 4(2y - 2) = 62

, which yields

y = 5

. Substituting these values into

3x + 4y

yields

3(6) + 4(5) = 38

.

Step-by-Step Solution

1

Formulate and solve the linear equation for financial modeling hours.

x = 6

6 Type A projects take

6x

hours and 4 Type B projects take

4(x + 3)

hours. Summing them yields

6x + 4x + 12 = 72

, which simplifies to

10x = 60

, giving

x = 6

.

2

Formulate and solve the linear equation for executive reporting hours.

y = 5

6 Type A projects take

6y

hours and 4 Type B projects take

4(2y - 2)

hours. Summing them yields

6y + 8y - 8 = 62

, which simplifies to

14y = 70

, giving

y = 5

.

3

Substitute

x = 6

and

y = 5

into the targeted expression

3x + 4y

.

38

3(6) + 4(5) = 18 + 20 = 38

.

Key Concept

Linear Equations in One and Two Variables

Question 22Question →

What is the sum of all valid real solutions to the equation $|2x + 15| = 1 - 5x$ ?

-2

\frac{10}{3}

\frac{16}{3}

-\frac{30}{7}

2

Show answer & explanation

Answer: $-2$

Answer

The sum of all valid real solutions is

-2

.

Solving the absolute value equation

|2x + 15| = 1 - 5x

yields two candidate values:

x = -2

and

x = \frac{16}{3}

. Substituting

x = -2

into the right-hand side gives

1 - 5(-2) = 11

, which is non-negative and matches

|2(-2) + 15| = 11

. Substituting

x = \frac{16}{3}

gives

1 - 5\left(\frac{16}{3}\right) = -\frac{77}{3}

, which is negative and therefore invalid. Thus,

x = -2

is the unique valid solution, making the sum

-2

.

Step-by-Step Solution

1

Set up the two cases for the absolute value equation

|2x + 15| = 1 - 5x

.

Case 1:

2x + 15 = 1 - 5x

; Case 2:

2x + 15 = -(1 - 5x)

.

By definition,

|u| = v

implies

u = v

or

u = -v

, provided

v \geq 0

.

2

Solve Case 1 for

x

.

7x = -14 \implies x = -2

.

Adding

5x

and subtracting

15

from both sides isolates

x

.

3

Solve Case 2 for

x

.

2x + 15 = -1 + 5x \implies 3x = 16 \implies x = \frac{16}{3}

.

Distributing the negative sign and combining like terms yields

x = \frac{16}{3}

.

4

Check candidate solutions against the non-negativity constraint

1 - 5x \geq 0

.

For

x = -2

:

1 - 5(-2) = 11 \geq 0

(Valid). For

x = \frac{16}{3}

:

1 - 5\left(\frac{16}{3}\right) = -\frac{77}{3} < 0

(Extraneous).

An absolute value cannot equal a negative number, so candidate solutions that make the right-hand side negative must be discarded.

5

Sum all valid real solutions.

The only valid solution is

x = -2

, so the sum is

-2

.

Extraneous solutions are excluded from the final sum.

Key Concept

Solving absolute value linear equations requires checking candidate solutions against domain constraints to filter out extraneous roots.

Estimated Time:2m 0s

Question 23Question →

A retail store sets the selling price, $P$ , of a custom item based on its wholesale cost, $C$ , according to the linear equation $P = 1.4C + 20$ . The store's profit on each item is defined as $P - C$ . If the profit on a certain item is also equal to $0.2P + 40$ , what is the wholesale cost, $C$ , of the item in dollars?

Show answer & explanation

Answer: 200

Answer

The wholesale cost

C

of the item is $200.

By writing profit both as

P - C = 0.4C + 20

and as

0.2P + 40 = 0.28C + 44

, we obtain a single linear equation in terms of

C

:

0.4C + 20 = 0.28C + 44

. Subtracting

0.28C

and

20

from both sides gives

0.12C = 24

, which simplifies to

C = 200

.

Step-by-Step Solution

1

Substitute

P = 1.4C + 20

into the standard profit expression

P - C

.

Profit =

0.4C + 20

This expresses the profit solely as a linear function of the wholesale cost

C

.

2

Substitute

P = 1.4C + 20

into the alternative profit expression

0.2P + 40

.

Profit =

0.28C + 44

This converts the given percentage-based profit condition into an expression dependent only on

C

.

3

Set the two profit expressions equal to each other and solve the resulting single-variable linear equation.

0.12C = 24 \implies C = 200

Equating two valid expressions for the same quantity allows solving for the unknown variable

C

.

Key Concept

Solving Systems of Linear Equations by Algebraic Substitution

Estimated Time:2m 0s

Question 24Question →

If $x$ is a real number that satisfies the equation $|5 - 2x| = 3x - 10$ , what is the value of $|x - 4| + 2x$ ?

5

7

11

14

18

Show answer & explanation

Answer: 11

Answer

11

Solving the equation

|5 - 2x| = 3x - 10

yields two potential values:

x = 3

and

x = 5

. Substituting

x = 3

into the original equation yields

|-1| = -1

, which is impossible because an absolute value cannot be negative. Therefore,

x = 3

is an extraneous root. Substituting

x = 5

yields

|-5| = 5

, which is true. Evaluating the requested expression

|x - 4| + 2x

at

x = 5

gives

|5 - 4| + 2(5) = 1 + 10 = 11

.

Step-by-Step Solution

1

Set up the two algebraic cases for the absolute value equation

|5 - 2x| = 3x - 10

.

Case 1:

5 - 2x = 3x - 10

when

5 - 2x \ge 0

(

x \le 2.5

).
Case 2:

-(5 - 2x) = 3x - 10

when

5 - 2x < 0

(

x > 2.5

).

An absolute value expression

|u|

equals

u

when

u \ge 0

and

-u

when

u < 0

.

2

Solve Case 1 algebraically.

5 - 2x = 3x - 10 \implies 15 = 5x \implies x = 3

.

Isolate the variable

x

on one side of the equation.

3

Solve Case 2 algebraically.

2x - 5 = 3x - 10 \implies 5 = x \implies x = 5$.

Simplify and isolate

x

.

4

Check both candidate solutions in the original equation to filter out extraneous roots.

For

x = 3

:

|5 - 2(3)| = 3(3) - 10 \implies |-1| = -1 \implies 1 = -1

(False, extraneous).
For

x = 5

:

|5 - 2(5)| = 3(5) - 10 \implies |-5| = 5 \implies 5 = 5

(True, valid).

Since the right-hand side

3x - 10

must be non-negative for the absolute value to hold, candidate solutions must be tested in the original equation.

5

Substitute the valid solution

x = 5

into the targeted expression

|x - 4| + 2x

.

|5 - 4| + 2(5) = |1| + 10 = 1 + 10 = 11

.

Evaluate the target expression using the single real value

x = 5

that satisfies the given linear absolute value equation.

Key Concept

Solving absolute value linear equations and identifying extraneous solutions

Estimated Time:2m 0s

Question 25Question →

An artisan workshop produces custom wooden chairs and tables. Each chair requires $3$ hours of carving and $2$ hours of finishing, while each table requires $5$ hours of carving and $4$ hours of finishing. If the workshop logged a total of $110$ hours of carving and $84$ hours of finishing last week, how many tables were produced?

10

14

16

18

22

Show answer & explanation

Answer: 16

Answer

The workshop produced 16 tables.

By defining

c

as the number of chairs and

t

as the number of tables, we can set up two linear equations representing total hours:

3c + 5t = 110

for carving and

2c + 4t = 84

for finishing. Multiplying the second equation by

1.5

yields

3c + 6t = 126

. Subtracting

3c + 5t = 110

from

3c + 6t = 126

leaves

t = 16

. Thus, the workshop produced 16 tables.

Step-by-Step Solution

1

Define variables and translate the word problem into a system of linear equations.

Let

c

be the number of chairs and

t

be the number of tables. Carving equation:

3c + 5t = 110

. Finishing equation:

2c + 4t = 84

.

The total hours for each activity equal the sum of hours spent on chairs and tables.

2

Eliminate variable

c

to solve for

t

.

Multiply the finishing equation by

1.5

to get

3c + 6t = 126

. Subtract the carving equation (

3c + 5t = 110

) from this equation:

(3c + 6t) - (3c + 5t) = 126 - 110 \implies t = 16

.

Aligning the coefficient of

c

in both equations allows direct elimination of

c

to isolate

t

.

3

Verify the solution by calculating

c

and checking both original equations.

Substitute

t = 16

into

2c + 4(16) = 84 \implies 2c + 64 = 84 \implies 2c = 20 \implies c = 10

. Check carving:

3(10) + 5(16) = 30 + 80 = 110

.

Ensures that

t = 16

and

c = 10

satisfy both resource constraints without calculation errors.

Key Concept

Setting up and solving a system of two linear equations in two variables

Hints

1

Set up two separate linear equations: one for total carving hours and one for total finishing hours.

2

Let

c

be the number of chairs and

t

be the number of tables. Your system is

3c + 5t = 110

and

2c + 4t = 84

.

3

Multiply

2c + 4t = 84

by

1.5

to get

3c + 6t = 126

, then subtract

3c + 5t = 110

to find

t

directly.

Practice More

Try solving a similar problem where the total revenue and total unit count are given to practice standard linear system modeling.

Alternative Method

Divide the finishing equation

2c + 4t = 84

by

2

to get

c + 2t = 42 \implies c = 42 - 2t

. Substitute this into the carving equation:

3(42 - 2t) + 5t = 110 \implies 126 - 6t + 5t = 110 \implies -t = -16 \implies t = 16

.

Estimated Time:1m 30s

Question 26Question →

A beverage producer creates two liquid mixtures, Mixture $A$ and Mixture $B$ . Mixture $A$ consists of $30\%$ fruit concentrate by volume, and Mixture $B$ consists of $70\%$ fruit concentrate by volume. A lab technician combines $x$ liters of Mixture $A$ with $y$ liters of Mixture $B$ to prepare an $80$ -liter batch that contains $45\%$ fruit concentrate by volume. What is the value of $x$ ?

Show answer & explanation

Answer: 50

Answer

The value of

x

is 50.

To find

x

, we construct two linear equations based on total liquid volume and total fruit concentrate. The total volume equation is

x + y = 80

, which gives

y = 80 - x

. The concentrate equation is

0.30x + 0.70y = 0.45(80) = 36

. Substituting

y = 80 - x

yields

0.30x + 0.70(80 - x) = 36

. Expanding gives

0.30x + 56 - 0.70x = 36

, so

-0.40x = -20

, which results in

x = 50

.

Step-by-Step Solution

1

Formulate a system of two linear equations representing total volume and total concentrate volume.

System equations:

x + y = 80

and

0.30x + 0.70y = 36

.

The sum of the component volumes equals the total mixture volume, and the sum of the pure concentrate from each component equals the total concentrate in the final mixture.

2

Substitute

y = 80 - x

into the concentrate equation to eliminate

y

.

0.30x + 0.70(80 - x) = 36

.

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

3

Simplify the single-variable linear equation and solve for

x

.

0.30x + 56 - 0.70x = 36 \implies -0.40x = -20 \implies x = 50

.

Combining like terms isolates the variable

x

.

Key Concept

Solving systems of two linear equations formed by weighted mixture word problems.

Question 27Question →

A web hosting company charges each enterprise client a one-time fixed setup fee plus a constant monthly maintenance fee per server. A client operating $4$ servers pays a total of $\$1,100$ for the setup fee and the first $6$ months of server maintenance. A client operating $9$ servers pays a total of $\$1,850$ for the setup fee and the first $6$ months of server maintenance. What is the one-time fixed setup fee, in dollars?

Show answer & explanation

Answer: 500

Answer

500

The fixed setup fee is

\$500

. Letting

S

represent the fixed setup fee and

M

represent the 6-month maintenance fee per server, the two given scenarios yield

S + 4M = 1100

and

S + 9M = 1850

. Subtracting the first equation from the second gives

5M = 750

, which simplifies to

M = 150

. Substituting

M = 150

into

S + 4(150) = 1100

gives

S + 600 = 1100

, so

S = 500

.

Step-by-Step Solution

1

Define variables for the unknown fixed cost and per-server cost, and construct the system of linear equations.

Let

S

be the fixed setup fee in dollars and

M

be the 6-month maintenance fee per server in dollars. The equations are

S + 4M = 1100

and

S + 9M = 1850

.

Modeling the situational relationships as a linear system allows isolated solution of each unknown.

2

Subtract the two linear equations to eliminate the fixed fee

S

and solve for

M

.

5M = 750 \implies M = 150

.

Since the coefficient of

S

is 1 in both equations, elimination by subtraction directly isolates

M

.

3

Substitute the value of

M

back into the first equation to solve for

S

.

S + 4(150) = 1100 \implies S + 600 = 1100 \implies S = 500

.

Replacing

M

with 150 yields a linear equation in one variable for the setup fee.

Key Concept

Linear Equations in One and Two Variables

Estimated Time:1m 30s

Question 28Question →

If

x

is a real number that satisfies the absolute value equation

|2x - 7| = 3x - 11

what is the value of

x^2 - 3x

?

2.16

4

16

28

270

Show answer & explanation

Answer: $4$

Answer

The value of

x^2 - 3x

is

4

.

Solving the absolute value equation

|2x - 7| = 3x - 11

yields two potential roots:

x = 4

and

x = 3.6

. Checking both in the original equation shows that

x = 3.6

makes the right-hand side negative (

-0.2

), making it an extraneous solution. The only valid solution is

x = 4

. Substituting

x = 4

into

x^2 - 3x

yields

4^2 - 3(4) = 16 - 12 = 4

.

Step-by-Step Solution

1

Set up equations based on the definition of absolute value

Case 1:

2x - 7 = 3x - 11

; Case 2:

2x - 7 = -(3x - 11)

An absolute value equation

|A| = B

splits into

A = B

or

A = -B

, with the requirement that

B \ge 0

.

2

Solve Case 1:

2x - 7 = 3x - 11

3x - 2x = 11 - 7 \implies x = 4

Isolate the variable

x

by algebraic rearrangement.

3

Solve Case 2:

2x - 7 = -3x + 11

5x = 18 \implies x = \frac{18}{5} = 3.6

Isolate

x

for the negative case.

4

Check both potential solutions in the original equation

|2x - 7| = 3x - 11

For

x = 4

:

|2(4) - 7| = |1| = 1

and

3(4) - 11 = 1

(Valid).
For

x = 3.6

:

|2(3.6) - 7| = |0.2| = 0.2

but

3(3.6) - 11 = -0.2

(Extraneous).

The right side

3x - 11

must be non-negative. Since

3(3.6) - 11 = -0.2 < 0

,

x = 3.6

is extraneous.

5

Evaluate the target expression

x^2 - 3x

using the valid root

x = 4

4^2 - 3(4) = 16 - 12 = 4

Substitute the single valid root into the requested expression.

Key Concept

Absolute Value Linear Equations and Extraneous Solution Checking

Question 29Question →

A software consultant charges an enterprise client a fixed setup fee of $200$ dollars plus an hourly rate of $r$ dollars for 1-on-1 technical training. For team training workshops, the consultant charges a fixed setup fee of $400$ dollars plus an hourly rate that is $50\%$ higher than the 1-on-1 rate. If the client was billed a total of $2,400$ dollars for $12$ hours of 1-on-1 training and $8$ hours of team training workshops, what is the value of $r$ ?

Show answer & explanation

Answer: 75

Answer

The individual hourly rate

r

is

75

dollars per hour.

The total cost consists of fixed setup fees (

200 + 400 = 600

) and variable hourly charges (

12r + 8 \times 1.5r = 24r

). Setting the total cost

600 + 24r

equal to

2,400

and solving for

r

yields

r = 75

.

Step-by-Step Solution

1

Formulate the total cost equation for the 1-on-1 technical training.

\text{Cost}_{\text{1-on-1}} = 200 + 12r

The client pays a

200

dollar setup fee plus

r

dollars for each of the

12

hours.

2

Determine the hourly rate and total cost equation for the team training workshops.

\text{Hourly Rate}_{\text{team}} = 1.5r

, so

\text{Cost}_{\text{team}} = 400 + 8(1.5r) = 400 + 12r

The team hourly rate is

50\%

greater than

r

, which equals

r + 0.5r = 1.5r

. Multiplying by

8

hours yields

12r

in hourly fees, plus the

400

dollar setup fee.

3

Combine the total costs to create a single linear equation in terms of

r

.

(200 + 12r) + (400 + 12r) = 2400 \implies 600 + 24r = 2400

The total amount billed across both services is given as

2,400

dollars.

4

Isolate the variable

r

and calculate its numerical value.

24r = 1800 \implies r = 75

Subtracting

600

from both sides gives

24r = 1800

, and dividing by

24

yields

r = 75

.

Key Concept

Linear Equations in One and Two Variables

Estimated Time:1m 30s

Linear Equations in One and Two Variables

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Key Concept

Common Mistakes

Hints

Practice More

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes

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Key Concept

Common Mistakes