Linear Equations in One Variable Practice Questions — GRE General Test — Page 2

For real constants

a

and

b

, consider the linear equation in one variable

x

:

\frac{a(x - 2)}{3} - \frac{2x + 1}{4} = \frac{(a - 3)x + b}{12}

If this equation has infinitely many solutions for

x

, which of the following statements must be true? Select all such statements.

Select all that apply

a + b = -10

ab < 0

2a - b = 13

a - b = -10

ab = 11

Show answer & explanation

Answer: $a + b = -10$ ; $ab < 0$ ; $2a - b = 13$

Answer

The statements

a + b = -10

,

ab < 0

, and

2a - b = 13

are all true.

Clearing denominators gives

4a(x - 2) - 3(2x + 1) = (a - 3)x + b

. Expanding both sides yields

(4a - 6)x - (8a + 3) = (a - 3)x + b

, which rearranges to

(3a - 3)x = 8a + b + 3

. For a linear equation in one variable to have infinitely many solutions, both the coefficient of

x

and the constant term must be zero (

0x = 0

). Setting

3a - 3 = 0

gives

a = 1

, and substituting

a = 1

into

8a + b + 3 = 0

yields

b = -11

. Evaluating the statements with

a = 1

and

b = -11

shows that

a + b = 1 + (-11) = -10

is true,

ab = -11 < 0

is true, and

2a - b = 2(1) - (-11) = 13

is true.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the equation by 12.

4a(x - 2) - 3(2x + 1) = (a - 3)x + b

Eliminating fractions simplifies expanding terms and combining like variables.

2

Expand all terms and group terms containing

x

on the left-hand side.

4ax - 8a - 6x - 3 = (a - 3)x + b \implies (4a - 6)x - (8a + 3) = (a - 3)x + b

Preparing the linear equation to be expressed in standard form

Ax = B

.

3

Rearrange into standard form

(A)x = B

.

[(4a - 6) - (a - 3)]x = 8a + b + 3 \implies (3a - 3)x = 8a + b + 3

A linear equation has infinitely many solutions if and only if

A = 0

and

B = 0

simultaneously.

4

Set the coefficient of

x

and the constant term equal to zero to determine

a

and

b

.

3a - 3 = 0 \implies a = 1

; then

8(1) + b + 3 = 0 \implies b + 11 = 0 \implies b = -11

The equation reduces to

0 \cdot x = 0

, which is satisfied by every real number

x

.

5

Evaluate the given statements using

a = 1

and

b = -11

.

a + b = 1 + (-11) = -10

(True);

ab = (1)(-11) = -11 < 0

(True);

2a - b = 2(1) - (-11) = 13

(True).

Direct substitution confirms which statements hold true.

Key Concept

Conditions for a linear equation in one variable to have infinitely many solutions

Estimated Time:2m 30s

Question 22Question →

Consider the linear equation in

x

shown below, where

k

is a real constant:

\frac{x - k}{2} - \frac{2x + 1}{3} = \frac{k(x + 1)}{6} - 1

Which of the following statements must be true? Select all such statements.

Select all that apply

If

k = 1

, then

x = 0

is the unique solution to the equation.

If

k = -1

, the equation has infinitely many solutions.

If

k = 0

, the solution to the equation is

x = 4

.

If

k = 3

, the solution to the equation is

x = 2

.

If

k = 2

, the solution to the equation is

x = -\frac{4}{3}

.

Show answer & explanation

Answer: If $k = 1$ , then $x = 0$ is the unique solution to the equation.; If $k = 0$ , the solution to the equation is $x = 4$ .; If $k = 2$ , the solution to the equation is $x = -\frac{4}{3}$ .

Answer

The correct statements are: if

k = 1

, then

x = 0

is the unique solution; if

k = 0

, the solution is

x = 4

; and if

k = 2

, the solution is

x = -\frac{4}{3}

.

Multiplying the equation by

6

yields

3(x - k) - 2(2x + 1) = k(x + 1) - 6

. Expanding and collecting terms in

x

leads to

(k + 1)x = 4(1 - k)

. Testing the proposed values of

k

shows that when

k = 1

,

2x = 0

gives

x = 0

; when

k = 0

,

x = 4

; and when

k = 2

,

3x = -4

gives

x = -\frac{4}{3}

. Therefore, the statements corresponding to

k = 1

,

k = 0

, and

k = 2

are all correct.

Step-by-Step Solution

1

Clear fractions by multiplying the entire equation by the least common denominator,

6

.

3(x - k) - 2(2x + 1) = k(x + 1) - 6

Eliminating fractions simplifies algebraic expansion and grouping of like terms.

2

Expand all expressions and combine like terms on both sides.

3x - 3k - 4x - 2 = kx + k - 6 \implies -x - 3k - 2 = kx + k - 6

Distribute terms carefully to prevent sign errors.

3

Isolate terms containing

x

on one side and parameter/constant terms on the other side.

-x - kx = k + 3k - 6 + 2 \implies -x(k + 1) = 4k - 4 \implies (k + 1)x = 4(1 - k)

Factoring out

x

provides the general solution form

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

for

k \neq -1

.

4

Evaluate each specified value of

k

against

(k + 1)x = 4(1 - k)

.

For

k = 1

:

2x = 0 \implies x = 0

.
For

k = -1

:

0x = 8 \implies

no solution.
For

k = 0

:

1x = 4 \implies x = 4

.
For

k = 3

:

4x = -8 \implies x = -2

.
For

k = 2

:

3x = -4 \implies x = -\frac{4}{3}

.

Direct substitution verifies which given statements are true.

Key Concept

Solving linear equations with parameters and analyzing existence and uniqueness of solutions.

Estimated Time:2m 30s

Question 23Question →

If

x

satisfies the linear equation

\frac{2(3x - 1)}{5} - \frac{4x - 3}{3} = \frac{x + 18}{15} - 2

what is the value of

3x - 5

?

7

16

-14

-12

40

Show answer & explanation

Answer: $16$

Answer

16

Multiplying the entire equation by the common denominator

15

clears all fractions, yielding

6(3x - 1) - 5(4x - 3) = (x + 18) - 30

. Expanding both sides produces

18x - 6 - 20x + 15 = x - 12

, which simplifies to

-2x + 9 = x - 12

. Rearranging terms yields

-3x = -21

, so

x = 7

. Substituting

x = 7

into

3x - 5

gives

3(7) - 5 = 16

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the equation by the least common multiple (LCM) of 5, 3, and 15, which is 15.

15 \cdot \left(\frac{2(3x - 1)}{5}\right) - 15 \cdot \left(\frac{4x - 3}{3}\right) = 15 \cdot \left(\frac{x + 18}{15}\right) - 15 \cdot 2

Eliminating fractions simplifies the linear equation into standard integer polynomial terms.

2

Simplify the products and distribute coefficients across parentheses.

3 \cdot 2(3x - 1) - 5(4x - 3) = (x + 18) - 30 \implies 6(3x - 1) - 5(4x - 3) = x - 12

Perform fractional reduction and simplify constants on the right side.

3

Expand both groupings and combine like terms on the left-hand side.

18x - 6 - 20x + 15 = x - 12 \implies -2x + 9 = x - 12

Ensure the negative sign is properly distributed to both terms inside

-5(4x - 3)

.

4

Isolate the variable

x

by subtracting

x

and

9

from both sides.

-3x = -21 \implies x = 7

Solve for the single variable

x

.

5

Substitute

x = 7

into the target expression

3x - 5

.

3(7) - 5 = 21 - 5 = 16

The question asks for the value of the algebraic expression

3x - 5

, not

x

alone.

Key Concept

Linear Equations in One Variable with Fractional Coefficients

Estimated Time:2m 0s

Question 24Question →

Two water pumps, Pump $A$ and Pump $B$ , operate at constant rates to fill a $910$ -gallon storage tank. The rate at which Pump $A$ fills the tank is $15$ gallons per hour more than twice the rate of Pump $B$ . Both pumps begin working together to fill the empty tank. After $4$ hours, Pump $A$ shuts down, and Pump $B$ continues working alone for another $5$ hours to finish filling the tank completely. What is the pumping rate of Pump $A$ , in gallons per hour?

50

85

100

115

130

Show answer & explanation

Answer: 115

Answer

115 gallons per hour

Letting

r

represent the rate of Pump B in gallons per hour, the rate of Pump A is

2r + 15

. During the first 4 hours, both pumps operate together at a combined rate of

(3r + 15)

gallons per hour, contributing

4(3r + 15) = 12r + 60

gallons. In the remaining 5 hours, Pump B alone contributes

5r

gallons. Setting the total amount equal to the tank capacity gives

12r + 60 + 5r = 910

, which simplifies to

17r = 850

, so

r = 50

. Substituting

r = 50

into

2r + 15

yields

115

gallons per hour.

Step-by-Step Solution

1

Define the variable for the unknown rate.

Let

r

be the pumping rate of Pump

B

in gallons per hour. Then the rate of Pump

A

is

2r + 15

gallons per hour.

Expressing both rates in terms of a single variable

r

allows us to formulate a linear equation in one variable.

2

Determine the combined pumping rate and the work done during the first phase.

Combined rate

= r + (2r + 15) = 3r + 15

gallons per hour. Work completed in 4 hours

= 4(3r + 15) = 12r + 60

gallons.

When both pumps work together, their individual rates add together.

3

Formulate the total work equation for filling the 910-gallon tank.

(12r + 60) + 5r = 910 \implies 17r + 60 = 910

The sum of the work done in the first 4 hours by both pumps and the work done in the next 5 hours by Pump B alone equals the total capacity.

4

Solve the linear equation for

r

.

17r = 850 \implies r = 50

gallons per hour.

Subtracting 60 from both sides and dividing by 17 gives the rate of Pump B.

5

Calculate the pumping rate of Pump A.

Rate of Pump

A = 2(50) + 15 = 115

gallons per hour.

Substitute

r = 50

into the expression for Pump A's rate (

2r + 15

).

Key Concept

Linear equation formulation from work and rate relationships

Question 25Question →

A technician charges a one-time fixed diagnostic fee of $\$35$ plus $\$25$ for each hour of repair work. If the total bill for a repair job was $\$160$ , how many hours of repair work were performed?

3

4

5

6

7

Show answer & explanation

Answer: $5$

Answer

5 hours

Let

h

represent the number of hours of repair work. The total charge is given by the sum of the fixed fee (

\$35

) and the hourly rate times hours worked (

25h

). Setting up the linear equation

35 + 25h = 160

and isolating

h

gives

25h = 125

, which simplifies to

h = 5

. Thus, 5 hours of work were performed.

Step-by-Step Solution

1

Define the variable and set up the linear equation

35 + 25h = 160

, where

h

is the number of repair hours.

The total cost is the sum of the fixed diagnostic fee and the hourly charge multiplied by hours worked.

2

Subtract the fixed fee from both sides of the equation

25h = 160 - 35 \implies 25h = 125

Isolate the variable term containing

h

.

3

Divide both sides by the hourly rate coefficient

h = \frac{125}{25} = 5

Solve for

h

.

Key Concept

Formulating and solving linear equations in one variable from real-world scenarios

Question 26Question →

If

x

satisfies the linear equation

\frac{3(x - 4)}{4} - \frac{2x + 1}{6} = \frac{x + 2}{3} - 5

what is the value of

2x + 5

?

-31

-23

-14

41

87

Show answer & explanation

Answer: $-23$

Answer

The value of

2x + 5

is

-23

.

Clearing denominators by multiplying the entire equation by

12

gives

9(x - 4) - 2(2x + 1) = 4(x + 2) - 60

. Expanding both sides yields

9x - 36 - 4x - 2 = 4x + 8 - 60

, which simplifies to

5x - 38 = 4x - 52

. Isolating

x

gives

x = -14

. Substituting

x = -14

into the target expression

2x + 5

yields

2(-14) + 5 = -23

.

Step-by-Step Solution

1

Clear the denominators by multiplying every term on both sides of the equation by the least common multiple of

4, 6,

and

3

, which is

12

.

12 \cdot \frac{3(x - 4)}{4} - 12 \cdot \frac{2x + 1}{6} = 12 \cdot \frac{x + 2}{3} - 12 \cdot 5

, leading to

9(x - 4) - 2(2x + 1) = 4(x + 2) - 60

.

Clearing denominators simplifies the multi-step fractional equation into an integer linear equation.

2

Expand all grouping symbols and combine like terms on both sides.

9x - 36 - 4x - 2 = 4x + 8 - 60

, which simplifies to

5x - 38 = 4x - 52

.

Distributing terms carefully ensures proper sign distribution, especially for negative signs across parentheses.

3

Isolate the variable

x

on one side of the equation.

5x - 4x = -52 + 38

, giving

x = -14

.

Subtracting

4x

and adding

38

isolates

x

.

4

Evaluate the target expression

2x + 5

using the solved value

x = -14

.

2(-14) + 5 = -28 + 5 = -23

.

The question asks for the value of

2x + 5

, not

x

.

Key Concept

Solving linear equations in one variable with fractional coefficients and evaluating algebraic expressions

Estimated Time:1m 30s

Question 27Question →

A water reservoir initially contains $450$ liters of water. Water drains out of the reservoir at a constant rate of $r$ liters per hour, while an inlet pipe supplies water at a constant rate of $18$ liters per hour. If the reservoir contains $390$ liters of water after $6$ hours, what is the value of $r$ ?

Show answer & explanation

Answer: 28

Answer

The value of

r

is

28

.

The reservoir starts with

450

liters. Over

6

hours, water enters at

18

liters/hour and leaves at

r

liters/hour, giving a net volume equation of

450 + 6(18 - r) = 390

. Simplifying yields

558 - 6r = 390

, which subtracts to

-6r = -168

, giving

r = 28

.

Step-by-Step Solution

1

Set up the linear equation representing the net change in water volume over time.

450 + 6(18 - r) = 390

The final volume equals the initial volume plus the net water added (inflow rate minus outflow rate multiplied by hours).

2

Expand and simplify the linear expression.

558 - 6r = 390

Distribute

6

across

(18 - r)

to obtain

108 - 6r

, then add to

450

.

3

Isolate the variable term

-6r

.

-6r = -168

Subtract

558

from both sides of the equation.

4

Solve for the rate

r

.

r = 28

Divide both sides by

-6

.

Key Concept

Formulating and solving a linear equation in one variable from a rate problem context.

Estimated Time:1m 30s

Question 28Question →

If

x

satisfies the linear equation

4(x - 2) - 15 = 2x + 3

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

Select all that apply

x

is a prime number

2x - 5

is equal to

21

x + 7

is a multiple of

5

x

is an even integer

\frac{x + 1}{2} = 6

Show answer & explanation

Answer: $x$ is a prime number; $2x - 5$ is equal to $21$ ; $x + 7$ is a multiple of $5$

Answer

The correct statements are that

x

is a prime number,

2x - 5

is equal to

21

, and

x + 7

is a multiple of

5

.

Solving the equation

4(x - 2) - 15 = 2x + 3

yields

x = 13

. Evaluating the choices with

x = 13

confirms that

13

is a prime number,

2(13) - 5 = 21

, and

13 + 7 = 20

(a multiple of

5

).

Step-by-Step Solution

1

Expand the left side of the linear equation by distributing the constant term.

4x - 8 - 15 = 2x + 3

Distribution removes parentheses so like terms can be combined.

2

Combine constant terms on the left side of the equation.

4x - 23 = 2x + 3

Simplifying

-8 - 15

yields

-23

.

3

Isolate variable terms on the left side and constant terms on the right side by subtracting

2x

and adding

23

to both sides.

2x = 26

Subtracting

2x

gives

2x

on the left, and adding

23

gives

26

on the right.

4

Divide both sides of the equation by

2

to solve for

x

.

x = 13

Dividing

26

by

2

determines the unique value of

x

.

5

Test each statement using

x = 13

.

13

is prime (True);

2(13) - 5 = 21

(True);

13 + 7 = 20

which is a multiple of

5

(True);

13

is even (False);

\frac{13 + 1}{2} = 7 \neq 6

(False).

Direct evaluation identifies which statements hold true.

Key Concept

Solving linear equations in one variable and evaluating numerical properties of the solution.

Estimated Time:1m 0s

Question 29Question →

An investor divided a total capital of $\$20,000$ between two accounts. Account A pays a simple annual interest rate of $6\%$ , and Account B pays a simple annual interest rate of $9\%$ . If the total interest earned from both accounts at the end of one year was $\$1,470$ , how much money was invested in Account A?

\$8,000

\$9,000

\$9,800

\$11,000

\$12,000

Show answer & explanation

Answer: $\$11,000$

Answer

The amount invested in Account A was

\$11,000

.

Let

x

represent the amount invested in Account A. The remaining capital,

20,000 - x

, is invested in Account B. The total interest earned in one year is given by

0.06x + 0.09(20,000 - x) = 1,470

. Expanding gives

0.06x + 1,800 - 0.09x = 1,470

, which simplifies to

-0.03x = -330

. Dividing by

-0.03

yields

x = 11,000

. Thus,

\$11,000

was invested in Account A.

Step-by-Step Solution

1

Define the variable for the unknown quantity.

Let

x

be the amount in dollars invested in Account A. Then

(20,000 - x)

is the amount invested in Account B.

Expressing both quantities in terms of a single variable allows setting up a single linear equation.

2

Formulate the total interest equation.

0.06x + 0.09(20,000 - x) = 1,470

Total interest is the sum of interest from Account A (

6\%

of

x

) and Account B (

9\%

of

20,000 - x

).

3

Expand and simplify the algebraic equation.

0.06x + 1,800 - 0.09x = 1,470 \implies -0.03x + 1,800 = 1,470

Distribute

0.09

across

(20,000 - x)

and combine like terms.

4

Isolate the variable

x

.

-0.03x = 1,470 - 1,800 \implies -0.03x = -330 \implies x = 11,000

Subtract

1,800

from both sides and divide by

-0.03

to find

x

.

Key Concept

Linear Equations in One Variable

Estimated Time:1m 30s

Question 30Question →

A plumbing service charges a flat diagnostic fee of $\$45$ plus $\$65$ for each hour of repair work. If the total bill for a repair job was $\$305$ , how many hours of repair work were performed?

Show answer & explanation

Answer: 4

Answer

The total number of hours of repair work performed was 4.

To determine the number of repair hours, represent the scenario with the linear equation

45 + 65h = 305

, where

h

is the number of hours. Subtracting 45 from both sides yields

65h = 260

. Dividing 260 by 65 gives

h = 4

.

Step-by-Step Solution

1

Formulate the linear equation from the word problem context.

45 + 65h = 305

The total charge consists of a one-time fixed fee of

45 plus a variable fee of

65 per hour

h

.

2

Isolate the variable term by subtracting 45 from both sides of the equation.

65h = 260

Subtracting the constant fee isolates the total cost incurred from hourly work.

3

Divide both sides by the coefficient of the variable to solve for

h

.

h = 4

Dividing the total labor charge (

260) by the hourly rate (

65) yields the number of hours worked.

Key Concept

Linear Equations in One Variable

Question 31Question →

A company has a total of $150$ employees assigned to either the Marketing department or the Development department. The number of employees in the Development department is $30$ more than $3$ times the number of employees in the Marketing department. How many employees work in the Marketing department?

25

30

40

45

120

Show answer & explanation

Answer: 30

Answer

The number of employees working in the Marketing department is 30.

Letting

m

represent the Marketing department employees, the Development department has

3m + 30

employees. Summing both yields

m + 3m + 30 = 150

, which simplifies to

4m + 30 = 150

. Subtracting 30 gives

4m = 120

, and dividing by 4 yields

m = 30

.

Step-by-Step Solution

1

Define variables for each department.

Let

m

be the number of employees in Marketing. Then the number of employees in Development is

3m + 30

.

Expressing both quantities in terms of a single variable simplifies setting up a linear equation.

2

Set up the linear equation representing total employees.

m + (3m + 30) = 150

The sum of employees in both departments equals the company total of 150.

3

Combine like terms.

4m + 30 = 150

Adding

m

and

3m

gives

4m

.

4

Isolate the variable term and solve for

m

.

4m = 120 \implies m = 30

Subtract 30 from both sides and then divide by 4.

Key Concept

Linear Equations in One Variable - Word Problem Modeling

Estimated Time:1m 0s

Question 32Question →

If $x$ satisfies the equation $\frac{3x - 1}{2} - \frac{x + 4}{5} = \frac{x + 7}{10} + 2$ , which of the following statements about $x$ must be true? Indicate all such statements.

Select all that apply

x > 3

3x

is an integer

x

is a solution to the equation

3x - 10 = 0

x

is an integer

x < 3.2

Show answer & explanation

Answer: $x > 3$ ; $3x$ is an integer; $x$ is a solution to the equation $3x - 10 = 0$

Answer

The statements that

x > 3

, that

3x

is an integer, and that

x

is a solution to the equation

3x - 10 = 0

are all correct.

Solving the equation yields

x = \frac{10}{3} \approx 3.33

. The statement

x > 3

is correct since

3.33 > 3

. The statement that

3x

is an integer is correct because

3 \times \frac{10}{3} = 10

. The statement that

x

is a solution to

3x - 10 = 0

is correct because

3\left(\frac{10}{3}\right) - 10 = 0

.

Step-by-Step Solution

1

Clear denominators by multiplying the entire linear equation by the least common multiple, 10.

5(3x - 1) - 2(x + 4) = (x + 7) + 20

Eliminating fractions simplifies the expression into a standard linear form.

2

Distribute terms across parentheses.

15x - 5 - 2x - 8 = x + 7 + 20

Expand both sides while correctly applying sign rules.

3

Combine like terms on each side.

13x - 13 = x + 27

Group variable terms together and constant terms together.

4

Isolate the variable

x

.

12x = 40 \implies x = \frac{40}{12} = \frac{10}{3}

Subtract

x

and add

13

on both sides, then divide by

12

.

5

Evaluate each given statement using

x = \frac{10}{3}

.

The value

\frac{10}{3} \approx 3.33

is greater than

3

;

3\left(\frac{10}{3}\right) = 10

is an integer;

3\left(\frac{10}{3}\right) - 10 = 0

is true;

\frac{10}{3}

is not an integer;

\frac{10}{3}

is not less than

3.2

.

Determine which logical assertions hold true for

x = \frac{10}{3}

.

Key Concept

Solving multi-step linear equations with fractions and evaluating properties of rational solutions.

Question 33Question →

If $x$ satisfies the linear equation $\frac{3(x - 1)}{2} - \frac{x + 5}{4} = \frac{2x - 3}{3}$ , what is the value of $4x + 7$ ?

15

19

23

27

31

Show answer & explanation

Answer: 19

Answer

19

Multiplying the entire equation by the least common denominator 12 yields

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

. Expanding the terms gives

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

, which simplifies to

15x - 33 = 8x - 12

. Subtracting

8x

and adding

33

gives

7x = 21

, so

x = 3

. Substituting

x = 3

into

4x + 7

produces

4(3) + 7 = 19

.

Step-by-Step Solution

1

Find the least common denominator (LCD) for all fractions in the equation.

The LCD of 2, 4, and 3 is 12.

Clearing denominators simplifies the linear equation to integer coefficients.

2

Multiply both sides of the equation by 12.

12 \cdot \left(\frac{3(x - 1)}{2}\right) - 12 \cdot \left(\frac{x + 5}{4}\right) = 12 \cdot \left(\frac{2x - 3}{3}\right) \implies 6 \cdot 3(x - 1) - 3(x + 5) = 4(2x - 3)

Multiplying each term by 12 eliminates all fractions.

3

Expand the terms on both sides of the equation.

18(x - 1) - 3(x + 5) = 8x - 12 \implies 18x - 18 - 3x - 15 = 8x - 12

Distribute the coefficients across each set of parentheses.

4

Combine like terms and solve for

x

.

15x - 33 = 8x - 12 \implies 7x = 21 \implies x = 3

Isolate the variable term

x

on one side of the equation.

5

Evaluate the requested expression

4x + 7

using

x = 3

.

4(3) + 7 = 12 + 7 = 19

Substitute the value of

x

into

4x + 7

to find the final requested value.

Key Concept

Solving multi-step linear equations in one variable with fractional coefficients by clearing denominators.

Estimated Time:1m 30s

Question 34Question →

A craft brewery blends two batches of cider. Batch A contains a $12\%$ sugar solution by volume, and Batch B contains a $20\%$ sugar solution by volume. The brewer mixes $15$ liters of Batch A with $x$ liters of Batch B to create a resulting mixture that is $17\%$ sugar by volume. What is the value of $x$ ?

Show answer & explanation

Answer: 25

Answer

The value of

x

is

25

.

Equating the total amount of pure sugar before and after mixing gives

0.12(15) + 0.20x = 0.17(15 + x)

. Simplifying yields

1.8 + 0.20x = 2.55 + 0.17x

, which reduces to

0.03x = 0.75

, giving

x = 25

.

Step-by-Step Solution

1

Calculate the volume of pure sugar contributed by Batch A

0.12 \times 15 = 1.8

liters of pure sugar

Batch A is

12\%

sugar by volume and has a total volume of

15

liters.

2

Express the total volume of pure sugar in terms of

x

Total sugar volume =

1.8 + 0.20x

liters

Batch B adds

x

liters of a

20\%

sugar solution.

3

Set up the mixture equation

1.8 + 0.20x = 0.17(15 + x)

The final mixture has a total volume of

(15 + x)

liters with a concentration of

17\%

sugar.

4

Expand and solve the linear equation for

x

1.8 + 0.20x = 2.55 + 0.17x \implies 0.03x = 0.75 \implies x = 25

Subtract

0.17x

and

1.8

from both sides, then divide by

0.03

.

Key Concept

Linear Equations in One Variable (Mixture Word Problems)

Estimated Time:1m 30s

Question 35Question →

A non-profit foundation distributes a total grant of $\$108,000$ among three research teams: Team A, Team B, and Team C. Team B receives $\$4,000$ more than Team A. Team C receives twice as much as the combined amount received by Team A and Team B. What amount does Team B receive?

\$16,000

\$20,000

\$24,000

\$28,000

\$72,000

Show answer & explanation

Answer: $\$20,000$

Answer

Team B receives

\$20,000

.

Defining Team A's share as

x

gives Team B a share of

x + 4,000

and Team C a share of

2(x + x + 4,000) = 4x + 8,000

. Combining these yields

6x + 12,000 = 108,000

, which solves to

x = 16,000

. Therefore, Team B receives

16,000 + 4,000 = \$20,000

.

Step-by-Step Solution

1

Define variables for each team's share in terms of a single unknown variable.

Let

x

be the amount received by Team A in dollars. Then Team B receives

x + 4,000

, and Team C receives

2 \cdot [x + (x + 4,000)] = 2(2x + 4,000) = 4x + 8,000

.

Expressing all quantities in terms of

x

reduces the problem to a linear equation in one variable.

2

Set up the linear equation representing the total grant amount.

x + (x + 4,000) + (4x + 8,000) = 108,000 \implies 6x + 12,000 = 108,000

.

The sum of the individual shares must equal the total grant of

\$108,000

.

3

Solve the linear equation for

x

.

6x = 96,000 \implies x = 16,000

.

Subtract

12,000

from both sides and divide by

6

.

4

Calculate Team B's share using the value of

x

.

Team B's share

= x + 4,000 = 16,000 + 4,000 = 20,000

.

The question specifically asks for Team B's share, not Team A's.

Key Concept

Formulating and solving a linear equation in one variable from a multi-step algebraic word problem.

Estimated Time:1m 30s

Question 36Question →

If $x$ satisfies the equation $\frac{5(x - 2)}{3} - \frac{2x - 1}{4} = \frac{x + 7}{6}$ , which of the following statements must be true? Select all that apply.

Select all that apply

x > 4

4x

is an integer

x

is an integer

2x + 1 > 10

3x - 2 = 10.75

Show answer & explanation

Answer: $x > 4$ ; $4x$ is an integer; $3x - 2 = 10.75$

Answer

The correct statements are that

x > 4

,

4x

is an integer, and

3x - 2 = 10.75

.

Solving the linear equation yields

x = 4.25

or

\frac{17}{4}

. Using this value:

4.25 > 4

is true;

4 \times \frac{17}{4} = 17

is an integer, so that statement is true; and

3(4.25) - 2 = 10.75

is also true.

Step-by-Step Solution

1

Find a common denominator to clear fractions in the given equation.

The least common multiple of denominators

3

,

4

, and

6

is

12

.

Clearing denominators simplifies multi-term fractional equations into standard linear form.

2

Multiply every term of the equation by

12

and expand numerators.

4 \cdot 5(x - 2) - 3 \cdot (2x - 1) = 2 \cdot (x + 7) \implies 20(x - 2) - 3(2x - 1) = 2(x + 7)

.

Multiplying each term by

12

eliminates all fraction bars.

3

Distribute the constants through the parentheses on both sides.

20x - 40 - 6x + 3 = 2x + 14

.

Applying the distributive property correctly accounts for negative signs across parentheses.

4

Combine like terms on the left side and solve for

x

.

14x - 37 = 2x + 14 \implies 12x = 51 \implies x = \frac{51}{12} = \frac{17}{4} = 4.25

.

Isolating

x

gives the exact rational value of the solution.

5

Evaluate each provided statement using

x = 4.25

.

The statement

x > 4

is true (

4.25 > 4

). The statement

4x

is an integer is true (

4 \times 4.25 = 17

). The statement

3x - 2 = 10.75

is true (

3 \times 4.25 - 2 = 10.75

). The other two statements are false.

Direct substitution verifies which conditions hold.

Key Concept

Solving multi-step linear equations in one variable by clearing denominators and combining variable terms.

Estimated Time:1m 30s

Question 37Question →

A technical consultant charges a one-time setup fee of $\$150$ plus a standard rate of $\$45$ per hour for the first $20$ hours of work on a project. For any hours worked beyond $20$ hours, the consultant charges an increased hourly rate that is $20\%$ higher than the standard rate. If the total bill for a project was $\$1,320$ , how many total hours did the consultant work on the project?

Show answer & explanation

Answer: 25

Answer

The consultant worked a total of 25 hours on the project.

To find the total hours worked, first subtract the setup fee (

150) and the cost of the first 20 hours (20 *

45 =

900) from the total bill of

1,320. This leaves

1,320 -

1,050 =

270. The hourly rate after 20 hours increases by 20% to

45 * 1.20 =

54 per hour. Dividing the remaining

270 by $54 gives 5 additional hours. Adding these 5 hours to the initial 20 hours gives a total of 25 hours.

Step-by-Step Solution

1

Determine the base cost for the setup fee and the initial 20 hours.

Base cost =

150 + (20 *

45) = $1,050.

The initial cost tier applies up to 20 hours of work.

2

Calculate the higher hourly rate applied to additional hours worked beyond 20.

Increased rate =

45 * 1.20 =

54 per hour.

The rate increases by 20% over the standard rate of $45 per hour.

3

Formulate and solve a linear equation for total hours

h

.

1050 + 54(h - 20) = 1320 \implies 54(h - 20) = 270 \implies h - 20 = 5 \implies h = 25

.

Subtracting the base cost leaves

270 for overtime hours, which divides by

54 per hour to yield 5 additional hours, for 25 hours total.

Key Concept

Linear Equations in One Variable

Question 38Question →

A non-profit organization hosted two fundraising events, Event A and Event B. Event A charged a ticket price of $\$30$ per person and collected an additional fixed donation of $\$250$ from a local sponsor. Event B charged a ticket price of $\$45$ per person and collected a fixed donation of $\$400$ from a corporate sponsor. The number of attendees at Event B was $10$ fewer than twice the number of attendees at Event A. If the total revenue raised from both events combined was $\$8,600$ , how many people attended Event B?

60

70

130

140

150

Show answer & explanation

Answer: 130

Answer

130 people attended Event B.

Setting up the linear equation for total revenue gives

(30x + 250) + (45(2x - 10) + 400) = 8,600

, where

x

is the attendance at Event A. Expanding and combining like terms yields

120x + 200 = 8,600

, which solves to

x = 70

. Substituting

x = 70

into the expression for Event B attendance (

2x - 10

) gives

2(70) - 10 = 130

.

Step-by-Step Solution

1

Define variables for the unknown quantities.

Let

x

be the number of attendees at Event A. The number of attendees at Event B is

2x - 10

.

Event B has 10 fewer attendees than twice Event A.

2

Express the revenue generated by each event in terms of

x

.

Event A revenue:

30x + 250

; Event B revenue:

45(2x - 10) + 400 = 90x - 450 + 400 = 90x - 50

.

Revenue equals ticket price times attendees plus fixed sponsor donations.

3

Set up and simplify the linear equation for total combined revenue.

(30x + 250) + (90x - 50) = 8,600 \implies 120x + 200 = 8,600

.

The sum of revenues from both events equals the total revenue of $8,600.

4

Solve the linear equation for

x

.

120x = 8,400 \implies x = 70

.

Subtract 200 from both sides and divide by 120 to isolate

x

.

5

Calculate the number of attendees at Event B.

Event B attendees

= 2(70) - 10 = 140 - 10 = 130

.

Substitute

x = 70

into the expression

2x - 10

.

Key Concept

Linear Equations in One Variable

Estimated Time:1m 30s

Question 39Question →

A bakery produces sourdough loaves using a mixture of two types of flour, Flour X and Flour Y. A standard batch requires a total of $84$ kilograms of flour. The amount of Flour X used is $12$ kilograms more than twice the amount of Flour Y used. If $y$ represents the amount of Flour Y, in kilograms, used in one standard batch, which of the following statements must be true? Select all such statements.

Select all that apply

y

is a positive integer multiple of

8

.

The ratio of the amount of Flour X to the amount of Flour Y used in one batch is

5:2

.

Flour X accounts for

75\%

of the total flour used in one batch.

The amount of Flour X used in one batch exceeds the amount of Flour Y used by

36

kilograms.

The combined amount of Flour Y used in

4

standard batches is

100

kilograms.

Show answer & explanation

Answer: $y$ is a positive integer multiple of $8$ .; The ratio of the amount of Flour X to the amount of Flour Y used in one batch is $5:2$ .; The amount of Flour X used in one batch exceeds the amount of Flour Y used by $36$ kilograms.

Answer

The correct statements are that

y

is a positive integer multiple of

8

, the ratio of Flour X to Flour Y used in one batch is

5:2

, and the amount of Flour X used in one batch exceeds the amount of Flour Y used by

36

kilograms.

Solving

y + (2y + 12) = 84

yields

y = 24

kilograms of Flour Y and

60

kilograms of Flour X. Since

24

is divisible by

8

, the statement asserting

y

is a multiple of

8

is true. The ratio of Flour X to Flour Y is

60:24 = 5:2

, making the ratio statement true. Finally,

60 - 24 = 36

, confirming that Flour X exceeds Flour Y by

36

kilograms.

Step-by-Step Solution

1

Set up the linear equation in terms of

y

.

Amount of Flour Y =

y

, Amount of Flour X =

2y + 12

. Total amount =

y + (2y + 12) = 84

.

The problem states Flour X is

12

kg more than twice Flour Y, and their sum equals

84

kg.

2

Solve the linear equation for

y

.

3y + 12 = 84 \implies 3y = 72 \implies y = 24

.

Subtract

12

from both sides and divide by

3

to isolate the variable

y

.

3

Calculate the amount of Flour X used.

Amount of Flour X =

2(24) + 12 = 48 + 12 = 60

kilograms.

Substitute

y = 24

back into the expression for Flour X.

4

Evaluate each given statement.

1)

y = 24

, which is a multiple of

8

(

24 = 8 \times 3

). (True)
2) Ratio of Flour X to Flour Y is

60 : 24 = 5 : 2

. (True)
3) Percentage of Flour X is

\frac{60}{84} = \frac{5}{7} \approx 71.43\% \neq 75\%

. (False)
4) Difference is

60 - 24 = 36

kg. (True)
5)

4

batches of Flour Y equal

4 \times 24 = 96

kg

\neq 100

kg. (False)

Direct calculation confirms which statements match the derived values.

Key Concept

Formulating and solving a linear equation in one variable from a word problem scenario, followed by logical evaluation of derived quantities.

Question 40Question →

A manufacturing plant operates two assembly lines, Line X and Line Y. Line X produces 15 units per hour, and Line Y produces 22 units per hour. On a certain day, Line X operated for 3 hours longer than Line Y did, and the two lines produced a total of 415 units. How many hours did Line Y operate?

Show answer & explanation

Answer: 10

Answer

10

Let

h

represent the number of hours Line Y operated. Because Line X operated for 3 hours longer than Line Y, Line X operated for

h + 3

hours. The total number of units produced by both lines is the sum of their individual outputs:

15(h + 3) + 22h = 415

. Distributing 15 yields

15h + 45 + 22h = 415

. Combining like terms gives

37h + 45 = 415

. Subtracting 45 from both sides yields

37h = 370

, and dividing by 37 gives

h = 10

. Therefore, Line Y operated for 10 hours.

Step-by-Step Solution

1

Define the variable representing Line Y's operating time in hours.

Let

h

represent the number of hours Line Y operated. Line X's operating time is

h + 3

hours.

Line X operated for 3 hours longer than Line Y.

2

Formulate a linear equation in one variable for total units produced.

15(h + 3) + 22h = 415

Total production is the sum of production from Line X (

15 \times (h + 3)

) and Line Y (

22 \times h

).

3

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

15h + 45 + 22h = 415 \implies 37h + 45 = 415

Apply the distributive property and combine variable terms.

4

Isolate the variable

h

.

37h = 370 \implies h = 10

Subtract 45 from both sides of the equation and divide by 37.

Key Concept

Linear Equations in One Variable

Estimated Time:1m 30s

Linear Equations in One Variable

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