Algebra and Functions Practice Questions — GMAT — Page 11

Question 201Question →

For all real numbers $x > 1$ , the expression $\frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} - \sqrt{x-1}}$ is equal to $x + \sqrt{x^2 - 1}$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because rationalizing the denominator of the algebraic radical expression simplifies directly to

x + \sqrt{x^2 - 1}

.

Rationalizing the radical expression by multiplying the numerator and denominator by the conjugate

(\sqrt{x+1} + \sqrt{x-1})

leads directly to the simplified form

x + \sqrt{x^2 - 1}

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator,

(\sqrt{x+1} + \sqrt{x-1})

.

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

Rationalizing the denominator removes radicals from the denominator using the difference of squares identity.

2

Expand the numerator using

(a+b)^2 = a^2 + 2ab + b^2

.

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

Simplifying sums of squared radicals and combining like terms.

3

Expand the denominator using

(a-b)(a+b) = a^2 - b^2

.

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

Evaluating the difference of squares of square root terms.

4

Divide the expanded numerator by the simplified denominator.

\frac{2x + 2\sqrt{x^2-1}}{2} = x + \sqrt{x^2-1}

Factoring out 2 from the numerator cancels the denominator of 2.

Key Concept

Rationalizing radical expressions using conjugates and difference of squares.

Question 202Question →

If $\frac{8^x \cdot 4^{x-1}}{16^{x+1}} = 32$ , what is the value of $x$ ?

Show answer & explanation

Answer: 11

Answer

The value of

x

is 11.

Rewriting all bases as powers of 2 yields

\frac{2^{3x} \cdot 2^{2x-2}}{2^{4x+4}} = 2^5

. Simplifying the left side gives

2^{x-6} = 2^5

. Equating exponents results in

x - 6 = 5

, giving

x = 11

.

Step-by-Step Solution

1

Express all bases as powers of 2.

8 = 2^3

,

4 = 2^2

,

16 = 2^4

, and

32 = 2^5

.

Converting to a common base allows exponents to be combined using standard exponent laws.

2

Simplify the numerator and denominator using the power rule

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

and product rule

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

.

Numerator:

2^{3x} \cdot 2^{2x-2} = 2^{5x-2}

. Denominator:

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

.

Distribute exponents carefully when multiplying powers with equal bases.

3

Apply the quotient rule

\frac{a^m}{a^n} = a^{m-n}

to combine the left side into a single power of 2.

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

.

Subtract the denominator's exponent from the numerator's exponent.

4

Equate exponents and solve for

x

.

2^{x-6} = 2^5 \implies x - 6 = 5 \implies x = 11

.

Since the bases are identical (

2

), the exponent expressions must be equal.

Key Concept

Solving exponential equations by expressing all numbers with a common base and applying exponent rules.

Question 203Question →

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

Question 204Question →

If $2^{a+b} = 12$ and $2^{a-b} = 3$ , what is the value of $2^{2a} - 2^{2b}$ ?

Show answer & explanation

Answer: 32

Answer

32

Multiplying the equations gives

(2^{a+b})(2^{a-b}) = 2^{2a} = 12 \times 3 = 36

. Dividing the first equation by the second gives

\frac{2^{a+b}}{2^{a-b}} = 2^{2b} = \frac{12}{3} = 4

. Subtracting the two values gives

2^{2a} - 2^{2b} = 36 - 4 = 32

.

Step-by-Step Solution

1

Multiply the given exponent expressions to evaluate

2^{2a}

.

2^{2a} = (2^{a+b})(2^{a-b}) = 12 \times 3 = 36

.

According to the product rule of exponents,

2^{x} \cdot 2^{y} = 2^{x+y}

. Adding

(a+b) + (a-b)

yields

2a

.

2

Divide the first exponent expression by the second to evaluate

2^{2b}

.

2^{2b} = \frac{2^{a+b}}{2^{a-b}} = \frac{12}{3} = 4

.

According to the quotient rule of exponents,

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

. Subtracting

(a+b) - (a-b)

yields

2b

.

3

Calculate the value of

2^{2a} - 2^{2b}

.

36 - 4 = 32

.

Substitute the evaluated components

2^{2a} = 36

and

2^{2b} = 4

into the expression.

Key Concept

Exponent Operations (Product and Quotient Rules)

Estimated Time:1m 30s

Question 205Question →

If $x$ is a positive real number such that $\sqrt{x} - \frac{2}{\sqrt{x}} = 3$ , what is the value of $x + \frac{4}{x}$ ?

5

7

9

11

13

Show answer & explanation

Answer: 13

Answer

The value of

x + \frac{4}{x}

is 13.

Squaring both sides of

\sqrt{x} - \frac{2}{\sqrt{x}} = 3

gives

(\sqrt{x})^2 - 2(\sqrt{x})(\frac{2}{\sqrt{x}}) + (\frac{2}{\sqrt{x}})^2 = 9

, which simplifies to

x - 4 + \frac{4}{x} = 9

. Adding 4 to both sides yields

x + \frac{4}{x} = 13

. Thus, the option with value 13 is correct.

Step-by-Step Solution

1

Square both sides of the given equation

\sqrt{x} - \frac{2}{\sqrt{x}} = 3

.

{\left(\sqrt{x} - \frac{2}{\sqrt{x}}\right)}^2 = 3^2

Squaring both sides relates the terms

\sqrt{x}

and

\frac{2}{\sqrt{x}}

to

x

and

\frac{4}{x}

.

2

Expand the left side using the perfect square identity

(a - b)^2 = a^2 - 2ab + b^2

.

(\sqrt{x})^2 - 2\left(\sqrt{x}\right)\left(\frac{2}{\sqrt{x}}\right) + \left(\frac{2}{\sqrt{x}}\right)^2 = 9

Applying the algebraic identity expands the squared expression correctly.

3

Simplify the terms on the left side.

x - 4 + \frac{4}{x} = 9

The product

\sqrt{x} \cdot \frac{2}{\sqrt{x}}

simplifies to

2

, making the middle term

-2 \cdot 2 = -4

.

4

Add 4 to both sides to isolate

x + \frac{4}{x}

.

x + \frac{4}{x} = 13

Isolating the target algebraic expression yields the final value.

Key Concept

Squaring binomial radical expressions using the identity

(a - b)^2 = a^2 - 2ab + b^2

.

Question 206Question →

A corporate catering service offers two lunch options: a Standard Meal and a Premium Meal. On Monday, an order of $40$ Standard Meals and $20$ Premium Meals cost a total of $\$1,300$ . On Tuesday, an order of $20$ Standard Meals and $30$ Premium Meals cost a total of $\$1,250$ . What is the total cost of an order consisting of $15$ Standard Meals and $15$ Premium Meals?

\$650.00

\$712.50

\$812.50

\$975.00

\$1,162.50

Show answer & explanation

Answer: $\$712.50$

Answer

The total cost of 15 Standard Meals and 15 Premium Meals is $712.50.

By setting up the system

40x + 20y = 1300

and

20x + 30y = 1250

, we simplify to

2x + y = 65

and

2x + 3y = 125

. Subtracting the first from the second gives

2y = 60

, so

y = 30

. Substituting

y = 30

into

2x + y = 65

yields

2x = 35

, so

x = 17.50

. The sum of one Standard Meal and one Premium Meal is

x + y = 47.50

. Therefore,

15

of each costs

15 \times 47.50 = \$712.50

. Alternatively, adding the two simplified equations directly yields

4x + 4y = 190 \implies x + y = 47.50

, so

15(x + y) = 15 \times 47.50 = \$712.50

.

Step-by-Step Solution

1

Set up a system of linear equations using variables

x

for the price of a Standard Meal and

y

for the price of a Premium Meal.

40x + 20y = 1300

and

20x + 30y = 1250

Translate the scenario information into algebraic equations.

2

Simplify both equations by dividing by their greatest common factors.

Equation 1:

2x + y = 65

; Equation 2:

2x + 3y = 125

Simplifying equations reduces computation complexity.

3

Subtract Equation 1 from Equation 2 to eliminate

x

and solve for

y

.

(2x + 3y) - (2x + y) = 125 - 65 \implies 2y = 60 \implies y = 30

Elimination isolates variable

y

.

4

Substitute

y = 30

back into Equation 1 to solve for

x

.

2x + 30 = 65 \implies 2x = 35 \implies x = 17.50

Finding the value of

x

gives the cost of one Standard Meal.

5

Calculate the target linear combination

15x + 15y = 15(x + y)

.

15(17.50 + 30.00) = 15(47.50) = 712.50

Multiply the combined price of one of each meal by 15.

Key Concept

Solving systems of two linear equations in two variables using elimination and finding linear combinations.

Estimated Time:1m 45s

Question 207Question →

A logistics company operates two types of delivery trucks: Model A and Model B. Model A consumes $0.25$ liters of fuel per mile and incurs $\$1.50$ in maintenance costs per mile. Model B consumes $0.40$ liters of fuel per mile and incurs $\$1.00$ in maintenance costs per mile. On a given day, the fleet consumed a total of $300$ liters of fuel and incurred a total of $\$1,100$ in maintenance costs. What was the total number of miles traveled by the Model A trucks on that day?

Show answer & explanation

Answer: 400

Answer

The Model A trucks traveled a total of 400 miles on that day.

Let

x

be the number of miles driven by Model A trucks and

y

be the number of miles driven by Model B trucks. Set up two equations:

0.25x + 0.40y = 300

(fuel) and

1.50x + 1.00y = 1100

(maintenance). Substituting

y = 1100 - 1.5x

into the fuel equation yields

0.25x + 0.40(1100 - 1.5x) = 300

, which simplifies to

-0.35x + 440 = 300

, leading directly to

x = 400

.

Step-by-Step Solution

1

Formulate the linear system

0.25x + 0.40y = 300

and

1.50x + 1.00y = 1100

, where

x

is miles driven by Model A and

y

is miles driven by Model B.

Translates the rates of fuel consumption and maintenance per mile into total daily figures.

2

Express

y

in terms of

x

y = 1100 - 1.5x

Isolating

y

makes substitution into the fuel equation straightforward.

3

Substitute

y

into the first equation and solve for

x

0.25x + 0.40(1100 - 1.5x) = 300 \implies -0.35x + 440 = 300 \implies -0.35x = -140 \implies x = 400

Reduces the system to a single linear equation in

x

to find the distance for Model A.

Key Concept

Setting up and solving a two-variable system of linear equations via substitution or elimination.

Question 208Question →

A financial advisory firm manages two types of investment accounts: Growth portfolios and Conservative portfolios. Each Growth portfolio generates $\$5,000$ in annual dividends and $\$12,000$ in annual capital gains. Each Conservative portfolio generates $\$8,000$ in annual dividends and $\$4,000$ in annual capital gains. Last year, the firm generated a total of $\$186,000$ in dividends and $\$188,000$ in capital gains from these portfolios. What was the total number of portfolios (Growth and Conservative combined) managed by the firm?

10

17

27

31

35

Show answer & explanation

Answer: $27$

Answer

The total number of portfolios managed by the firm is

27

.

Setting up the system of equations with

x

as Growth portfolios and

y

as Conservative portfolios gives

5x + 8y = 186

and

12x + 4y = 188

. Multiplying the second equation by

2

yields

24x + 8y = 376

. Subtracting the first equation gives

19x = 190

, so

x = 10

. Substituting

x = 10

gives

y = 17

. The total number of portfolios is

10 + 17 = 27

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations in thousands of dollars.

Let

x

be the number of Growth portfolios and

y

be the number of Conservative portfolios.
Dividends equation:

5x + 8y = 186

Capital gains equation:

12x + 4y = 188

Dividing all dollar amounts by

1,000

simplifies the coefficients.

2

Solve for

x

using the elimination method.

Multiply the capital gains equation by

2

:

2(12x + 4y) = 2(188) \implies 24x + 8y = 376

Subtract the dividends equation (

5x + 8y = 186

) from this new equation:

(24x + 8y) - (5x + 8y) = 376 - 186

19x = 190 \implies x = 10

Eliminating

y

isolates

x

to determine the number of Growth portfolios.

3

Substitute

x = 10

into the capital gains equation to solve for

y

.

Divide

12(10) + 4y = 188

simplify to

120 + 4y = 188 \implies 4y = 68 \implies y = 17

Finds the number of Conservative portfolios.

4

Calculate the total number of portfolios.

Total portfolios =

x + y = 10 + 17 = 27

The question asks for the sum of both types of portfolios.

Key Concept

Solving Systems of Two Linear Equations with Word Problem Constraints

Estimated Time:1m 30s

Question 209Question →

If $x > 1$ is an integer such that $\sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = 6$ , what is the value of $x$ ?

7

9

10

18

37

Show answer & explanation

Answer: 10

Answer

10

By rewriting the expressions under the radicals using the perfect square identity

(\sqrt{x-1} \pm 1)^2 = (x-1) \pm 2\sqrt{x-1} + 1 = x \pm 2\sqrt{x-1}

, the equation simplifies directly to

(\sqrt{x-1} + 1) + (\sqrt{x-1} - 1) = 6

. Adding the terms gives

2\sqrt{x-1} = 6

, which leads to

\sqrt{x-1} = 3

, and squaring both sides yields

x - 1 = 9

, so

x = 10

.

Step-by-Step Solution

1

Recognize the nested radical identity structure

Observe that

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

and

x - 2\sqrt{x-1} = (\sqrt{x-1} - 1)^2

.

Rewriting the terms under the outer square roots as perfect squares eliminates the outer radicals.

2

Simplify the square roots

Since

x > 1

,

\sqrt{x-1} + 1 > 0

and

\sqrt{x-1} - 1 \ge 0

for

x \ge 2

. Thus,

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

.

The principal square root of

a^2

is

|a|

, which is positive.

3

Set the simplified expression equal to 6 and solve for

x

2\sqrt{x-1} = 6 \implies \sqrt{x-1} = 3 \implies x - 1 = 9 \implies x = 10

.

Isolating the radical and squaring both sides yields the exact integer value of

x

.

Key Concept

Simplifying nested radicals using algebraic identities

Question 210Question →

A boutique hotel offers two types of rooms: Executive Suites and Deluxe Suites. During a convention weekend, booking $12$ Executive Suites and $18$ Deluxe Suites generated total revenue of $\$7{}800$ . The following weekend, booking $15$ Executive Suites and $10$ Deluxe Suites generated total revenue of $\$7{}250$ . If the nightly rate for each type of suite remained constant, what was the nightly rental rate, in dollars, of an Executive Suite?

Show answer & explanation

Answer: 350

Answer

The nightly rental rate of an Executive Suite was 350 dollars.

Formulating the revenue statements as linear equations gives

12x + 18y = 7,800

and

15x + 10y = 7,250

. Simplifying these by dividing by

6

and

5

yields

2x + 3y = 1,300

and

3x + 2y = 1,450

. Multiplying the first simplified equation by

2

(

4x + 6y = 2,600

) and the second by

3

(

9x + 6y = 4,350

) aligns the

y

-coefficients. Subtracting the equations leads to

5x = 1,750

, giving

x = 350

.

Step-by-Step Solution

1

Define variables and construct the system of equations.

12x + 18y = 7,800

and

15x + 10y = 7,250

, where

x

represents the Executive Suite rate and

y

represents the Deluxe Suite rate.

Translating word problem information into algebraic equations.

2

Simplify the system equations.

Dividing the first equation by

6

gives

2x + 3y = 1,300

. Dividing the second equation by

5

gives

3x + 2y = 1,450

.

Reducing the coefficients minimizes computation steps and errors.

3

Eliminate variable

y

to solve for

x

.

Multiply

2x + 3y = 1,300

by

2

to get

4x + 6y = 2,600

. Multiply

3x + 2y = 1,450

by

3

to get

9x + 6y = 4,350

. Subtracting gives

5x = 1,750

, so

x = 350

.

Eliminating

y

yields the requested value of

x

directly.

Key Concept

Solving Systems of Two Linear Equations via Elimination

Estimated Time:2m 0s

Question 211Question →

A technology company purchased a total of $35$ computers for its new office, consisting of standard desktop units costing $\$800$ each and high-performance workstation units costing $\$1,400$ each. If the total expenditure for all $35$ computers was $\$34,000$ , how many standard desktop units did the company purchase?

Show answer & explanation

Answer: 25

Answer

The company purchased 25 standard desktop units.

Let

x

be the number of standard desktop units and

y

be the number of workstation units. The situation yields two equations:

x + y = 35

and

800x + 1400y = 34000

. Simplifying the second equation gives

4x + 7y = 170

. Multiplying the first equation by

7

gives

7x + 7y = 245

. Subtracting the simplified cost equation from this product yields

3x = 75

, so

x = 25

.

Step-by-Step Solution

1

Define variables and establish equations from the problem statement.

System of equations:

x + y = 35

and

800x + 1400y = 34000

, where

x

represents standard desktop units and

y

represents workstation units.

Linear systems represent total quantity and total cost constraints simultaneously.

2

Simplify the linear cost equation by dividing by the greatest common divisor (

200

).

Simplified equation:

4x + 7y = 170

.

Simplifying coefficients reduces arithmetic complexity and potential errors.

3

Solve for

x

using substitution or elimination.

-3x = -75 \implies x = 25

.

Isolating

x

directly calculates the required count of standard desktop units.

Key Concept

Solving a system of two linear equations in two variables using substitution or elimination

Estimated Time:1m 30s

Question 212Question →

An art gallery placed two separate orders for exhibition lighting fixtures from the same supplier. The first order consisted of $5$ LED spotlights and $3$ halogen floodlights for a total cost of $\$840$ . The second order consisted of $3$ LED spotlights and $5$ halogen floodlights for a total cost of $\$760$ . What is the cost of $1$ LED spotlight?

\$80

\$100

\$120

\$150

\$200

Show answer & explanation

Answer: $\$120$

Answer

The cost of 1 LED spotlight is $120.

Setting up the system of equations

5x + 3y = 840

and

3x + 5y = 760

, adding both equations yields

8x + 8y = 1600

, which simplifies to

x + y = 200

. Subtracting the second equation from the first yields

2x - 2y = 80

, which simplifies to

x - y = 40

. Adding

x + y = 200

and

x - y = 40

gives

2x = 240

, so

x = 120

. Thus,

1

LED spotlight costs

\$120

.

Step-by-Step Solution

1

Define variables and set up the system of linear equations based on the two orders.

Let

x

be the cost of

1

LED spotlight and

y

be the cost of

1

halogen floodlight.
Equation (1):

5x + 3y = 840

Equation (2):

3x + 5y = 760

Translating the verbal statements into algebraic equations.

2

Add Equation (1) and Equation (2) to find the combined linear expression

(x + y)

.

(5x + 3y) + (3x + 5y) = 840 + 760 \implies 8x + 8y = 1600 \implies x + y = 200

Simplifying the system using linear combination.

3

Subtract Equation (2) from Equation (1) to find

(x - y)

.

(5x + 3y) - (3x + 5y) = 840 - 760 \implies 2x - 2y = 80 \implies x - y = 40

Determining the difference between the costs of the two fixtures.

4

Solve the reduced two-variable system for

x

.

Add

(x + y = 200)

and

(x - y = 40)

:

2x = 240 \implies x = 120

Eliminating

y

to determine the exact unit price of

1

LED spotlight.

Key Concept

Solving Systems of Linear Equations via Linear Combination

Estimated Time:1m 30s

Question 213Question →

A university admissions committee evaluates applicants from two distinct pools, Pool $A$ and Pool $B$ . In the first evaluation phase, accepting $40\%$ of the applicants from Pool $A$ and $60\%$ of the applicants from Pool $B$ results in a total of $180$ accepted students. In the second phase, accepting $70\%$ of the applicants from Pool $A$ and $30\%$ of the applicants from Pool $B$ results in a total of $210$ accepted students. What is the total number of applicants across both pools combined?

240

380

390

420

195

Show answer & explanation

Answer: 380

Answer

The total number of applicants across both pools combined is 380.

Translating the conditions into linear equations gives

0.40A + 0.60B = 180

and

0.70A + 0.30B = 210

. Multiplying both equations to clear decimals yields

2A + 3B = 900

and

7A + 3B = 2100

. Subtracting the first from the second eliminates

B

, giving

5A = 1200

, so

A = 240

. Substituting

A = 240

into

2A + 3B = 900

gives

B = 140

. The sum

A + B = 240 + 140 = 380

.

Step-by-Step Solution

1

Set up a system of linear equations based on the percentage of accepted applicants from Pool A (

A

) and Pool B (

B

).

Equation 1:

0.40A + 0.60B = 180

; Equation 2:

0.70A + 0.30B = 210

.

Converting percentages to decimals forms linear equations relating the total applicants in each pool to the accepted totals.

2

Simplify the equations by multiplying by 10 and dividing by common factors.

Equation 1 becomes

2A + 3B = 900

; Equation 2 becomes

7A + 3B = 2100

.

Simplifying coefficients facilitates easier variable elimination.

3

Subtract the simplified Equation 1 from Equation 2 to eliminate

B

and solve for

A

.

(7A + 3B) - (2A + 3B) = 2100 - 900 \implies 5A = 1200 \implies A = 240

.

Eliminating

B

yields a single linear equation in

A

.

4

Substitute

A = 240

back into

2A + 3B = 900

to solve for

B

.

2(240) + 3B = 900 \implies 480 + 3B = 900 \implies 3B = 420 \implies B = 140

.

Determines the population of Pool B.

5

Calculate the combined total of applicants from both pools.

A + B = 240 + 140 = 380

.

The question asks for the sum of applicants in both pools.

Key Concept

Solving Systems of Two-Variable Linear Equations using Elimination

Question 214Question →

A beverage manufacturer creates a functional tea blend by combining Extract $A$ , which contains $12\%$ polyphenols by volume, and Extract $B$ , which contains $28\%$ polyphenols by volume. If the manufacturer needs to produce a $200$ -liter batch of tea with a total polyphenol concentration of $18\%$ by volume, how many liters of Extract $A$ must be included in the mixture?

Show answer & explanation

Answer: 125

Answer

The manufacturer must include

125

liters of Extract

A

in the mixture.

By setting up the linear system

x + y = 200

and

0.12x + 0.28y = 36

, substituting

y = 200 - x

leads directly to

-0.16x = -20

, giving

x = 125

liters.

Step-by-Step Solution

1

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

x + y = 200

and

0.12x + 0.28y = 36

The sum of the individual extract volumes equals the total batch volume, and the sum of the pure polyphenols from each extract equals the total pure polyphenols in the final mixture.

2

Substitute

y = 200 - x

into the polyphenol equation to solve for

x

.

0.12x + 0.28(200 - x) = 36 \implies 0.12x + 56 - 0.28x = 36

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

x

.

3

Combine like terms and solve for

x

.

-0.16x = -20 \implies x = \frac{-20}{-0.16} = 125

Isolating

x

yields the required volume of Extract

A

.

Key Concept

Systems of Linear Equations in Applied Mixture Problems

Question 215Question →

A freight logistics company charges fixed rates per unit for two types of cargo containers: Type $X$ and Type $Y$ . Shipping $4$ Type $X$ containers and $3$ Type $Y$ containers costs a total of $\$290$ . Shipping $6$ Type $X$ containers and $5$ Type $Y$ containers costs a total of $\$460$ . Based on these rates, what is the total cost to ship $7$ Type $X$ containers and $6$ Type $Y$ containers?

$375

$495

$545

$630

$750

Show answer & explanation

Answer: $545

Answer

$545

By solving the system

4x + 3y = 290

and

6x + 5y = 460

, subtracting the first equation from the second yields

2x + 2y = 170

, so

x + y = 85

. Adding

x + y = 85

to

6x + 5y = 460

gives

7x + 6y = 545

. Alternatively, solving individually yields

x = 35

and

y = 50

, giving

7(35) + 6(50) = 545

.

Step-by-Step Solution

1

Set up a system of linear equations representing the given shipping costs.

Let

x

be the cost per Type

X

container and

y

be the cost per Type

Y

container. Equation (1):

4x + 3y = 290

. Equation (2):

6x + 5y = 460

.

Translate the word problem statements into linear equations.

2

Subtract Equation (1) from Equation (2) to find a linear combination.

(6x + 5y) - (4x + 3y) = 460 - 290 \implies 2x + 2y = 170 \implies x + y = 85.

Determining the combined cost of one container of each type simplifies calculating larger combinations.

3

Add the expression for

x + y

to Equation (2) to obtain

7x + 6y

.

(6x + 5y) + (x + y) = 460 + 85 \implies 7x + 6y = 545.

Combining

6x + 5y

and

x + y

directly gives the cost for

7

Type

X

and

6

Type

Y

containers without requiring individual variable solving.

Key Concept

Linear combinations and systems of linear equations in two variables

Estimated Time:1m 30s

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