All practice questions

Question 281Question →

If the roots of the quadratic equation $x^2 - (k - 2)x + (k + 1) = 0$ have a sum that is equal to twice their product, what is the value of $k$ ?

Show answer & explanation

Answer: -4

Answer

The value of

k

is

-4

.

By Vieta's formulas, the sum of the roots of

x^2 - (k-2)x + (k+1) = 0

is

k-2

and the product of the roots is

k+1

. Equating the sum to twice the product gives

k - 2 = 2(k + 1)

, which simplifies to

k - 2 = 2k + 2

and yields

k = -4

. Evaluating the discriminant with

k = -4

gives

\Delta = (-6)^2 - 4(-3) = 48 > 0

, ensuring real roots exist.

Step-by-Step Solution

1

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

k

.

Sum of roots

= -\frac{b}{a} = k - 2

, and product of roots

= \frac{c}{a} = k + 1

.

For any quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is

-\frac{b}{a}

and the product of the roots is

\frac{c}{a}

.

2

Set up an equation based on the condition that the sum of the roots equals twice their product.

k - 2 = 2(k + 1)

This directly translates the verbal condition given in the problem.

3

Solve the linear equation for

k

.

k - 2 = 2k + 2 \implies -4 = k

Subtracting

k

and 2 from both sides isolates

k

.

4

Check the discriminant to ensure real roots exist when

k = -4

.

\Delta = b^2 - 4ac = (-6)^2 - 4(1)(-3) = 36 + 12 = 48 > 0

A positive discriminant guarantees that the quadratic equation has two distinct real roots.

Key Concept

Vieta's Formulas for Quadratic Equations

Question 282Question →

A glass bowl contains 6 red candies and $N$ green candies, where $N$ is a positive integer. If two candies are selected at random one after another without replacement, the probability that both candies selected are red is $\frac{1}{3}$ . What is the value of $N$ ?

Show answer & explanation

Answer: 4

Answer

4

The correct value of

N

is 4. Because the selection is performed without replacement, the probability of drawing the first red candy is

\frac{6}{N+6}

and the second red candy is

\frac{5}{N+5}

. Setting their product equal to

\frac{1}{3}

yields

\frac{30}{(N+6)(N+5)} = \frac{1}{3}

, which expands to

N^2 + 11N - 60 = 0

. Factoring gives

(N-4)(N+15) = 0

, yielding

N = 4

as the only positive solution.

Step-by-Step Solution

1

Define total outcomes and event probabilities for dependent selection

Initial total candies =

N + 6

. Probability of first red =

\frac{6}{N+6}

. Probability of second red given first is red =

\frac{5}{N+5}

.

Since selection is without replacement, the total count and red count both decrease by 1 after the first draw, making the events dependent.

2

Set up the probability multiplication equation

P(\text{both red}) = \frac{6}{N+6} \times \frac{5}{N+5} = \frac{30}{(N+6)(N+5)} = \frac{1}{3}

.

The probability of joint dependent events

A

and

B

occurring is

P(A \cap B) = P(A) \times P(B|A)

.

3

Solve the equation for

N

(N+6)(N+5) = 90 \Rightarrow N^2 + 11N - 60 = 0 \Rightarrow (N-4)(N+15) = 0

.

Cross-multiplying gives a quadratic equation in terms of

N

.

4

Select the valid positive integer root

N = 4

.

The count of candies must be a positive number, so

N = -15

is discarded.

Key Concept

Probability of Dependent Events (Sampling without Replacement)

Question 283Question →

A cyclist and a runner start simultaneously from Point $A$ and travel along a straight path toward Point $B$ . The cyclist travels at a constant speed, and the runner travels at a lower constant speed. When the cyclist reaches Point $B$ , she immediately turns around and travels back toward Point $A$ at her original speed. She meets the runner at a location $12$ miles from Point $B$ .

If the cyclist had instead increased her speed by $20\%$ and the runner had decreased his speed by $10\%$ , the cyclist would have met the runner on her return trip at a location $20$ miles from Point $B$ .

What is the distance, in miles, between Point $A$ and Point $B$ ?

Show answer & explanation

Answer: 60

Answer

The distance between Point A and Point B is 60 miles.

The correct answer of 60 miles is obtained by setting up the ratio of distances traveled by the cyclist and the runner for both scenarios. Since time elapsed until they meet is the same for both people in each scenario, the ratio of distances equals the ratio of speeds. Setting the scaled speed ratio (4/3 times the original ratio) equal to the new distance ratio (D + 20)/(D - 20) yields the quadratic equation D^2 - 56D - 240 = 0, whose unique positive solution is D = 60.

Step-by-Step Solution

1

Set up the relative distance ratio for the first meeting condition.

v_c / v_r = (D + 12) / (D - 12)

Because time is equal for both participants, distance traveled is directly proportional to speed.

2

Adjust the speed ratio based on the given percentage changes.

v_c' / v_r' = (1.20 / 0.90) * (v_c / v_r) = (4/3) * (D + 12) / (D - 12)

Increasing speed by 20% scales it by 1.20, and decreasing speed by 10% scales it by 0.90.

3

Equate the adjusted speed ratio to the distance ratio for the second meeting condition.

(D + 20) / (D - 20) = 4(D + 12) / [3(D - 12)]

In the second scenario, the cyclist covers D + 20 miles while the runner covers D - 20 miles.

4

Expand and solve the resulting quadratic equation for D.

D^2 - 56D - 240 = 0, yielding D = 60 miles.

Cross-multiplying and simplifying yields a quadratic equation with positive root D = 60.

Key Concept

Distance-Speed Proportionality for Simultaneous Travel and Relative Motion

Estimated Time:3m 0s

Question 284Question →

A specialty tea manufacturer blends three varieties of tea leaves—Darjeeling, Assam, and Ceylon. Initially, the weights of Darjeeling, Assam, and Ceylon in a batch are in the ratio $3 : 4 : 5$ , respectively. After $20$ kilograms of Ceylon tea leaves are removed from the batch and replaced with $20$ kilograms of Darjeeling tea leaves, the new ratio of Darjeeling to Assam to Ceylon becomes $5 : 4 : 3$ . What was the total weight, in kilograms, of the initial tea batch?

Show answer & explanation

Answer: 120

Answer

The initial total weight of the tea batch was 120 kg.

By assigning

x

as the multiplier for the initial ratio

3 : 4 : 5

, the initial total weight is

12x

. The swap of

20

kg changes Darjeeling to

3x + 20

and Ceylon to

5x - 20

, while Assam remains

4x

. Setting

\frac{3x + 20}{4x} = \frac{5}{4}

gives

x = 10

, making the initial total weight

12 \times 10 = 120

kg.

Step-by-Step Solution

1

Define initial quantities using a common ratio multiplier

x

.

Darjeeling =

3x

, Assam =

4x

, Ceylon =

5x

. Total weight =

12x

.

Expressing quantities in terms of a single variable

x

allows setting up algebraic equations based on the changed quantities.

2

Adjust quantities after the exchange of tea leaves.

New Darjeeling =

3x + 20

, New Assam =

4x

, New Ceylon =

5x - 20

.

Replacing

20

kg of Ceylon with

20

kg of Darjeeling increases Darjeeling by

20

kg and decreases Ceylon by

20

kg while leaving Assam unchanged.

3

Set up a proportion equation comparing the new ratio of Darjeeling to Assam.

\frac{3x + 20}{4x} = \frac{5}{4}

The new ratio of Darjeeling to Assam is given as

5 : 4

.

4

Solve the linear equation for

x

.

x = 10

Cross-multiplying yields

12x + 80 = 20x

, which simplifies to

8x = 80

.

5

Compute the total initial weight.

12 \times 10 = 120

kg

The initial total weight is

12x

, so substituting

x = 10

gives the final answer of

120

kg.

Key Concept

Setting up algebraic equations for 3-part ratios under quantity addition/subtraction.

Estimated Time:1m 30s

Question 285Question →

An oceanographic research station uses two deep-sea water sampling pumps, Pump $A$ and Pump $B$ , to fill a designated storage tank with seawater. Working alone at its constant rate, Pump $A$ can fill the tank in $10$ hours. Working alone at its constant rate, Pump $B$ can fill the tank in $15$ hours. If Pump $A$ begins filling the empty tank alone and operates for $4$ hours before shutting down, after which Pump $B$ operates alone to finish filling the tank, how many total hours does it take to fill the tank?

Show answer & explanation

Answer: 13

Answer

The total time required to fill the storage tank is 13 hours.

Pump A completes 2/5 of the tank during its 4 hours of operation. The remaining 3/5 of the tank is filled by Pump B in 9 hours. Summing the 4 hours from Pump A and the 9 hours from Pump B gives a total time of 13 hours.

Step-by-Step Solution

1

Calculate individual work rates

Pump A rate = 1/10 tank/hour; Pump B rate = 1/15 tank/hour

Work rate is equal to the fraction of the job completed per unit of time (1 divided by total time).

2

Calculate work completed during the first stage

Work done by Pump A = 4 * (1/10) = 2/5 of the tank

Work completed equals work rate multiplied by operating time.

3

Determine the remaining work

Remaining work = 1 - 2/5 = 3/5 of the tank

The entire job is represented by 1 whole unit.

4

Calculate time required for the second stage

Time for Pump B = (3/5) / (1/15) = 9 hours

Time taken equals remaining work divided by the rate of Pump B.

5

Sum the time for both stages

Total time = 4 + 9 = 13 hours

The question asks for the total elapsed time from the start of filling to completion.

Key Concept

Work Rate and Combined Work

Estimated Time:2m 0s

Question 286Question →

An investment portfolio consists of three asset classes: Equities, Bonds, and Real Estate. The total monetary value of Equities to Bonds is in the ratio $3 : 2$ , and the total value of Bonds to Real Estate is in the ratio $4 : 5$ . Over the past year, Equities yielded an average return of $12\%$ . If the overall weighted average return of the entire portfolio was $10.4\%$ , and the average percentage return of Bonds was $25\%$ higher than the average percentage return of Real Estate, what was the average percentage return of Bonds?

Show answer & explanation

Answer: 10.5

Answer

The average percentage return of Bonds was 10.5%.

The correct return of Bonds is obtained by first unifying the given ratios into a continuous ratio of weights

6 : 4 : 5

for Equities, Bonds, and Real Estate respectively. Substituting these weights and the algebraic relation between Real Estate return (

r

) and Bond return (

1.25r

) into the weighted average expression yields

r = 8.4\%

. Multiplying this by

1.25

gives the bond return of

10.5\%

.

Step-by-Step Solution

1

Determine the combined ratio of portfolio weights for all three asset classes

The ratio Equities : Bonds : Real Estate is 6 : 4 : 5, giving a total weight sum of 15 units.

Since Equities to Bonds is 3:2 (or 6:4) and Bonds to Real Estate is 4:5, the common term Bonds allows combining the ratios into 6 : 4 : 5.

2

Express the unknown asset returns in terms of a single variable

Return of Real Estate =

r\%

, Return of Bonds =

1.25r\%

.

The problem states the return of Bonds is 25% higher than that of Real Estate (

r + 0.25r = 1.25r

).

3

Set up and solve the weighted average formula for the portfolio

r = 8.4\%

.

Using

\frac{6(12) + 4(1.25r) + 5r}{15} = 10.4

, we get

72 + 10r = 156

, leading to

10r = 84

.

4

Calculate the targeted return for Bonds

Return of Bonds =

10.5\%

.

Multiplying

1.25

by

8.4

yields

10.5

.

Key Concept

Combining multiple ratios to establish weighted average component weights and solving multi-variable weighted average equations

Estimated Time:2m 30s

Question 287Question →

A cargo logistics warehouse handles three types of freight: Standard, Express, and Fragile. At the start of the week, $\frac{2}{5}$ of the total shipments in the warehouse were Standard freight, $0.35$ of the total shipments were Express freight, and the remaining shipments were Fragile freight. During a midweek inventory adjustment, the number of Standard freight shipments increased by $25\%$ , the number of Express freight shipments decreased by $20\%$ , and the number of Fragile freight shipments remained unchanged at $125$ units. What was the total number of shipments in the warehouse after the midweek inventory adjustment?

Show answer & explanation

Answer: 515

Answer

515 shipments

Converting

\frac{2}{5}

to a decimal gives

0.40

. The initial proportion of Fragile freight is

1 - 0.40 - 0.35 = 0.25

. Since Fragile freight equals

125

shipments, the initial total number of shipments is

\frac{125}{0.25} = 500

. Prior to adjustment, the warehouse has

0.40 \times 500 = 200

Standard shipments and

0.35 \times 500 = 175

Express shipments. After the adjustment, Standard shipments increase to

200 \times 1.25 = 250

, Express shipments decrease to

175 \times 0.80 = 140

, and Fragile shipments remain

125

. Summing these post-adjustment counts yields

250 + 140 + 125 = 515

shipments.

Step-by-Step Solution

1

Determine the initial fraction representing Fragile freight.

Fragile freight constitutes

0.25

(or

25\%

) of the total initial shipments.

Converting

\frac{2}{5}

to a decimal yields

0.40

. Subtracting the Standard (

0.40

) and Express (

0.35

) proportions from the total (

1.00

) leaves

1.00 - 0.40 - 0.35 = 0.25

.

2

Calculate the total initial number of shipments in the warehouse.

Initial total shipments =

500

.

Since

25\%

of the total initial shipments equals

125

Fragile units, the initial total is

\frac{125}{0.25} = 500

.

3

Calculate the initial quantities of Standard and Express shipments.

Initial Standard shipments =

200

, Initial Express shipments =

175

.

Standard shipments =

0.40 \times 500 = 200

, and Express shipments =

0.35 \times 500 = 175

.

4

Apply the inventory percentage changes to each category.

New Standard shipments =

250

, New Express shipments =

140

, Fragile shipments =

125

.

A

25\%

increase in Standard shipments gives

200 \times 1.25 = 250

. A

20\%

decrease in Express shipments gives

175 \times 0.80 = 140

. Fragile shipments remain

125

.

5

Sum the adjusted shipment counts to determine the net total.

Final total shipments =

515

.

Adding the three updated categories gives

250 + 140 + 125 = 515

.

Key Concept

Fractions, Decimals, and Percents Arithmetic

Estimated Time:1m 30s

Question 288Question →

A café sells small cold brew coffees for $\$4$ each and large cold brew coffees for $\$6$ each. On Monday morning, the café sold a total of $40$ cold brew coffees for $\$196$ . How many large cold brew coffees were sold?

Show answer & explanation

Answer: 18

Answer

The café sold 18 large cold brew coffees.

Setting up the system

s + l = 40

and

4s + 6l = 196

, substituting

s = 40 - l

yields

4(40 - l) + 6l = 196

. Simplifying gives

160 + 2l = 196

, so

2l = 36

and

l = 18

.

Step-by-Step Solution

1

Set up equations based on total count and total cost.

Let

s

= number of small coffees and

l

= number of large coffees. The system is

s + l = 40

and

4s + 6l = 196

.

The problem gives two distinct relationships between the quantity sold and the total dollar value.

2

Substitute

s = 40 - l

into the second equation.

4(40 - l) + 6l = 196 \implies 160 + 2l = 196

.

Substituting reduces the system from two variables to a single equation in

l

.

3

Solve the linear equation for

l

.

2l = 36 \implies l = 18

.

Subtracting 160 from both sides and dividing by 2 yields the number of large cold brews.

Key Concept

Solving systems of linear equations in two variables using substitution.

Question 289Question →

A financial analyst models a company's total annual revenue, $R(t)$ , and total annual expenses, $E(t)$ , in thousands of dollars, as linear functions of time $t$ , where $t$ represents the number of years since 2020. In 2022 ( $t = 2$ ), the annual revenue was $\$850,000$ and annual expenses were $\$610,000$ . In 2025 ( $t = 5$ ), the annual revenue reached $\$1,120,000$ while annual expenses reached $\$700,000$ . Assuming these linear trends continue, in which calendar year will the company's annual profit (defined as total annual revenue minus total annual expenses) reach exactly $\$540,000$ ?

Show answer & explanation

Answer: 2027

Answer

The company's annual profit will reach $540,000 in the year 2027.

The linear profit function increases at a constant rate of

60,000 per year from a base profit of

240,000 in 2022 (

t = 2

). Solving

240 + 60(t - 2) = 540

yields

t = 7

, which corresponds to the calendar year 2027.

Step-by-Step Solution

1

Calculate the annual profit in thousands of dollars for the known years 2022 (

t = 2

) and 2025 (

t = 5

).

In 2022 (

t = 2

), profit

P(2) = 850 - 610 = 240

thousand dollars. In 2025 (

t = 5

), profit

P(5) = 1120 - 700 = 420

thousand dollars.

Profit is defined as revenue minus expenses.

2

Determine the linear profit function

P(t) = mt + b

.

The rate of change of profit per year is

m = \frac{420 - 240}{5 - 2} = \frac{180}{3} = 60

thousand dollars per year.

Since both revenue and expenses are linear functions of time

t

, their difference

P(t) = R(t) - E(t)

is also a linear function of

t

.

3

Set up the linear equation for profit using point-slope form.

P(t) - 240 = 60(t - 2) \implies P(t) = 60t + 120

.

Using the point

(2, 240)

and slope

m = 60

establishes the complete linear equation for annual profit.

4

Solve the linear equation for

t

when profit

P(t) = 540

thousand dollars.

60t + 120 = 540 \implies 60t = 420 \implies t = 7

.

Setting the profit equal to 540 gives the value of

t

years after 2020.

5

Convert the value of

t

into the target calendar year.

Calendar Year

= 2020 + 7 = 2027

.

Since

t

represents years elapsed since 2020,

t = 7

corresponds to calendar year 2027.

Key Concept

Linear Modeling and Linear Equations in Two Variables

Question 290Question →

How many integer values of $x$ satisfy both $|x^2 - 10| \le 6$ and $|x - 1| > 2$ ?

Show answer & explanation

Answer: 4

Answer

4 integer values satisfy both inequalities:

-4

,

-3

,

-2

, and

4

.

Solving

|x^2 - 10| \le 6

gives

4 \le x^2 \le 16

, which yields 6 integer solutions:

\{-4, -3, -2, 2, 3, 4\}

. Applying the condition

|x - 1| > 2

requires

x < -1

or

x > 3

. Only

-4, -3, -2

, and

4

satisfy both conditions, giving exactly 4 valid integer values.

Step-by-Step Solution

1

Solve the quadratic absolute value inequality

|x^2 - 10| \le 6

.

-6 \le x^2 - 10 \le 6 \implies 4 \le x^2 \le 16

.

Removing the absolute value bars gives the bounded inequality

-6 \le x^2 - 10 \le 6

. Adding 10 across all parts isolates

x^2

.

2

Identify all candidate integer values of

x

for which

4 \le x^2 \le 16

.

x \in \{-4, -3, -2, 2, 3, 4\}

.

Taking the square root yields

-4 \le x \le -2

or

2 \le x \le 4

. The integer solutions are

-4, -3, -2, 2, 3, 4

.

3

Solve the linear absolute value inequality

|x - 1| > 2

.

x > 3

or

x < -1

.

An absolute value distance greater than 2 expands into two disjunctive linear inequalities:

x - 1 > 2

or

x - 1 < -2

.

4

Intersect the candidate integer set

\{-4, -3, -2, 2, 3, 4\}

with the domain

x < -1

or

x > 3

.

The valid integers are

-4, -3, -2,

and

4

. The values

2

and

3

fail the inequality since they fall inside the excluded interval

[-1, 3]

.

Testing each candidate shows that

-4, -3, -2

are strictly less than

-1

, and

4

is strictly greater than

3

.

Key Concept

Solving compound absolute value inequalities and finding integer solution set intersections.

Question 291Question →

What is the remainder when the integer $N = 3^{83} + 7^{83}$ is divided by $100$ ?

Show answer & explanation

Answer: 70

Answer

The remainder when

N = 3^{83} + 7^{83}

is divided by

100

is 70.

Finding the remainder when an integer is divided by

100

is equivalent to finding its last two digits. Powers of

3

repeat their last two digits in cycles of

20

, and

3^{20} \equiv 1 \pmod{100}

. Thus

3^{83} \equiv 3^3 = 27 \pmod{100}

. Powers of

7

repeat their last two digits in cycles of

4

, with

7^4 = 2401 \equiv 1 \pmod{100}

. Thus

7^{83} \equiv 7^3 = 343 \equiv 43 \pmod{100}

. Combining these gives

27 + 43 = 70

.

Step-by-Step Solution

1

Find the remainder of

3^{83}

modulo

100

by identifying the cyclicity pattern of powers of

3

.

3^{20} \equiv 1 \pmod{100}

, which implies

3^{83} = (3^{20})^4 \cdot 3^3 \equiv 1^4 \cdot 27 = 27 \pmod{100}

.

Since

3^{20} \equiv 1 \pmod{100}

, exponent

83

can be reduced to

83 \pmod{20} = 3

.

2

Find the remainder of

7^{83}

modulo

100

by identifying the cyclicity pattern of powers of

7

.

7^4 = 2401 \equiv 1 \pmod{100}

, which implies

7^{83} = (7^4)^{20} \cdot 7^3 \equiv 1^{20} \cdot 343 \equiv 43 \pmod{100}

.

Since

7^4 \equiv 1 \pmod{100}

, exponent

83

can be reduced to

83 \pmod{4} = 3

.

3

Sum the modular remainders to obtain the final remainder of

N

.

N = 3^{83} + 7^{83} \equiv 27 + 43 = 70 \pmod{100}

.

The remainder of a sum is equal to the sum of the individual remainders modulo

100

.

Key Concept

Modular Arithmetic and Cyclicity of Last Two Digits

Question 292Question →

What is the smallest positive integer $m$ such that the remainder when $6^m$ is divided by $11$ is equal to $8$ ?

Show answer & explanation

Answer: 7

Answer

The smallest positive integer

m

is 7.

Evaluating powers of 6 modulo 11 sequentially gives:

6^1 \equiv 6

,

6^2 \equiv 3

,

6^3 \equiv 7

,

6^4 \equiv 9

,

6^5 \equiv 10

,

6^6 \equiv 5

, and

6^7 \equiv 8 \pmod{11}

. Therefore,

m = 7

is the smallest positive integer satisfying the condition.

Step-by-Step Solution

1

Calculate the remainder of powers of 6 when divided by 11 starting from

m = 1

.

6^1 \equiv 6 \pmod{11}

,

6^2 \equiv 3 \pmod{11}

,

6^3 \equiv 7 \pmod{11}

,

6^4 \equiv 9 \pmod{11}

,

6^5 \equiv 10 \pmod{11}

,

6^6 \equiv 5 \pmod{11}

,

6^7 \equiv 8 \pmod{11}

.

Modular multiplication allows computing powers step-by-step by multiplying the previous remainder by 6 and finding the remainder modulo 11.

2

Identify the smallest exponent

m

that yields a remainder of 8.

m = 7

At

m = 7

,

6^7 = 279,936 = 11 \times 25,448 + 8

, which leaves a remainder of 8.

Key Concept

Modular Arithmetic and Cyclicity of Powers

Question 293Question →

A production facility uses two automated machines, Alpha and Beta, which operate at constant hourly production rates. Machine Alpha produces $x$ units per hour and Machine Beta produces $y$ units per hour. When both machines operate together for 4 hours, followed by Machine Alpha operating alone for 3 hours, the facility produces a total of 310 units. When Machine Beta operates alone for 2 hours, followed by both machines operating together for 5 hours, the facility produces a total of 325 units. If Machine Alpha's hourly rate is increased by $20\%$ and Machine Beta's hourly rate is decreased by $20\%$ , how many total units will both machines produce if they work together for 8 hours at their new rates?

Show answer & explanation

Answer: 448

Answer

The total number of units produced by both machines working together for 8 hours under the modified rates is 448.

Setting up the system of equations based on individual hours worked yields

7x + 4y = 310

and

5x + 7y = 325

. Solving this system gives original rates

x = 30

units/hr and

y = 25

units/hr. Applying the

20\%

increase to Alpha (

36

units/hr) and

20\%

decrease to Beta (

20

units/hr) gives a combined rate of

56

units/hr. Multiplying by

8

hours gives the final answer of

448

units.

Step-by-Step Solution

1

Translate the given operational scenarios into a system of two linear equations.

Equation 1:

7x + 4y = 310

; Equation 2:

5x + 7y = 325

.

Operating both machines for 4 hours and Alpha alone for 3 hours means Alpha works

4 + 3 = 7

hours while Beta works

4

hours, giving

7x + 4y = 310

. Operating Beta alone for 2 hours and both for 5 hours means Alpha works

5

hours while Beta works

2 + 5 = 7

hours, giving

5x + 7y = 325

.

2

Solve the system of linear equations for variables

x

and

y

.

x = 30

and

y = 25

.

Eliminating

y

by multiplying the first equation by 7 and the second by 4 yields

49x - 20x = 2170 - 1300 \implies 29x = 870 \implies x = 30

. Substituting

x = 30

back into

7x + 4y = 310

gives

210 + 4y = 310 \implies 4y = 100 \implies y = 25

.

3

Calculate the modified production rates after percentage adjustments.

New rate for Alpha is

36

units/hr; new rate for Beta is

20

units/hr.

A

20\%

increase on

x = 30

yields

30 \times 1.20 = 36

. A

20\%

decrease on

y = 25

yields

25 \times 0.80 = 20

.

4

Compute total combined output over 8 hours.

Total units produced =

448

.

Combined modified rate is

36 + 20 = 56

units per hour. Total production over 8 hours is

56 \times 8 = 448

.

Key Concept

Linear Equations in One and Two Variables

Estimated Time:2m 30s

Question 294Question →

A positive integer $n$ has exactly three distinct prime factors. If $n^2$ has exactly $147$ positive divisors, what is the minimum possible value of $n$ ?

Show answer & explanation

Answer: 1080

Answer

The minimum possible value of

n

is 1080.

By prime factorizing

n = p^a q^b r^c

, the number of divisors of

n^2

is given by

(2a+1)(2b+1)(2c+1) = 147

. Decomposing 147 into three factors greater than 1 gives

7 \times 7 \times 3

, which determines the exponents

a=3

,

b=3

, and

c=1

. Minimizing

n

by assigning the smallest primes (2, 3, and 5) to these exponents yields

2^3 \times 3^3 \times 5^1 = 1080

.

Step-by-Step Solution

1

Set up the prime factorization of

n

and

n^2

Let

n = p^a q^b r^c

, where

p < q < r

are prime numbers and

a, b, c \ge 1

. Then

n^2 = p^{2a} q^{2b} r^{2c}

.

Any positive integer can be uniquely represented as a product of prime powers.

2

Formulate the divisor counting equation for

n^2

The total number of positive divisors of

n^2

is

(2a+1)(2b+1)(2c+1) = 147

.

If an integer has prime factorization

\prod p_i^{e_i}

, its number of positive divisors is

\prod (e_i + 1)

.

3

Factor 147 into three odd factors greater than 1

The prime factorization of 147 is

3 \times 7^2

. The only factorization into three integers greater than 1 is

7 \times 7 \times 3

.

Since

a, b, c \ge 1

, we have

2a+1 \ge 3

,

2b+1 \ge 3

, and

2c+1 \ge 3

.

4

Solve for the exponents

a, b, c

Setting

2a+1=7

,

2b+1=7

, and

2c+1=3

yields

a=3

,

b=3

, and

c=1

.

Equating the factor values allows us to determine the required exponents of the prime factors of

n

.

5

Minimize

n

by choosing prime bases and assigning exponents

Choose the three smallest primes

p=2, q=3, r=5

and assign exponents

3, 3, 1

to get

n = 2^3 \times 3^3 \times 5^1 = 1080

.

To make the product as small as possible, smaller prime bases must be paired with larger exponents.

Key Concept

Determining the prime factorization and number of divisors of prime power expressions

Question 295Question →

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 296Question →

What is the sum of all the distinct prime factors of the integer $6^8 - 6^6$ ?

Show answer & explanation

Answer: 17

Answer

The sum of the distinct prime factors of

6^8 - 6^6

is 17.

Factoring out

6^6

yields

6^6(6^2 - 1) = 6^6 \times 35

. The prime factors of

6^6

are 2 and 3, while the prime factors of 35 are 5 and 7. The distinct prime factors are 2, 3, 5, and 7. Summing these prime numbers gives

2 + 3 + 5 + 7 = 17

.

Step-by-Step Solution

1

Factor out the common term

6^6

6^8 - 6^6 = 6^6(6^2 - 1) = 6^6 \times 35

Factoring simplifies the large exponent expression into a product of smaller integers.

2

Break down each base into its prime factors

6^6 = (2 \times 3)^6 = 2^6 \times 3^6

and

35 = 5 \times 7

Prime factorization requires expressing all bases as prime numbers.

3

List the distinct prime bases and compute their sum

Distinct prime factors: 2, 3, 5, 7. Sum =

2 + 3 + 5 + 7 = 17

.

Exponents do not affect which prime numbers are factors, only how many times they divide the number.

Key Concept

Prime Factorization of Difference of Exponents

Question 297Question →

If $n$ is an integer such that $|3 - 2n| \le 9$ and $|n - 1| \ge 2$ , what is the sum of all possible values of $n$ ?

Show answer & explanation

Answer: 12

Answer

The sum of all possible integer values of

n

is 12.

Solving

|3 - 2n| \le 9

gives

-3 \le n \le 6

. Solving

|n - 1| \ge 2

restricts

n

to

n \le -1

or

n \ge 3

. Taking the intersection yields the integer set

\{-3, -2, -1, 3, 4, 5, 6\}

, whose sum equals 12.

Step-by-Step Solution

1

Unpack and solve

|3 - 2n| \le 9

-3 \le n \le 6

Removing absolute value yields

-9 \le 3 - 2n \le 9

. Dividing by

-2

requires reversing the inequality direction.

2

Unpack and solve

|n - 1| \ge 2

n \le -1

or

n \ge 3

An absolute value greater than or equal to 2 implies distance from 1 is at least 2 units in either direction.

3

Determine the intersection set of integers

\{-3, -2, -1, 3, 4, 5, 6\}

Filters out integers 0, 1, and 2 from the continuous range

[-3, 6]

.

4

Calculate the sum of the valid integers

12

Summing

(-3) + (-2) + (-1) + 3 + 4 + 5 + 6 = 12

.

Key Concept

Combining system of absolute value inequalities and handling inequality sign flips when dividing by negative quantities.

Question 298Question →

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 299Question →

An artisanal coffee roasting company purchases unroasted green coffee beans at a cost of $\$12.00$ per kilogram. During the roasting process, moisture loss reduces the total weight of the coffee beans by $20\%$ . To set the regular retail price per kilogram of roasted beans, the company marks up the effective cost price per kilogram of roasted beans by $50\%$ . If the company sells the roasted beans at a promotional discount of $10\%$ off the regular retail price, what is the net profit percentage earned on the total cost of the unroasted green coffee beans?

Show answer & explanation

Answer: 35

Answer

The net profit percentage earned on the total cost of the unroasted green coffee beans is 35%.

The correct calculation yields a total revenue of

16.20 from 0.80 kg of roasted coffee produced from 1 kg of green coffee purchased for

12.00. This results in a net profit of

4.20, which is exactly 35% of the initial

12.00 cost.

Step-by-Step Solution

1

Calculate the effective cost per kilogram of roasted beans after weight loss.

Effective cost per kg of roasted beans =

12.00 / (1 - 0.20) =

15.00 per kg.

Because weight decreases by 20%, 1 kg of green beans yields 0.80 kg of roasted beans, raising the per-unit cost.

2

Calculate the regular retail price per kilogram of roasted beans.

Regular retail price =

15.00 * (1 + 0.50) =

22.50 per kg.

The 50% markup is applied to the effective cost price of the roasted beans.

3

Calculate the promotional selling price after a 10% discount.

Discounted selling price =

22.50 * (1 - 0.10) =

20.25 per kg.

The discount reduces the regular retail price by 10%.

4

Find total revenue generated from the yield of 1 kg of green coffee beans.

Total revenue = 0.80 kg *

20.25/kg =

16.20.

1 kg of initial green beans produces 0.80 kg of sellable roasted product.

5

Calculate the net profit percentage based on the initial cost.

Net profit percentage = ((

16.20 -

12.00) /

12.00) * 100% = (

4.20 / $12.00) * 100% = 35%.

Profit percentage is the net profit divided by the original total cost.

Key Concept

Multi-step profit calculation involving shrink/yield loss, markup, successive discounts, and base identification.

Question 300Question →

If $k$ is an integer such that $|2k - 7| \le 11$ and $|k + 2| > 4$ , how many possible values of $k$ exist?

Show answer & explanation

Answer: 7

Answer

The total number of possible integer values of

k

is 7.

Unfolding

|2k - 7| \le 11

gives

-2 \le k \le 9

, representing 12 possible integers. Unfolding

|k + 2| > 4

gives

k > 2

or

k < -6

. Intersecting these two regions for real integers eliminates

k < -6

(since

k \ge -2

) and restricts the set to

3 \le k \le 9

. The integers in this range are

3, 4, 5, 6, 7, 8,

and

9

, yielding a total of 7 valid values.

Step-by-Step Solution

1

Solve the first absolute value inequality

|2k - 7| \le 11

.

-2 \le k \le 9

Expanding the absolute value gives

-11 \le 2k - 7 \le 11

. Adding 7 to all parts yields

-4 \le 2k \le 18

, and dividing by 2 results in

-2 \le k \le 9

.

2

Solve the second absolute value inequality

|k + 2| > 4

.

k > 2

or

k < -6

Expanding the strict absolute value inequality gives two cases:

k + 2 > 4 \implies k > 2

, or

k + 2 < -4 \implies k < -6

.

3

Find the intersection of the two solution sets for integer values of

k

.

3 \le k \le 9

Since the first inequality requires

k \ge -2

, no integer can satisfy both

k \ge -2

and

k < -6

. Thus,

k

must satisfy

-2 \le k \le 9

and

k > 2

, which reduces to

3 \le k \le 9

.

4

Count the number of integers in the range

3 \le k \le 9

.

7

The valid integers in this range are

3, 4, 5, 6, 7, 8,

and

9

. The total count is

9 - 3 + 1 = 7

.

Key Concept

Solving compound absolute value inequalities for integer solution counts

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