All practice questions

Question 21Question →

A commercial bakery produces artisan cakes at a base cost of $\$40$ per cake. The bakery marks up the base cost by $75\%$ to set the catalog price. To encourage bulk corporate orders, the bakery offers a $20\%$ discount off the catalog price. If the bakery incurs an additional packaging and delivery expense of $\$6$ per cake, what is the bakery's net profit, in dollars, on each corporate cake sold?

Show answer & explanation

Answer: 10

Answer

The bakery earns a net profit of $10 on each corporate cake sold.

The catalog price is

40 \times 1.75 =

70. Applying a 20% discount gives a discounted selling price of

70 \times 0.80 =

56. The total cost to produce and package each cake is

40 +

6 =

46. Subtracting total cost from selling price yields a net profit of

56 -

46 =

10.

Step-by-Step Solution

1

Calculate the catalog price using the base cost and markup percentage.

Catalog price =

40 \times (1 + 0.75) =

70

The catalog price is established by marking up the base production cost by 75%.

2

Calculate the selling price after applying the discount.

Selling price =

70 \times (1 - 0.20) =

56

The 20% promotional discount applies to the marked-up catalog price.

3

Calculate the total cost incurred by the bakery for each cake.

Total cost = Base cost + Overhead expense =

40 +

6 = $46

Net profit calculations must account for both production base cost and direct selling/packaging expenses.

4

Subtract total cost from selling price to determine net profit.

Net profit =

56 -

46 = $10

Net profit equals total revenue per unit minus total costs per unit.

Key Concept

Profit, Loss, and Markup with Successive Adjustments and Overhead Expenses

Question 22Question →

A dataset consists of five integers: $2, 5, 8, 11,$ and $x$ . The variance of the dataset is $10$ . If the median of the dataset is $8$ , what is the value of $x$ ?

Show answer & explanation

Answer: 9

Answer

The value of

x

is 9.

By setting up the equation for variance in terms of

x

, we obtain the quadratic equation

x^2 - 13x + 36 = 0

, which yields

x = 4

or

x = 9

. Arranging the set in ascending order for

x = 9

gives

\{2, 5, 8, 9, 11\}

, where the middle number (median) is 8, satisfying all conditions.

Step-by-Step Solution

1

Calculate the mean of the dataset in terms of

x

.

The mean is

\mu = \frac{26 + x}{5}

.

The mean of a dataset is the sum of all elements divided by the number of elements.

2

Set up the variance equation using the definition of population variance.

Variance

\sigma^2 = \frac{(2-\mu)^2 + (5-\mu)^2 + (8-\mu)^2 + (11-\mu)^2 + (x-\mu)^2}{5} = 10

.

Variance measures the average squared deviation from the mean.

3

Substitute

\mu = \frac{26+x}{5}

, clear denominators, expand the algebraic expression, and set up the quadratic equation.

x^2 - 13x + 36 = 0

.

Multiplying through by 25 and simplifying the quadratic terms yields a standard quadratic form.

4

Solve the quadratic equation for

x

.

x = 4

or

x = 9

.

Factoring

(x-4)(x-9) = 0

gives two potential solutions.

5

Evaluate the median condition for both potential values of

x

.

For

x = 9

, the ordered dataset is

\{2, 5, 8, 9, 11\}

, which has a median of 8.

For

x = 4

, the ordered set is

\{2, 4, 5, 8, 11\}

with median 5, which fails the given condition.

Key Concept

Variance calculation and dataset order statistics

Question 23Question →

A delivery van leaves a warehouse traveling along a straight highway at a constant speed of $45$ miles per hour. Exactly $40$ minutes later, a courier motorcycle leaves the same warehouse along the exact same route, traveling at a constant speed of $60$ miles per hour. How many hours after the courier motorcycle departs will it overtake the delivery van?

Show answer & explanation

Answer: 2

Answer

The courier motorcycle will overtake the delivery van 2 hours after the motorcycle departs.

The delivery van travels for

\frac{2}{3}

hour before the motorcycle begins moving, establishing a initial lead of

45 \times \frac{2}{3} = 30

miles. Since both vehicles travel in the same direction, the motorcycle gains on the van at a rate equal to the difference of their speeds,

60 - 45 = 15

miles per hour. To cover the 30-mile gap, it takes

\frac{30}{15} = 2

hours after the motorcycle departs.

Step-by-Step Solution

1

Convert the van's head-start time from minutes to hours.

Head-start time is

\frac{40}{60} = \frac{2}{3}

hours.

Speed is given in miles per hour, so time units must be converted to hours for consistency.

2

Calculate the lead distance gained by the van before the motorcycle starts moving.

Lead distance =

45 \times \frac{2}{3} = 30

miles.

Distance equals rate multiplied by time (

d = r \cdot t

).

3

Calculate the relative speed at which the motorcycle gains on the van.

Relative speed =

60 - 45 = 15

miles per hour.

When two objects move in the same direction, the rate at which the distance between them decreases is the difference between their individual speeds.

4

Determine the time needed for the motorcycle to eliminate the 30-mile gap.

Time =

\frac{30}{15} = 2

hours.

Time equals distance divided by relative speed (

t = \frac{d}{r_{\text{rel}}}

).

Key Concept

Catch-up scenarios and relative speed in same-direction movement.

Alternative Method

Equate the distance expressions for both vehicles at the moment of overtaking. Let

t

represent the motorcycle's travel time in hours. The van's travel time is

t + \frac{2}{3}

hours. Since both travel equal distances from the warehouse:

60t = 45\left(t + \frac{2}{3}\right) \implies 60t = 45t + 30 \implies 15t = 30 \implies t = 2

hours.

Estimated Time:1m 30s

Question 24Question →

At the beginning of the year, a cloud storage service had a total of $S$ active subscribers. During the first quarter, the number of subscribers increased by $20\%$ . In the second quarter, the number of subscribers decreased by $10\%$ . In the third quarter, the number of subscribers increased by $25\%$ . If the total net increase in subscribers from the beginning of the year to the end of the third quarter was $7{}000$ , what was the value of $S$ ?

Show answer & explanation

Answer: 20000

Answer

The initial number of subscribers

S

was

20{}000

.

Each quarter's percent change applies to the balance at the start of that quarter. An increase of

20\%

multiplies the quantity by

1.20

. A decrease of

10\%

multiplies it by

0.90

. A subsequent increase of

25\%

multiplies it by

1.25

. The cumulative factor is

1.20 \times 0.90 \times 1.25 = 1.35

. Since the final subscriber total is

1.35S

, the net gain is

0.35S

. Given that

0.35S = 7{}000

, dividing

7{}000

by

0.35

yields

S = 20{}000

.

Step-by-Step Solution

1

Calculate the composite growth factor after three quarters of successive percentage changes.

Growth factor

= (1.20) \times (0.90) \times (1.25) = 1.35

Successive percentage changes multiply together, each using the previous period's value as its new base.

2

Express the net increase in terms of the initial amount

S

.

Net increase

= 1.35S - S = 0.35S

The net change is the final value minus the starting value.

3

Set the net increase equal to

7{}000

and solve for

S

.

S = \frac{7{}000}{0.35} = 20{}000

Dividing the absolute net increase by the net percentage increase rate yields the original base value.

Key Concept

Successive Percent Change and Base Value Calculation

Estimated Time:1m 30s

Question 25Question →

A municipal water treatment facility utilizes three primary purification units—Unit P, Unit Q, and Unit R—to process large water batches. Operating alone at its constant rate, Unit P can process a full batch in $12$ hours, Unit Q can process a full batch in $18$ hours, and Unit R can process a full batch in $36$ hours. All three units begin processing a batch together. After $3$ hours, Unit P stops operating due to scheduled maintenance. Unit Q and Unit R continue operating together for an additional $2$ hours, after which Unit Q is shut down. Unit R then completes the remainder of the batch working alone. How many total hours does it take from start to finish to process the complete batch?

Show answer & explanation

Answer: 17

Answer

The total time required to process the complete batch from start to finish is 17 hours.

To find the total time from start to finish, analyze the work completed in each of the three stages:

1. In Stage 1 (3 hours), all three units operate together. Their combined rate is

\frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{3 + 2 + 1}{36} = \frac{6}{36} = \frac{1}{6}

batch per hour. In

3

hours, they complete

3 \times \frac{1}{6} = \frac{1}{2}

of the total batch.

2. In Stage 2 (2 hours), Unit P stops and Units Q and R continue together. Their combined rate is

\frac{1}{18} + \frac{1}{36} = \frac{2 + 1}{36} = \frac{3}{36} = \frac{1}{12}

batch per hour. In

2

hours, they complete

2 \times \frac{1}{12} = \frac{1}{6}

of the total batch.

3. After Stage 2, the total fraction of work completed is

\frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}

of the batch. The remaining fraction of work is

1 - \frac{2}{3} = \frac{1}{3}

of the batch.

4. In Stage 3, Unit R completes the remaining

\frac{1}{3}

batch alone at its rate of

\frac{1}{36}

batch per hour. The time taken by Unit R is

\frac{1/3}{1/36} = 12

hours.

Adding the durations of all three stages gives

3 + 2 + 12 = 17

total hours.

Step-by-Step Solution

1

Calculate individual work rates

Rate of P =

\frac{1}{12}

batch/hr, Rate of Q =

\frac{1}{18}

batch/hr, Rate of R =

\frac{1}{36}

batch/hr

Work rate is the reciprocal of the total time required to complete one entire job.

2

Calculate work done during Stage 1 (first 3 hours)

Combined rate of P, Q, R =

\frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{6}{36} = \frac{1}{6}

batch/hr. Work done =

3 \times \frac{1}{6} = \frac{1}{2}

batch

All three units operate simultaneously for 3 hours.

3

Calculate work done during Stage 2 (next 2 hours)

Combined rate of Q and R =

\frac{1}{18} + \frac{1}{36} = \frac{3}{36} = \frac{1}{12}

batch/hr. Work done =

2 \times \frac{1}{12} = \frac{1}{6}

batch

Unit P stops working, leaving Q and R to operate together for 2 hours.

4

Determine remaining work and time required for Stage 3

Remaining work =

1 - \left(\frac{1}{2} + \frac{1}{6}\right) = \frac{1}{3}

batch. Time for R alone =

\frac{1/3}{1/36} = 12

hours

Unit R must complete the remaining fraction of the batch operating alone at its constant rate.

5

Calculate total elapsed time

3\text{ hours} + 2\text{ hours} + 12\text{ hours} = 17\text{ hours}

Sum the durations of all three distinct operational stages.

Key Concept

Combined Work Rate in Multi-Stage Work Scenarios

Question 26Question →

What is the product of all real solutions to the equation

(x^2 - 6x + 10)^2 - 2(x^2 - 6x + 10) - 35 = 0?

Show answer & explanation

Answer: 3

Answer

The product of all real solutions to the equation is 3.

Substituting

u = x^2 - 6x + 10

converts the given equation into

u^2 - 2u - 35 = 0

, which factors as

(u - 7)(u + 5) = 0

. Substituting back yields two quadratic equations:

x^2 - 6x + 3 = 0

and

x^2 - 6x + 15 = 0

. Checking the discriminants reveals that

x^2 - 6x + 3 = 0

has

D = 24 > 0

, producing two real roots with product

\frac{c}{a} = 3

, whereas

x^2 - 6x + 15 = 0

has

D = -24 < 0

, producing no real roots. Therefore, the product of all real solutions is 3.

Step-by-Step Solution

1

Substitute a dummy variable to simplify the equation structure.

Let

u = x^2 - 6x + 10

. The equation becomes

u^2 - 2u - 35 = 0

.

Recognizing the repeated quadratic expression allows transforming a fourth-degree equation into a standard quadratic form.

2

Factor the quadratic equation in terms of

u

.

(u - 7)(u + 5) = 0 \implies u = 7 \text{ or } u = -5

.

Finding the values of

u

establishes the intermediate equations for

x

.

3

Analyze the first case

u = 7

for real solutions.

x^2 - 6x + 3 = 0

. Discriminant

D_1 = 36 - 12 = 24 > 0

. Product of real roots is

\frac{3}{1} = 3

.

A positive discriminant guarantees two real roots, and Vieta's formulas give their product directly.

4

Analyze the second case

u = -5

for real solutions.

x^2 - 6x + 15 = 0

. Discriminant

D_2 = 36 - 60 = -24 < 0

. No real roots.

A negative discriminant indicates complex conjugate roots, which must be excluded when finding the product of real solutions.

5

Combine results to find the final product of all real solutions.

Product = 3.

Only the two roots from the first case are real, so their product is the product of all real solutions.

Key Concept

Solving disguised quadratics via algebraic substitution and testing discriminants to filter out non-real roots before applying Vieta's formulas.

Question 27Question →

A manufacturing plant has a fixed daily setup cost of $\$1,500$ . For a specific product, the production cost for each of the first $50$ units is $\$60$ per unit, while each additional unit produced beyond the first $50$ costs $\$30$ per unit. If the average total cost per unit produced on a given day was $\$45$ , how many total units were produced on that day?

Show answer & explanation

Answer: 200

Answer

The total number of units produced on that day was 200.

Setting up the piecewise linear total cost expression

C(n) = 1,500 + 50(60) + (n-50)30 = 3,000 + 30n

and equating average cost

\frac{3,000 + 30n}{n}

to

45

yields

15n = 3,000

, which solves to

n = 200

.

Step-by-Step Solution

1

Model total daily production cost as a piecewise linear algebraic equation.

C(n) = 1,500 + (50 \times 60) + (n - 50) \times 30 = 3,000 + 30n

for

n > 50

.

Total cost combines fixed setup fees, cost of the initial 50 units, and tier-2 cost for units exceeding 50.

2

Formulate the equation for average cost per unit.

\frac{C(n)}{n} = \frac{3,000 + 30n}{n} = 45

Average cost is total daily cost divided by total quantity produced, given as $45 per unit.

3

Solve the algebraic equation for n.

3,000 + 30n = 45n \implies 15n = 3,000 \implies n = 200

Isolating n yields the exact volume of units needed to satisfy the average cost target.

Key Concept

Algebraic Equation Modeling with Piecewise Cost Functions

Estimated Time:2m 0s

Question 28Question →

Four distinct books—two mathematics books and two history books—are to be arranged in a single line on a shelf. If the two mathematics books must stand next to each other, in how many different linear arrangements can the four books be placed?

Show answer & explanation

Answer: 12

Answer

12 linear arrangements

To find the total number of linear arrangements where the two mathematics books are adjacent, treat the two mathematics books as a single combined block. This leaves 3 items to arrange (the mathematics block and the 2 individual history books), which can be arranged in

3! = 6

ways. Within the block, the 2 mathematics books can be ordered in

2! = 2

ways. Multiplying the independent arrangements yields

6 \times 2 = 12

total arrangements.

Step-by-Step Solution

1

Group the adjacent items into a single unit and count total units to arrange

3 units (1 grouped math unit + 2 individual history books), which can be arranged in

3! = 6

ways

Treating items that must be adjacent as a single block simplifies the linear arrangement into distinct available positions.

2

Determine internal permutations within the grouped unit

2! = 2

ways to order the two mathematics books inside their block

The mathematics books can switch positions with each other within their designated block.

3

Multiply external and internal arrangements

6 \times 2 = 12

total linear arrangements

By the Fundamental Counting Principle, total arrangements equal the product of unit arrangements and internal arrangements.

Key Concept

Linear Permutations with Adjacent Restrictions (Tie-Together Method)

Question 29Question →

A sequence of positive integers $a_n$ is defined by $a_n = 2^n + 5^n$ for all integers $n \ge 1$ . What is the remainder when $a_{100}$ is divided by $7$ ?

Show answer & explanation

Answer: 4

Answer

The remainder when

a_{100}

is divided by

7

is

4

.

By reducing

5 \equiv -2 \pmod 7

, we observe that

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even. The powers of

2 \pmod 7

follow a 3-step cycle (

2, 4, 1

). Dividing the exponent

100

by

3

leaves a remainder of

1

, meaning

2^{100} \equiv 2^1 = 2 \pmod 7

. Adding the remainders for both terms yields

2 + 2 = 4

.

Step-by-Step Solution

1

Use modular arithmetic to simplify the base

5 \pmod 7

.

5 \equiv -2 \pmod 7

, which implies

5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7

because the exponent

100

is even.

Converting

5

to

-2

allows both terms to be expressed using powers of

2

.

2

Determine the remainder cycle of powers of

2

when divided by

7

.

2^1 \equiv 2 \pmod 7

,

2^2 \equiv 4 \pmod 7

, and

2^3 = 8 \equiv 1 \pmod 7

. The pattern repeats every

3

powers.

Finding the period of cyclicity simplifies evaluating large powers.

3

Divide the exponent

100

by the cycle length

3

.

100 = 3 \times 33 + 1

, leaving a remainder of

1

. Thus,

2^{100} \equiv 2^1 = 2 \pmod 7

.

The remainder of the exponent modulo the cycle length determines the equivalent reduced power.

4

Combine the remainders for

2^{100}

and

5^{100}

.

a_{100} = 2^{100} + 5^{100} \equiv 2 + 2 = 4 \pmod 7

.

Adding the individual modular values gives the overall remainder.

Key Concept

Modular arithmetic cyclicity and negative congruences

Question 30Question →

A set $S$ consists of consecutive integers. The arithmetic mean of all the positive integers in set $S$ is $18.5$ , and the arithmetic mean of all the negative integers in set $S$ is $-12$ . How many integers are in set $S$ ?

Show answer & explanation

Answer: 60

Answer

The total number of integers in set

S

is

60

.

The positive integers in set

S

are

1, 2, \dots, 36

, which have an arithmetic mean of

\frac{1 + 36}{2} = 18.5

. The negative integers in set

S

are

-23, -22, \dots, -1

, which have an arithmetic mean of

\frac{-23 + (-1)}{2} = -12

. Since set

S

consists of consecutive integers spanning from

-23

to

36

, it includes

23

negative integers,

36

positive integers, and the integer

0

. The total number of elements is

23 + 1 + 36 = 60

.

Step-by-Step Solution

1

Find the largest positive integer in set

S

.

The largest positive integer is

36

, meaning there are

36

positive integers in

S

.

The positive integers in

S

must form a consecutive sequence starting at

1

up to some maximum integer

m

. The average of consecutive integers from

1

to

m

is

\frac{1 + m}{2}

. Setting

\frac{1 + m}{2} = 18.5

yields

1 + m = 37

, so

m = 36

.

2

Find the smallest negative integer in set

S

.

The smallest negative integer is

-23

, meaning there are

23

negative integers in

S

.

The negative integers in

S

must form a consecutive sequence ending at

-1

down to some minimum integer

-k

. The average of consecutive integers from

-k

to

-1

is

\frac{-k + (-1)}{2}

. Setting

\frac{-k - 1}{2} = -12

yields

-k - 1 = -24

, so

k = 23

.

3

Calculate the total number of elements in set

S

.

Set

S

contains

60

integers.

Because set

S

contains consecutive integers ranging from negative to positive values, it must also contain

0

. The total count is

23\text{ (negative integers)} + 1\text{ (the integer zero)} + 36\text{ (positive integers)} = 60

.

Key Concept

Evenly spaced set averages and classification of zero in consecutive integer sets.

Estimated Time:2m 0s

Question 31Question →

A manufacturing plant uses three assembly lines, $L_1$ , $L_2$ , and $L_3$ , to produce three custom components, $X$ , $Y$ , and $Z$ .

- Producing one unit of component $X$ requires 2 hours on $L_1$ , 1 hour on $L_2$ , and 3 hours on $L_3$ .
- Producing one unit of component $Y$ requires 3 hours on $L_1$ , 4 hours on $L_2$ , and 2 hours on $L_3$ .
- Producing one unit of component $Z$ requires 1 hour on $L_1$ , 2 hours on $L_2$ , and 4 hours on $L_3$ .

During a given week, assembly lines $L_1$ , $L_2$ , and $L_3$ were operated for a total of 140 hours, 165 hours, and 235 hours, respectively, with zero idle time. Assuming full capacity utilization, how many units of component $Z$ were produced during that week?

Show answer & explanation

Answer: 30

Answer

30 units of component Z were produced.

Translating the assembly line operational hours into a 3x3 system of linear equations yields

2X + 3Y + Z = 140

,

X + 4Y + 2Z = 165

, and

3X + 2Y + 4Z = 235

. Eliminating

X

results in two equations in

Y

and

Z

:

5Y + 3Z = 190

and

5Y + Z = 130

. Subtracting these equations gives

2Z = 60

, so

Z = 30

.

Step-by-Step Solution

1

Formulate a system of 3 linear equations representing total hours logged on each assembly line.

Line 1:

2X + 3Y + Z = 140

; Line 2:

X + 4Y + 2Z = 165

; Line 3:

3X + 2Y + 4Z = 235

.

Each component requires specific line processing time, and total time per line equals total available capacity.

2

Eliminate variable

X

by substituting

X = 165 - 4Y - 2Z

into the other two equations.

Equation A:

5Y + 3Z = 190

and Equation B:

5Y + Z = 130

.

Reducing a 3-variable system to a 2-variable system simplifies linear elimination.

3

Subtract Equation B from Equation A.

2Z = 60, so Z = 30.

Since the coefficients of

Y

in both reduced equations are identical (

5Y

), subtraction directly isolates

Z

.

Key Concept

Solving Systems of Three Linear Equations via Variable Substitution and Elimination

Question 32Question →

Set $S$ consists of $n$ consecutive even integers. The arithmetic mean of all the integers in set $S$ is $45$ . If the difference between the largest integer and the smallest integer in set $S$ is $34$ , what is the value of the largest integer in set $S$ ?

Show answer & explanation

Answer: 62

Answer

The largest integer in set

S

is

62

.

In any set of evenly spaced numbers, such as consecutive even integers, the arithmetic mean is equal to the average of the smallest term and the largest term. Since the arithmetic mean is 45, the sum of the smallest term and the largest term must be

45 \times 2 = 90

. Combining this with the given fact that the difference between the largest term and the smallest term is 34 creates a system of equations: Smallest + Largest = 90 and Largest - Smallest = 34. Adding these two equations cancels out the smallest term, resulting in 2 * Largest = 124, which gives 62 for the largest integer.

Step-by-Step Solution

1

Express the arithmetic mean of the set in terms of the smallest element (

F

) and largest element (

L

).

F + L = 90

For any evenly spaced set of numbers, the arithmetic mean is equal to the average of the first and last elements:

\frac{F + L}{2} = 45

.

2

Set up the equation for the difference between the largest and smallest elements.

L - F = 34

The question specifies that the largest integer exceeds the smallest integer by 34.

3

Solve the system of two linear equations for

L

.

L = 62

Adding

(F + L) + (L - F) = 90 + 34

eliminates

F

, leaving

2L = 124

, which yields

L = 62

.

Key Concept

Equivalence of arithmetic mean to the average of the first and last terms in an evenly spaced set

Question 33Question →

A software analytics company offers three annual subscription plans: Standard, Professional, and Enterprise.

• A client purchasing 3 Standard, 2 Professional, and 1 Enterprise plan pays a total of $1,110. �� A client purchasing 1 Standard, 4 Professional, and 2 Enterprise plans pays a total of$ 1,620.
• A client purchasing 4 Standard, 1 Professional, and 3 Enterprise plans pays a total of $1,730.

What is the cost, in dollars, of 1 Enterprise plan?

Show answer & explanation

Answer: 350

Answer

The cost of 1 Enterprise plan is 350 dollars.

Setting up equations for each purchase tier gives

3x + 2y + z = 1110

,

x + 4y + 2z = 1620

, and

4x + y + 3z = 1730

. Isolating

x

in the second equation gives

x = 1620 - 4y - 2z

. Substituting

x

into the first and third equations yields

2y + z = 750

and

3y + z = 950

, respectively. Subtracting these reduced equations gives

y = 200

, which upon back-substitution into

2y + z = 750

reveals

z = 350

. Thus, 1 Enterprise plan costs 350 dollars.

Step-by-Step Solution

1

Formulate the linear system of equations from the given conditions.

Let

x

be the price of a Standard plan,

y

the price of a Professional plan, and

z

the price of an Enterprise plan:
(1)

3x + 2y + z = 1110

(2)

x + 4y + 2z = 1620

(3)

4x + y + 3z = 1730

Translating the verbal conditions into algebraic equations creates a solvable system.

2

Isolate variable

x

in equation (2) and substitute it into equations (1) and (3).

From (2),

x = 1620 - 4y - 2z

.
Substituting into (1):

3(1620 - 4y - 2z) + 2y + z = 1110 \implies 4860 - 12y - 6z + 2y + z = 1110 \implies 10y + 5z = 3750 \implies 2y + z = 750

(Equation 4)

Substituting into (3):

4(1620 - 4y - 2z) + y + 3z = 1730 \implies 6480 - 16y - 8z + y + 3z = 1730 \implies 15y + 5z = 4750 \implies 3y + z = 950

(Equation 5)

Eliminating

x

reduces the system to two linear equations with two variables.

3

Solve the 2x2 system of equations for

y

and

z

.

Subtracting Equation (4) from Equation (5):

(3y + z) - (2y + z) = 950 - 750 \implies y = 200

.

Substitute

y = 200

back into Equation (4):

2(200) + z = 750 \implies 400 + z = 750 \implies z = 350

.

Solving the reduced system yields the exact values of

y

and

z

.

Key Concept

Solving Systems of Three Linear Equations via Gaussian Elimination / Variable Substitution

Question 34Question →

At an investment firm with $200$ portfolio managers, $110$ managers invest in Equities, $95$ invest in Fixed Income, and $80$ invest in Commodities. Furthermore, $45$ managers invest in both Equities and Fixed Income, $35$ invest in both Equities and Commodities, and $30$ invest in both Fixed Income and Commodities. If $15$ managers do not invest in any of these three asset classes, how many portfolio managers invest in all three asset classes?

Show answer & explanation

Answer: 10

Answer

The correct answer is 10 portfolio managers.

Subtracting the 15 managers who do not invest in any asset class from the total 200 yields 185 managers in at least one asset class. Applying the 3-set inclusion-exclusion formula

185 = (110 + 95 + 80) - (45 + 35 + 30) + x

simplifies to

185 = 175 + x

, which gives

x = 10

for the number of managers investing in all three asset classes.

Step-by-Step Solution

1

Determine the total number of portfolio managers investing in at least one asset class.

185 managers

Subtracting the 15 managers who invest in none of the three asset classes from the total of 200 gives the size of the union of the three sets:

200 - 15 = 185

.

2

Set up the Inclusion-Exclusion formula for three overlapping sets.

|E \cup F \cup C| = |E| + |F| + |C| - (|E \cap F| + |E \cap C| + |F \cap C|) + |E \cap F \cap C|

To avoid overcounting elements present in multiple set intersections, pairwise intersections are subtracted and the triple intersection is added back.

3

Substitute known values into the equation and solve for the unknown triple intersection

x

.

x = 10

185 = 110 + 95 + 80 - (45 + 35 + 30) + x \implies 185 = 285 - 110 + x \implies 185 = 175 + x \implies x = 10

.

Key Concept

Three-Set Inclusion-Exclusion Principle

Question 35Question →

An antique dealer purchased a vintage writing desk for $\$600$ . To prepare the desk for sale, the dealer spent $\$40$ on restoration. The dealer then marked up the initial purchase price of $\$600$ by $50\%$ to establish the list price. During a seasonal promotion, the desk was sold at a $20\%$ discount off the list price. What was the dealer's profit as a percentage of the total expenditure (purchase price plus restoration cost)?

Show answer & explanation

Answer: 12.5

Answer

The dealer's profit as a percentage of the total expenditure is

12.5\%

.

To find the net profit percentage on total expenditure, first compute total cost (

\$600 + \$40 = \$640

). Next, compute list price by applying a

50\%

markup to the initial purchase price (

\$600 \times 1.50 = \$900

). Then compute the selling price after a

20\%

discount on list price (

\$900 \times 0.80 = \$720

). Net profit is selling price minus total cost (

\$720 - \$640 = \$80

). Finally, express net profit as a percentage of total cost:

(\$80 / \$640) \times 100\% = 12.5\%

.

Step-by-Step Solution

1

Calculate the total cost (expenditure)

Total cost =

\$600 + \$40 = \$640

The total investment includes both the purchase price of the desk and the mandatory restoration fee incurred before sale.

2

Determine the list price

List price =

\$600 \times 1.50 = \$900

The markup of

50\%

is applied specifically to the initial purchase price of

\$600

.

3

Determine the selling price

Selling price =

\$900 \times 0.80 = \$720

The

20\%

discount reduces the list price of

\$900

.

4

Calculate the net dollar profit

Net profit =

\$720 - \$640 = \$80

Profit is calculated as total revenue (selling price) minus total cost.

5

Calculate the net profit percentage

Net profit percentage =

(\$80 / \$640) \times 100\% = 12.5\%

The percentage return must be evaluated using total expenditure (

\$640

) as the base.

Key Concept

Calculating successive percentage markups and discounts with secondary cost components

Estimated Time:1m 45s

Question 36Question →

A logistics company maintains a fleet consisting of three types of delivery vehicles: Vans, Medium Trucks, and Heavy Trucks. The ratio of the number of Vans to Medium Trucks is $3 : 2$ , and the ratio of the number of Medium Trucks to Heavy Trucks is $4 : 1$ . If the average fuel efficiency is $24$ miles per gallon for the Vans and $15$ miles per gallon for the Medium Trucks, and the average fuel efficiency for the entire combined fleet is $19$ miles per gallon, what is the average fuel efficiency, in miles per gallon, of the Heavy Trucks?

Show answer & explanation

Answer: 5

Answer

The average fuel efficiency of the Heavy Trucks is 5 miles per gallon.

Combining the given vehicle ratios gives a unified proportion of 6 Vans to 4 Medium Trucks to 1 Heavy Truck (total 11 units). The total weighted efficiency sum is

6(24) + 4(15) + 1(x) = 144 + 60 + x = 204 + x

. Setting this equal to the combined fleet total of

11 \times 19 = 209

yields

x = 5

miles per gallon.

Step-by-Step Solution

1

Unify the two relative ratios into a single compound ratio for all three vehicle groups.

Vans : Medium Trucks : Heavy Trucks = 6 : 4 : 1, giving a total of 11 ratio units.

Both individual ratios share Medium Trucks as a common element (

2

units in the first ratio and

4

units in the second). Multiplying the first ratio by

2

aligns the Medium Truck count.

2

Set up the weighted mean equation using group weights and individual group averages.

Weighted Sum = 6(24) + 4(15) + 1(x) = 204 + x, where total elements = 11.

The sum of (group quantity × group mean) divided by total quantity equals the combined average.

3

Equate to total fleet mean and solve for the unknown average fuel efficiency x.

204 + x = 11 × 19 = 209, so x = 5 miles per gallon.

Multiplying total items (11) by total mean (19) gives total combined miles per gallon units (209).

Key Concept

Weighted Average of Multiple Combined Subgroups

Question 37Question →

A boutique bookstore sells hardcover books for $\$25$ each and paperback books for $\$12$ each. On a certain day, the store sold a total of $85$ books and generated a total revenue of $\$1,579$ . How many hardcover books were sold on that day?

Show answer & explanation

Answer: 43

Answer

The total number of hardcover books sold on that day is 43.

By setting up the two linear equations

h + p = 85

(quantity) and

25h + 12p = 1579

(revenue), eliminating

p

yields

13h = 559

, which gives

h = 43

.

Step-by-Step Solution

1

Define variables and construct the system of linear equations

Let

h

be the number of hardcover books and

p

be the number of paperback books. The equations are

h + p = 85

and

25h + 12p = 1579

.

The total quantity of items sold gives a sum equation, and the individual prices multiplied by their quantities yield the total revenue equation.

2

Use substitution or elimination to solve for

h

Multiplying the total books equation by 12 yields

12h + 12p = 1020

. Subtracting this from

25h + 12p = 1579

gives

13h = 559

.

Eliminating

p

directly isolates the variable

h

representing hardcover books.

3

Calculate the value of

h

h = \frac{559}{13} = 43

.

Dividing the remaining total revenue by the coefficient of

h

yields the precise quantity.

Key Concept

Solving two-variable systems of linear equations using elimination or substitution

Question 38Question →

A commuter travels to work along a straight route at a constant speed of $40$ miles per hour. On the return trip along the exact same route, traffic delays reduce the average speed to $24$ miles per hour. If the return trip takes $40$ minutes longer than the trip to work, what is the total distance, in miles, traveled by the commuter for the round trip?

Show answer & explanation

Answer: 80

Answer

The total distance traveled by the commuter for the round trip is 80 miles.

The total distance for the round trip is 80 miles. By expressing time as distance divided by speed, the outbound trip takes

\frac{D}{40}

hours and the return trip takes

\frac{D}{24}

hours. Since the return trip takes

40

minutes (

\frac{2}{3}

hours) longer, setting up the equation

\frac{D}{24} - \frac{D}{40} = \frac{2}{3}

yields

D = 40

miles for one way. The round-trip distance is

2 \times 40 = 80

miles.

Step-by-Step Solution

1

Define variables for the one-way distance and expressions for the travel time of each leg.

Let

D

represent the one-way distance in miles. Outbound time

t_1 = \frac{D}{40}

hours; return time

t_2 = \frac{D}{24}

hours.

Using the relation

\text{Time} = \frac{\text{Distance}}{\text{Speed}}

expresses time in terms of a single unknown variable.

2

Convert the given time difference from minutes to hours.

40 \text{ minutes} = \frac{40}{60} \text{ hours} = \frac{2}{3} \text{ hours}

.

Speeds are given in miles per hour, so time units must be in hours for dimensional consistency.

3

Formulate the equation representing the time difference and solve for

D

.

\frac{D}{24} - \frac{D}{40} = \frac{2}{3} \implies \frac{2D}{120} = \frac{2}{3} \implies \frac{D}{60} = \frac{2}{3} \implies D = 40 \text{ miles}

.

Subtracting the faster outbound time from the slower return time equals the time difference.

4

Compute the total distance for the entire round trip.

\text{Total distance} = 2 \times D = 2 \times 40 = 80 \text{ miles}

.

The question asks for the round-trip distance, which includes both the outbound and return journeys (

2D

).

Key Concept

Rate, Time, and Distance Problems

Question 39Question →

Let $d(k)$ denote the number of positive divisors of a positive integer $k$ . What is the smallest positive integer $n$ that is a multiple of $72$ and satisfies $d(n) = 35$ ?

Show answer & explanation

Answer: 5184

Answer

The smallest positive integer

n

that satisfies all conditions is 5184.

The prime factorization of

72

is

2^3 \times 3^2

. Any multiple

n

of

72

must be written as

2^a \times 3^b \times dots

with

a \ge 3

and

b \ge 2

. The total number of positive divisors

d(n) = (a+1)(b+1) dots = 35

. The integer

35

can be factored into integers greater than

1

only as

7 \times 5

. Thus,

n

must have exactly two distinct prime factors, which must be

2

and

3

. The required exponents are

a, b \in \{6, 4\}

. To minimize

n

, we assign the larger exponent

6

to the smaller base

2

, yielding

n = 2^6 \times 3^4 = 64 \times 81 = 5184

.

Step-by-Step Solution

1

Express 72 in prime factor form to determine minimum required prime exponents.

72 = 2^3 \times 3^2

, meaning

n

must have prime factorization

2^a \times 3^b \dots

where

a \ge 3

and

b \ge 2

.

Any multiple of 72 must contain at least three factors of 2 and two factors of 3.

2

Analyze the divisor count condition

d(n) = 35

.

Since

35 = 7 \times 5

, the number of prime factors of

n

can be at most 2, with exponent increments

(a+1)(b+1) = 7 \times 5

.

The number of positive divisors for

p_1^{e_1} p_2^{e_2} \dots

is

(e_1+1)(e_2+1)\dots

, and 35 factors into integers greater than 1 only as 35 or

7 \times 5

.

3

Determine the required exponents for the prime factors of

n

.

The exponents of the prime factors must be

7-1 = 6

and

5-1 = 4

.

Since

n

must contain both 2 and 3 as prime factors,

n

cannot have only 1 prime factor, so it has exactly two prime factors (2 and 3).

4

Optimize the exponent assignment to minimize

n

.

Assign exponent 6 to base 2 and exponent 4 to base 3, giving

n = 2^6 \times 3^4 = 64 \times 81 = 5184

.

Assigning the larger exponent to the smaller prime base minimizes the total product while meeting the conditions

a \ge 3

and

b \ge 2

.

Key Concept

Divisor counting formula and prime factor exponent allocation under divisibility constraints

Estimated Time:2m 0s

Question 40Question →

An investor deposits $\$800$ into a savings account that pays a simple annual interest rate of $5\%$ . If no additional deposits or withdrawals are made, what is the total interest, in dollars, earned on the account after $3$ years?

Show answer & explanation

Answer: 120

Answer

The total interest earned after 3 years is 120 dollars.

The total interest earned is computed using the simple interest formula

I = P \times r \times t

. Substituting

P = 800

,

r = 0.05

, and

t = 3

yields

I = 800 \times 0.05 \times 3 = 120

dollars.

Step-by-Step Solution

1

Identify the principal, rate, and time from the problem stem.

Principal

P = \$800

, interest rate

r = 5\% = 0.05

, and time

t = 3

years.

These values are required to apply the simple interest formula.

2

Calculate simple interest using

I = P \times r \times t

.

I = 800 \times 0.05 \times 3 = 120

.

Simple interest accumulates linearly over time based on the initial principal.

Key Concept

Simple Interest Calculation

Estimated Time:45s

Answer

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