Word Problems and Applied Math Practice Questions — GMAT — Page 6

Question 101Question →

A wholesale distributor purchased a shipment of 50 units of Model A laptops and 30 units of Model B laptops for a total cost of $4,000. To set the list price for each model, the distributor marked up the unit cost of Model A by 40% and the unit cost of Model B by 50%. During a clearance promotion, Model A was sold at a 10% discount off its list price, and Model B was sold at a 20% discount off its list price. If the distributor earned a total profit of$ 995 on the entire shipment of 80 laptops, what was the original purchase cost of one Model A laptop?

$25

$50

$60.50

$65

$75

Show answer & explanation

Answer: $65

Answer

The original purchase cost of one Model A laptop was $65.

The original purchase cost of one Model A laptop is found by setting up a system of two linear equations: one for total cost (

50x + 30y = 4,000

, or

5x + 3y = 400

) and one for total profit (

50(0.26x) + 30(0.20y) = 995

, or

13x + 6y = 995

). Solving this system yields

x = 65

, which accurately reflects the unit cost of Model A.

Step-by-Step Solution

1

Define variables and set up the total cost equation.

Let

x

be the unit cost of Model A and

y

be the unit cost of Model B. The equation is

50x + 30y = 4,000

, which simplifies to

5x + 3y = 400

.

The distributor purchased 50 units of Model A and 30 units of Model B for a total cost of $4,000.

2

Calculate the selling price and profit per unit for each model.

For Model A:

\text{List Price} = 1.40x

,

\text{Selling Price} = 1.40x \times 0.90 = 1.26x

, so

\text{Profit per unit} = 1.26x - x = 0.26x

.
For Model B:

\text{List Price} = 1.50y

,

\text{Selling Price} = 1.50y \times 0.80 = 1.20y

, so

\text{Profit per unit} = 1.20y - y = 0.20y

.

Selling prices are derived by applying the discount to the list price, and unit profit is selling price minus cost price.

3

Set up the total profit equation.

Total profit is

50(0.26x) + 30(0.20y) = 995

, which simplifies to

13x + 6y = 995

.

Summing the total profit from all 50 units of Model A and 30 units of Model B equals the given total profit of $995.

4

Solve the system of linear equations for

x

.

Multiply

5x + 3y = 400

by 2 to get

10x + 6y = 800

. Subtract this from

13x + 6y = 995

to obtain

(13x - 10x) + (6y - 6y) = 995 - 800 \implies 3x = 195 \implies x = 65

.

Eliminating

y

isolates the variable

x

, representing the unit purchase cost of Model A.

Key Concept

Profit, Loss, and Markup with Successive Percentages

Question 102Question →

A wholesaler purchases a desk for $\$250$ and marks up the cost price by $20\%$ to establish the retail price. If the desk is sold at the retail price, what is the profit, in dollars, earned on the sale?

Show answer & explanation

Answer: 50

Answer

The profit earned on the sale of the desk is

\$50

.

The profit is calculated by finding

20\%

of the cost price of

\$250

, which is

0.20 \times 250 = 50

.

Step-by-Step Solution

1

Identify the cost price and markup rate from the problem statement.

Cost price

= \$250

; Markup rate

= 20\%

.

Markup is calculated directly as a percentage of the original cost price.

2

Multiply the cost price by the markup percentage to find the profit.

\text{Profit} = 0.20 \times 250 = 50

.

Taking

20\%

of the

\$250

purchase price gives the exact dollar markup/profit.

Key Concept

Dollar profit is calculated as the markup percentage multiplied by the cost price.

Question 103Question →

A chemical storage vessel contains three industrial solvents—Solvent A, Solvent B, and Solvent C—blended in the initial ratio of $2 : 5 : 7$ , respectively. After $24$ liters of Solvent A and $12$ liters of Solvent B are added to the vessel, the ratio of Solvent A to Solvent B becomes $4 : 7$ . What was the initial volume, in liters, of Solvent C in the vessel?

Show answer & explanation

Answer: 140

Answer

The initial volume of Solvent C in the vessel was 140 liters.

Defining the initial amounts of Solvents A, B, and C as

2x

,

5x

, and

7x

, respectively, the addition of

24

liters of Solvent A and

12

liters of Solvent B yields the ratio equation

\frac{2x + 24}{5x + 12} = \frac{4}{7}

. Cross-multiplying gives

14x + 168 = 20x + 48

, which simplifies to

6x = 120

and

x = 20

. The initial volume of Solvent C is therefore

7 \times 20 = 140

liters.

Step-by-Step Solution

1

Express the initial quantities of the three solvents in terms of a common ratio multiplier

x

.

Solvent A =

2x

, Solvent B =

5x

, Solvent C =

7x

.

A ratio of

2 : 5 : 7

implies the actual quantities are proportional to 2, 5, and 7 by a factor

x

.

2

Set up an equation representing the modified ratio of Solvent A to Solvent B.

\frac{2x + 24}{5x + 12} = \frac{4}{7}

Adding 24 liters to Solvent A yields

2x + 24

liters, and adding 12 liters to Solvent B yields

5x + 12

liters.

3

Solve the algebraic equation for

x

.

7(2x + 24) = 4(5x + 12) \implies 14x + 168 = 20x + 48 \implies 6x = 120 \implies x = 20

.

Cross-multiplication converts the ratio comparison into a linear equation.

4

Compute the initial volume of Solvent C.

Initial volume of Solvent C =

7x = 7(20) = 140

liters.

Substitute

x = 20

into the expression

7x

for the original amount of Solvent C.

Key Concept

Ratio scaling and algebraic modeling of partial additions

Question 104Question →

At a research institute, $120$ scientists are each assigned to at least one of three interdisciplinary projects: Alpha, Beta, and Gamma. The project assignments meet the following conditions:

- Exactly $70$ scientists work on Project Alpha.
- Exactly $65$ scientists work on Project Gamma.
- Exactly $25$ scientists work on both Project Alpha and Project Beta.
- Exactly $30$ scientists work on both Project Beta and Project Gamma.
- Exactly $20$ scientists work on both Project Alpha and Project Gamma.
- The number of scientists working on all three projects is equal to half the number of scientists who work on Project Gamma only.

How many scientists work on Project Beta only?

5

10

15

25

30

Show answer & explanation

Answer: $5$

Answer

Exactly

5

scientists work on Project Beta only.

The correct answer is

5

. By using the 3-set Venn diagram equation for Project Gamma, we find the number of scientists in all three projects is

x = 15

. Decomposing all disjoint regions gives Alpha only

= 40

, Gamma only

= 30

, and two-project overlaps of

10

,

15

, and

5

. Subtracting the sum of these six regions (

115

) from the total universe (

120

) leaves exactly

5

scientists working on Project Beta only.

Step-by-Step Solution

1

Define variables for the regions of the 3-set Venn diagram.

Let

x

be the number of scientists working on all three projects (Alpha

\cap

Beta

\cap

Gamma).

Setting the central intersection as

x

allows expressing all pairwise overlaps in terms of

x

.

2

Express the two-project-only regions and Project Gamma only in terms of

x

.

Alpha & Beta only

= 25 - x

; Beta & Gamma only

= 30 - x

; Alpha & Gamma only

= 20 - x

; Gamma only

= 2x

.

Each given two-set intersection includes the triple intersection

x

. Gamma only is twice

x

per the stem.

3

Set up an equation for the total number of scientists in Project Gamma (

65

) to solve for

x

.

\text{Gamma Total} = (2x) + (20 - x) + (30 - x) + x = x + 50 = 65 \implies x = 15

.

Summing all four regions comprising Project Gamma equates to the total given as

65

.

4

Calculate the numerical values of the intersection regions and Gamma only.

All three

= 15

; Gamma only

= 30

; Alpha & Beta only

= 10

; Beta & Gamma only

= 15

; Alpha & Gamma only

= 5

.

Substituting

x = 15

back into the expressions from Step 2 gives exact region values.

5

Calculate the number of scientists working on Project Alpha only.

\text{Alpha Total} = 70 = \text{Alpha only} + 10 + 5 + 15 \implies \text{Alpha only} = 40

.

Project Alpha is composed of Alpha only plus its three overlapping regions (

10

,

5

, and

15

).

6

Use the total universe of

120

scientists to solve for Project Beta only.

120 = \text{Alpha only} + \text{Beta only} + \text{Gamma only} + (A \cap B)_{\text{only}} + (B \cap G)_{\text{only}} + (A \cap G)_{\text{only}} + \text{All three} \implies 120 = 40 + \text{Beta only} + 30 + 10 + 15 + 5 + 15 = 115 + \text{Beta only} \implies \text{Beta only} = 5

.

Since every scientist is on at least one project, the sum of all 7 disjoint region counts equals the total population of

120

.

Key Concept

Three-Set Overlapping Sets and Venn Diagram Analysis

Question 105Question →

Three automated server clusters—Cluster $X$ , Cluster $Y$ , and Cluster $Z$ —process large-scale data analytics tasks. Working alone at their respective constant rates, Cluster $X$ can complete a standard workload in $12$ hours, Cluster $Y$ can complete it in $20$ hours, and Cluster $Z$ can complete it in $30$ hours. Cluster $X$ and Cluster $Y$ begin processing a standard workload together. After $4$ hours, Cluster $X$ goes offline due to maintenance while Cluster $Y$ continues working alone. Exactly $2$ hours after Cluster $X$ goes offline, Cluster $Z$ is brought online to assist Cluster $Y$ . How many additional hours will it take Cluster $Y$ and Cluster $Z$ working together to complete the remaining portion of the workload?

4.4

hours

5.6

hours

2.8

hours

18.33

hours

6.2

hours

Show answer & explanation

Answer: $4.4$ hours

Answer

The additional time required for Cluster Y and Cluster Z to finish the remaining workload is

4.4

hours.

The combined rate of Cluster X and Cluster Y is

\frac{1}{12} + \frac{1}{20} = \frac{2}{15}

job/hr. In

4

hours, they complete

4 \times \frac{2}{15} = \frac{8}{15}

of the job, leaving

\frac{7}{15}

. Cluster Y then works alone for

2

hours, completing

2 \times \frac{1}{20} = \frac{1}{10}

of the job. The remaining work is

\frac{7}{15} - \frac{1}{10} = \frac{11}{30}

. Finally, Cluster Y and Cluster Z work together at a combined rate of

\frac{1}{20} + \frac{1}{30} = \frac{1}{12}

job/hr. The time needed to finish is

\frac{11/30}{1/12} = 4.4

hours.

Step-by-Step Solution

1

Calculate individual work rates for each cluster per hour.

r_X = \frac{1}{12}

,

r_Y = \frac{1}{20}

, and

r_Z = \frac{1}{30}

of the total workload per hour.

Work rate is the reciprocal of the total time required to complete one full workload.

2

Calculate the work completed in Stage 1 when Cluster X and Cluster Y work together for 4 hours.

Combined rate

r_{X+Y} = \frac{1}{12} + \frac{1}{20} = \frac{5 + 3}{60} = \frac{8}{60} = \frac{2}{15}

. Work done in 4 hours =

4 \times \frac{2}{15} = \frac{8}{15}

. Remaining work =

1 - \frac{8}{15} = \frac{7}{15}

.

Multiply the combined rate of Clusters X and Y by 4 hours to find the fraction of work completed.

3

Calculate the work completed in Stage 2 when Cluster Y works alone for 2 hours.

Work done by Y in 2 hours =

2 \times \frac{1}{20} = \frac{1}{10} = \frac{3}{30}

. Remaining work =

\frac{7}{15} - \frac{1}{10} = \frac{14}{30} - \frac{3}{30} = \frac{11}{30}

.

Subtract the work done by Cluster Y during its solo 2-hour interval from the remaining work.

4

Calculate the time required for Cluster Y and Cluster Z working together to complete the remaining

11/30

of the work.

Combined rate

r_{Y+Z} = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12}

. Time

t = \frac{11/30}{1/12} = \frac{11}{30} \times 12 = \frac{132}{30} = 4.4

hours.

Divide the remaining workload fraction by the combined hourly work rate of Clusters Y and Z.

Key Concept

Combined Work Rate in Multi-Stage Problems

Estimated Time:2m 0s

Question 106Question →

A chemist prepares a mixture by combining $20$ liters of a $10\%$ saline solution with $30$ liters of a $20\%$ saline solution. What is the saline concentration, as a percentage, of the resulting mixture?

Show answer & explanation

Answer: 16

Answer

The concentration of the resulting mixture is

16\%

.

The total amount of pure saline in the mixture is

8

liters out of a total solution volume of

50

liters. The overall concentration is

\frac{8}{50} \times 100\% = 16\%

.

Step-by-Step Solution

1

Calculate the total amount of pure saline solute contributed by both solutions.

20 \text{ L} \times 0.10 + 30 \text{ L} \times 0.20 = 2 + 6 = 8 \text{ liters}

The total solute is the sum of the solute amounts in each component solution.

2

Calculate the total volume of the combined mixture.

20 + 30 = 50 \text{ liters}

The combined volume is the sum of the individual volumes.

3

Compute the weighted average concentration of the mixture.

\frac{8}{50} = 0.16 = 16\%

The weighted average concentration equals total solute divided by total solution volume.

Key Concept

Weighted Average in Mixture Problems

Question 107Question →

A pharmaceutical facility uses two automated purification columns, Column $X$ and Column $Y$ , to process standard batches of a synthetic compound. Working alone at its constant rate, Column $X$ can purify a full batch in $15$ hours, while Column $Y$ can purify a full batch in $10$ hours working alone at its constant rate. Column $X$ begins purifying a batch alone. After $3$ hours, Column $Y$ is turned on to work alongside Column $X$ , and both columns work together at their respective constant rates until the batch is completely purified. What is the total time, in hours, from the moment Column $X$ started until the batch is completely purified?

4.8 hours

7.2 hours

7.8 hours

9.6 hours

11.0 hours

Show answer & explanation

Answer: 7.8 hours

Answer

7.8 hours

The correct answer of 7.8 hours is obtained by evaluating the work completed in stages. In the first 3 hours, Column X completes 3/15 or 1/5 of the batch, leaving 4/5 of the batch remaining. When Column Y joins Column X, their combined rate is 1/15 + 1/10 = 1/6 batch per hour. Completing the remaining 4/5 of the batch at a rate of 1/6 per hour requires (4/5) / (1/6) = 4.8 hours. Adding the initial 3 hours yields a total time of 7.8 hours.

Step-by-Step Solution

1

Determine the individual hourly work rates for each column

Rate of Column

X = \frac{1}{15}

batch per hour; Rate of Column

Y = \frac{1}{10}

batch per hour.

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

2

Calculate the fraction of work completed during the initial stage

Work done by

X = 3 \times \frac{1}{15} = \frac{3}{15} = \frac{1}{5}

of the batch.

Column

X

works alone for the first 3 hours.

3

Find the remaining fraction of work to be completed

Remaining work

= 1 - \frac{1}{5} = \frac{4}{5}

of the batch.

Subtract the completed fraction from the total job (1 full batch).

4

Determine the combined work rate of both columns working together

Combined rate

= \frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30} = \frac{1}{6}

batch per hour.

Rates add when two entities work simultaneously.

5

Calculate the time required for both columns to finish the remaining work

Time together

= \frac{\text{Remaining Work}}{\text{Combined Rate}} = \frac{4/5}{1/6} = \frac{4}{5} \times 6 = \frac{24}{5} = 4.8

hours.

Time equals work divided by rate.

6

Calculate the total elapsed time from the start

Total time

= 3 \text{ hours (alone)} + 4.8 \text{ hours (together)} = 7.8

hours.

Add the initial duration to the combined working duration.

Key Concept

Work Rate and Combined Work

Question 108Question →

A store owner purchases a bicycle for $150. To establish the list price, the store owner marks up the cost price by 40%. During a promotional sale, the bicycle is sold at a 20% discount off the list price. What is the store owner's net profit from the sale of the bicycle?

$18

$30

$42

$60

$168

Show answer & explanation

Answer: $18

Answer

$18

The correct answer of

18 is obtained by first calculating the list price (

150 + 40% of

150 =

210), then computing the selling price (

210 - 20% of

210 =

168), and finally finding the profit (

168 -

150 =

18).

Step-by-Step Solution

1

Calculate the list price after a 40% markup on the cost price of $150.

List Price =

150 × 1.40 =

210

Markup is applied directly to the cost price.

2

Calculate the selling price after a 20% discount on the list price of $210.

Selling Price =

210 × (1 - 0.20) =

210 × 0.80 = $168

Discounts are calculated based on the list price, not the cost price.

3

Subtract the original cost price from the final selling price to find the net profit.

Net Profit =

168 -

150 = $18

Profit is the difference between total revenue (selling price) and total cost.

Key Concept

Successive percentage changes with different bases (cost price vs. list price)

Question 109Question →

An international airline surveyed $360$ frequent flyers regarding three premium services they utilized during the past year: First-Class Lounge Access, Priority Baggage Handling, and In-Flight Wi-Fi. The survey revealed the following data:

- Every surveyed passenger utilized at least one of the three services.
- $220$ passengers utilized First-Class Lounge Access.
- $180$ passengers utilized Priority Baggage Handling.
- $140$ passengers utilized In-Flight Wi-Fi.
- $60$ passengers utilized all three services.

If the ratio of the number of passengers who utilized exactly one service to the number of passengers who utilized exactly two services is $4 : 1$ , how many passengers utilized exactly two services?

48

60

75

120

240

Show answer & explanation

Answer: $60$

Answer

The number of passengers who utilized exactly two services is 60.

By breaking the total set of passengers into disjoint groups—those using exactly one service (

E_1

), exactly two services (

E_2

), all three services (

E_3 = 60

), and none (

\text{None} = 0

)—we set up the equation

360 = E_1 + E_2 + 60 + 0

. Subtracting

60

from both sides gives

E_1 + E_2 = 300

. Given the ratio

E_1 : E_2 = 4 : 1

, we substitute

E_1 = 4E_2

, yielding

5E_2 = 300

, so

E_2 = 60

.

Step-by-Step Solution

1

Set up the fundamental 3-set total elements equation in terms of disjoint regions.

\text{Total} = E_1 + E_2 + E_3 + \text{None}

, where

E_1

is exactly 1 service,

E_2

is exactly 2 services, and

E_3

is all 3 services.

Categorizing members into mutually exclusive regions simplifies group accounting without double-counting.

2

Substitute the known values into the total elements formula.

360 = E_1 + E_2 + 60 + 0 \implies E_1 + E_2 = 300

.

Since every passenger used at least one service,

\text{None} = 0

, leaving

300

passengers who used either exactly 1 or exactly 2 services.

3

Apply the given ratio of

E_1 : E_2 = 4 : 1

to solve for

E_2

.

E_1 = 4E_2 \implies 4E_2 + E_2 = 300 \implies 5E_2 = 300 \implies E_2 = 60

.

The ratio implies that out of

5

total parts for

E_1 + E_2

,

1

part represents

E_2

.

Key Concept

3-Set Overlapping Sets Disjoint Regions Formula

Estimated Time:2m 0s

Question 110Question →

A municipal water treatment facility uses three pumps—Pump $P$ , Pump $Q$ , and Pump $R$ —to fill a main reservoir. Working alone at their respective constant rates, Pump $P$ can fill the reservoir in 20 hours, Pump $Q$ in 30 hours, and Pump $R$ in 40 hours.

All three pumps begin filling the empty reservoir simultaneously. After 4 hours, Pump $P$ shuts off. Pumps $Q$ and $R$ continue operating together for another 8 hours before Pump $Q$ is also shut off. Pump $R$ continues to run alone until the reservoir is completely filled.

How many total hours does it take to fill the reservoir from the start of the process until it is completely filled?

Show answer & explanation

Answer: 16

Answer

The total time required to fill the reservoir from start to finish is 16 hours.

Converting the individual completion times into work rates per hour (1/20, 1/30, and 1/40 of the reservoir per hour) allows us to determine the combined output per phase. In the first 4 hours, all three pumps fill 52/120 of the reservoir. In the next 8 hours, Pumps Q and R fill 56/120 of the reservoir. This leaves 12/120 (or 1/10) of the job remaining. Pump R, working at a rate of 1/40 per hour, takes 4 hours to complete the final 1/10. Adding all three durations (4 + 8 + 4) gives a total of 16 hours.

Step-by-Step Solution

1

Determine the individual hourly rates of work for each pump.

Pump P completes

\frac{1}{20}

of the job per hour, Pump Q completes

\frac{1}{30}

of the job per hour, and Pump R completes

\frac{1}{40}

of the job per hour.

Work rate is the reciprocal of the total time required to complete one whole task.

2

Calculate the work completed during Stage 1 (4 hours with all 3 pumps working).

Combined rate =

\frac{1}{20} + \frac{1}{30} + \frac{1}{40} = \frac{6 + 4 + 3}{120} = \frac{13}{120}

. Work done =

4 \times \frac{13}{120} = \frac{52}{120}

.

Work done equals combined rate multiplied by time spent working together.

3

Calculate the work completed during Stage 2 (8 hours with Pumps Q and R working).

Combined rate of Q and R =

\frac{1}{30} + \frac{1}{40} = \frac{4 + 3}{120} = \frac{7}{120}

. Work done =

8 \times \frac{7}{120} = \frac{56}{120}

.

Only Pumps Q and R are active during this second period.

4

Calculate the remaining fraction of work after Stage 1 and Stage 2.

Total work done so far =

\frac{52}{120} + \frac{56}{120} = \frac{108}{120} = \frac{9}{10}

. Remaining work =

1 - \frac{9}{10} = \frac{1}{10}

.

Subtracting completed work from the whole (1) leaves the remaining work to be completed.

5

Calculate the time required for Pump R to finish the remaining work alone.

Time =

\frac{1/10}{1/40} = \frac{40}{10} = 4

hours.

Time equals remaining work divided by the individual rate of Pump R.

6

Sum the time spent across all stages.

Total time =

4 \text{ hours (Stage 1)} + 8 \text{ hours (Stage 2)} + 4 \text{ hours (Stage 3)} = 16

hours.

The total elapsed time is the sum of the durations of each distinct phase.

Key Concept

Work Rate and Combined Work in Multi-Stage Processes

Alternative Method

Assume a convenient total reservoir volume equal to the least common multiple of the times: 120 units. Pump P produces 6 units/hr, Pump Q produces 4 units/hr, and Pump R produces 3 units/hr. Stage 1 (4 hrs): 4 × (6 + 4 + 3) = 52 units. Stage 2 (8 hrs): 8 × (4 + 3) = 56 units. Total filled = 108 units. Remaining = 12 units. Stage 3 (Pump R alone): 12 / 3 = 4 hrs. Total time = 4 + 8 + 4 = 16 hours.

Estimated Time:2m 0s

Question 111Question →

A cyclist travels from Town A to Town B, a distance of $60$ miles, at a constant speed of $30$ miles per hour. On the return trip from Town B to Town A along the exact same route, she travels at a constant speed of $20$ miles per hour. What is the cyclist's average speed, in miles per hour, for the entire round trip?

25 miles per hour

24 miles per hour

50 miles per hour

12 miles per hour

10 miles per hour

Show answer & explanation

Answer: 24 miles per hour

Answer

24 miles per hour

To find the average speed for any multi-leg trip, divide total distance by total time. The first leg of

60

miles at

30

mph takes

2

hours. The return leg of

60

miles at

20

mph takes

3

hours. The total distance is

120

miles and total time is

5

hours. Therefore, the average speed is

\frac{120}{5} = 24

miles per hour.

Step-by-Step Solution

1

Calculate the time taken for the trip from Town A to Town B.

t_1 = \frac{60}{30} = 2

hours

Time equals distance divided by speed.

2

Calculate the time taken for the return trip from Town B to Town A.

t_2 = \frac{60}{20} = 3

hours

Time equals distance divided by speed.

3

Find the total distance and total time for the round trip.

Total Distance =

60 + 60 = 120

miles; Total Time =

2 + 3 = 5

hours

Average speed is based on the entire journey's distance and duration.

4

Compute the average speed.

\text{Average Speed} = \frac{120}{5} = 24\text{ miles per hour}

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

Key Concept

Average Speed for a Multi-Leg Trip

Estimated Time:1m 0s

Question 112Question →

Train X and Train Y depart simultaneously from Station A and Station B, respectively, heading toward each other along a straight, 360-mile track. Train X travels at a constant speed of 40 miles per hour for the first 2 hours, after which its speed increases by 50 percent. Train Y travels at a constant speed of 50 miles per hour for the first hour, stops for 30 minutes due to a signal delay, and then resumes its journey at a constant speed of 70 miles per hour. How many hours after their simultaneous departure will the two trains meet?

Show answer & explanation

Answer: 3.5

Answer

3.5 hours

Evaluating distance piecewise up to

t = 2

hours shows Train X covers 80 miles and Train Y covers 85 miles, leaving 195 miles between them. After

t = 2

hours, their relative rate of convergence is

60 + 70 = 130

mph. Dividing 195 miles by 130 mph yields 1.5 additional hours, bringing the total time since departure to 3.5 hours.

Step-by-Step Solution

1

Determine distance traveled by Train X in the first 2 hours and its new speed.

Train X distance =

40 \text{ mph} \times 2 \text{ hours} = 80 \text{ miles}

. New speed =

40 \times (1 + 0.50) = 60 \text{ mph}

.

Train X increases its speed by 50% after the 2-hour mark.

2

Determine distance traveled by Train Y during the first 2 hours.

In hour 1 (

t=0

to

t=1

):

50 \text{ mph} \times 1 \text{ h} = 50 \text{ miles}

. During delay (

t=1

to

t=1.5

):

0 \text{ miles}

. From

t=1.5

to

t=2.0

:

70 \text{ mph} \times 0.5 \text{ h} = 35 \text{ miles}

. Total distance =

50 + 0 + 35 = 85 \text{ miles}

.

Accounting for Train Y's initial leg, 30-minute delay, and new speed up to

t = 2

hours.

3

Find the remaining distance separating the two trains at

t = 2

hours.

Total distance covered by both trains =

80 + 85 = 165 \text{ miles}

. Remaining distance =

360 - 165 = 195 \text{ miles}

.

Subtracting total distance covered by both trains from the track length of 360 miles.

4

Calculate the time required to cover the remaining distance using relative speed.

Relative closing speed for

t > 2

is

60 + 70 = 130 \text{ mph}

. Additional time =

\frac{195}{130} = 1.5 \text{ hours}

.

When two objects move toward each other, their rates add together to find the closing rate.

5

Calculate total elapsed time from departure.

Total time =

2.0 + 1.5 = 3.5 \text{ hours}

.

Combining the initial 2-hour evaluation window with the additional 1.5 hours required to meet.

Key Concept

Relative speed and piecewise distance calculations for converging vehicles with variable speeds and delays.

Estimated Time:3m 0s

Question 113Question →

At a culinary institute, a cohort of $250$ professional chefs was evaluated on three specialized culinary skills: Molecular Gastronomy, Pastry Arts, and Artisanal Baking. Each chef in the cohort possessed proficiency in at least one of the three skills. According to the evaluation report:

- $150$ chefs were proficient in Molecular Gastronomy
- $140$ chefs were proficient in Pastry Arts
- $130$ chefs were proficient in Artisanal Baking
- The number of chefs proficient in all three skills was $30$

How many chefs in the cohort were proficient in EXACTLY ONE of the three skills?

80

90

110

140

170

Show answer & explanation

Answer: 110

Answer

110 chefs were proficient in exactly one of the three skills.

The correct answer of 110 is derived by setting up two fundamental 3-set relationships. First, the total number of chefs is the sum of those proficient in exactly one skill (

x

), exactly two skills (

y

), and all three skills (

z = 30

), giving

x + y + 30 = 250

, so

x + y = 220

. Second, the sum of individual set sizes counts elements according to how many sets they belong to:

150 + 140 + 130 = x + 2y + 3(30)

, which simplifies to

x + 2y = 330

. Subtracting the first equation from the second yields

y = 110

, and substituting

y = 110

back into

x + y = 220

gives

x = 110

.

Step-by-Step Solution

1

Define variables for the regions of the 3-set Venn diagram

Let

x

be the number of chefs proficient in exactly 1 skill,

y

be the number proficient in exactly 2 skills, and

z

be the number proficient in all 3 skills (

z = 30

).

Categorizing elements by the exact number of sets they belong to simplifies the set inclusion-exclusion equations.

2

Set up the total population equation

x + y + z = 250 \implies x + y + 30 = 250 \implies x + y = 220

Since every chef is proficient in at least one skill, the sum of all distinct regions equals the total number of chefs (

250

).

3

Set up the sum of individual set sizes equation

150 + 140 + 130 = x + 2y + 3z \implies 420 = x + 2y + 3(30) \implies x + 2y = 330

Summing the individual set totals counts elements in exactly one set once, elements in exactly two sets twice, and elements in all three sets three times.

4

Solve the system of two linear equations for

x

and

y

Subtracting Equation 1 from Equation 2 gives

(x + 2y) - (x + y) = 330 - 220 \implies y = 110

. Substituting

y = 110

back into Equation 1 gives

x + 110 = 220 \implies x = 110

.

Finding

x

directly answers the question regarding chefs proficient in exactly one skill.

Key Concept

Overlapping Sets (3-Set Inclusion-Exclusion and Region Partitioning)

Estimated Time:2m 0s

Question 114Question →

A financial firm consists of three departments: Analytics, Trading, and Research. The ratio of the number of analysts in the Analytics department to traders in the Trading department is $2:3$ . The average annual bonus is $\$90,000$ for an analyst in the Analytics department, $\$60,000$ for a trader in the Trading department, and $\$45,000$ for a researcher in the Research department. If the Research department has $40$ researchers and the average annual bonus across all employees in all three departments combined is $\$64,000$ , how many analysts are in the Analytics department?

19

38

40

57

95

Show answer & explanation

Answer: 38

Answer

The Analytics department has 38 analysts.

Using the given ratio

2:3

, let the number of analysts be

2k

and traders be

3k

. The weighted average bonus for these two groups is

\frac{2k(90,000) + 3k(60,000)}{5k} = \$72,000

. Combining this sub-group with the 40 researchers earning

\$45,000

gives an overall average of

\frac{72,000(5k) + 45,000(40)}{5k + 40} = 64,000

. Solving

\frac{360k + 1800}{5k + 40} = 64

yields

360k + 1800 = 320k + 2560

, so

40k = 760

and

k = 19

. The number of analysts is

2(19) = 38

.

Step-by-Step Solution

1

Express the number of employees in Analytics and Trading in terms of a variable

k

.

Let

N_{\text{Analytics}} = 2k

and

N_{\text{Trading}} = 3k

. The total number of employees in these two departments is

5k

.

The ratio of analysts to traders is given as

2:3

.

2

Calculate the combined average bonus for the Analytics and Trading departments.

Combined average =

\frac{2k(90,000) + 3k(60,000)}{5k} = \frac{180,000k + 180,000k}{5k} = \frac{360,000k}{5k} = \$72,000

.

Weighting each department's average bonus by its relative ratio simplifies the calculation.

3

Set up the weighted average equation combining all three departments.

\frac{72,000(5k) + 45,000(40)}{5k + 40} = 64,000

The total bonus pool divided by the total number of employees across all three departments equals the overall average bonus of

\$64,000

.

4

Solve for

k

and find the number of analysts.

Dividing all terms by

1,000

yields

\frac{360k + 1800}{5k + 40} = 64 \implies 360k + 1800 = 320k + 2560 \implies 40k = 760 \implies k = 19

. Therefore,

N_{\text{Analytics}} = 2k = 2(19) = 38

.

Linear algebraic simplification isolates

k

to compute the requested quantity.

Key Concept

Weighted Averages with Compound Ratios

Alternative Method

Use the scale-balance method for weighted averages: The combined average of Analytics and Trading is

\$72,000

. The Research average is

\$45,000

, and the target combined average is

\$64,000

. The distances from the target are

(72,000 - 64,000) = 8,000

and

(64,000 - 45,000) = 19,000

. The ratio of group sizes

(N_{\text{Analytics}+\text{Trading}} : N_{\text{Research}})

is inversely proportional to these distances:

5k : 40 = 19,000 : 8,000 = 19 : 8

. Cross-multiplying gives

40k = 760 \implies k = 19

, so analysts

= 2k = 38

.

Estimated Time:2m 0s

Question 115Question →

A container holds a solution of acid and water that is $50\%$ acid by volume. First, $20$ liters of the solution are removed and replaced with $20$ liters of pure water, resulting in a solution that is $40\%$ acid by volume. Next, $25$ liters of this $40\%$ solution are removed and replaced with $25$ liters of pure acid. What is the percentage of acid, by volume, in the final solution?

47.5\%

52.5\%

55\%

65\%

70\%

Show answer & explanation

Answer: $55\%$

Answer

The final concentration of acid by volume is

55\%

.

The solution requiring

55\%

concentration is correct. From the first replacement step, we establish that the total container volume is

100

liters (since removing

20

liters of

50\%

solution leaves

0.50V - 10

liters of acid, and

\frac{0.50V - 10}{V} = 0.40

yields

V = 100

). Before the second replacement, the container holds

40

liters of acid. Removing

25

liters of this

40\%

solution removes

10

liters of acid, leaving

30

liters of acid in

75

liters of solution. Adding

25

liters of pure acid increases the acid amount to

55

liters in a total volume of

100

liters, producing a final concentration of

55\%

.

Step-by-Step Solution

1

Determine the total volume of the container (

V

) using the first replacement step.

Initial acid volume is

0.50V

. Removing

20

liters of solution removes

0.50 \times 20 = 10

liters of acid. Replacing with

20

liters of pure water restores total volume to

V

with an acid content of

0.50V - 10

. Setting up the new concentration equation:

\frac{0.50V - 10}{V} = 0.40 \implies 0.50V - 10 = 0.40V \implies 0.10V = 10 \implies V = 100

liters.

Finding the fixed total volume of the container is required to calculate the exact amounts of acid removed and added in subsequent steps.

2

Calculate the amount of acid remaining after removing

25

liters of the

40\%

solution.

Before the second step, the container has

100

liters of

40\%

acid solution, which contains

0.40 \times 100 = 40

liters of acid. Removing

25

liters of this solution removes

0.40 \times 25 = 10

liters of acid. Acid remaining in the container

= 40 - 10 = 30

liters.

When solution is removed, both solute (acid) and solvent (water) are removed proportional to their concentration.

3

Calculate the final acid concentration after adding

25

liters of pure acid.

Adding

25

liters of pure (

100\%

) acid adds exactly

25

liters of acid. Total final acid

= 30 + 25 = 55

liters. Total final volume

= (100 - 25) + 25 = 100

liters. Final concentration

= \frac{55}{100} \times 100\% = 55\%

.

The new concentration is the total volume of acid divided by the total volume of the solution.

Key Concept

Multi-stage removal and replacement in mixture problems

Estimated Time:2m 0s

Question 116Question →

A craft brewery blends three types of hops—Cascade, Centennial, and Mosaic—in a master batch. Initially, the weights of Cascade, Centennial, and Mosaic hops are in the ratio $3 : 4 : 5$ , respectively. After $12$ kilograms of Cascade hops and $12$ kilograms of Centennial hops are added to the batch, the new ratio of Cascade to Centennial to Mosaic becomes $5 : 6 : 5$ . What was the initial weight, in kilograms, of Centennial hops in the master batch?

6

18

24

30

36

Show answer & explanation

Answer: $24$

Answer

The initial weight of Centennial hops in the master batch was

24

kilograms.

Let the initial weights of Cascade, Centennial, and Mosaic hops be

3x

,

4x

, and

5x

kilograms, respectively. Adding

12

kg to Cascade and Centennial gives new weights of

3x + 12

kg and

4x + 12

kg, while Mosaic stays at

5x

kg. Equating the ratio of New Cascade to New Mosaic (

\frac{3x + 12}{5x}

) to the new ratio term (

\frac{5}{5} = 1

) yields

3x + 12 = 5x

, so

2x = 12

and

x = 6

. Therefore, the initial weight of Centennial hops was

4x = 4(6) = 24

kg.

Step-by-Step Solution

1

Define initial quantities using a common ratio multiplier

x

.

Cascade =

3x

kg, Centennial =

4x

kg, Mosaic =

5x

kg.

An initial ratio of

3 : 4 : 5

means the actual weights are multiples of

x

.

2

Express the new quantities after adding the specified weights.

New Cascade =

3x + 12

, New Centennial =

4x + 12

, New Mosaic =

5x

.

12

kg of Cascade and

12

kg of Centennial were added, while Mosaic remained unchanged.

3

Set up a proportion using the unchanged quantity and the new ratio.

Since Mosaic quantity remains

5x

and its component in the new ratio

5 : 6 : 5

is

5

, the multiplier for the new ratio is also

x

.

Comparing New Cascade to New Mosaic:

\frac{3x + 12}{5x} = \frac{5}{5} = 1 \implies 3x + 12 = 5x \implies 2x = 12 \implies x = 6

.

4

Calculate the initial weight of Centennial hops.

Initial Centennial weight =

4x = 4 \times 6 = 24

kilograms.

Multiplying the ratio part for Centennial (

4

) by the multiplier (

x = 6

) gives the initial weight.

Key Concept

Ratio scaling and setting up algebraic equations for altered multi-part ratios.

Estimated Time:2m 0s

Question 117Question →

A commuter drives from home to work at a constant speed of $45$ miles per hour and arrives $10$ minutes late. On the following day, driving along the exact same route at a constant speed of $60$ miles per hour, the commuter arrives $5$ minutes early. What is the distance, in miles, from the commuter's home to work?

Show answer & explanation

Answer: 45

Answer

The distance from the commuter's home to work is 45 miles.

Let

d

be the distance in miles between home and work. At

45

miles per hour, the time taken is

\frac{d}{45}

hours. At

60

miles per hour, the time taken is

\frac{d}{60}

hours. The difference between being

10

minutes late and

5

minutes early is

15

minutes, or

\frac{15}{60} = \frac{1}{4}

hour. Setting up the equation

\frac{d}{45} - \frac{d}{60} = \frac{1}{4}

and finding a common denominator of

180

gives

\frac{4d - 3d}{180} = \frac{1}{4}

, which simplifies to

\frac{d}{180} = \frac{1}{4}

. Multiplying both sides by

180

yields

d = 45

miles.

Step-by-Step Solution

1

Calculate the difference in travel time between the two trips in hours.

The time difference is 10 minutes late - (-5 minutes early) = 15 minutes = 1/4 hour.

Since speeds are given in miles per hour, time must be expressed in hours.

2

Set up an equation relating the two travel times using the formula time = distance / speed.

d/45 - d/60 = 1/4, where d represents the distance in miles.

The slower speed (45 mph) takes 1/4 hour longer than the faster speed (60 mph).

3

Solve the equation for the distance variable d.

(4d - 3d)/180 = 1/4 => d/180 = 1/4 => d = 45.

Find a common denominator for 45 and 60, which is 180, and solve for d.

Key Concept

Rate, Time, and Distance Problems

Question 118Question →

At a commercial bank, an audit of $300$ business loan applications revealed that each application underwent at least one of three specialized risk evaluations: Credit Risk, Market Risk, and Operational Risk. Exactly $180$ applications underwent Credit Risk evaluation, $150$ underwent Market Risk evaluation, and $135$ underwent Operational Risk evaluation. If exactly $95$ applications underwent exactly two of the three risk evaluations, how many loan applications underwent all three risk evaluations?

Show answer & explanation

Answer: 35

Answer

35 loan applications underwent all three risk evaluations.

The total number of applications is

300

. The sum of individual evaluations is

180 + 150 + 135 = 465

. Using the Venn diagram region formula,

\text{Sum of individual sets} - \text{Total} = \text{Exactly } 2 + 2 \times (\text{Exactly } 3)

. Substituting the known values gives

465 - 300 = 95 + 2x

, which simplifies to

165 = 95 + 2x

. Solving for

x

gives

2x = 70

, so

x = 35

.

Step-by-Step Solution

1

State the fundamental overlapping set region equation for 3 sets.

\text{Total} = \text{Exactly } 1 + \text{Exactly } 2 + \text{Exactly } 3 + \text{Neither}

Every application falls into exactly one of these non-overlapping region categories.

2

Calculate the sum of the individual set counts.

180 + 150 + 135 = 465

Summing individual set counts counts items in 1 set once, 2 sets twice, and 3 sets three times.

3

Formulate the algebraic identity linking individual sums to region totals.

\text{Sum of Individual Sets} = \text{Exactly } 1 + 2(\text{Exactly } 2) + 3(\text{Exactly } 3)

This accounts for the multiple counting of overlapping regions.

4

Subtract the Total equation from the Sum equation to eliminate the 'Exactly 1' term.

465 - 300 = \text{Exactly } 2 + 2(\text{Exactly } 3) \Rightarrow 165 = 95 + 2(\text{Exactly } 3)

Subtracting

(\text{Exactly } 1 + \text{Exactly } 2 + \text{Exactly } 3)

from

(\text{Exactly } 1 + 2\cdot\text{Exactly } 2 + 3\cdot\text{Exactly } 3)

leaves

1\cdot\text{Exactly } 2 + 2\cdot\text{Exactly } 3

.

5

Solve the linear equation for the 'Exactly 3' intersection.

2(\text{Exactly } 3) = 165 - 95 = 70 \Rightarrow \text{Exactly } 3 = 35

Dividing the remaining difference of 70 by 2 gives the number of applications in all three sets.

Key Concept

Three-Set Venn Diagram Region Decomposition

Question 119Question →

An agricultural facility uses two automated loading conveyors, Conveyor $A$ and Conveyor $B$ , to fill a grain storage silo. Working alone at its constant rate, Conveyor $A$ can fill the silo in 10 hours. Working alone at its constant rate, Conveyor $B$ can fill the silo in 15 hours. Conveyor $A$ begins filling the empty silo alone. After 4 hours, Conveyor $B$ is turned on, and both conveyors work together at their respective constant rates until the silo is completely full. How many total hours does it take, from the moment Conveyor $A$ starts, to completely fill the silo?

Show answer & explanation

Answer: 7.6

Answer

The total time required to fill the silo is 7.6 hours.

Conveyor A fills 1/10 of the silo per hour and works alone for 4 hours, completing 2/5 of the total job. The remaining 3/5 of the job is completed by both conveyors working together at a combined rate of 1/10 + 1/15 = 1/6 silo per hour. Dividing 3/5 by 1/6 gives 3.6 hours for the second stage. Adding the initial 4 hours yields a total time of 7.6 hours.

Step-by-Step Solution

1

Determine the individual hourly rates of Conveyor A and Conveyor B.

Conveyor A rate =

1/10

silo per hour; Conveyor B rate =

1/15

silo per hour.

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

2

Calculate the fraction of the silo filled by Conveyor A working alone for 4 hours.

Work completed =

4 \times \frac{1}{10} = \frac{2}{5}

of the silo.

Work done equals rate multiplied by time.

3

Determine the remaining fraction of work to be completed.

Remaining work =

1 - \frac{2}{5} = \frac{3}{5}

of the silo.

Subtract the completed fraction from the total job (1).

4

Calculate the combined work rate when both conveyors operate together.

Combined rate =

\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}

silo per hour.

When working together, individual rates add up.

5

Determine the time required for both conveyors to finish the remaining work.

Time =

\frac{3/5}{1/6} = \frac{18}{5} = 3.6

hours.

Divide remaining work by the combined work rate.

6

Sum the time spent in both stages to find the total time.

Total time =

4 + 3.6 = 7.6

hours.

The total duration includes 4 hours of Conveyor A operating alone plus 3.6 hours of combined operation.

Key Concept

Work Rate and Combined Work

Question 120Question →

A merchant purchased a shipment of 200 identical items for a total cost of $8,000. The merchant marked up the cost price of each item by$ P $percent to establish its list price. During the first month of sales, 60 percent of the items were sold at the list price. During the second month, 50 percent of the remaining items were sold at a 20 percent discount off the list price. The remaining items were then sold during a clearance sale at a 50 percent discount off the list price. If the merchant made an overall net profit of 29 percent on the total purchase cost of the shipment, what is the value of$ P$?

Show answer & explanation

Answer: 50

Answer

The value of

P

is 50.

The unit cost of each item is

\frac{\$8,000}{200} = \$40

. To make an overall net profit of 29%, total revenue must equal

\$8,000 \times 1.29 = \$10,320

. Sales are divided into three tiers: 120 items sold at list price

L

, 40 items sold at

0.80L

, and 40 items sold at

0.50L

. Summing the revenue yields

120L + 32L + 20L = 172L

. Solving

172L = 10,320

gives

L = 60

. The percentage markup over cost price is

\frac{60 - 40}{40} \times 100 = 50\%

.

Step-by-Step Solution

1

Calculate the unit cost price and required total revenue.

Unit cost

C = \frac{\$8,000}{200} = \$40

. Total target revenue

R = \$8,000 \times (1 + 0.29) = \$10,320

.

Determining the total required revenue sets the benchmark to solve for the list price.

2

Determine the quantity of items sold at each price tier.

Month 1 sales:

0.60 \times 200 = 120

items at list price

L

.
Month 2 sales:

0.50 \times (200 - 120) = 40

items at discounted price

0.80L

.
Clearance sales:

200 - 120 - 40 = 40

items at clearance price

0.50L

.

Categorizing quantities by price tier enables building the total revenue expression.

3

Express total revenue in terms of list price

L

and solve for

L

.

Total Revenue =

120L + 40(0.80L) + 40(0.50L) = 120L + 32L + 20L = 172L

.
Setting

172L = 10,320 \implies L = \frac{10,320}{172} = 60

.

Summing revenue from all tiers and equating to target revenue yields the list price per item.

4

Calculate the markup percentage

P

.

P = \frac{L - C}{C} \times 100 = \frac{60 - 40}{40} \times 100 = 50\%

.

Percentage markup is measured relative to the initial cost price per unit.

Key Concept

Multi-tiered inventory pricing and percentage markup calculation

Estimated Time:2m 30s

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