Word Problems and Applied Math Practice Questions — GMAT — Page 7

Question 121Question →

A global logistics firm consists of three distinct divisions: Express, Freight, and Solutions. In a given fiscal year, the annual revenue per employee was $\$120,000$ for the Express division, $\$180,000$ for the Freight division, and $\$240,000$ for the Solutions division. The number of employees in the Freight division was $25\%$ greater than the number of employees in the Express division. If the overall weighted average revenue per employee across all three divisions combined was $\$180,000$ , and the total annual revenue generated by the Solutions division was $\$14,400,000$ , what was the total number of employees across all three divisions?

Show answer & explanation

Answer: 195

Answer

The total number of employees across all three divisions was 195.

By setting up the weighted average equation based on employee counts, we establish that the Solutions division headcount equals the Express division headcount (

L = E = 60

). Incorporating the Freight division headcount (

F = 1.25 \times 60 = 75

) gives a total of

60 + 75 + 60 = 195

employees.

Step-by-Step Solution

1

Relate the number of employees in the Freight and Express divisions

F = 1.25E

The Freight division employs 25% more people than the Express division.

2

Set up the weighted average formula and solve for the relationship between Express (

E

) and Solutions (

L

) headcount

L = E

Setting the overall average revenue per employee to

180,000 yields

34.5E + 24L = 40.5E + 18L

, which simplifies to

6L = 6E$.

3

Calculate the exact employee count of the Solutions division

L = 60

Dividing the total revenue of

\$14,400,000

by the revenue per employee of

\$240,000

gives 60 employees.

4

Determine employee counts for all divisions and calculate the total

Total employees =

60 + 75 + 60 = 195

Since

E = L = 60

and

F = 1.25(60) = 75

, the total employee count is

60 + 75 + 60 = 195

.

Key Concept

Weighted Average in Combined Sets

Question 122Question →

An athlete swims across a lake at a constant rate of $10$ miles per hour and swims back to the starting point along the exact same path at a constant rate of $15$ miles per hour. What is the athlete's average speed, in miles per hour, for the entire trip?

11.5

12

12.5

13

25

Show answer & explanation

Answer: $12$

Answer

12

miles per hour

To find the average speed for a round trip, divide the total distance by the total time. If the distance of one leg is

d

, the time for the first leg is

\frac{d}{10}

and for the second leg is

\frac{d}{15}

. Total distance is

2d

, and total time is

\frac{d}{10} + \frac{d}{15} = \frac{d}{6}

. Dividing total distance by total time yields

\frac{2d}{\frac{d}{6}} = 12

miles per hour.

Step-by-Step Solution

1

Assign a convenient distance for one leg of the trip.

Let the distance across the lake be

d = 30

miles (the least common multiple of

10

and

15

).

Choosing a concrete distance simplifies calculating travel times.

2

Calculate the time spent on each leg.

Outbound time

t_1 = \frac{30}{10} = 3

hours; return time

t_2 = \frac{30}{15} = 2

hours.

Time equals distance divided by rate.

3

Calculate the total distance and total time.

Total distance

= 30 + 30 = 60

miles; total time

= 3 + 2 = 5

hours.

The round trip consists of two equal legs.

4

Compute the average speed.

Average speed

= \frac{60}{5} = 12

miles per hour.

Average speed is defined as total distance divided by total time.

Key Concept

Average speed for a multi-leg trip is equal to the total distance divided by the total time, not the simple arithmetic mean of the individual speeds.

Estimated Time:1m 0s

Question 123Question →

A ferry travels from Port Alpha to Port Beta across a bay, a distance of $30$ miles, at a constant speed of $15$ miles per hour. On the return trip from Port Beta to Port Alpha along the exact same route, the ferry travels at a constant speed of $30$ miles per hour. What is the ferry's average speed, in miles per hour, for the entire round trip?

Show answer & explanation

Answer: 20

Answer

The ferry's average speed for the entire round trip is

20

miles per hour.

Average speed is found by dividing the total distance traveled by the total time taken. The round trip consists of two 30-mile legs, making the total distance 60 miles. The outbound leg takes 2 hours (30 / 15), while the return leg takes 1 hour (30 / 30), giving a total time of 3 hours. Dividing 60 miles by 3 hours yields an average speed of 20 miles per hour.

Step-by-Step Solution

1

Calculate outbound trip duration

2 hours

Dividing distance by rate gives time:

30 / 15 = 2

hours.

2

Calculate return trip duration

1 hour

Dividing distance by rate gives time:

30 / 30 = 1

hour.

3

Calculate total distance and total time

60 miles in 3 hours

Total distance is

30 + 30 = 60

miles, and total time is

2 + 1 = 3

hours.

4

Calculate overall average speed

20 mph

Average speed is total distance divided by total time:

60 / 3 = 20

mph.

Key Concept

Average speed for any multi-leg or round-trip journey is defined as Total Distance divided by Total Time, rather than the simple arithmetic mean of the speeds.

Question 124Question →

A piece of industrial machinery valued at $\$5,000$ depreciates in value by $20\%$ during its first year of operation, and by $10\%$ of its reduced value during its second year. What is the value of the machinery, in dollars, at the end of the second year?

Show answer & explanation

Answer: 3600

Answer

3600 dollars

Applying the successive percent reductions sequentially gives

\$5,000 \times 0.80 = \$4,000

at the end of Year 1, and

\$4,000 \times 0.90 = \$3,600

at the end of Year 2.

Step-by-Step Solution

1

Calculate the value after Year 1 depreciation

$4,000

Depreciation of 20% on $5,000 leaves 80% of the original value.

2

Calculate the value after Year 2 depreciation

$3,600

Depreciation of 10% on the new base of $4,000 leaves 90% of the Year 1 value.

Key Concept

Successive Percent Change

Question 125Question →

A solar-powered desalination facility uses two purification units, Unit $X$ and Unit $Y$ , to process standard batches of seawater. Working alone at its constant rate, Unit $X$ can desalinate a full batch of seawater in $12$ hours. Working alone at its constant rate, Unit $Y$ can desalinate an identical batch of seawater in $18$ hours. Unit $X$ begins processing a batch alone and operates for $3$ hours. Then, Unit $Y$ is also turned on, and both units work together at their respective constant rates to complete the remaining portion of the batch. What is the total number of hours required to process the entire batch from the time Unit $X$ was first turned on?

5.4

7.2

8.4

14.25

25.5

Show answer & explanation

Answer: $8.4$

Answer

8.4 hours

Unit X completes 1/12 of the batch per hour, so in 3 hours it processes 3/12 = 1/4 of the batch. This leaves 3/4 of the batch unfinished. Working together, Unit X and Unit Y have a combined rate of 1/12 + 1/18 = 5/36 batch per hour. Dividing the remaining 3/4 batch by 5/36 gives 27/5 = 5.4 hours for the joint operation phase. Summing the initial 3 hours and the 5.4 hours yields a total processing time of 8.4 hours.

Step-by-Step Solution

1

Calculate individual hourly work rates for each unit

Unit X rate =

\frac{1}{12}

batch/hour, Unit Y rate =

\frac{1}{18}

batch/hour

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

2

Determine the fraction of work completed during the first 3 hours

Work completed by Unit X =

3 \times \frac{1}{12} = \frac{1}{4}

of the batch

Unit X operates alone for 3 hours at a rate of

\frac{1}{12}

batch per hour.

3

Calculate the remaining fraction of work to be done

Remaining work =

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

of the batch

Subtract the completed fraction from the total job equal to 1 whole batch.

4

Find the combined hourly work rate of both units

Combined rate =

\frac{1}{12} + \frac{1}{18} = \frac{3}{36} + \frac{2}{36} = \frac{5}{36}

batch/hour

When entities work together, their individual rates add up.

5

Calculate the time required for both units to complete the remaining work

Time together =

\frac{\frac{3}{4}}{\frac{5}{36}} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4

hours

Time equals remaining work divided by combined rate.

6

Add the initial single-unit operating time to the combined operating time

Total time =

3 + 5.4 = 8.4

hours

The question asks for the total elapsed time from when Unit X first started.

Key Concept

Work Rate and Combined Work

Question 126Question →

Tank X initially contains $50$ liters of a solvent solution that is $20\%$ chemical compound by volume. Tank Y initially contains $30$ liters of a solvent solution that is $70\%$ of the same chemical compound by volume. First, $v$ liters of the solution in Tank X are transferred to Tank Y and thoroughly mixed. Then, $v$ liters of the resulting mixture in Tank Y are transferred back to Tank X. If the final concentration of the chemical compound in Tank X is $32\%$ by volume, what is the value of $v$ ?

Show answer & explanation

Answer: 20

Answer

The value of

v

is 20.

By setting up the conservation of solute equation for Tank X, the initial

10

liters of compound minus the

0.20v

liters transferred to Tank Y, plus the return portion

v \cdot \frac{21 + 0.20v}{30 + v}

equals the final

16

liters of compound. Solving this relation yields

v = 20

.

Step-by-Step Solution

1

Calculate the initial volume of pure chemical compound in each tank.

Tank X initially contains

50 \times 0.20 = 10

liters of compound. Tank Y initially contains

30 \times 0.70 = 21

liters of compound.

Establishing initial solute quantities is essential for building the mass balance equations.

2

Determine the concentration of Tank Y after the first transfer of

v

liters from Tank X.

Tank Y contains

30 + v

total liters of solution and

21 + 0.20v

liters of compound. Its concentration becomes

C_Y = \frac{21 + 0.20v}{30 + v}

.

The solution transferred from Tank X carries a 20% concentration into Tank Y, altering Tank Y's volume and concentration.

3

Formulate the total compound equation for Tank X after transferring

v

liters back from Tank Y.

Remaining compound in X after first transfer =

10 - 0.20v

. Compound returned from Y =

v \cdot C_Y = v \cdot \frac{21 + 0.20v}{30 + v}

. Final compound in X =

50 \times 0.32 = 16

liters.

Tank X ends with its original total volume (50 liters) at a 32% concentration.

4

Solve the algebraic equation for

v

.

(10 - 0.20v) + \frac{21v + 0.20v^2}{30 + v} = 16

-0.20v + \frac{21v + 0.20v^2}{30 + v} = 6

Multiply by

(30 + v)

:

-0.20v(30 + v) + 21v + 0.20v^2 = 6(30 + v)

-6v - 0.20v^2 + 21v + 0.20v^2 = 180 + 6v

15v = 180 + 6v \implies 9v = 180 \implies v = 20

The quadratic terms cancel out cleanly, leaving a simple linear equation.

Key Concept

Two-Stage Transfer and Replacement Mixture Balance

Question 127Question →

A financial consulting firm audited $500$ corporate investment portfolios for exposure to three alternative asset classes: Private Equity ( $P$ ), Venture Capital ( $V$ ), and Infrastructure ( $I$ ). Every audited portfolio contained at least one of the three asset classes.

The audit revealed the following:
- Exactly $270$ portfolios contained Private Equity.
- Exactly $250$ portfolios contained Venture Capital.
- Exactly $220$ portfolios contained Infrastructure.
- The ratio of the number of portfolios containing both Private Equity and Venture Capital to those containing both Venture Capital and Infrastructure to those containing both Private Equity and Infrastructure was $11 : 9 : 10$ , respectively.
- The number of portfolios containing all three asset classes was $60$ .

How many portfolios contained exactly two of the three asset classes?

Show answer & explanation

Answer: 120

Answer

The total number of portfolios containing exactly two of the three asset classes is 120.

Using the 3-set inclusion-exclusion formula, we establish that

500 = 270 + 250 + 220 - (11k + 9k + 10k) + 60

, which simplifies to

30k = 300

, so

k = 10

. The sum of the pairwise intersections is

110 + 90 + 100 = 300

. Since each pairwise intersection includes the 60 portfolios that contain all three asset classes, the number of portfolios containing exactly two asset classes is

(110-60) + (90-60) + (100-60) = 50 + 30 + 40 = 120

.

Step-by-Step Solution

1

Set up the 3-Set Inclusion-Exclusion Principle equation

\text{Total} = |P| + |V| + |I| - (|P \cap V| + |V \cap I| + |P \cap I|) + |P \cap V \cap I|

This formula accounts for all region overlaps without double-counting or triple-counting.

2

Define variables using the given ratio for pairwise intersections

|P \cap V| = 11k

,

|V \cap I| = 9k

, and

|P \cap I| = 10k

Expressing the pairwise intersection sizes with a common ratio constant

k

allows us to substitute them into a single-variable linear equation.

3

Substitute the known numerical values and solve for

k

500 = 270 + 250 + 220 - (11k + 9k + 10k) + 60 \implies 500 = 800 - 30k \implies k = 10

Using algebraic simplification to solve for

k

gives the scaling factor for the pairwise intersections.

4

Calculate the actual sizes of the pairwise intersections

|P \cap V| = 110

,

|V \cap I| = 90

, and

|P \cap I| = 100

; Sum of pairwise intersections =

300

Multiplying each ratio term by

k=10

gives the exact count of portfolios containing at least the respective pairs.

5

Calculate portfolios containing exactly two asset classes

(110 - 60) + (90 - 60) + (100 - 60) = 50 + 30 + 40 = 120

The triple intersection (

60

) must be subtracted from each pairwise overlap to isolate the regions corresponding to portfolios holding exactly two asset classes.

Key Concept

Three-Set Overlapping Venn Diagrams and Inclusion-Exclusion Principle

Estimated Time:2m 0s

Question 128Question →

Two automated document redaction pipelines, Pipeline $A$ and Pipeline $B$ , process digital legal archives. Working alone at its constant rate, Pipeline $A$ can redact a standard archive in $8$ hours. Working alone at its constant rate, Pipeline $B$ can redact the same standard archive in $12$ hours. Both pipelines begin processing a standard archive together. After $3$ hours of joint operation, Pipeline $A$ encounters a system error and stops. Pipeline $B$ continues working alone at its constant rate until the entire archive is redacted. How many total hours does it take from start to finish to complete the redaction of the archive?

Show answer & explanation

Answer: 7.5

Answer

The total time required to redact the entire archive from start to finish is 7.5 hours.

To find the total time required, first determine the combined rate of Pipeline A and Pipeline B:

\frac{1}{8} + \frac{1}{12} = \frac{5}{24}

archive per hour. In the first 3 hours of joint work, they complete

3 \times \frac{5}{24} = \frac{5}{8}

of the job, leaving

1 - \frac{5}{8} = \frac{3}{8}

of the archive remaining. Pipeline B processes the remaining

\frac{3}{8}

at its individual rate of

\frac{1}{12}

per hour, taking

\frac{3/8}{1/12} = 4.5

hours. Adding the initial 3 hours to the 4.5 additional hours yields a total of 7.5 hours.

Step-by-Step Solution

1

Determine the individual work rate for each pipeline.

Pipeline A's rate is

\frac{1}{8}

archive per hour; Pipeline B's rate is

\frac{1}{12}

archive per hour.

Work rate is the fraction of job completed per hour (

1 / \text{time required alone}

).

2

Calculate the combined work rate during joint operation.

Combined rate is

\frac{1}{8} + \frac{1}{12} = \frac{5}{24}

archive per hour.

When entities work concurrently, their rates are added together.

3

Determine the amount of work completed in the first 3 hours.

Work completed is

3 \times \frac{5}{24} = \frac{5}{8}

of the total archive.

Work done equals rate multiplied by time.

4

Find the fraction of the archive remaining to be redacted.

Remaining work is

1 - \frac{5}{8} = \frac{3}{8}

of the archive.

The full task represents 1 complete unit of work.

5

Calculate how long Pipeline B needs to finish the remaining work alone.

Solo time for Pipeline B is

\frac{3/8}{1/12} = 4.5

hours.

Time equals remaining work divided by Pipeline B's individual rate.

6

Sum the joint operation time and the solo operation time.

Total time is

3 + 4.5 = 7.5

hours.

The overall duration consists of two sequential periods: 3 hours working together plus 4.5 hours for Pipeline B working alone.

Key Concept

Combined Work Rates and Multi-Stage Work Scenarios

Question 129Question →

A motorboat travels downstream along a straight river from Port Alpha to Port Beta, a distance of $120$ miles, completing the downstream leg in exactly $3$ hours. At the moment the motorboat departs from Port Alpha, an unpowered raft is released from Port Alpha and drifts downstream driven solely by the river's constant current. Upon reaching Port Beta, the motorboat immediately turns around and travels upstream toward Port Alpha. If the motorboat meets the drifting raft at a point $30$ miles downstream from Port Alpha, and the motorboat's speed relative to the water remains constant throughout the journey, what is the motorboat's speed in still water, in miles per hour?

20

25

32

35

45

Show answer & explanation

Answer: 35

Answer

35 miles per hour

The correct answer is 35 miles per hour. Since the motorboat covers the 120-mile downstream distance in 3 hours, its downstream rate is 40 mph, which means the boat's speed in still water

v

plus the current speed

c

equals 40. The raft drifts 30 miles at speed

c

, requiring

\frac{30}{c}

hours. In that same total time, the boat spends 3 hours going downstream and then travels 90 miles upstream at speed

v - c = 40 - 2c

. Equating the two time expressions

\frac{30}{c} = 3 + \frac{90}{40-2c}

produces the quadratic equation

c^2 - 45c + 200 = 0

. The valid root is

c = 5

mph, yielding

v = 40 - 5 = 35

mph.

Step-by-Step Solution

1

Determine the downstream speed of the motorboat and express the still-water speed

v

in terms of the current

c

.

The downstream speed is

\frac{120 \text{ miles}}{3 \text{ hours}} = 40

mph. Therefore,

v + c = 40

, which implies

v = 40 - c

mph and upstream speed

v - c = 40 - 2c

mph.

Downstream speed combines the motorboat's still-water speed and the river's current speed.

2

Formulate expressions for the time elapsed for both the raft and the motorboat until they meet.

The raft drifts

30

miles at speed

c

, so raft time

T = \frac{30}{c}

hours. The motorboat travels

120

miles downstream in

3

hours, and then

120 - 30 = 90

miles upstream at speed

(40 - 2c)

mph. So motorboat time

T = 3 + \frac{90}{40 - 2c}

hours.

Both vessels start at the exact same moment from Port Alpha, so their total travel times until they meet are equal.

3

Equate total times and solve the resulting algebraic equation for

c

.

Setting

\frac{30}{c} = 3 + \frac{90}{40 - 2c}

and dividing by

3

yields

\frac{10}{c} = 1 + \frac{15}{20 - c} = \frac{35 - c}{20 - c}

. Cross-multiplying gives

10(20 - c) = c(35 - c) \implies c^2 - 45c + 200 = 0

. Factoring gives

(c - 5)(c - 40) = 0

. Since

c < 40

,

c = 5

mph.

Solving the quadratic equation isolates the valid physical value for the current speed.

4

Calculate the motorboat's speed in still water,

v

.

v = 40 - 5 = 35

mph.

Subtracting the current speed from the downstream speed yields the speed in still water.

Key Concept

Relative speed in current and multi-leg journey time equilibrium

Question 130Question →

An industrial facility uses two automated assembly units, Unit $X$ and Unit $Y$ , to fabricate custom components. Working alone at its constant rate, Unit $X$ can complete a batch of components in $6$ hours, while Unit $Y$ can complete an identical batch working alone at its constant rate in $12$ hours. Unit $X$ begins working on a batch alone. After $2$ hours, Unit $Y$ joins Unit $X$ , and both units work together at their respective constant rates until the batch is complete. What is the total time, in hours, required to complete the batch from start to finish?

2\frac{2}{3}

3\frac{1}{2}

4\frac{2}{3}

6

8

Show answer & explanation

Answer: $4\frac{2}{3}$

Answer

4\frac{2}{3}

hours

The correct answer is

4\frac{2}{3}

hours. Unit

X

works alone for

2

hours, completing

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

of the job. This leaves

\frac{2}{3}

of the job remaining. Working together, Unit

X

and Unit

Y

have a combined rate of

\frac{1}{6} + \frac{1}{12} = \frac{1}{4}

of the job per hour. The time spent working together is

\frac{2/3}{1/4} = \frac{8}{3}

hours. Adding the initial

2

hours gives a total duration of

2 + \frac{8}{3} = 4\frac{2}{3}

hours.

Step-by-Step Solution

1

Calculate the individual work rates of Unit

X

and Unit

Y

.

Unit

X

's rate is

\frac{1}{6}

batch per hour, and Unit

Y

's rate is

\frac{1}{12}

batch per hour.

Work rate is defined as

\text{Rate} = \frac{1}{\text{Time}}

for completing one full job.

2

Determine the fraction of the batch completed by Unit

X

during its

2

hours of solo work.

Unit

X

completes

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

of the batch.

Work done equals

\text{Rate} \times \text{Time}

.

3

Find the remaining fraction of work to be completed.

The remaining work is

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

of the batch.

Subtract the completed fraction from the total job (

1

).

4

Calculate the combined work rate when both units operate together.

Combined rate =

\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}

batch per hour.

Simultaneous work rates add linearly.

5

Calculate the time needed for both units to finish the remaining work.

Time together =

\frac{2/3}{1/4} = \frac{2}{3} \times 4 = \frac{8}{3} = 2\frac{2}{3}

hours.

Time equals

\frac{\text{Remaining Work}}{\text{Combined Rate}}

.

6

Add the solo phase duration to the combined phase duration to get total elapsed time.

Total time =

2 + \frac{8}{3} = \frac{14}{3} = 4\frac{2}{3}

hours.

Total time is the sum of the times from each sequential stage.

Key Concept

Work Rate and Combined Work

Estimated Time:2m 0s

Question 131Question →

An art gallery displays works in three mediums: oil paintings, sculptures, and watercolors. Initially, the ratio of oil paintings to sculptures to watercolors is $3 : 4 : 5$ , respectively. After the gallery acquires 12 additional oil paintings and sells 6 watercolors, the ratio of oil paintings to watercolors becomes $1 : 1$ . If no sculptures were acquired or sold, how many sculptures are currently on display at the gallery?

27

36

45

48

108

Show answer & explanation

Answer: 36

Answer

36

By defining the initial quantities of oil paintings, sculptures, and watercolors as

3x

,

4x

, and

5x

respectively, we update the modified amounts to

3x + 12

for oil paintings and

5x - 6

for watercolors. Equating these gives

3x + 12 = 5x - 6

, leading to

2x = 18

and

x = 9

. The number of sculptures, which remained unchanged, is

4x = 4 \times 9 = 36

.

Step-by-Step Solution

1

Define variables based on the initial ratio

Let the multiplier be

x

. Initial quantities are Oil Paintings

= 3x

, Sculptures

= 4x

, and Watercolors

= 5x

.

Ratios expressed as

3 : 4 : 5

can be represented using a common multiplier

x

to find actual quantities.

2

Set up an equation incorporating the changes in inventory

New Oil Paintings

= 3x + 12

; New Watercolors

= 5x - 6

. The ratio between them is

1 : 1

, so

3x + 12 = 5x - 6

.

A

1 : 1

ratio means the updated quantity of oil paintings equals the updated quantity of watercolors.

3

Solve for the common multiplier

x

5x - 3x = 12 + 6 \implies 2x = 18 \implies x = 9

.

Linear algebraic manipulation isolates

x

.

4

Calculate the quantity of sculptures

Number of Sculptures

= 4x = 4 \times 9 = 36

.

Substitute the value of

x

back into the expression for sculptures.

Key Concept

Ratio modification and single-variable setting for multi-part ratios

Estimated Time:2m 0s

Question 132Question →

A laboratory container initially holds $120$ liters of a chemical solution that is $40\%$ acid by volume. A chemist removes $x$ liters of the solution and replaces it with an equal volume of pure acid. Next, the chemist removes $x$ liters of the resulting mixture and replaces it with an equal volume of pure water. If the final concentration of acid in the container is $40\%$ by volume, what is the value of $x$ ?

20

30

40

50

60

Show answer & explanation

Answer: $40$

Answer

The value of

x

is

40

.

The initial amount of acid is

120 \times 0.40 = 48

liters. After replacing

x

liters with pure acid, the amount of acid becomes

48 - 0.40x + x = 48 + 0.60x

liters. Removing

x

liters of this intermediate solution removes a fraction

\frac{x}{120}

of its acid, leaving

(48 + 0.60x)\left(1 - \frac{x}{120}\right)

liters. Adding pure water does not add acid. Setting this equal to the final acid quantity of

48

liters yields the equation

(48 + 0.60x)(120 - x) = 5760

, which simplifies to

24x - 0.60x^2 = 0

. Solving for non-zero

x

gives

x = 40

.

Step-by-Step Solution

1

Calculate initial acid quantity and acid quantity after the first replacement.

Initial acid =

120 \times 0.40 = 48

liters. Removing

x

liters of

40\%

solution removes

0.40x

liters of acid. Adding

x

liters of pure acid gives a new acid quantity of

48 - 0.40x + x = 48 + 0.60x

liters.

Pure acid contains

100\%

acid, so replacing

x

liters adds

1.00x

acid while removing

0.40x

acid.

2

Determine the acid quantity after the second replacement.

The solution concentration after Step 1 is

\frac{48 + 0.60x}{120}

. Removing

x

liters removes a fraction

\frac{x}{120}

of the solution, leaving a fraction

\left(1 - \frac{x}{120}\right)

. Replacing with pure water adds

0

acid, so final acid quantity is

(48 + 0.60x)\left(1 - \frac{x}{120}\right)

.

Pure water contributes

0\%

acid to the total solute count.

3

Set up and solve the equation for the final acid amount.

(48 + 0.60x)\left(\frac{120 - x}{120}\right) = 48 \implies (48 + 0.60x)(120 - x) = 5760

. Expanding gives

5760 - 48x + 72x - 0.60x^2 = 5760 \implies 24x - 0.60x^2 = 0

. Since

x > 0

,

0.60x = 24 \implies x = 40

.

The final concentration is given as

40\%

, which corresponds to

48

liters of acid in

120

liters total volume.

Key Concept

Multi-stage mixture dilution and replacement equations using solute mass balance.

Estimated Time:2m 0s

Question 133Question →

An investor deposits a principal amount into a savings account that earns interest at a constant annual compound rate of $r\%$ . At the end of the first year, immediately after annual interest is credited, the investor withdraws $20\%$ of the interest earned during that first year, leaving the original principal and all remaining interest in the account. At the end of the second year, the interest earned in the second year alone is $10\%$ greater than the interest earned in the first year alone. What is the value of $r$ ?

10.0\%

12.5\%

15.0\%

25.0\%

37.5\%

Show answer & explanation

Answer: $12.5\%$

Answer

12.5%

The option specifying 12.5% is correct. Let

P

be the initial principal. The interest earned in the first year is

I_1 = P r

. After withdrawing

20\%

of this interest,

80\%

of the interest remains in the account, making the new principal for Year 2 equal to

P + 0.8Pr = P(1 + 0.8r)

. The interest earned in Year 2 is

I_2 = P(1 + 0.8r)r

. Since Year 2 interest is

10\%

greater than Year 1 interest,

I_2 = 1.10 I_1

. Equating the two expressions gives

P(1 + 0.8r)r = 1.10 Pr

. Dividing both sides by

Pr

yields

1 + 0.8r = 1.10

, which simplifies to

0.8r = 0.10

, so

r = 0.125

or

12.5\%

.

Step-by-Step Solution

1

Express the interest earned in Year 1 in terms of principal

P

and interest rate

r

.

Year 1 interest

I_1 = P \times r

.

Simple/compound interest earned over the first single compounding period on initial principal

P

is

P \times r

.

2

Determine the account balance at the start of Year 2 after the partial interest withdrawal.

Balance at start of Year 2 =

P + 0.80(P \times r) = P(1 + 0.8r)

.

The investor keeps the principal

P

and retains

80\%

of Year 1 interest (

I_1 - 0.20 I_1 = 0.80 I_1

).

3

Express the interest earned in Year 2 (

I_2

) and set up the equation

I_2 = 1.10 \times I_1

.

P(1 + 0.8r) \times r = 1.10 \times (P \times r)

.

Year 2 interest is calculated on the updated balance at rate

r

, and is given as

10\%

greater than Year 1 interest.

4

Simplify the equation to solve for

r

.

1 + 0.8r = 1.10 \implies 0.8r = 0.10 \implies r = \frac{0.10}{0.80} = 0.125 = 12.5\%

.

Divide both sides by the non-zero quantity

P \times r

, then solve linear equation for

r

.

Key Concept

Percent Change and Compound Interest Base Tracking

Estimated Time:2m 0s

Question 134Question →

An investment fund allocated its total capital into three distinct asset classes: Class X, Class Y, and Class Z. Class X yielded an annual return of $12\%$ , Class Y yielded an annual return of $18\%$ , and Class Z yielded an annual return of $6\%$ . The capital invested in Class X was exactly twice the capital invested in Class Y. If the overall weighted annual return for the entire fund was $11\%$ , what percentage of the total capital was invested in Class Z?

Show answer & explanation

Answer: 37.5

Answer

37.5%

Using the given relationship

W_X = 2W_Y

and total weight constraint

W_X + W_Y + W_Z = 1

, Class Z's weight can be written as

1 - 3W_Y

. Plugging these into the overall weighted return formula yields

12(2W_Y) + 18W_Y + 6(1 - 3W_Y) = 11

, which solves to

W_Y = 5/24

. Substituting back gives

W_Z = 9/24 = 37.5\%

.

Step-by-Step Solution

1

Define fractional weight variables for each asset class.

Let

W_X

,

W_Y

, and

W_Z

be the proportions of total capital invested in Class X, Class Y, and Class Z, such that

W_X + W_Y + W_Z = 1

.

Establishing a standard framework for weighted average calculation.

2

Express

W_X

and

W_Z

in terms of

W_Y

.

W_X = 2W_Y

, so

2W_Y + W_Y + W_Z = 1 \implies W_Z = 1 - 3W_Y

.

Reducing the number of unknown variables to one.

3

Formulate the weighted average equation using component returns.

12W_X + 18W_Y + 6W_Z = 11 \implies 12(2W_Y) + 18W_Y + 6(1 - 3W_Y) = 11

.

The fund's overall return equals the weighted sum of individual returns.

4

Solve the algebraic equation for

W_Y

.

24W_Y + 18W_Y + 6 - 18W_Y = 11 \implies 24W_Y + 6 = 11 \implies 24W_Y = 5 \implies W_Y = \frac{5}{24}

.

Determining the exact weight of Class Y.

5

Calculate the weight percentage for Class Z.

W_Z = 1 - 3\left(\frac{5}{24}\right) = 1 - \frac{15}{24} = \frac{9}{24} = \frac{3}{8} = 0.375 \text{ or } 37.5\%

.

Evaluating the target component weight requested by the question.

Key Concept

Weighted Average in Combined Financial Assets

Question 135Question →

A merchant blends two types of coffee beans: Type A, which costs $\$12$ per kilogram and contains $15\%$ caffeine by weight, and Type B, which costs $\$18$ per kilogram and contains $25\%$ caffeine by weight. The merchant mixes Type A and Type B to create a $60$ -kilogram batch of Blend C. If the total cost of the $60$ -kilogram batch of Blend C is $\$936$ , what is the percentage of caffeine by weight in Blend C?

Show answer & explanation

Answer: 21

Answer

The percentage of caffeine by weight in Blend C is 21%.

Solving the linear system for total mass (

A + B = 60

) and total cost (

12A + 18B = 936

) reveals that Blend C consists of 24 kg of Type A beans and 36 kg of Type B beans. Type A contributes

24 \times 0.15 = 3.6

kg of caffeine, and Type B contributes

36 \times 0.25 = 9.0

kg of caffeine. The total caffeine mass of 12.6 kg out of 60 kg total mass equals a concentration of

12.6 / 60 = 21\%

.

Step-by-Step Solution

1

Formulate linear system for bean masses based on total weight and cost

A + B = 60

and

12A + 18B = 936

, solving to

A = 24

kg and

B = 36

kg

The weight of each constituent bean type must be determined before total caffeine content can be evaluated.

2

Calculate caffeine mass from each component

Type A provides

3.6

kg (

24 \times 0.15

) and Type B provides

9.0

kg (

36 \times 0.25

)

Solute mass is found by multiplying total component weight by its percentage concentration.

3

Sum caffeine masses to find total caffeine in the mixture

Total caffeine =

3.6 + 9.0 = 12.6

kg

The total amount of caffeine present in Blend C is the combined sum from both bean types.

4

Determine caffeine concentration percentage of Blend C

\frac{12.6}{60} \times 100 = 21\%

Concentration is calculated as the ratio of total solute mass to total solution mass expressed as a percentage.

Key Concept

Weighted concentration combined with multi-variable cost systems

Question 136Question →

A manufacturing plant produces electronic components across three assembly lines: Line 1, Line 2, and Line 3. Line 1 accounts for $40\%$ of the plant's total production and has a defect rate of $2.0\%$ . Line 2 accounts for $35\%$ of total production and has a defect rate of $4.0\%$ . Line 3 produces the remaining components. If the overall defect rate for the entire manufacturing plant is $3.1\%$ , what is the defect rate of Line 3?

3.0\%

3.2\%

3.3\%

3.6\%

4.1\%

Show answer & explanation

Answer: $3.6\%$

Answer

3.6\%

The correct answer is

3.6\%

. Since Line 1 accounts for

40\%

of production at a

2.0\%

defect rate (

0.8\%

overall contribution) and Line 2 accounts for

35\%

at a

4.0\%

defect rate (

1.4\%

overall contribution), their combined contribution is

2.2\%

. Line 3 represents the remaining

25\%

(

0.25

) of production. Setting up

2.2\% + 0.25 \times d = 3.1\%

gives

0.25 d = 0.9\%

, yielding

d = 3.6\%

.

Step-by-Step Solution

1

Determine the proportion of total production for Line 3

The proportion for Line 3 is

100\% - (40\% + 35\%) = 25\%

(or

0.25

).

The sum of the production proportions of all three lines must equal

100\%

of total production.

2

Set up the weighted average equation for the overall defect rate

0.40(2.0\%) + 0.35(4.0\%) + 0.25(d\%) = 3.1\%

, where

d\%

represents the defect rate of Line 3.

The overall weighted average defect rate is the sum of each component's defect rate multiplied by its relative production weight.

3

Calculate the defect contributions from Line 1 and Line 2

Contribution of Line 1 =

0.40 \times 2.0\% = 0.8\%

; Contribution of Line 2 =

0.35 \times 4.0\% = 1.4\%

. Combined contribution =

0.8\% + 1.4\% = 2.2\%

.

Multiplying individual weights by their defect rates yields their weighted contribution to the overall plant defect rate.

4

Solve for the defect rate of Line 3 (

d

)

2.2\% + 0.25d\% = 3.1\% \implies 0.25d\% = 0.9\% \implies d = \frac{0.9}{0.25} = 3.6

. Thus, the defect rate of Line 3 is

3.6\%

.

Isolating the variable

d

gives the exact required defect rate for Line 3.

Key Concept

Weighted Average in Combined Sets

Estimated Time:2m 0s

Question 137Question →

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 138Question →

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 139Question →

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 140Question →

A tech hub tracked $150$ start-up companies to determine their use of three cloud services: Storage ( $S$ ), Analytics ( $A$ ), and Machine Learning ( $M$ ). Every company uses at least one of the three services. Exactly $95$ companies use Storage, $80$ use Analytics, and $65$ use Machine Learning. If exactly $20$ companies use all three services, and $45$ companies use exactly two of the services, how many companies use exactly one of the three cloud services?

65

75

85

95

105

Show answer & explanation

Answer: $75$

Answer

The number of companies that use exactly one of the three cloud services is

75

.

Using the region-based formula for three overlapping sets, the total number of elements is equal to the sum of elements in exactly one set, elements in exactly two sets, elements in all three sets, and elements in none of the sets. Here, Total

= 150

, Exactly Two

= 45

, Exactly Three

= 20

, and Neither

= 0

. Substituting these values gives

150 = \text{Exactly One} + 45 + 20 + 0

, which simplifies to

\text{Exactly One} = 75

.

Step-by-Step Solution

1

State the standard 3-set overlapping sets formula in terms of exact regions.

Total

= (\text{Exactly 1}) + (\text{Exactly 2}) + (\text{Exactly 3}) + (\text{Neither})

Dividing the Venn diagram into mutually exclusive regions simplifies calculating specific counts.

2

Substitute the known values into the regional total equation.

150 = (\text{Exactly 1}) + 45 + 20 + 0

Since every company uses at least one service,

\text{Neither} = 0

.

3

Solve for the number of companies using exactly one service.

\text{Exactly 1} = 150 - 65 = 75

Subtracting the sum of companies using exactly two services (

45

) and all three services (

20

) from the total (

150

) yields

75

.

Key Concept

Three-Set Overlapping Sets Regional Decomposition

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