Algebraic Word Problems and Modeling Practice Questions — GRE General Test — Page 2

Question 21Question →

A manufacturing facility uses Machine A and Machine B to process standard orders of raw materials. Operating independently at its constant rate, Machine A requires $x$ hours to process one standard order, where $x > 0$ . Machine B operates at a constant rate and requires $x + 2$ hours to process one standard order. When both machines operate simultaneously at their respective constant rates for $6$ hours, the total number of standard orders processed is $2$ fewer than the number of standard orders Machine A would process operating alone for $18$ hours. What is the value of $x$ ?

2

3

4

6

8

Show answer & explanation

Answer: 4

Answer

4

The rate of Machine A is

\frac{1}{x}

orders per hour, and the rate of Machine B is

\frac{1}{x+2}

orders per hour. In 18 hours, Machine A processes

\frac{18}{x}

orders. Working together for 6 hours, both machines process

6\left(\frac{1}{x} + \frac{1}{x+2}\right)

orders. Setting up the difference:

\frac{18}{x} - 6\left(\frac{1}{x} + \frac{1}{x+2}\right) = 2

, which simplifies to

\frac{12}{x} - \frac{6}{x+2} = 2

. Multiplying both sides by

x(x+2)

yields

12(x+2) - 6x = 2x^2 + 4x

, leading to

2x^2 - 2x - 24 = 0

or

x^2 - x - 12 = 0

. Factoring gives

(x-4)(x+3) = 0

. Because time must be positive,

x = 4

.

Step-by-Step Solution

1

Express the individual work rates of Machine A and Machine B.

Machine A completes

\frac{1}{x}

orders per hour; Machine B completes

\frac{1}{x+2}

orders per hour.

Work rate is the reciprocal of the total time required to complete one unit of work.

2

Formulate the equation based on the total orders processed in the given time frames.

18 \cdot \frac{1}{x} - 6\left(\frac{1}{x} + \frac{1}{x+2}\right) = 2

Machine A alone in 18 hours processes

\frac{18}{x}

orders. Together in 6 hours, they process

6\left(\frac{1}{x} + \frac{1}{x+2}\right)

orders, which is 2 orders less.

3

Simplify the algebraic equation.

\frac{12}{x} - \frac{6}{x+2} = 2

Subtracting

6 \cdot \frac{1}{x}

from

18 \cdot \frac{1}{x}

yields

\frac{12}{x}

.

4

Clear the denominators by multiplying through by

x(x+2)

and solve the resulting quadratic equation.

12(x+2) - 6x = 2x(x+2) \implies 6x + 24 = 2x^2 + 4x \implies 2x^2 - 2x - 24 = 0 \implies x^2 - x - 12 = 0 \implies (x-4)(x+3) = 0

Clearing denominators transforms the rational equation into a standard quadratic equation.

5

Select the physically meaningful solution for time

x

.

x = 4

hours (rejecting

x = -3

since

x > 0

).

Time must be positive.

Key Concept

Algebraic Work-Rate Modeling and Quadratic Solution

Estimated Time:2m 0s

Question 22Question →

A water reservoir is filled by Pipe A and Pipe B operating simultaneously at their respective constant rates. Operating together at their original rates, the two pipes can fill the empty reservoir completely in $12$ hours. On a certain day, both pipes begin filling the empty reservoir together at their original rates. After $4$ hours, Pipe A's rate decreases by $25\%$ , while Pipe B's rate increases by $50\%$ . Operating at these new constant rates, the two pipes require an additional $7$ hours to fill the remainder of the reservoir. How many hours would it take Pipe A, operating alone at its original rate, to fill the entire reservoir?

Show answer & explanation

Answer: 25.2

Answer

It would take Pipe A 25.2 hours operating alone at its original rate to fill the entire reservoir.

By defining the original work rates

a

and

b

in reservoirs per hour, the initial condition yields

a + b = \frac{1}{12}

. In the first 4 hours,

\frac{1}{3}

of the job is completed, leaving

\frac{2}{3}

. Setting up the equation for the remaining job with modified rates

0.75a

and

1.5b

over 7 hours produces

7(0.75a + 1.5b) = \frac{2}{3}

. Solving this system of two linear equations yields

a = \frac{5}{126}

reservoirs per hour. Taking the reciprocal gives the time required for Pipe A alone to fill the reservoir, which is

25.2

hours.

Step-by-Step Solution

1

Set up equations for the original rates of Pipe A (

a

) and Pipe B (

b

).

The combined original rate is

a + b = \frac{1}{12}

reservoir per hour.

Together they complete

1

reservoir in

12

hours.

2

Determine the fraction of the reservoir filled in the first 4 hours and the remaining fraction.

Work completed =

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

; Remaining work =

\frac{2}{3}

.

The pipes worked at their original combined rate for 4 hours.

3

Set up an equation for the work done during the remaining 7 hours at the adjusted rates.

7 \left(0.75a + 1.5b\right) = \frac{2}{3} \implies 5.25a + 10.5b = \frac{2}{3} \implies 63a + 126b = 8

.

Pipe A's rate decreases by

25\%

to

0.75a

, and Pipe B's rate increases by

50\%

to

1.5b

.

4

Solve the system of linear equations for

a

.

a = \frac{5}{126}

reservoir per hour.

Multiplying

a + b = \frac{1}{12}

by

126

yields

126a + 126b = 10.5

. Subtracting

63a + 126b = 8

gives

63a = 2.5

, so

a = \frac{2.5}{63} = \frac{5}{126}

.

5

Calculate the time for Pipe A alone to fill the entire reservoir.

Time

= \frac{1}{a} = \frac{126}{5} = 25.2

hours.

Time equals total work divided by individual rate.

Key Concept

Algebraic modeling of combined work and rates with mid-process rate modifications

Question 23Question →

A total initial capital of $\$100,000$ is split between Fund X and Fund Y. Fund X yields an annual simple interest rate of $r\%$ , while Fund Y yields an annual simple interest rate of $(r + 2)\%$ . Under the original capital allocation, the ratio of the annual interest earned from Fund X to the annual interest earned from Fund Y is $15 : 14$ . If the initial allocation to Fund X had been increased by $25\%$ by transferring funds directly from Fund Y, the total annual interest earned from both funds combined would have been $\$5,500$ . What was the original amount allocated to Fund X?

\$40,000

\$50,000

\$60,000

\$64,000

\$75,000

Show answer & explanation

Answer: $\$60,000$

Answer

The original amount allocated to Fund X was

\$60,000

.

The correct option is

\$60,000

. Substituting

X = 60,000

gives Fund Y allocation

Y = 40,000

. From the modified interest condition,

100,000r - 2.5(60,000) = 350,000

, giving

r = 5\%

. Fund X yields

60,000 \times 0.05 = \$3,000

and Fund Y yields

40,000 \times 0.07 = \$2,800

, which satisfies the ratio

3,000 : 2,800 = 15 : 14

. Under the modified allocation (

75,000

in X and

25,000

in Y), interest is

75,000(0.05) + 25,000(0.07) = 3,750 + 1,750 = \$5,500

.

Step-by-Step Solution

1

Formulate variables and express initial conditions.

Let

X

be the initial capital in Fund X and

Y = 100,000 - X

be the initial capital in Fund Y. The interest rates are

\frac{r}{100}

and

\frac{r+2}{100}

respectively.

Establish a single-variable representation for the fund allocations.

2

Set up the interest ratio equation.

\frac{X \cdot r}{(100,000 - X)(r+2)} = \frac{15}{14} \implies 14 X r = 15(100,000 - X)(r+2).$

Relate the original interest outputs according to the

15:14

ratio.

3

Model the modified allocation scenario.

Fund X becomes

1.25X

and Fund Y becomes

100,000 - 1.25X

. Total interest equation:

(1.25X)\left(\frac{r}{100}\right) + (100,000 - 1.25X)\left(\frac{r+2}{100}\right) = 5,500.

Express total combined interest under the hypothetical

25\%

transfer.

4

Simplify the total interest equation to express

r

in terms of

X

.

1.25Xr + 100,000r + 200,000 - 1.25Xr - 2.5X = 550,000 \implies 100,000r - 2.5X = 350,000 \implies r = 3.5 + 0.000025X.$

Eliminate the

Xr

product term to solve for

r

linearly.

5

Substitute

r

into the ratio equation and solve for

X

.

Expanding

29Xr + 30X = 1,500,000r + 3,000,000

with

r = 3.5 + 0.000025X

yields

0.000725X^2 + 94X - 8,250,000 = 0

, which factors to give

X = 60,000

.

Determine the exact value for the initial allocation to Fund X.

Key Concept

Algebraic modeling of multi-variable financial rate and allocation systems.

Estimated Time:3m 0s

Question 24Question →

Two automated data processing algorithms, Algorithm X and Algorithm Y, operate at their respective constant rates. Algorithm X takes $x$ hours to process a full dataset when working alone. Algorithm Y operates at a constant rate that is $50\%$ faster than Algorithm X.

In Scenario 1, Algorithm X processes the dataset alone for $2$ hours, after which Algorithm Y joins, and both algorithms work together for an additional $t$ hours to complete the dataset.

In Scenario 2, Algorithm Y processes the dataset alone for $3$ hours, after which Algorithm X joins, and both algorithms complete the remaining work together in $\frac{4}{5}t$ hours.

Which of the following statements must be true? Indicate all such statements.

Select all that apply

Algorithm X operating alone requires 14 hours and 30 minutes to process one full dataset.

Algorithm Y operating alone requires 10 hours and 15 minutes to process one full dataset.

If both algorithms operate together from the beginning, they will complete one full dataset in 5 hours and 48 minutes.

The total time spent by Algorithm X working across Scenario 1 is 5 hours.

Algorithm Y processes exactly 60 percent of the dataset in Scenario 2.

Show answer & explanation

Answer: Algorithm X operating alone requires 14 hours and 30 minutes to process one full dataset.; If both algorithms operate together from the beginning, they will complete one full dataset in 5 hours and 48 minutes.

Answer

The true statements are that Algorithm X operating alone requires 14 hours and 30 minutes to process one full dataset, and that operating together from the start, both algorithms complete one full dataset in 5 hours and 48 minutes.

The statement specifying that Algorithm X takes 14 hours and 30 minutes alone is correct because solving the system yields

x = 14.5

hours. The statement regarding the combined time of 5 hours and 48 minutes is correct because

\frac{1}{R_X + R_Y} = 5.8

hours, which equals 5 hours and 48 minutes.

Step-by-Step Solution

1

Define individual and combined work rates in terms of

x

.

Rate of Algorithm X is

R_X = \frac{1}{x}

datasets/hr. Since Algorithm Y is

50\%

faster, its rate is

R_Y = 1.5 R_X = \frac{3}{2x}

datasets/hr. Their combined rate is

R_{X+Y} = \frac{1}{x} + \frac{3}{2x} = \frac{5}{2x}

datasets/hr.

Establishing accurate rate expressions is essential for modeling combined work scenarios.

2

Formulate equations for both scenarios and solve for

t

and

x

.

From Scenario 1:

2 \cdot \frac{1}{x} + t \cdot \frac{5}{2x} = 1 \implies 2 + 2.5t = x

.
From Scenario 2:

3 \cdot \frac{3}{2x} + \frac{4}{5}t \cdot \frac{5}{2x} = 1 \implies 4.5 + 2t = x

.
Equating the two expressions:

2 + 2.5t = 4.5 + 2t \implies 0.5t = 2.5 \implies t = 5

hours.
Substituting

t = 5

yields

x = 2 + 2.5(5) = 14.5

hours (14 hours 30 minutes).

Setting work completed equal to 1 dataset in both scenarios yields a solvable system of equations.

3

Evaluate the given statement choices using the derived parameters.

1. Algorithm X alone time:

x = 14.5

hours = 14 hours 30 minutes (True).
2. Algorithm Y alone time:

\frac{1}{R_Y} = \frac{2x}{3} = \frac{29}{3} = 9

hours 40 minutes (False).
3. Combined time from start:

\frac{1}{R_{X+Y}} = \frac{2x}{5} = \frac{29}{5} = 5.8

hours = 5 hours 48 minutes (True).
4. Total time Algorithm X works in Scenario 1:

2 + t = 2 + 5 = 7

hours (False).
5. Work fraction by Algorithm Y in Scenario 2:

(3 + \frac{4}{5}(5)) \cdot \frac{3}{2x} = 7 \cdot \frac{3}{29} = \frac{21}{29} \approx 72.4\%

(False).

Direct numerical verification reveals which statements hold true.

Key Concept

Combined work rate equations with variable initial delays and relative rate multipliers.

Question 25Question →

A catering service charges a fixed base setup fee of $\$150$ plus $\$25$ per guest for a private dinner event. If a client was billed a total of $\$900$ for an event, how many guests attended?

24

30

34

36

42

Show answer & explanation

Answer: $30$

Answer

30 guests attended the event.

The total billing model is represented by the equation

150 + 25g = 900

, where

g

is the number of guests. Subtracting the fixed setup fee of

\$150

gives

25g = 750

. Dividing

750

by

25

yields

g = 30

, showing that exactly

30

guests attended.

Step-by-Step Solution

1

Formulate a linear modeling equation for total cost

\text{Total Cost} = \text{Base Fee} + (\text{Rate per Guest} \times g)

, where

g

represents the number of guests. Thus,

150 + 25g = 900

.

The total charge consists of a fixed constant component plus a variable component that depends on the number of guests.

2

Isolate the variable term by subtracting the fixed setup fee from both sides

25g = 900 - 150 \implies 25g = 750

.

Subtracting the fixed fee isolates the portion of the total cost spent strictly on guests.

3

Solve for

g

by dividing by the per-guest rate

g = \frac{750}{25} = 30

.

Dividing the variable total cost by the rate per guest yields the exact number of attending guests.

Key Concept

Linear Modeling and Single-Variable Word Problems

Estimated Time:1m 0s

Question 26Question →

A retailer purchases items for $x$ dollars each and sets the regular selling price at a $40\%$ markup above the purchase price. During a clearance event, the retailer discounts the regular selling price by $15\%$ . Which of the following statements regarding the pricing of an item must be true? Select all such statements.

Select all that apply

The regular selling price of an item is

1.40x

dollars.

The clearance sale price of an item is

1.19x

dollars.

The clearance sale price represents a

25\%

markup above the original purchase price

x

.

The dollar profit earned on each item sold during the clearance event is

0.25x

dollars.

Show answer & explanation

Answer: The regular selling price of an item is $1.40x$ dollars.; The clearance sale price of an item is $1.19x$ dollars.

Answer

The statements confirming that the regular selling price is

1.40x

dollars and that the clearance sale price is

1.19x

dollars are correct.

The regular selling price is computed by increasing the cost

x

by 40%, yielding

1.40x

. The clearance price is found by reducing

1.40x

by 15%, which equals

1.40x \times 0.85 = 1.19x

. Therefore, both the statement that the regular price is

1.40x

dollars and the statement that the clearance price is

1.19x

dollars are correct.

Step-by-Step Solution

1

Model the regular selling price using the original purchase price

x

and the

40\%

markup.

Regular price

= x \times (1 + 0.40) = 1.40x

.

A 40% markup increases the base cost

x

by 0.40 times its original value.

2

Model the clearance sale price by applying a

15\%

discount to the regular selling price.

Clearance price

= 1.40x \times (1 - 0.15) = 1.40x \times 0.85 = 1.19x

.

A 15% discount reduces the regular price to 85% of its value.

3

Calculate the effective markup percentage and profit per item to evaluate all statements.

Profit

= 1.19x - x = 0.19x

, which is a

19\%

net markup over the purchase price

x

.

Comparing

1.19x

to the base cost

x

shows a net gain of

0.19x

per unit.

Key Concept

Applying successive percent changes in algebraic modeling by using correct base values.

Estimated Time:1m 0s

Question 27Question →

Container X contains a liquid solution that is $80\%$ alcohol by volume, and Container Y contains a liquid solution that is $25\%$ alcohol by volume. A chemist removes a specific volume of solution from Container X and mixes it with a solution from Container Y to produce $100$ liters of a new mixture that is $58\%$ alcohol by volume. Following this removal, $20$ liters of pure alcohol are added to the liquid remaining in Container X. If Container X initially held $120$ liters of solution, what is the concentration of alcohol, by volume, in Container X after the pure alcohol is added?

56.67\%

68\%

70\%

80\%

85\%

Show answer & explanation

Answer: $85\%$

Answer

85%

To find the final concentration, we first determine the volume of solution removed from Container X. Using the weighted average for the 100-liter mixture:

0.80 V_x + 0.25 (100 - V_x) = 58

, which simplifies to

0.55 V_x = 33

, so

V_x = 60

liters. Container X originally held 120 liters, so removing 60 liters leaves 60 liters of solution containing

80\% \times 60 = 48

liters of alcohol. Adding 20 liters of pure alcohol increases the total alcohol to

48 + 20 = 68

liters and the total volume to

60 + 20 = 80

liters. The final concentration is

\frac{68}{80} = 85\%

.

Step-by-Step Solution

1

Find the volume of solution

V_x

removed from Container X to make the 100-liter mixture.

V_x = 60

liters.

Let

V_x

be the volume from X and

100 - V_x

be the volume from Y. Setting up the alcohol concentration equation:

0.80 V_x + 0.25 (100 - V_x) = 0.58(100) \implies 0.55 V_x + 25 = 58 \implies 0.55 V_x = 33 \implies V_x = 60

liters.

2

Determine the remaining solution volume and alcohol volume in Container X after removing 60 liters.

Remaining solution =

60

liters; Remaining alcohol =

48

liters.

Container X initially had

120

liters. Removing

60

liters leaves

120 - 60 = 60

liters. Since the mixture is homogeneous, the remaining liquid retains an

80\%

alcohol concentration, yielding

0.80 \times 60 = 48

liters of alcohol.

3

Calculate the total alcohol volume and total solution volume in Container X after adding 20 liters of pure alcohol.

New alcohol volume =

68

liters; New total solution volume =

80

liters.

Adding

20

liters of pure alcohol increases both the alcohol amount (

48 + 20 = 68

liters) and the total liquid volume (

60 + 20 = 80

liters).

4

Calculate the final concentration of alcohol in Container X.

Concentration =

\frac{68}{80} = 0.85 = 85\%

.

The final alcohol concentration is the ratio of final alcohol volume to final total liquid volume.

Key Concept

Algebraic Modeling of Multi-Step Mixture Solutions

Estimated Time:2m 30s

Question 28Question →

Three water pumps, A, B, and C, are used to drain a large industrial reservoir. Operating alone at its constant rate, Pump A can drain the reservoir in $x$ hours, where $x > 0$ . Pump B operating alone takes $x + 4$ hours to drain the reservoir. When operating together for 2 hours, Pump A and Pump B complete the exact same fraction of the total job that Pump C completes operating alone in 3 hours. If all three pumps working simultaneously at their respective constant rates can drain the entire reservoir in $\frac{9}{4}$ hours, what is the value of $x$ ?

4

6

8

10

12

Show answer & explanation

Answer: 6

Answer

The value of

x

is 6.

The correct answer 6 is derived by properly formulating the work rates

R_A = \frac{1}{x}

,

R_B = \frac{1}{x+4}

, and

R_C = \frac{4(x+2)}{3x(x+4)}

. Setting their sum

\frac{10(x+2)}{3x(x+4)}

equal to the combined rate

\frac{4}{9}

forms the quadratic

2x^2 - 7x - 30 = 0

, which yields the unique positive solution

x = 6

.

Step-by-Step Solution

1

Express the individual work rates of Pump A and Pump B in terms of

x

.

R_A = \frac{1}{x}

and

R_B = \frac{1}{x+4}

. Combined rate

R_{A+B} = \frac{1}{x} + \frac{1}{x+4} = \frac{2x+4}{x(x+4)}

.

Work rate is defined as the fraction of the job completed per hour.

2

Determine the work rate of Pump C using the given relationship.

Work done by A and B in 2 hours is

2 \cdot \frac{2x+4}{x(x+4)} = \frac{4x+8}{x(x+4)}

. Since Pump C does this in 3 hours,

R_C = \frac{1}{3} \cdot \frac{4x+8}{x(x+4)} = \frac{4(x+2)}{3x(x+4)}

.

Pump C's hourly rate is one-third of the total work completed by A and B in 2 hours.

3

Sum all three rates to find the total combined rate and set it equal to the given combined rate.

R_{total} = \frac{2x+4}{x(x+4)} + \frac{4x+8}{3x(x+4)} = \frac{3(2x+4) + 4x+8}{3x(x+4)} = \frac{10(x+2)}{3x(x+4)}

. Given total time is

\frac{9}{4}

hours, total rate is

\frac{4}{9}

. Thus,

\frac{10(x+2)}{3x(x+4)} = \frac{4}{9}

.

The sum of individual rates equals the inverse of total time required when working simultaneously.

4

Solve the algebraic equation for

x

.

Multiply both sides by 9 to get

\frac{90(x+2)}{3x(x+4)} = 4 \implies \frac{30(x+2)}{x(x+4)} = 4 \implies 15(x+2) = 2x(x+4) \implies 15x + 30 = 2x^2 + 8x \implies 2x^2 - 7x - 30 = 0

. Factoring gives

(2x + 5)(x - 6) = 0

. Since

x > 0

,

x = 6

.

Solving the quadratic yields the valid positive real root for time

x

.

Key Concept

Work Rate Modeling and Rational Equation Systems

Question 29Question →

An executive chartered a private aircraft to complete a trip between two cities separated by a non-stop distance of $600$ miles. On the return flight along the exact same route, strong headwinds reduced the aircraft's average ground speed by $50$ miles per hour compared to its outbound speed. As a result, the return flight took $2$ hours longer than the outbound flight. What was the average speed, in miles per hour, of the aircraft on the outbound flight?

Show answer & explanation

Answer: 150

Answer

150 miles per hour

By setting the outbound speed to

v

and return speed to

v - 50

, the relationship between outbound flight time

\frac{600}{v}

and return flight time

\frac{600}{v - 50}

yields the rational equation

\frac{600}{v - 50} - \frac{600}{v} = 2

. Solving this equation gives the quadratic

v^2 - 50v - 15,000 = 0

, which factors as

(v - 150)(v + 100) = 0

. The positive root gives an outbound speed of 150 miles per hour.

Step-by-Step Solution

1

Define variables for the unknown outbound rate and express travel times for both legs.

Let

v

be the outbound speed in miles per hour. Outbound time is

\frac{600}{v}

hours and return time is

\frac{600}{v - 50}

hours.

Distance equals speed multiplied by time (

d = v \cdot t

), so time equals distance divided by speed.

2

Formulate the algebraic equation using the given difference in flight durations.

\frac{600}{v - 50} - \frac{600}{v} = 2

The return flight took

2

hours longer than the outbound flight.

3

Clear denominators and simplify into standard quadratic form.

v^2 - 50v - 15,000 = 0

Multiplying both sides by

v(v - 50)

yields

600v - 600v + 30,000 = 2(v^2 - 50v)

, which simplifies to

2v^2 - 100v - 30,000 = 0

or

v^2 - 50v - 15,000 = 0

.

4

Solve the quadratic equation for

v

.

v = 150

Factoring

(v - 150)(v + 100) = 0

gives solutions

v = 150

or

v = -100

. Physical speed must be positive.

Key Concept

Distance-Rate-Time Quadratic Algebraic Modeling

Question 30Question →

A community library purchases two types of books: hardcover books for $\$24$ each and paperback books for $\$15$ each. The library spends a total of $\$360$ on these books and purchases at least one book of each type. Which of the following could be the total number of books purchased? Select all such numbers.

Select all that apply

15

18

20

21

24

Show answer & explanation

Answer: 18; 21

Answer

18 and 21

The linear modeling equation

24x + 15y = 360

simplifies to

8x + 5y = 120

. Since

x

and

y

must be positive integers,

x

must be a multiple of 5. The only valid solutions satisfying

x \ge 1

and

y \ge 1

are

(x=5, y=16)

and

(x=10, y=8)

, which yield total book counts of 21 and 18, respectively.

Step-by-Step Solution

1

Set up the linear equation from the given word problem context.

24x + 15y = 360

, where

x \ge 1

is the number of hardcover books and

y \ge 1

is the number of paperback books, with

x, y \in \mathbb{Z}^+

.

Total expenditure is the sum of cost per hardcover times number of hardcovers plus cost per paperback times number of paperbacks.

2

Simplify the equation by dividing both sides by the greatest common divisor, 3.

8x + 5y = 120

Simplifying coefficients reduces arithmetic complexity and isolates integer conditions.

3

Express

y

in terms of

x

to identify valid integer pairs

(x, y)

.

y = \frac{120 - 8x}{5} = 24 - \frac{8x}{5}

For

y

to be an integer,

8x

must be divisible by 5, meaning

x

must be a positive multiple of 5.

4

Test valid positive integer values for

x

such that

y \ge 1

.

If

x = 5

,

y = 24 - 8 = 16

, giving total books

x + y = 21

. If

x = 10

,

y = 24 - 16 = 8

, giving total books

x + y = 18

. If

x \ge 15

,

y \le 0

, which is invalid.

These are the only integer solutions satisfying

x \ge 1

and

y \ge 1

.

Key Concept

Linear Diophantine Equations in Word Problems

Estimated Time:1m 30s

Question 31Question →

Two automated assembly lines, Line A and Line B, produce components at constant individual rates. Under normal operating conditions, Line A operating for $3$ hours and Line B operating for $4$ hours together produce a combined total of $1,400$ units. Under adjusted operating conditions, Line A operates at a rate $20\%$ higher than its normal rate, while Line B operates at a rate $10\%$ lower than its normal rate. Operating together under these adjusted conditions for $5$ hours, the two lines produce a total of $2,100$ units. What is the normal rate of Line A, in units per hour?

160

200

240

250

300

Show answer & explanation

Answer: $200$

Answer

The normal rate of Line A is

200

units per hour.

Let

r_A

and

r_B

represent the normal production rates in units per hour for Line A and Line B, respectively. From the first condition,

3r_A + 4r_B = 1400

. From the second condition, operating for

5

hours at rates

1.20r_A

and

0.90r_B

yields

5(1.20r_A + 0.90r_B) = 2100

, which simplifies to

1.20r_A + 0.90r_B = 420

, or

4r_A + 3r_B = 1400

. Subtracting

3r_A + 4r_B = 1400

from

4r_A + 3r_B = 1400

gives

r_A - r_B = 0

, meaning

r_A = r_B

. Substituting

r_B = r_A

into

3r_A + 4r_A = 1400

gives

7r_A = 1400

, so

r_A = 200

units per hour.

Step-by-Step Solution

1

Define variables and set up the equation for normal operating conditions.

3r_A + 4r_B = 1400

Line A operates for

3

hours at rate

r_A

and Line B operates for

4

hours at rate

r_B

to produce

1,400

units.

2

Set up the equation for adjusted operating conditions.

5(1.20r_A + 0.90r_B) = 2100 \implies 1.20r_A + 0.90r_B = 420

Line A's rate increases by

20\%

(

1.20r_A

) and Line B's rate decreases by

10\%

(

0.90r_B

). Divided by

5

hours, their combined hourly adjusted rate is

420

units per hour.

3

Multiply the simplified adjusted equation by

10

to clear decimals.

12r_A + 9r_B = 4200 \implies 4r_A + 3r_B = 1400

Dividing all terms by

3

simplifies the linear equation for easier elimination.

4

Solve the system of equations for

r_A

.

r_A = 200

From Step 1,

4r_B = 1400 - 3r_A \implies r_B = 350 - 0.75r_A

. Substituting into

4r_A + 3(350 - 0.75r_A) = 1400

gives

4r_A + 1050 - 2.25r_A = 1400 \implies 1.75r_A = 350 \implies r_A = 200

.

Key Concept

Linear Modeling of Combined Work and Rates

Question 32Question →

A coffee roaster creates a custom blend by combining Grade X beans costing $\$p$ per pound with Grade Y beans costing $\$q$ per pound, where $0 < p < q$ . The batch contains a total of $M$ pounds, consisting of $x$ pounds of Grade X and $y$ pounds of Grade Y. The total cost of the batch is $C$ dollars, and the average cost per pound of the blend is $A$ dollars. Which of the following algebraic relationships MUST be true? Indicate all such statements.

Select all that apply

The weight of Grade X beans in pounds is given by

x = \frac{qM - C}{q - p}

.

The weight of Grade X beans in pounds is given by

x = \frac{C - pM}{q - p}

.

The average cost per pound of the blend satisfies

A = q - \frac{x(q - p)}{M}

.

If the weight of Grade X in the blend is increased by

20\%

while keeping the weight of Grade Y constant, the total cost of the batch increases by

0.20C

dollars.

The ratio of the weight of Grade X to the weight of Grade Y in the blend is equal to

\frac{q - A}{A - p}

.

Show answer & explanation

Answer: The weight of Grade X beans in pounds is given by $x = \frac{qM - C}{q - p}$ .; The average cost per pound of the blend satisfies $A = q - \frac{x(q - p)}{M}$ .; The ratio of the weight of Grade X to the weight of Grade Y in the blend is equal to $\frac{q - A}{A - p}$ .

Answer

The true statements are that the weight of Grade X beans is

x = \frac{qM - C}{q - p}

, the average cost per pound satisfies

A = q - \frac{x(q - p)}{M}

, and the ratio of Grade X weight to Grade Y weight is

\frac{q - A}{A - p}

.

The correct statements correctly model the system of equations

x + y = M

and

C = px + qy

. Solving for

x

in terms of total cost gives

x = \frac{qM - C}{q - p}

. Dividing the expression for total cost by total mass

M

yields

A = q - \frac{x(q - p)}{M}

. Finally, setting total cost

C = A(x+y) = px + qy

and rearranging gives the ratio

\frac{x}{y} = \frac{q - A}{A - p}

.

Step-by-Step Solution

1

Set up equations for total mass

M

and total cost

C

using individual weights

x

and

y

.

x + y = M \implies y = M - x

and

C = px + qy = px + q(M - x)

.

Relating the two variables through total weight eliminates

y

to express cost purely in terms of

x

.

2

Solve the total cost equation for

x

.

C = qM - (q - p)x \implies (q - p)x = qM - C \implies x = \frac{qM - C}{q - p}

.

Isolating

x

confirms the valid algebraic formula for the weight of Grade X.

3

Calculate average cost

A = \frac{C}{M}

by substituting the simplified expression for

C

.

A = \frac{qM - (q - p)x}{M} = q - \frac{x(q - p)}{M}

.

Dividing total cost by total weight

M

yields the weighted average cost per pound.

4

Determine the ratio of component weights

\frac{x}{y}

in terms of unit costs

p, q

and average cost

A

.

A(x + y) = px + qy \implies Ax + Ay = px + qy \implies y(q - A) = x(A - p) \implies \frac{x}{y} = \frac{q - A}{A - p}

.

Rearranging the weighted average equation isolates the ratio of the quantities of the two components.

Key Concept

Linear weighted averages and multi-variable system modeling in mixture word problems.

Estimated Time:2m 30s

Question 33Question →

A municipal utility company charges residential customers according to two seasonal rate structures. During the peak summer season, electricity costs $\$0.24$ per kilowatt-hour ( $\text{kWh}$ ). During the off-peak season, electricity costs $\$0.15$ per $\text{kWh}$ . A homeowner consumed a total of $14,000\text{ kWh}$ of electricity over the entire year, resulting in a total annual bill of $\$2,640$ . How many kilowatt-hours of electricity did the homeowner consume during the peak summer season?

6,000

8,000

2,250

11,000

7,000

Show answer & explanation

Answer: $6,000$

Answer

6,000\text{ kWh}

The correct answer of

6,000\text{ kWh}

is found by setting up a linear model for total energy consumption (

x + y = 14,000

) and total cost (

0.24x + 0.15y = 2,640

). Substituting

y = 14,000 - x

gives

0.09x = 540

, which solves to

x = 6,000\text{ kWh}

.

Step-by-Step Solution

1

Define variables for the two seasons.

Let

x

be the number of kilowatt-hours consumed during the peak summer season, and let

y

be the number of kilowatt-hours consumed during the off-peak season.

Establishing clear variables allows modeling the problem as a system of linear equations.

2

Set up the total consumption equation and express

y

in terms of

x

.

x + y = 14,000 \implies y = 14,000 - x

The total consumption across both seasons equals

14,000\text{ kWh}

.

3

Set up the total cost equation using the respective seasonal rates.

0.24x + 0.15y = 2,640

Peak electricity costs

\$0.24/\text{kWh}

and off-peak costs

\$0.15/\text{kWh}

, summing to the total annual cost of

\$2,640

.

4

Substitute

y = 14,000 - x

into the cost equation and solve for

x

.

0.24x + 0.15(14,000 - x) = 2,640 \implies 0.24x + 2,100 - 0.15x = 2,640 \implies 0.09x = 540 \implies x = 6,000

Simplifying the single-variable linear equation isolates the peak season consumption.

Key Concept

Linear Modeling and Systems of Linear Equations

Estimated Time:1m 30s

Question 34Question →

An industrial facility has three supply pipes—Pipe A, Pipe B, and Pipe C—that can fill a storage reservoir. Pipe A operating alone can fill the empty reservoir in $a$ hours. Pipe B operating alone takes $50\%$ longer than Pipe A to fill the empty reservoir. Pipe C operating alone fills the empty reservoir at a rate equal to the combined filling rate of Pipe A and Pipe B.

Initially, the reservoir is empty. Pipe A and Pipe B are opened simultaneously. After $2$ hours, Pipe B is closed and Pipe C is opened, while Pipe A remains open. The reservoir becomes completely full exactly $4$ hours after Pipe A and Pipe B were initially opened.

Which of the following statements must be true? Select all such statements.

Select all that apply

Pipe A operating alone would fill the empty reservoir in

8

hours and

40

minutes.

During the first

2

hours of the process, exactly

\frac{5}{13}

of the total capacity of the reservoir is filled.

Pipe C operating alone would fill the empty reservoir in

5

hours and

12

minutes.

Pipe B operating alone would fill the empty reservoir in

11

hours and

30

minutes.

If all three pipes were opened simultaneously from the beginning, the empty reservoir would be filled in

3

hours and

15

minutes.

Show answer & explanation

Answer: Pipe A operating alone would fill the empty reservoir in $8$ hours and $40$ minutes.; During the first $2$ hours of the process, exactly $\frac{5}{13}$ of the total capacity of the reservoir is filled.; Pipe C operating alone would fill the empty reservoir in $5$ hours and $12$ minutes.

Answer

The statements asserting that Pipe A operating alone takes 8 hours and 40 minutes, that 5/13 of the capacity is filled in the first 2 hours, and that Pipe C operating alone takes 5 hours and 12 minutes are all true.

Using the rate relationships

R_A = \frac{1}{a}

,

R_B = \frac{2}{3a}

, and

R_C = \frac{5}{3a}

, the equation

2(R_A + R_B) + 2(R_A + R_C) = 1

simplifies to

\frac{26}{3a} = 1

, giving

a = \frac{26}{3}

hours. This verifies that Pipe A takes 8 hours 40 minutes alone, Phase 1 fills 5/13 of the reservoir capacity, and Pipe C takes 5.2 hours (5 hours 12 minutes) alone.

Step-by-Step Solution

1

Express the individual work rates of Pipe A, Pipe B, and Pipe C in terms of parameter

a

.

R_A = \frac{1}{a}

,

R_B = \frac{1}{1.5a} = \frac{2}{3a}

, and

R_C = R_A + R_B = \frac{1}{a} + \frac{2}{3a} = \frac{5}{3a}

.

Work rate is defined as the reciprocal of the time required to complete one unit of work.

2

Formulate an equation for total work completed over the two 2-hour phases.

2(R_A + R_B) + 2(R_A + R_C) = 1 \implies 2\left(\frac{5}{3a}\right) + 2\left(\frac{8}{3a}\right) = 1 \implies \frac{10}{3a} + \frac{16}{3a} = 1

.

Pipes A and B operate for the first 2 hours, followed by Pipes A and C operating for the next 2 hours to complete 1 full reservoir.

3

Solve the work equation for

a

and determine Pipe A's solo time.

\frac{26}{3a} = 1 \implies a = \frac{26}{3} = 8\frac{2}{3} \text{ hours} = 8 \text{ hours } 40 \text{ minutes}.

Fractional hours are converted to minutes by multiplying

\frac{2}{3}

by

60

.

4

Evaluate the volume filled in Phase 1 and the solo filling time for Pipe C.

\text{Phase 1 volume} = \frac{10}{3(26/3)} = \frac{5}{13}. \text{ Pipe C solo time} = \frac{1}{R_C} = \frac{3a}{5} = \frac{26}{5} = 5.2 \text{ hours} = 5 \text{ hours } 12 \text{ minutes}.

Substituting

a = \frac{26}{3}

into the respective rate expressions yields the exact time and volume parameters.

Key Concept

Formulating combined rate models and solving sequential multi-stage work problems.

Question 35Question →

Courier A departs from Warehouse X heading toward Warehouse Y at 8:00 AM traveling at a constant speed of $40$ miles per hour. Courier B departs from Warehouse Y heading toward Warehouse X along the same straight route at 9:00 AM traveling at a constant speed of $60$ miles per hour. If the total distance between Warehouse X and Warehouse Y is $190$ miles, at what time will the two couriers meet?

9:54 AM

10:00 AM

10:30 AM

11:00 AM

11:15 AM

Show answer & explanation

Answer: 10:30 AM

Answer

10:30 AM

The correct answer is 10:30 AM. Between 8:00 AM and 9:00 AM, Courier A travels 40 miles alone. At 9:00 AM, the distance remaining between them is 150 miles. Because they travel toward each other, their speeds combine to 100 mph (40 + 60). Dividing 150 miles by 100 mph gives 1.5 hours (1 hour and 30 minutes). Adding 1 hour and 30 minutes to 9:00 AM gives a meeting time of 10:30 AM.

Step-by-Step Solution

1

Calculate the distance traveled by Courier A before Courier B starts moving.

From 8:00 AM to 9:00 AM (1 hour), Courier A travels

40 \text{ mph} \times 1 \text{ hour} = 40 \text{ miles}

.

Courier A has a 1-hour head start.

2

Determine the remaining distance to be covered between the two couriers at 9:00 AM.

Remaining distance

= 190 - 40 = 150 \text{ miles}

.

Subtract Courier A's distance from the total distance of 190 miles.

3

Calculate the combined rate of both couriers and solve for elapsed time after 9:00 AM.

Combined rate

= 40 + 60 = 100 \text{ mph}

. Elapsed time

t = \frac{150}{100} = 1.5 \text{ hours} = 1 \text{ hour } 30 \text{ minutes}

.

Since they move toward each other, their speeds add up to close the gap.

4

Add the elapsed combined time to 9:00 AM to find the meeting time.

9:00 AM

+ 1 \text{ hour } 30 \text{ minutes} = 10:30 \text{ AM}

.

The combined movement began at 9:00 AM when Courier B started traveling.

Key Concept

Distance-rate-time relationship with staggered start times

Estimated Time:1m 30s

Question 36Question →

Two automated assembly robots, Robot P and Robot Q, produce identical components. Robot P operates at a constant rate of $p$ components per hour, and Robot Q operates at a constant rate of $q$ components per hour, where $p > q > 0$ . During a shift, Robot P worked for $4$ hours and Robot Q worked for $6$ hours to produce a combined total of $360$ components. If $T$ represents the total number of components produced when Robot P works for $7$ hours and Robot Q works for $3$ hours, which of the following statements must be true? Select all such statements.

Select all that apply

The rate of Robot P,

p

, must be greater than

36

components per hour.

The rate of Robot Q,

q

, must be less than

36

components per hour.

The total number of components

T

must be greater than

360

and less than

630

.

The rate of Robot P,

p

, must be less than

60

components per hour.

The total number of components

T

could be equal to

320

.

Show answer & explanation

Answer: The rate of Robot P, $p$ , must be greater than $36$ components per hour.; The rate of Robot Q, $q$ , must be less than $36$ components per hour.; The total number of components $T$ must be greater than $360$ and less than $630$ .

Answer

The correct statements are: the rate of Robot P,

p

, must be greater than

36

components per hour; the rate of Robot Q,

q

, must be less than

36

components per hour; and the total number of components

T

must be greater than

360

and less than

630

.

The system of equations

2p + 3q = 180

combined with

p > q > 0

strictly bounds

p

between

36

and

90

, and

q

between

0

and

36

. Substituting these boundary conditions into the total expression

T = 5p + 180

yields the strict range

360 < T < 630

. Consequently, the three statements asserting

p > 36

,

q < 36

, and

360 < T < 630

are all mathematically required.

Step-by-Step Solution

1

Set up the linear equation from the initial production shift and simplify.

4p + 6q = 360 \implies 2p + 3q = 180 \implies q = 60 - \frac{2}{3}p

This establishes the exact relationship between the production rates

p

and

q

.

2

Apply the given constraints

p > q > 0

to determine the domain bounds for

p

and

q

.

p > 60 - \frac{2}{3}p \implies \frac{5}{3}p > 60 \implies p > 36

. Also,

q > 0 \implies 60 - \frac{2}{3}p > 0 \implies p < 90

. Thus,

36 < p < 90

and

0 < q < 36

.

Determining extreme bounds for

p

automatically constrains both individual rates.

3

Formulate

T

in terms of

p

and evaluate its numerical boundaries.

T = 7p + 3q = 7p + (180 - 2p) = 5p + 180

. Substituting

36 < p < 90

gives

360 < T < 630

.

Substituting

3q = 180 - 2p

simplifies

T

into a single-variable linear modeling equation.

Key Concept

Linear word problem modeling, variable elimination, and system inequality constraint analysis

Estimated Time:1m 45s

Question 37Question →

Two cyclists, Clara and Dan, start simultaneously from opposite ends of a $120$ -mile trail and ride directly toward each other along the same path. Clara rides at a constant speed of $c$ miles per hour, and Dan rides at a constant speed of $d$ miles per hour. If they maintain their initial speeds, they will meet in $3$ hours. However, if Clara increases her speed by $20\%$ and Dan increases his speed by $33\frac{1}{3}\%$ , they will meet $36$ minutes earlier. What is Clara's original speed $c$ , in miles per hour?

15

20

22.5

25

30

Show answer & explanation

Answer: 25

Answer

Clara's original speed

c

is 25 miles per hour.

The correct answer is 25 miles per hour. Initial combined speed is

c + d = \frac{120}{3} = 40

mph. Meeting 36 minutes earlier means a new duration of 2.4 hours, giving a new combined speed of

\frac{120}{2.4} = 50

mph. Setting up the equation

\frac{6}{5}c + \frac{4}{3}(40 - c) = 50

yields

c = 25

.

Step-by-Step Solution

1

Determine the initial combined rate of Clara and Dan.

Since they cover

120

miles in

3

hours, their combined speed is

c + d = \frac{120}{3} = 40

miles per hour.

Distance equals rate multiplied by time (

D = r \cdot t

). When moving toward each other, combined rate is

c + d

.

2

Calculate the new travel time and the new combined rate.

The new travel time is

3\text{ hours} - 36\text{ minutes} = 3 - 0.6 = 2.4

hours (or

\frac{12}{5}

hours). The new combined rate is

\frac{120}{2.4} = 50

miles per hour.

36 minutes is equivalent to

\frac{36}{60} = 0.6

hours.

3

Express the modified individual rates algebraically.

Clara's new speed is

1.2c = \frac{6}{5}c

. Dan's new speed is

\left(1 + \frac{1}{3}\right)d = \frac{4}{3}d

.

Increasing by

20\%

multiplies rate by

1.2

, and increasing by

33\frac{1}{3}\%

multiplies rate by

\frac{4}{3}

.

4

Formulate and solve the system of linear equations.

We have the system:
1)

c + d = 40 \implies d = 40 - c

2)

\frac{6}{5}c + \frac{4}{3}d = 50

Substituting

d = 40 - c

into equation 2:

\frac{6}{5}c + \frac{4}{3}(40 - c) = 50

Multiplying the entire equation by

15

to eliminate denominators:

18c + 20(40 - c) = 750

18c + 800 - 20c = 750

-2c = -50 \implies c = 25

Solving for

c

yields Clara's original speed in miles per hour.

Key Concept

Distance-Rate-Time modeling using systems of linear equations with percentage changes in individual rates

Estimated Time:2m 30s

Question 38Question →

An electronics manufacturer models the monthly demand $N$ for a specific model of tablet as $N(P) = 1,200 - 3P$ , where $P$ is the selling price per tablet in dollars. The monthly total cost of producing these tablets consists of a fixed overhead cost of $\$15,000$ plus a variable cost of $\$100$ per tablet produced. If the manufacturer earned a monthly net profit of $\$45,000$ , what is the smaller of the two possible selling prices, in dollars, that could yield this profit?

Show answer & explanation

Answer: 200

Answer

The smaller of the two possible selling prices is 200 dollars.

Total revenue is given by

R(P) = P \cdot N(P) = P(1,200 - 3P) = 1,200P - 3P^2

, and total cost is given by

C(P) = 15,000 + 100(1,200 - 3P) = 135,000 - 300P

. The profit equation

\text{Profit}(P) = R(P) - C(P) = 45,000

yields

-3P^2 + 1,500P - 135,000 = 45,000

. Rearranging gives

-3P^2 + 1,500P - 180,000 = 0

, which simplifies to

P^2 - 500P + 60,000 = 0

. Factoring gives

(P - 200)(P - 300) = 0

, leading to prices of

\$200

and

\$300

. The smaller price is

\$200

.

Step-by-Step Solution

1

Formulate algebraic functions for monthly revenue and monthly total cost in terms of price

P

.

Revenue

R(P) = P(1,200 - 3P) = 1,200P - 3P^2

and Total Cost

C(P) = 15,000 + 100(1,200 - 3P) = 135,000 - 300P

.

Total revenue equals unit price multiplied by demand quantity, while total cost equals fixed overhead plus variable cost per unit multiplied by demand quantity.

2

Formulate the net profit function

\text{Profit}(P) = R(P) - C(P)

and equate it to the target profit of

\$45,000

.

(1,200P - 3P^2) - (135,000 - 300P) = 45,000

, which simplifies to

-3P^2 + 1,500P - 180,000 = 0$.

Net profit is defined as total revenue minus total cost.

3

Divide the resulting quadratic equation by

-3

to normalize coefficients and solve for

P

.

P^2 - 500P + 60,000 = 0 \implies (P - 200)(P - 300) = 0

, giving roots

P = 200

and

P = 300$.

Simplifying the quadratic equation reveals two factored linear roots.

4

Identify the smaller of the two valid price solutions.

P = 200 dollars.

The question specifically requests the smaller of the two possible selling prices.

Key Concept

Algebraic Modeling of Revenue, Cost, and Profit Functions using Quadratic Equations

Question 39Question →

A solar power facility operates two types of solar panel arrays: Array Alpha and Array Beta. When operational, Array Beta produces electricity at a constant hourly rate that is $25\%$ greater than the constant hourly rate of Array Alpha. On a clear day, Array Alpha operated for $8$ hours and Array Beta operated for $6$ hours, together generating a total of $3,100$ kilowatt-hours (kWh) of electricity. What was the hourly production rate of Array Alpha, in kWh per hour?

160

193.75

200

248

250

Show answer & explanation

Answer: 200

Answer

200 kWh per hour

Let the hourly rate of Array Alpha be

r

kWh per hour. Since Array Beta produces at a rate

25\%

greater, its hourly rate is

1.25r

kWh per hour. Multiply each rate by the respective number of hours operated: Array Alpha produced

8r

kWh and Array Beta produced

6 \times 1.25r = 7.5r

kWh. Combining these gives

8r + 7.5r = 15.5r = 3,100

. Solving for

r

yields

r = 200

kWh per hour.

Step-by-Step Solution

1

Define variables for the hourly rates of Array Alpha and Array Beta.

Let

r

be the hourly rate of Array Alpha in kWh per hour. Since Array Beta's rate is

25\%

greater, Array Beta's rate is

r + 0.25r = 1.25r

kWh per hour.

Establishing the linear relationship between the two unknown rates.

2

Set up the total electricity output equation using rate times time for each array.

\text{Total Energy} = (8 \text{ hours} \times r) + (6 \text{ hours} \times 1.25r) = 3,100

Total production is the sum of production from Array Alpha and Array Beta.

3

Simplify the algebraic equation and solve for

r

.

8r + 7.5r = 3,100 \implies 15.5r = 3,100 \implies r = \frac{3,100}{15.5} = 200

Isolating

r

gives the hourly rate of Array Alpha.

Key Concept

Linear Algebraic Modeling of Combined Rates and Percentages

Estimated Time:1m 30s

Question 40Question →

Two industrial pumps, Pump A and Pump B, are used to fill a storage tank. Working alone at its constant standard operating rate, Pump A can fill the empty tank in $12$ hours. Working alone at its constant standard operating rate, Pump B can fill the empty tank in $18$ hours.

To fill the tank, both pumps begin operating simultaneously at their standard rates. After $4$ hours of joint operation, Pump A undergoes maintenance that reduces its operating rate by $25\%$ , while Pump B continues operating at its standard rate. Exactly $2$ hours after Pump A's rate is reduced, Pump B's operating rate is increased by $50\%$ above its standard rate due to a valve adjustment. If both pumps continue operating at these adjusted rates until the tank is full, how many total hours from the initial start does it take to completely fill the tank?

6\frac{2}{3}

hours

7\frac{1}{7}

hours

7\frac{3}{7}

hours

8\frac{1}{7}

hours

8\frac{2}{3}

hours

Show answer & explanation

Answer: $7\frac{3}{7}$ hours

Answer

The total time required to completely fill the tank from the initial start is

7\frac{3}{7}

hours.

The solution proceeds in three stages. In the first 4 hours, both pumps at standard rates completed 5/9 of the job. In the next 2 hours, Pump A operated at 1/16 tank/hr and Pump B at 1/18 tank/hr, completing an additional 17/72 of the job, bringing total completed work to 19/24 of the tank. For the final 5/24 of the tank, Pump A (1/16 tank/hr) and Pump B (1/12 tank/hr) worked at a combined rate of 7/48 tank/hr, taking 10/7 (or 1 3/7) hours. The total time is 4 + 2 + 1 3/7 = 7 3/7 hours.

Step-by-Step Solution

1

Determine individual standard rates and combined initial rate.

Pump A rate

r_A = \frac{1}{12}

tank/hr; Pump B rate

r_B = \frac{1}{18}

tank/hr. Initial combined rate

r_1 = \frac{1}{12} + \frac{1}{18} = \frac{5}{36}

tank/hr.

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

2

Calculate work completed in Phase 1 (first 4 hours).

Work completed

= 4 \times \frac{5}{36} = \frac{20}{36} = \frac{5}{9}

of the tank. Remaining work

= 1 - \frac{5}{9} = \frac{4}{9}

of the tank.

Both pumps work at standard combined rate for 4 hours.

3

Adjust rates for Phase 2 (hours 4 to 6) and calculate work done.

Pump A reduced rate

= 0.75 \times \frac{1}{12} = \frac{1}{16}

tank/hr. Combined rate

r_2 = \frac{1}{16} + \frac{1}{18} = \frac{17}{144}

tank/hr. Work done in 2 hours

= 2 \times \frac{17}{144} = \frac{17}{72}

. Cumulative work

= \frac{5}{9} + \frac{17}{72} = \frac{57}{72} = \frac{19}{24}

of the tank.

Phase 2 lasts 2 hours with Pump A operating at

75\%

efficiency and Pump B at standard rate.

4

Adjust rates for Phase 3 (from hour 6 onwards) and calculate remaining time.

Pump B increased rate

= 1.50 \times \frac{1}{18} = \frac{1}{12}

tank/hr. Combined rate

r_3 = \frac{1}{16} + \frac{1}{12} = \frac{7}{48}

tank/hr. Remaining work

= 1 - \frac{19}{24} = \frac{5}{24}

. Additional time required

= \frac{5/24}{7/48} = \frac{5}{24} \times \frac{48}{7} = \frac{10}{7} = 1\frac{3}{7}

hours.

Divide remaining fractional work by the new combined rate.

5

Calculate total elapsed time.

Total time

= 4 + 2 + 1\frac{3}{7} = 7\frac{3}{7}

hours.

Sum the durations of all three phases.

Key Concept

Multi-stage work-rate problems with variable individual rates and fractional job completion tracking.

Alternative Method

Define the tank capacity as 144 units (the LCM of 12, 18, 16, 48). Pump A standard rate = 12 units/hr; Pump B standard rate = 8 units/hr. Phase 1 (4 hrs): combined rate = 20 units/hr, work done = 80 units. Phase 2 (2 hrs): Pump A rate = 9 units/hr, Pump B rate = 8 units/hr, combined rate = 17 units/hr, work done = 34 units. Total work done in 6 hrs = 114 units. Remaining work = 30 units. Phase 3: Pump A rate = 9 units/hr, Pump B rate = 12 units/hr, combined rate = 21 units/hr. Additional time = 30/21 = 10/7 hrs. Total time = 6 + 10/7 = 7 3/7 hrs.

Estimated Time:2m 30s

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