Algebraic Word Problems and Modeling

62 soru

Soru 1Soru

A commercial coffee roaster creates a custom blend by mixing two existing bean blends: Blend X and Blend Y. Blend X consists of 60% Arabica beans and 40% Robusta beans by weight, whereas Blend Y consists of 30% Arabica beans and 70% Robusta beans by weight. The roaster mixes a quantity of Blend X with a quantity of Blend Y to produce a total of 50 pounds of a new mixture that is 42% Arabica beans by weight. How many pounds of Blend X are in the final mixture?

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Cevap: 20

Cevap

20 pounds
Let xx represent the number of pounds of Blend X. The remaining weight of the mixture, (50x)(50 - x) pounds, comes from Blend Y. Setting up the equation for the total weight of Arabica beans gives 0.60x+0.30(50x)=0.42(50)0.60x + 0.30(50 - x) = 0.42(50). Simplifying this expression yields 0.60x+150.30x=210.60x + 15 - 0.30x = 21, which reduces to 0.30x=60.30x = 6. Dividing by 0.300.30 gives x=20x = 20. Therefore, 20 pounds of Blend X were used.

Adım Adım Çözüm

1
Define variables for component weights
Let xx be the pounds of Blend X. The weight of Blend Y used is 50x50 - x pounds.
The total combined weight of the mixture is given as 50 pounds.
2
Formulate an equation for the total weight of Arabica beans
0.60x+0.30(50x)=0.42(50)0.60x + 0.30(50 - x) = 0.42(50), which simplifies to 0.60x+150.30x=210.60x + 15 - 0.30x = 21.
The sum of Arabica beans contributed by each blend must equal the total weight of Arabica beans in the combined mixture.
3
Solve the linear equation for xx
0.30x+15=21    0.30x=6    x=200.30x + 15 = 21 \implies 0.30x = 6 \implies x = 20.
Subtract 15 from both sides to isolate the variable term, then divide by 0.30.

Anahtar Kavram

Linear Modeling and Mixture Problems
Tahmini Süre:1m 30s
Soru 2Soru

A manufacturing company produces two models of office chairs, Model P and Model Q. Producing each Model P chair requires 33 hours of assembly and 11 hour of finishing. Producing each Model Q chair requires 22 hours of assembly and 22 hours of finishing. On a given day, the assembly department was scheduled for 9696 total hours of work and the finishing department was scheduled for 4848 total hours of work. If both departments operated at full capacity and used all scheduled hours, how many Model P chairs were produced on that day?

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Cevap: 24

Cevap

24
Letting pp be the number of Model P chairs and qq be the number of Model Q chairs, the hours used by the assembly department give 3p+2q=963p + 2q = 96 and the hours used by the finishing department give p+2q=48p + 2q = 48. Subtracting the finishing equation from the assembly equation eliminates 2q2q, resulting in 2p=482p = 48, or p=24p = 24.

Adım Adım Çözüm

1
Define variables for the unknown quantities.
Let pp equal the number of Model P chairs produced and qq equal the number of Model Q chairs produced.
Assigning variables to the unknown quantities enables the construction of algebraic equations.
2
Formulate a system of linear equations representing total department hours.
Assembly department equation: 3p+2q=963p + 2q = 96; Finishing department equation: p+2q=48p + 2q = 48.
The sum of hours required for both models in each department must equal that department's total scheduled hours.
3
Solve the system of equations for pp using the elimination method.
Subtracting p+2q=48p + 2q = 48 from 3p+2q=963p + 2q = 96 yields 2p=482p = 48, which gives p=24p = 24.
Eliminating qq directly provides the value for pp, which corresponds to the target quantity requested in the problem.

Anahtar Kavram

Modeling real-world resource allocation using systems of linear equations
Soru 3Soru

A hotel renovated its guest rooms and purchased a total of 100 lighting fixtures, consisting of wall sconces costing $45\$45 each and ceiling pendants costing $70\$70 each. If the average (arithmetic mean) cost per fixture for the entire purchase was $52\$52, how many wall sconces were purchased?

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Cevap: 72

Cevap

72
Setting up the weighted total cost equation 45x+70(100x)=520045x + 70(100 - x) = 5200 simplifies to 25x=1800-25x = -1800, giving x=72x = 72 wall sconces.

Adım Adım Çözüm

1
Define the variable for the unknown quantity
Let xx represent the number of wall sconces purchased. The number of ceiling pendants is 100x100 - x.
Defining the target quantity as xx allows the problem to be modeled using a single-variable linear equation.
2
Set up the algebraic equation for total cost
45x+70(100x)=52×10045x + 70(100 - x) = 52 \times 100
The total cost of all fixtures is the sum of the total sconce cost and total pendant cost, which equals the overall average cost multiplied by the total number of fixtures.
3
Expand and solve the linear equation
45x+700070x=5200    25x=1800    x=7245x + 7000 - 70x = 5200 \implies -25x = -1800 \implies x = 72
Simplifying algebraic terms isolates xx to find the exact number of wall sconces.

Anahtar Kavram

Linear Algebraic Modeling and Weighted Averages
Soru 4Soru

A food processing plant packages grain using two automated lines, Line A and Line B. Line A operates at a constant rate of 4040 bags per minute, and Line B operates at a constant rate of 6060 bags per minute. Line A begins filling a batch of 4,2004,200 bags alone. After 1515 minutes, Line A's rate decreases by 25%25\%, and Line B is turned on to work alongside Line A until the batch is complete. How many total minutes does Line A operate to complete the batch?

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Cevap: 55 minutes

Cevap

55 minutes
In the first 15 minutes, Line A fills 40×15=60040 \times 15 = 600 bags, leaving 4,200600=3,6004,200 - 600 = 3,600 bags remaining. Line A's rate then drops to 40×0.75=3040 \times 0.75 = 30 bags per minute. Working together with Line B (6060 bags per minute), their combined rate is 30+60=9030 + 60 = 90 bags per minute. Filling the remaining 3,6003,600 bags takes 3,60090=40\frac{3,600}{90} = 40 minutes. Thus, Line A operates for a total of 15+40=5515 + 40 = 55 minutes.

Adım Adım Çözüm

1
Calculate the number of bags filled by Line A during the first 15 minutes.
Line A filled 40 bags/min×15 min=600 bags40 \text{ bags/min} \times 15 \text{ min} = 600 \text{ bags}.
Line A worked alone for 15 minutes at its initial constant rate.
2
Determine the remaining number of bags to be filled.
Remaining bags = 4,200600=3,600 bags4,200 - 600 = 3,600 \text{ bags}.
Subtract the bags already completed from the total batch size.
3
Calculate Line A's reduced rate and the new combined rate of both lines.
Line A's new rate = 40×(10.25)=30 bags/min40 \times (1 - 0.25) = 30 \text{ bags/min}. Combined rate = 30+60=90 bags/min30 + 60 = 90 \text{ bags/min}.
Line A's filling rate drops by 25%, and Line B contributes 60 bags per minute simultaneously.
4
Calculate the time needed for both lines to finish the remaining bags.
Time for second phase = 3,600 bags90 bags/min=40 minutes\frac{3,600 \text{ bags}}{90 \text{ bags/min}} = 40 \text{ minutes}.
Divide remaining work by the combined filling rate.
5
Calculate the total time Line A was operating.
Total operating time = 15 min+40 min=55 minutes15 \text{ min} + 40 \text{ min} = 55 \text{ minutes}.
Line A operated during both the initial 15-minute period and the subsequent 40-minute period.

Anahtar Kavram

Linear Modeling and Piecewise Work-Rate Scenarios
Tahmini Süre:2m 0s
Soru 5Soru

A university archive uses two high-speed scanners, Machine A and Machine B, to digitize manuscript pages. Machine A operates at a constant rate of 6060 pages per hour, and Machine B operates at a constant rate of 8080 pages per hour. On Monday, Machine A starts scanning at 8:00 AM. At 9:30 AM, Machine B is brought online to work alongside Machine A. If both machines continue scanning together at their respective constant rates until a total of 510510 pages have been scanned, at what time will the scanning project be completed?

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Cevap: 12:30 PM

Cevap

12:30 PM
The option stating 12:30 PM is correct. Machine A operates alone from 8:00 AM to 9:30 AM (1.5 hours), completing 60×1.5=9060 \times 1.5 = 90 pages. This leaves 51090=420510 - 90 = 420 pages to be scanned. From 9:30 AM onward, both machines work together at a combined rate of 60+80=14060 + 80 = 140 pages per hour. The remaining 420 pages require 420140=3\frac{420}{140} = 3 hours. Adding 3 hours to 9:30 AM gives a completion time of 12:30 PM.

Adım Adım Çözüm

1
Calculate the number of pages scanned by Machine A during its head start
60 pages/hr×1.5 hours=90 pages60 \text{ pages/hr} \times 1.5 \text{ hours} = 90 \text{ pages}
Machine A operates alone from 8:00 AM to 9:30 AM, which is 1.51.5 hours.
2
Determine the remaining number of pages to be scanned
51090=420 pages510 - 90 = 420 \text{ pages}
Subtract the pages already completed from the total requirement of 510 pages.
3
Calculate the combined scanning rate of both machines
60+80=140 pages/hr60 + 80 = 140 \text{ pages/hr}
When both machines operate simultaneously, their individual rates add together.
4
Calculate the time required to scan the remaining pages
420 pages140 pages/hr=3 hours\frac{420 \text{ pages}}{140 \text{ pages/hr}} = 3 \text{ hours}
Divide the remaining work by the combined rate.
5
Determine the final completion time
9:30 AM+3 hours=12:30 PM9:30 \text{ AM} + 3 \text{ hours} = 12:30 \text{ PM}
Add the combined work duration to the time when both machines began working together (9:30 AM).

Anahtar Kavram

Combined Work Rates with Staggered Start Times
Tahmini Süre:1m 30s
Soru 6Soru

A IT consultant charges a flat setup fee of $150\$150 plus a standard hourly rate of $80\$80 for regular hours worked on a project. For any rush hours worked, the hourly rate increases by 50%50\%. On a recently completed project, the consultant worked a total of tt hours, of which rr hours were rush hours (where 0rt0 \leq r \leq t). Which of the following statements must be true regarding the total charge CC, in dollars, for this project? Select all such statements.

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Cevap: The total charge, in dollars, can be expressed as C=150+80t+40rC = 150 + 80t + 40r.; If the consultant worked a total of 2020 hours and the total charge was $2,150\$2,150, then exactly 1010 of those hours were rush hours.; If no rush hours were worked (r=0r = 0), the average charge per hour, including the setup fee, is 150t+80\frac{150}{t} + 80 dollars per hour.

Cevap

The correct statements are those expressing the total charge as C=150+80t+40rC = 150 + 80t + 40r, determining that 1010 rush hours were worked when total cost is $2,150\$2,150 for 2020 hours, and calculating the average hourly cost as 150t+80\frac{150}{t} + 80 dollars per hour when no rush hours are worked.
The model correctly partitions total hours tt into (tr)(t - r) regular hours at $80\$80/hr and rr rush hours at $120\$120/hr, which simplifies algebraically to C=150+80t+40rC = 150 + 80t + 40r. Solving this equation for t=20t = 20 and C=2150C = 2150 yields r=10r = 10. Furthermore, when r=0r = 0, dividing total charge 150+80t150 + 80t by tt gives 150t+80\frac{150}{t} + 80.

Adım Adım Çözüm

1
Calculate the hourly rush rate from the given percentage increase.
The standard rate is $80\$80/hr. The rush rate is $80×(1+0.50)=$120\$80 \times (1 + 0.50) = \$120/hr.
Rush hours cost 50%50\% more than standard regular hours.
2
Set up the algebraic model for total charge CC using standard hours (tr)(t - r) and rush hours rr.
C=150+80(tr)+120r=150+80t80r+120r=150+80t+40rC = 150 + 80(t - r) + 120r = 150 + 80t - 80r + 120r = 150 + 80t + 40r.
This combines the fixed setup fee with variable costs from regular and rush hours.
3
Evaluate the specific case where total time t=20t = 20 and total cost C=2150C = 2150.
2150=150+80(20)+40r    2150=1750+40r    40r=400    r=102150 = 150 + 80(20) + 40r \implies 2150 = 1750 + 40r \implies 40r = 400 \implies r = 10.
Plugging given values into the algebraic model allows solving for the unknown number of rush hours.
4
Calculate average hourly cost when r=0r = 0.
Average cost =150+80tt=150t+80= \frac{150 + 80t}{t} = \frac{150}{t} + 80.
Average hourly rate is total charge divided by total hours worked.

Anahtar Kavram

Linear algebraic modeling of rate problems with multiple rate components
Soru 7Soru

A boutique perfume workshop creates a signature fragrance blend by combining two fragrance oils, Oil A and Oil B. Oil A costs 12perounceandcontains4012 per ounce and contains 40% pure essential oil by volume. Oil B costs 20 per ounce and contains 80% pure essential oil by volume. The perfumer creates a 30-ounce batch of the signature blend at a total cost that averages exactly $15 per ounce. What is the total volume, in ounces, of pure essential oil contained in this 30-ounce blend?

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Cevap: 16.5

Cevap

The total volume of pure essential oil contained in the 30-ounce blend is 16.5 ounces.
To find the total essential oil content, first set up a system of equations for the volumes of Oil A (xx) and Oil B (yy): x+y=30x + y = 30 and 12x+20y=45012x + 20y = 450. Solving this system yields x=18.75x = 18.75 ounces and y=11.25y = 11.25 ounces. Multiplying each by its respective essential oil concentration gives 0.40×18.75=7.50.40 \times 18.75 = 7.5 ounces from Oil A and 0.80×11.25=90.80 \times 11.25 = 9 ounces from Oil B, totaling 16.5 ounces of essential oil.

Adım Adım Çözüm

1
Set up linear equations representing the total volume and total cost of the mixture.
x+y=30x + y = 30 and 12x+20y=45012x + 20y = 450, where xx is ounces of Oil A and yy is ounces of Oil B.
The total cost of the 30-ounce blend at 15perounceis15 per ounce is 15 \times 30 = 450$ dollars.
2
Solve for the quantities of Oil A and Oil B used in the mixture.
x=18.75x = 18.75 ounces of Oil A and y=11.25y = 11.25 ounces of Oil B.
Substituting y=30xy = 30 - x into 12x+20(30x)=45012x + 20(30 - x) = 450 yields 8x=150-8x = -150, giving x=18.75x = 18.75.
3
Compute the amount of pure essential oil contributed by each component and sum them.
0.40(18.75)+0.80(11.25)=7.5+9=16.50.40(18.75) + 0.80(11.25) = 7.5 + 9 = 16.5 ounces.
Oil A contains 40% essential oil by volume and Oil B contains 80% essential oil by volume.

Anahtar Kavram

Systems of Linear Equations and Mixture Modeling
Tahmini Süre:2m 0s
Soru 8Soru

A logistics company calculates the shipping cost CC, in dollars, for a parcel of weight ww pounds shipped over a distance of dd miles using the following piecewise pricing model:
- A fixed base fee of $10\$10.
- An additional weight surcharge of $1.50\$1.50 per pound for every pound (or fraction thereof) exceeding 55 pounds.
- An additional distance surcharge of $0.05\$0.05 per mile for every mile exceeding 100100 miles.
- A flat combined heavy-distance surcharge of $25\$25 applied if and only if w>10w > 10 AND d>500d > 500.

A customer ships two parcels, Parcel 1 and Parcel 2:
- Parcel 1 weighs w1w_1 pounds (where w1w_1 is an integer greater than 10) and is shipped a distance of d1d_1 miles (where d1d_1 is an integer greater than 500).
- Parcel 2 weighs w2=w14w_2 = w_1 - 4 pounds and is shipped a distance of d2=d1+200d_2 = d_1 + 200 miles.

Let C1C_1 be the shipping cost of Parcel 1 and C2C_2 be the shipping cost of Parcel 2.

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

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Cevap: If w1=12w_1 = 12, then Parcel 1 costs exactly $21\$21 more to ship than Parcel 2.; If w1>14w_1 > 14, Parcel 2 always costs exactly $4\$4 more to ship than Parcel 1.

Cevap

The correct statements are the ones indicating that if the weight of Parcel 1 is 12 pounds, Parcel 1 costs exactly 21morethanParcel2,andiftheweightofParcel1isgreaterthan14pounds,Parcel2costsexactly21 more than Parcel 2, and if the weight of Parcel 1 is greater than 14 pounds, Parcel 2 costs exactly 4 more than Parcel 1.
The cost functions must be broken down by conditional threshold. For Parcel 1, both weight (>10) and distance (>500) thresholds are met, giving C_1 = 1.5w_1 + 0.05d_1 + 22.5. For Parcel 2, d_2 = d_1 + 200 > 700 is always > 500, but w_2 = w_1 - 4 exceeds 10 if and only if w_1 > 14. If w_1 = 12, w_2 = 8, so Parcel 2 gets no 25surcharge,resultinginC1C2=21.Ifw1>14,w2>10,soParcel2getsthe25 surcharge, resulting in C_1 - C_2 = 21. If w_1 > 14, w_2 > 10, so Parcel 2 gets the 25 surcharge, giving C_2 - C_1 = 4.

Adım Adım Çözüm

1
Formulate the algebraic cost function C(w, d) for any parcel.
C(w, d) = 10 + 1.50(w - 5) + 0.05(d - 100) + S(w, d), where S(w, d) = 25 if w > 10 and d > 500, else S(w, d) = 0.
This establishes the piecewise linear equations governing shipping costs.
2
Express C_1 for Parcel 1 given w_1 > 10 and d_1 > 500.
Since w_1 > 10 and d_1 > 500, S(w_1, d_1) = 25. Thus, C_1 = 10 + 1.50(w_1 - 5) + 0.05(d_1 - 100) + 25 = 1.50 w_1 + 0.05 d_1 + 22.50.
Parcel 1 always meets both conditions for the $25 heavy-distance surcharge.
3
Analyze Parcel 2 when 10 < w_1 <= 14 (meaning 6 < w_2 <= 10).
For w_2 <= 10, S(w_2, d_2) = 0 even though d_2 > 500. C_2 = 10 + 1.50(w_1 - 9) + 0.05(d_1 + 100) + 0 = 1.50 w_1 + 0.05 d_1 + 1.50. Therefore, C_1 - C_2 = (1.50 w_1 + 0.05 d_1 + 22.50) - (1.50 w_1 + 0.05 d_1 + 1.50) = $21.00.
This confirms that for w_1 = 12, Parcel 1 costs exactly $21 more than Parcel 2.
4
Analyze Parcel 2 when w_1 > 14 (meaning w_2 > 10).
For w_2 > 10 and d_2 > 700 > 500, S(w_2, d_2) = 25. Thus, C_2 = 1.50 w_1 + 0.05 d_1 + 26.50. Subtracting C_1 gives C_2 - C_1 = (1.50 w_1 + 0.05 d_1 + 26.50) - (1.50 w_1 + 0.05 d_1 + 22.50) = $4.00.
This proves Parcel 2 costs exactly $4 more than Parcel 1 whenever w_1 > 14.

Anahtar Kavram

Piecewise linear modeling and conditional surcharge threshold evaluations
Soru 9Soru

A reservoir initially contains 4,8004,800 gallons of water. A drainage pump operates at a constant rate of 120120 gallons per minute to empty the reservoir. How many minutes of continuous pumping will it take for the amount of water remaining in the reservoir to be reduced to 1,2001,200 gallons?

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Cevap: 3030

Cevap

30 minutes
To find the time needed for 1,2001,200 gallons to remain, subtract 1,2001,200 from the initial 4,8004,800 gallons to determine that 3,6003,600 gallons of water must be removed. Dividing the 3,6003,600 gallons by the drainage rate of 120120 gallons per minute gives exactly 3030 minutes.

Adım Adım Çözüm

1
Calculate the total volume of water that must be drained from the reservoir.
4,800 gallons1,200 gallons=3,600 gallons4,800 \text{ gallons} - 1,200 \text{ gallons} = 3,600 \text{ gallons}
The question asks for the time until 1,2001,200 gallons remain, so the amount removed is the difference between the starting volume and ending volume.
2
Set up the linear rate equation to solve for elapsed time tt in minutes.
Rate×t=Volume Drained    120t=3,600\text{Rate} \times t = \text{Volume Drained} \implies 120t = 3,600
The rate of drainage is constant at 120120 gallons per minute.
3
Solve for tt.
t=3,600120=30 minutest = \frac{3,600}{120} = 30 \text{ minutes}
Dividing total gallons to drain by the rate gives the required time.

Anahtar Kavram

Linear Modeling and Distance-Rate-Time / Work-Rate Relationships
Tahmini Süre:1m 0s
Soru 10Soru

A train travels from Station A to Station B, a distance of 180 miles, at a constant speed of vv miles per hour. On the return trip from Station B to Station A, the train travels the first half of the distance at a constant speed that is 20%20\% less than vv, and the remaining half of the distance at a constant speed that is 25%25\% greater than vv. If the total time for the return trip is 6 minutes longer than the total time for the trip from Station A to Station B, what is the value of vv?

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Cevap: 45

Cevap

The value of vv is 4545.
The outbound travel time for 180 miles at speed vv is 180v\frac{180}{v} hours. On the return trip, the first 90 miles at speed 0.80v0.80v require 900.80v=112.5v\frac{90}{0.80v} = \frac{112.5}{v} hours, while the second 90 miles at speed 1.25v1.25v require 901.25v=72v\frac{90}{1.25v} = \frac{72}{v} hours. The total return duration is 112.5+72v=184.5v\frac{112.5 + 72}{v} = \frac{184.5}{v} hours. Setting the difference between the return time and outbound time equal to 6 minutes (0.10.1 hours) gives 184.5v180v=0.1\frac{184.5}{v} - \frac{180}{v} = 0.1, which simplifies to 4.5v=0.1\frac{4.5}{v} = 0.1, yielding v=45v = 45.

Adım Adım Çözüm

1
Write the expression for the outbound trip duration in terms of vv.
Toutbound=180vT_{\text{outbound}} = \frac{180}{v} hours.
Time is equal to total distance divided by constant speed.
2
Calculate the duration for each half of the return trip in terms of vv.
The first 90 miles take 900.80v=112.5v\frac{90}{0.80v} = \frac{112.5}{v} hours, and the second 90 miles take 901.25v=72v\frac{90}{1.25v} = \frac{72}{v} hours, giving a total return duration of 184.5v\frac{184.5}{v} hours.
The return trip consists of two 90-mile segments driven at 0.80v0.80v and 1.25v1.25v respectively.
3
Equate the difference between return and outbound times to 0.1 hours and solve for vv.
184.5v180v=0.1    4.5v=0.1    v=45\frac{184.5}{v} - \frac{180}{v} = 0.1 \implies \frac{4.5}{v} = 0.1 \implies v = 45.
The time difference of 6 minutes is equal to 660=0.1\frac{6}{60} = 0.1 hours.

Anahtar Kavram

Distance, Rate, and Time Modeling with Piecewise Speed Changes
Soru 11Soru

Pipeline A operating alone can fill a storage tank in xx hours, whereas Pipeline B operating alone takes 50%50\% longer than Pipeline A to fill the same tank. Drainage Pipe C operating alone can empty a full tank in 2x2x hours. If all three pipes are opened simultaneously when the tank is empty, the tank becomes completely full in 1212 hours. What is the value of xx?

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Cevap: 1414

Cevap

The value of xx is 1414.
To find xx, calculate the hourly rate of each pipe: Pipeline A fills 1x\frac{1}{x} of the tank per hour, Pipeline B takes 1.5x=32x1.5x = \frac{3}{2}x hours so it fills 23x\frac{2}{3x} per hour, and Pipe C empties 12x\frac{1}{2x} per hour. The combined rate is 1x+23x12x=76x\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{7}{6x}. Since the tank fills in 1212 hours, the net rate is 112\frac{1}{12}. Setting 76x=112\frac{7}{6x} = \frac{1}{12} gives 6x=846x = 84, so x=14x = 14.

Adım Adım Çözüm

1
Express the individual work rates per hour in terms of xx.
Pipeline A rate = 1x\frac{1}{x}, Pipeline B time = 1.5x=32x    1.5x = \frac{3}{2}x \implies Pipeline B rate = 23x\frac{2}{3x}, Pipe C rate = 12x-\frac{1}{2x}.
Work rate is the reciprocal of the total time required to complete the job alone, with drainage represented as a negative rate.
2
Formulate the equation for the combined rate of all three pipes operating together.
Combined Rate = 1x+23x12x=1x(1+2312)=1x(6+436)=76x\frac{1}{x} + \frac{2}{3x} - \frac{1}{2x} = \frac{1}{x} \left(1 + \frac{2}{3} - \frac{1}{2}\right) = \frac{1}{x} \left(\frac{6 + 4 - 3}{6}\right) = \frac{7}{6x}.
Simultaneous operation means summing the individual rates.
3
Equate the combined rate to the overall rate required to fill the tank in 12 hours and solve for xx.
\frac{7}{6x} = \frac{1}{12} \implies 6x = 84 \implies x = 14.
The tank is filled in 12 hours, so the net rate per hour is 112\frac{1}{12}.

Anahtar Kavram

Combined Work Rates with Inflow and Outflow
Tahmini Süre:2m 0s
Soru 12Soru

A pharmaceutical laboratory produces a custom solution in a mixing tank using two automated pumps, Pump A and Pump B. Initially, the tank contains 600600 liters of a solution that is 20%20\% active reagent by volume. Pump A adds pure active reagent to the tank at a constant rate of 1515 liters per minute, while Pump B simultaneously adds distilled water (0%0\% active reagent) to the tank at a constant rate of 2525 liters per minute. At the exact instant when the mixture in the tank reaches 35%35\% active reagent by volume, Pump B is turned off while Pump A continues to add pure active reagent at 1515 liters per minute. How many total minutes, from the moment both pumps were initially started, does it take for the solution in the tank to reach 50%50\% active reagent by volume?

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Cevap: 174

Cevap

The total elapsed time is 174 minutes.
The problem requires setting up a two-stage rational rate model. In the first stage, initial reagent volume is 120120 liters (20%20\% of 600600). With Pump A adding 1515 L/min of reagent and Pump B adding 2525 L/min of water, the mixture concentration after t1t_1 minutes is 120+15t1600+40t1=0.35\frac{120 + 15t_1}{600 + 40t_1} = 0.35, yielding t1=90t_1 = 90 minutes. At t1=90t_1 = 90, the total volume is 4,2004,200 liters and the reagent volume is 1,4701,470 liters. In the second stage, only Pump A is active, so concentration after t2t_2 additional minutes is 1470+15t24200+15t2=0.50\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50, yielding t2=84t_2 = 84 minutes. Summing both intervals gives 90+84=17490 + 84 = 174 total minutes.

Adım Adım Çözüm

1
Determine initial quantities of active reagent and total solution.
Initial total volume V0=600V_0 = 600 liters. Initial volume of active reagent A0=0.20×600=120A_0 = 0.20 \times 600 = 120 liters.
Establishing the baseline volume and solute concentration before any pumps are turned on.
2
Model Phase 1 where both pumps operate concurrently for t1t_1 minutes to reach 35%35\% concentration.
Total volume equation: V(t1)=600+(15+25)t1=600+40t1V(t_1) = 600 + (15 + 25)t_1 = 600 + 40t_1. Reagent volume equation: A(t1)=120+15t1A(t_1) = 120 + 15t_1. Set concentration to 0.350.35: 120+15t1600+40t1=0.35    t1=90\frac{120 + 15t_1}{600 + 40t_1} = 0.35 \implies t_1 = 90 minutes.
Both pumps add volume, but only Pump A adds active reagent. Solving 120+15t1600+40t1=720\frac{120 + 15t_1}{600 + 40t_1} = \frac{7}{20} gives 20(120+15t1)=7(600+40t1)    2400+300t1=4200+280t1    20t1=1800    t1=9020(120 + 15t_1) = 7(600 + 40t_1) \implies 2400 + 300t_1 = 4200 + 280t_1 \implies 20t_1 = 1800 \implies t_1 = 90 minutes.
3
Calculate the accumulated total volume and reagent volume at the end of Phase 1.
Total volume V1=600+40(90)=4,200V_1 = 600 + 40(90) = 4,200 liters. Active reagent volume A1=120+15(90)=1,470A_1 = 120 + 15(90) = 1,470 liters.
These amounts serve as the starting parameters for Phase 2.
4
Model Phase 2 where only Pump A operates for t2t_2 additional minutes to reach 50%50\% concentration.
In Phase 2, volume increases at 1515 L/min and reagent increases at 1515 L/min. Set concentration to 0.500.50: 1470+15t24200+15t2=0.50    t2=84\frac{1470 + 15t_2}{4200 + 15t_2} = 0.50 \implies t_2 = 84 minutes.
Solving 2(1470+15t2)=4200+15t2    2940+30t2=4200+15t2    15t2=1260    t2=842(1470 + 15t_2) = 4200 + 15t_2 \implies 2940 + 30t_2 = 4200 + 15t_2 \implies 15t_2 = 1260 \implies t_2 = 84 minutes.
5
Sum the durations of Phase 1 and Phase 2 to find the total time.
Total time T=t1+t2=90+84=174T = t_1 + t_2 = 90 + 84 = 174 minutes.
The question asks for the total minutes from the initial start of both pumps.

Anahtar Kavram

Multi-Phase Algebraic Rate and Mixture Modeling
Soru 13Soru

A coffee shop owner creates a 3030-pound blend of coffee by mixing Bean X, which costs $8\$8 per pound, with Bean Y, which costs $14\$14 per pound. If the final mixture costs $10\$10 per pound, how many pounds of Bean X are in the mixture?

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Cevap: 20

Cevap

The mixture contains 20 pounds of Bean X.
The quantity of Bean X required is 20 pounds. Setting up the cost equation 8x+14(30x)=3008x + 14(30 - x) = 300 simplifies to 6x=120-6x = -120, giving x=20x = 20.

Adım Adım Çözüm

1
Define the unknown variables representing the quantities of each component.
Let xx be the weight of Bean X in pounds. The weight of Bean Y is (30x)(30 - x) pounds.
Since the total weight of the mixture is 30 pounds, expressing Bean Y in terms of xx creates a single-variable system.
2
Set up a linear equation based on total financial value.
8x+14(30x)=10×308x + 14(30 - x) = 10 \times 30
The total cost of Bean X plus the total cost of Bean Y equals the total cost of the combined 30-pound mixture.
3
Solve the algebraic equation for xx.
8x+42014x=300    6x=120    x=208x + 420 - 14x = 300 \implies -6x = -120 \implies x = 20
Combining like terms isolates the variable xx to yield the required quantity of Bean X.

Anahtar Kavram

Linear mixture modeling and single-variable algebraic modeling
Soru 14Soru

Working alone at a constant rate, Pump A can drain a full water reservoir in 66 hours, while Pump B working alone at a constant rate can drain the same full reservoir in 44 hours. Which of the following statements must be true? Select all such statements.

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Cevap: Working together, Pump A and Pump B can drain the entire reservoir in 2.42.4 hours.; In 11 hour, Pump A and Pump B working together drain 512\frac{5}{12} of the total reservoir volume.

Cevap

The statements confirming that the two pumps together drain the reservoir in 2.4 hours and that they drain 5/12 of the reservoir in 1 hour are correct.
The rate for Pump A is 16\frac{1}{6} job/hr and for Pump B is 14\frac{1}{4} job/hr. Their combined rate is 16+14=512\frac{1}{6} + \frac{1}{4} = \frac{5}{12} of the reservoir per hour, which directly validates the statement regarding 1 hour of combined work. Dividing 11 full reservoir by 512\frac{5}{12} yields 2.42.4 hours, which validates the total elapsed time statement.

Adım Adım Çözüm

1
Calculate individual work rates
Pump A rate = 16\frac{1}{6} reservoir/hr, Pump B rate = 14\frac{1}{4} reservoir/hr
Work rate is the reciprocal of time required to complete the job.
2
Calculate combined work rate
Combined rate = 16+14=212+312=512\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} reservoir/hr
Rates add when workers/pumps perform simultaneously.
3
Calculate time needed for full task
Total time = 1Combined Rate=125=2.4\frac{1}{\text{Combined Rate}} = \frac{12}{5} = 2.4 hours
Total work (11) divided by combined rate gives total elapsed time.

Anahtar Kavram

Combined Work Rates and Inverse Time Relationships
Tahmini Süre:1m 0s
Soru 15Soru

A car rental agency charges a flat daily rate of 45plus45 plus 0.20 per mile driven. If a customer rented a car for one day and the total rental cost before taxes was $75, how many miles did the customer drive?

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Cevap: 150

Cevap

150 miles
Subtracting the 45fixedfeefromthetotalchargeof45 fixed fee from the total charge of 75 leaves 30attributabletomileage.Dividing30 attributable to mileage. Dividing 30 by the variable rate of $0.20 per mile gives 150 miles.

Adım Adım Çözüm

1
Set up the linear equation representing total cost
45+0.20m=7545 + 0.20m = 75, where mm represents the number of miles driven.
The total cost consists of a fixed fee plus the variable per-mile charge.
2
Isolate the variable term by subtracting the fixed fee from both sides
0.20m=7545    0.20m=300.20m = 75 - 45 \implies 0.20m = 30
This determines the portion of the total cost accrued strictly from mileage.
3
Solve for mm by dividing by the per-mile rate
m=300.20=150m = \frac{30}{0.20} = 150
Dividing total mileage cost by the cost per mile gives the total miles driven.

Anahtar Kavram

Linear Modeling and Single-Variable Equations
Tahmini Süre:45s
Soru 16Soru

Working independently at their respective constant rates, Alex and Blair can complete a certain job together. Alex works alone for 44 hours, after which Blair joins Alex, and together they work for an additional 66 hours to finish the entire job. If Alex takes strictly less time to complete the job working alone than Blair takes working alone, which of the following statements must be true? Select all that apply.

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Cevap: Alex working alone would take less than 1616 hours to complete the job.; The time required for Alex and Blair to complete the job working together from start to finish is strictly between 88 hours and 1010 hours.; Alex completes more than 60%60\% of the entire job.

Cevap

The true statements are that Alex working alone would take less than 16 hours, the combined time to finish the job working together is between 8 and 10 hours, and Alex completes more than 60 percent of the entire job.
By setting up the total work equation 10rA+6rB=110 r_A + 6 r_B = 1 and using the inequality rA>rB>0r_A > r_B > 0, we find that 1/16<rA<1/101/16 < r_A < 1/10. This implies Alex's solo time is strictly less than 16 hours. The combined time TtogetherT_{together} is constrained between 8 and 10 hours because 6/Ttogether=14rA6/T_{together} = 1 - 4 r_A. Additionally, Alex's total work contribution is 10rA>10/16=62.5%10 r_A > 10/16 = 62.5\%, which is strictly greater than 60%.

Adım Adım Çözüm

1
Set up the work equation using individual rates rAr_A and rBr_B.
4rA+6(rA+rB)=1    10rA+6rB=14 r_A + 6(r_A + r_B) = 1 \implies 10 r_A + 6 r_B = 1.
Alex works alone for 4 hours and then both Alex and Blair work together for 6 hours to complete 1 unit of work.
2
Apply the condition that Alex takes strictly less time alone than Blair (rA>rB>0r_A > r_B > 0).
10rA+6rA>10rA+6rB=1    16rA>1    rA>11610 r_A + 6 r_A > 10 r_A + 6 r_B = 1 \implies 16 r_A > 1 \implies r_A > \frac{1}{16}.
Since Alex's rate rAr_A is strictly greater than Blair's rate rBr_B, replacing rBr_B with rAr_A gives an upper bound on Alex's solo time TA=1/rA<16T_A = 1/r_A < 16 hours.
3
Determine the feasible range for Alex's rate rAr_A.
116<rA<110\frac{1}{16} < r_A < \frac{1}{10}.
From 6rB=110rA>06 r_B = 1 - 10 r_A > 0, we get rA<1/10r_A < 1/10. Combined with rA>1/16r_A > 1/16, we have 1/16<rA<1/101/16 < r_A < 1/10.
4
Calculate the combined time Ttogether=1rA+rBT_{together} = \frac{1}{r_A + r_B}.
8<Ttogether<108 < T_{together} < 10.
Since 6(rA+rB)=14rA6(r_A + r_B) = 1 - 4 r_A, substituting 1/16<rA<1/101/16 < r_A < 1/10 gives 3/5<6(rA+rB)<3/43/5 < 6(r_A + r_B) < 3/4, which simplifies to 8<Ttogether<108 < T_{together} < 10.
5
Calculate the fraction of total work performed by Alex.
Alex performs 10rA>10×116=0.625=62.5%10 r_A > 10 \times \frac{1}{16} = 0.625 = 62.5\% of the total work.
Alex works for a total of 10 hours (4+64 + 6). Since rA>1/16r_A > 1/16, Alex completes over 62.5%62.5\% of the job, which is greater than 60%60\%.

Anahtar Kavram

Formulating algebraic inequalities for work rate problems when relative individual speeds are given.
Soru 17Soru

Train XX departs from a station traveling due east at a constant speed of 5050 miles per hour. Exactly 11 hour later, Express Train YY departs from the same station along the same track, traveling due east at a constant speed of 7575 miles per hour. How many hours after Express Train YY departs will it catch up to Train XX?

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Cevap: 22

Cevap

Express Train YY will catch up to Train XX exactly 22 hours after Express Train YY departs.
Let tt represent the number of hours Express Train YY travels. Because Train XX departed 11 hour earlier, it has been traveling for t+1t + 1 hours. For Express Train YY to catch up to Train XX, both trains must cover the exact same distance from the starting station. Setting up the distance equation 75t=50(t+1)75t = 50(t + 1) gives 75t=50t+5075t = 50t + 50, which simplifies to 25t=5025t = 50, yielding t=2t = 2 hours.

Adım Adım Çözüm

1
Define variables for the time traveled by each train.
Let tt be the time in hours that Express Train YY travels. Since Train XX departed 11 hour earlier, Train XX travels for t+1t + 1 hours.
Train XX has a 11-hour head start.
2
Express the distance traveled by each train using Distance=Rate×Time\text{Distance} = \text{Rate} \times \text{Time}.
Distance of Train X=50(t+1)X = 50(t + 1) miles; Distance of Express Train Y=75tY = 75t miles.
Both trains travel at constant rates along the same path.
3
Equate the two distance expressions to solve for tt.
75t=50(t+1)    75t=50t+50    25t=50    t=275t = 50(t + 1) \implies 75t = 50t + 50 \implies 25t = 50 \implies t = 2.
Express Train YY catches Train XX when both have covered the exact same distance.

Anahtar Kavram

Distance, Rate, and Time Modeling for Catch-up Scenarios
Soru 18Soru

An express train travels along a straight track between Station A and Station B. For the first 40%40\% of the total distance, the train travels at a constant speed of vv miles per hour. For the next 50%50\% of the remaining distance, due to track maintenance, it travels at a constant speed that is 25%25\% slower than vv. For the final leg of the journey, the train increases its speed to a constant rate that is 20%20\% faster than vv. If the average speed for the entire journey from Station A to Station B is 6060 miles per hour, what is the value of vv, in miles per hour?

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Cevap: 63.063.0

Cevap

The base speed vv is 63.063.0 miles per hour.
The correct answer is derived by setting up a model for distance, speed, and time across all three segments. The first leg covers 0.4D0.4D at speed vv (time =0.4D/v= 0.4D/v). The second leg covers 50%50\% of the remaining 0.6D0.6D, which is 0.3D0.3D, at speed 0.75v0.75v (time =0.3D/(0.75v)=0.4D/v= 0.3D/(0.75v) = 0.4D/v). The third leg covers the remaining 0.3D0.3D at speed 1.2v1.2v (time =0.3D/(1.2v)=0.25D/v= 0.3D/(1.2v) = 0.25D/v). Summing these gives total time T=1.05D/vT = 1.05D/v. The average speed is D/(1.05D/v)=v/1.05=20v/21D / (1.05D/v) = v / 1.05 = 20v/21. Setting 20v/21=6020v/21 = 60 yields v=63v = 63.

Adım Adım Çözüm

1
Define distances for each leg of the trip in terms of total distance DD.
Leg 1 distance d1=0.40Dd_1 = 0.40D. Remaining distance is D0.40D=0.60DD - 0.40D = 0.60D. Leg 2 distance d2=0.50×0.60D=0.30Dd_2 = 0.50 \times 0.60D = 0.30D. Leg 3 distance d3=0.60D0.30D=0.30Dd_3 = 0.60D - 0.30D = 0.30D.
The problem specifies percentages of remaining distance, requiring step-by-step subtraction of completed distance.
2
Determine the speed for each leg in terms of vv.
Leg 1 speed v1=vv_1 = v. Leg 2 speed v2=v(10.25)=0.75vv_2 = v(1 - 0.25) = 0.75v. Leg 3 speed v3=v(1+0.20)=1.20vv_3 = v(1 + 0.20) = 1.20v.
Percentage increases and decreases are applied to the base rate vv.
3
Calculate time spent on each leg (t=dvt = \frac{d}{v}) and sum for total time TT.
t1=0.40Dvt_1 = \frac{0.40D}{v}, t2=0.30D0.75v=0.40Dvt_2 = \frac{0.30D}{0.75v} = \frac{0.40D}{v}, t3=0.30D1.20v=0.25Dvt_3 = \frac{0.30D}{1.20v} = \frac{0.25D}{v}. Total time T=0.40D+0.40D+0.25Dv=1.05Dv=21D20vT = \frac{0.40D + 0.40D + 0.25D}{v} = \frac{1.05D}{v} = \frac{21D}{20v}.
Average speed requires total distance divided by total time.
4
Set up the average speed equation and solve for vv.
Average speed =DT=D21D20v=20v21=60    20v=1260    v=63= \frac{D}{T} = \frac{D}{\frac{21D}{20v}} = \frac{20v}{21} = 60 \implies 20v = 1260 \implies v = 63.
Equating the algebraic average speed expression to the given value of 60 mph yields the value of vv.

Anahtar Kavram

Weighted Average Speed in Multi-Phase Motion Problems
Soru 19Soru

A company allocates a total budget of $4,000\$4,000 between its marketing and research departments. The amount allocated to marketing is $400\$400 more than three times the amount allocated to research. How many dollars are allocated to research?

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Cevap: 900

Cevap

The amount allocated to research is 900900 dollars.
If xx represents the research budget, the marketing budget is 3x+4003x + 400. Summing both department allocations gives x+(3x+400)=4,000x + (3x + 400) = 4,000. Simplifying this linear equation gives 4x+400=4,0004x + 400 = 4,000, leading to 4x=3,6004x = 3,600 and x=900x = 900.

Adım Adım Çözüm

1
Define the unknown variable and express both allocations algebraically.
Let xx be the research budget. The marketing budget is 3x+4003x + 400.
The marketing allocation is defined relative to the research allocation.
2
Formulate a linear equation representing the combined budget.
x+(3x+400)=4000x + (3x + 400) = 4000
The total budget allocated across both departments is $4,000\$4,000.
3
Solve the linear equation for xx.
4x+400=4000    4x=3600    x=9004x + 400 = 4000 \implies 4x = 3600 \implies x = 900
Combine like terms, isolate the variable term by subtracting 400400, and divide by 44.

Anahtar Kavram

Setting up and solving a linear equation in one variable from a real-life word problem context.
Soru 20Soru

A bookstore sells hardcover books for $15\$15 each and paperback books for $10\$10 each. A customer purchased a total of 1212 books and spent $150\$150 in total. Let hh represent the number of hardcover books purchased and pp represent the number of paperback books purchased. Which of the following equations correctly model this situation? Select all such equations.

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Cevap: h+p=12h + p = 12; 15h+10p=15015h + 10p = 150; 3h+2p=303h + 2p = 30

Cevap

The correct equations are h+p=12h + p = 12, 15h+10p=15015h + 10p = 150, and 3h+2p=303h + 2p = 30.
The system of equations that models this situation requires one equation for the total count of books (h+p=12h + p = 12) and one for the total monetary expenditure (15h+10p=15015h + 10p = 150). Dividing the cost equation 15h+10p=15015h + 10p = 150 by 5 gives another mathematically equivalent relation 3h+2p=303h + 2p = 30. Thus, all three of these statements accurately represent the given conditions.

Adım Adım Çözüm

1
Formulate the total quantity equation
h+p=12h + p = 12
The total number of books bought is 12, which is the sum of hardcover books hh and paperback books pp.
2
Formulate the total cost equation
15h+10p=15015h + 10p = 150
Hardcover books cost $15\$15 each and paperbacks cost $10\$10 each, yielding a total cost of $150\$150.
3
Simplify the total cost equation by dividing by the greatest common divisor
3h+2p=303h + 2p = 30
Dividing all coefficients in 15h+10p=15015h + 10p = 150 by 55 produces an equivalent simplified linear equation.

Anahtar Kavram

Linear Modeling and Equivalent Equations
Tahmini Süre:1m 0s
Sayfa 1 / 4Sonraki