Word Problems and Applied Math

188 questions

Question 141Question

An express delivery truck travels from Warehouse A to Warehouse B along a 180180-mile route. For the first 6060 miles, the truck maintains a constant speed of vv miles per hour. For the remaining 120120 miles, heavy traffic reduces its constant speed to v2\frac{v}{2} miles per hour. If the average speed of the truck for the entire 180180-mile trip was 3636 miles per hour, what was the truck's speed vv, in miles per hour, during the first 6060 miles?

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Answer: 6060

Answer

The speed vv during the first leg of the trip was 6060 miles per hour.
The correct answer is 6060 miles per hour. Total trip time is the sum of time on the first leg (60v\frac{60}{v} hours) and time on the second leg (120v/2=240v\frac{120}{v/2} = \frac{240}{v} hours), giving a total time of 300v\frac{300}{v} hours. Since average speed equals total distance divided by total time, 180300/v=3v5=36\frac{180}{300/v} = \frac{3v}{5} = 36, solving to v=60v = 60.

Step-by-Step Solution

1
Express the time spent on each leg of the trip in terms of vv.
Leg 1 time: t1=60vt_1 = \frac{60}{v} hours. Leg 2 time: t2=120v/2=240vt_2 = \frac{120}{v/2} = \frac{240}{v} hours.
Time equals distance divided by rate. The rate for the second leg is half of vv.
2
Calculate the total time for the entire 180180-mile trip in terms of vv.
Total time T=t1+t2=60v+240v=300vT = t_1 + t_2 = \frac{60}{v} + \frac{240}{v} = \frac{300}{v} hours.
Summing the travel times of both legs gives the total trip duration.
3
Set up the average speed equation using total distance divided by total time.
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{180}{\frac{300}{v}} = \frac{180v}{300} = \frac{3v}{5}$.
Average speed for any multi-leg trip is defined as total distance divided by total time.
4
Equate the average speed expression to the given value of 3636 mph and solve for vv.
\frac{3v}{5} = 36 \implies 3v = 180 \implies v = 60$ miles per hour.
Multiplying both sides by 55 and dividing by 33 isolates vv.

Key Concept

Average speed for a multi-leg journey is always calculated as total distance divided by total time, not by taking the arithmetic mean of the speeds.
Estimated Time:2m 0s
Question 142Question

A commuter drives from home to work along a straight route at a constant speed of vv miles per hour. On the return trip along the exact same route, heavy traffic causes the drive to take 50%50\% longer than the outbound trip. If the commuter's average speed for the entire round trip was 4848 miles per hour, what was the value of vv, in miles per hour?

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Answer: 60

Answer

60 miles per hour
The total distance for the round trip is 2d2d. If outbound time is t1=dvt_1 = \frac{d}{v}, the return time taking 50%50\% longer is 1.5t1=1.5dv1.5t_1 = \frac{1.5d}{v}. The total time is 2.5t1=5d2v2.5t_1 = \frac{5d}{2v}. Dividing total distance 2d2d by total time 5d2v\frac{5d}{2v} gives average speed 4v5\frac{4v}{5}. Setting 4v5=48\frac{4v}{5} = 48 gives v=60v = 60 miles per hour.

Step-by-Step Solution

1
Define variables for distance and outbound time.
Let the one-way distance be dd miles. Outbound time is t1=dvt_1 = \frac{d}{v} hours.
Relating time to distance and speed using the standard formula t=dvt = \frac{d}{v}.
2
Express return time and return speed in terms of outbound parameters.
Return time is t2=1.5t1=1.5dvt_2 = 1.5 t_1 = \frac{1.5d}{v} hours.
The return trip takes 50% longer, so t2=t1+0.5t1=1.5t1t_2 = t_1 + 0.5 t_1 = 1.5 t_1.
3
Calculate total distance and total time for the round trip.
Total distance =2d= 2d. Total time =t1+t2=t1+1.5t1=2.5t1=2.5dv=5d2v= t_1 + t_2 = t_1 + 1.5 t_1 = 2.5 t_1 = \frac{2.5d}{v} = \frac{5d}{2v} hours.
Average speed requires total distance divided by total time.
4
Set up the average speed equation and solve for vv.
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{2d}{\frac{5d}{2v}} = \frac{4v}{5}.Givenaveragespeedis48mph:. Given average speed is 48 mph: \frac{4v}{5} = 48 \implies 4v = 240 \implies v = 60$.
Solving the linear algebraic equation yields the exact outbound speed.

Key Concept

Average speed for multi-leg journeys must always be calculated as Total Distance divided by Total Time.
Estimated Time:2m 0s
Question 143Question

An artisanal coffee roasting company purchases unroasted green coffee beans at a cost of $12.00\$12.00 per kilogram. During the roasting process, moisture loss reduces the total weight of the coffee beans by 20%20\%. To set the regular retail price per kilogram of roasted beans, the company marks up the effective cost price per kilogram of roasted beans by 50%50\%. If the company sells the roasted beans at a promotional discount of 10%10\% off the regular retail price, what is the net profit percentage earned on the total cost of the unroasted green coffee beans?

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Answer: 35

Answer

The net profit percentage earned on the total cost of the unroasted green coffee beans is 35%.
The correct calculation yields a total revenue of 16.20from0.80kgofroastedcoffeeproducedfrom1kgofgreencoffeepurchasedfor16.20 from 0.80 kg of roasted coffee produced from 1 kg of green coffee purchased for 12.00. This results in a net profit of 4.20,whichisexactly354.20, which is exactly 35% of the initial 12.00 cost.

Step-by-Step Solution

1
Calculate the effective cost per kilogram of roasted beans after weight loss.
Effective cost per kg of roasted beans = 12.00/(10.20)=12.00 / (1 - 0.20) = 15.00 per kg.
Because weight decreases by 20%, 1 kg of green beans yields 0.80 kg of roasted beans, raising the per-unit cost.
2
Calculate the regular retail price per kilogram of roasted beans.
Regular retail price = 15.00(1+0.50)=15.00 * (1 + 0.50) = 22.50 per kg.
The 50% markup is applied to the effective cost price of the roasted beans.
3
Calculate the promotional selling price after a 10% discount.
Discounted selling price = 22.50(10.10)=22.50 * (1 - 0.10) = 20.25 per kg.
The discount reduces the regular retail price by 10%.
4
Find total revenue generated from the yield of 1 kg of green coffee beans.
Total revenue = 0.80 kg * 20.25/kg=20.25/kg = 16.20.
1 kg of initial green beans produces 0.80 kg of sellable roasted product.
5
Calculate the net profit percentage based on the initial cost.
Net profit percentage = ((16.2016.20 - 12.00) / 12.00)10012.00) * 100% = ( 4.20 / $12.00) * 100% = 35%.
Profit percentage is the net profit divided by the original total cost.

Key Concept

Multi-step profit calculation involving shrink/yield loss, markup, successive discounts, and base identification.
Question 144Question

An jeweler creates a custom silver alloy by melting together two available alloys. Alloy X is 40%40\% silver by weight, and Alloy Y is 70%70\% silver by weight. If the jeweler uses 1515 grams more of Alloy Y than Alloy X to produce a final alloy mixture that is 60%60\% silver by weight, what is the total weight, in grams, of the final alloy mixture?

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Answer: 4545 grams

Answer

4545 grams
The total weight of the mixture is 4545 grams. Letting xx represent the grams of Alloy X, the weight of Alloy Y is x+15x + 15. Equating the pure silver content gives 0.40x+0.70(x+15)=0.60(2x+15)0.40x + 0.70(x + 15) = 0.60(2x + 15). Solving yields x=15x = 15 grams for Alloy X and 3030 grams for Alloy Y, making the total final mixture weight 15+30=4515 + 30 = 45 grams.

Step-by-Step Solution

1
Define variables for the weights of the alloys.
Let the weight of Alloy X be xx grams. Since Alloy Y has 1515 grams more than Alloy X, the weight of Alloy Y is x+15x + 15 grams. The total weight of the mixture is x+(x+15)=2x+15x + (x + 15) = 2x + 15 grams.
Establishing explicit algebraic terms for component and total weights is essential for setting up the mixture equation.
2
Set up the equation for total silver content.
0.40x+0.70(x+15)=0.60(2x+15)0.40x + 0.70(x + 15) = 0.60(2x + 15)
The total amount of pure silver from Alloy X and Alloy Y combined must equal the amount of pure silver in the final mixture.
3
Solve the algebraic equation for xx.
0.40x+0.70x+10.5=1.20x+91.10x+10.5=1.20x+90.10x=1.5x=150.40x + 0.70x + 10.5 = 1.20x + 9 \Rightarrow 1.10x + 10.5 = 1.20x + 9 \Rightarrow 0.10x = 1.5 \Rightarrow x = 15.
Isolating xx determines the weight of Alloy X used.
4
Calculate the total weight of the final alloy mixture.
Total weight = 2x+15=2(15)+15=452x + 15 = 2(15) + 15 = 45 grams.
The question asks for the total weight of the mixture, requiring substitution of x=15x = 15 back into the total weight expression.

Key Concept

Weighted average and concentration conservation in mixtures
Question 145Question

Car 1 departs from City A at 8:00 AM and travels along a straight highway toward City B at a constant speed of 5050 miles per hour. At 8:30 AM, Car 2 departs from City B and travels along the same highway toward City A at a constant speed of 6060 miles per hour. At 9:15 AM, Car 1 encounters heavy traffic and reduces its speed by 2020 percent, maintaining this reduced speed for the rest of its journey, while Car 2 continues at its original speed. If the two cars pass each other at exactly 10:00 AM, what is the distance, in miles, between City A and City B?

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Answer: 182.5

Answer

The distance between City A and City B is 182.5 miles.
The total distance between the two cities is the sum of the distances traveled by both cars prior to their meeting point at 10:00 AM. Car 1 travels for 1.25 hours at 50 mph (62.5 miles) and 0.75 hours at 40 mph (30 miles), totaling 92.5 miles. Car 2 travels for 1.5 hours at 60 mph (90 miles). Summing these gives 92.5 + 90 = 182.5 miles.

Step-by-Step Solution

1
Determine the distance traveled by Car 1 prior to its speed reduction
Car 1 traveled 62.5 miles between 8:00 AM and 9:15 AM
Car 1 drove for 1 hour and 15 minutes (1.25 hours) at 50 mph, so distance = 50 × 1.25 = 62.5 miles.
2
Determine the distance traveled by Car 1 after its speed reduction
Car 1 traveled 30 miles between 9:15 AM and 10:00 AM
Car 1 reduced its speed by 20%, bringing its rate to 50 × 0.80 = 40 mph. It traveled at 40 mph for 45 minutes (0.75 hours), so distance = 40 × 0.75 = 30 miles.
3
Determine the distance traveled by Car 2
Car 2 traveled 90 miles between 8:30 AM and 10:00 AM
Car 2 drove from 8:30 AM to 10:00 AM, which is 1 hour and 30 minutes (1.5 hours), at a constant speed of 60 mph, so distance = 60 × 1.5 = 90 miles.
4
Calculate the total distance between City A and City B
The total distance is 182.5 miles
Because the two cars traveled toward each other from opposite cities until they met, the total distance between the cities equals the sum of the individual distances: 62.5 + 30 + 90 = 182.5 miles.

Key Concept

Converging motion with variable rates and time offsets
Question 146Question

A market research firm surveyed 200200 smartphone owners regarding two security features enabled on their devices: Facial Recognition and Fingerprint Scanning. Exactly 140140 owners enabled Facial Recognition, 110110 owners enabled Fingerprint Scanning, and 3030 owners enabled neither feature. How many of the surveyed smartphone owners enabled Facial Recognition ONLY?

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Answer: 60

Answer

60 smartphone owners enabled Facial Recognition only.
Out of 200200 owners, 3030 used neither feature, leaving 170170 owners who enabled at least one feature. Using the overlapping set formula At Least One=Facial+FingerprintBoth\text{At Least One} = \text{Facial} + \text{Fingerprint} - \text{Both}, we get 170=140+110Both170 = 140 + 110 - \text{Both}, so Both=80\text{Both} = 80. To find the number of owners who enabled Facial Recognition ONLY, we subtract the overlap from the total Facial Recognition count: 14080=60140 - 80 = 60.

Step-by-Step Solution

1
Calculate the total number of owners who enabled at least one security feature.
20030=170200 - 30 = 170 owners.
Owners who used neither feature must be subtracted from the total universe of surveyed owners.
2
Apply the Overlapping Sets formula to find the number of owners who enabled both features.
Total At Least One=Facial+FingerprintBoth \text{Total At Least One} = \text{Facial} + \text{Fingerprint} - \text{Both}
170=140+110Both    Both=250170=80170 = 140 + 110 - \text{Both} \implies \text{Both} = 250 - 170 = 80.
Summing individual sets double-counts the intersection area.
3
Subtract the intersection from the total Facial Recognition set to find Facial Recognition only.
14080=60140 - 80 = 60 owners.
The Facial Recognition group consists of those who enabled Facial Recognition only plus those who enabled both features.

Key Concept

Two-Set Overlapping Sets Formula: Total = Set A + Set B - Both + Neither
Question 147Question

A passenger train and a cargo train both travel along parallel tracks on a 120120-mile route between Station P and Station Q. The passenger train travels at a constant speed of 5050 miles per hour for the entire trip. The cargo train completes the first 60%60\% of the total distance at a constant speed of 3030 miles per hour and the remaining 40%40\% of the distance at a constant speed of 6060 miles per hour. By what percentage is the average speed of the passenger train for the entire trip greater than the average speed of the cargo train for the entire trip?

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Answer: 3313%33\frac{1}{3}\%

Answer

The average speed of the passenger train is 3313%33\frac{1}{3}\% greater than the average speed of the cargo train.
The correct answer accurately determines the total travel time of the cargo train (3.23.2 hours) to find its true average speed (37.537.5 mph). Comparing the passenger train's speed (5050 mph) against this baseline gives 5037.537.5=13\frac{50 - 37.5}{37.5} = \frac{1}{3}, which corresponds to 3313%33\frac{1}{3}\%.

Step-by-Step Solution

1
Calculate the travel time and average speed of the passenger train.
The passenger train travels 120120 miles at 5050 mph, so its average speed is 5050 mph.
The passenger train maintains a single constant speed throughout the trip.
2
Calculate the distance and travel time for each leg of the cargo train's trip.
Leg 1 distance: 60%60\% of 120=72120 = 72 miles; Leg 1 time: 7230=2.4\frac{72}{30} = 2.4 hours. Leg 2 distance: 40%40\% of 120=48120 = 48 miles; Leg 2 time: 4860=0.8\frac{48}{60} = 0.8 hours.
Time equals distance divided by speed for each specific segment.
3
Determine the total travel time and overall average speed of the cargo train.
Total time = 2.4+0.8=3.22.4 + 0.8 = 3.2 hours. Cargo train average speed = 120 miles3.2 hours=37.5\frac{120\text{ miles}}{3.2\text{ hours}} = 37.5 mph.
Average speed for a multi-leg journey is strictly defined as total distance divided by total time.
4
Calculate by what percentage the passenger train's average speed is greater than the cargo train's average speed.
Percentage increase = 5037.537.5×100%=12.537.5×100%=13×100%=3313%\frac{50 - 37.5}{37.5} \times 100\% = \frac{12.5}{37.5} \times 100\% = \frac{1}{3} \times 100\% = 33\frac{1}{3}\%.
The baseline for 'percent greater than the cargo train' is the cargo train's speed (37.537.5 mph).

Key Concept

Average Speed for Multi-Leg Journeys and Percentage Difference Baseline
Question 148Question

A commercial retailer purchased 100 identical smartwatches for a total cost of $10,000\$10,000. Before any sales were made, 10 smartwatches were damaged and could not be sold. The retailer marked up the unit cost of each remaining smartwatch by PP percent to determine its list price. During a promotional period, 50 smartwatches were sold at the full list price. The remaining 40 undamaged smartwatches were subsequently sold during a clearance sale at a 20 percent discount off the list price. If the retailer earned a total net profit of 23 percent on the initial investment of $10,000\$10,000, what is the value of PP?

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Answer: 50

Answer

50
The unit cost of each smartwatch is $10,000/100=$100\$10,000 / 100 = \$100. A markup of P%P\% sets the list price to L=100+PL = 100 + P. The retailer sells 50 units at full list price LL and 40 units at 0.80L0.80L, earning total revenue of 50L+32L=82L=82(100+P)=8,200+82P50L + 32L = 82L = 82(100 + P) = 8,200 + 82P. To earn a 23% profit on the entire $10,000\$10,000 investment, total revenue must equal $12,300\$12,300. Solving 8,200+82P=12,3008,200 + 82P = 12,300 yields 82P=4,10082P = 4,100, so P=50P = 50.

Step-by-Step Solution

1
Calculate the unit cost and total required revenue.
Unit cost c=$10,000100=$100c = \frac{\$10,000}{100} = \$100. Required revenue R=$10,000×(1+0.23)=$12,300R = \$10,000 \times (1 + 0.23) = \$12,300.
Net profit is 23% of the total initial investment of $10,000\$10,000.
2
Express the list price LL and clearance price in terms of PP.
List price L=100×(1+P100)=100+PL = 100 \times \left(1 + \frac{P}{100}\right) = 100 + P. Clearance price =0.80×L=0.80(100+P)=80+0.8P= 0.80 \times L = 0.80(100 + P) = 80 + 0.8P.
Markup P%P\% is calculated relative to the unit cost of $100\$100, and clearance discount is 20% off list price.
3
Set up the total revenue equation based on sold units.
Total Revenue =50L+40(0.80L)=50L+32L=82L=82(100+P)=8,200+82P= 50L + 40(0.80L) = 50L + 32L = 82L = 82(100 + P) = 8,200 + 82P.
10 units are unsellable ($0 revenue), 50 units generate full list price, and 40 units generate 80% of list price.
4
Solve for PP.
8,200+82P=12,300    82P=4,100    P=508,200 + 82P = 12,300 \implies 82P = 4,100 \implies P = 50.
Equating total generated revenue to the required revenue of $12,300\$12,300.

Key Concept

Profit, Loss, and Markup with Damaged Inventory and Tiered Discounts
Question 149Question

Container X initially contains 3030 liters of a solution that is 20%20\% acid by volume. Container Y contains VV liters of a solution that is 50%50\% acid by volume. If 1010 liters of pure water are added to Container X and the entire contents of Container X are then mixed with Container Y, the resulting mixture is 30%30\% acid by volume. What is the value of VV, in liters?

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Answer: 3030

Answer

3030 liters
The option stating 3030 liters is correct because Container X initially contains 66 liters of acid (0.20×300.20 \times 30). After adding 1010 liters of water, Container X holds 4040 liters of solution with 66 liters of acid. Container Y contributes 0.50V0.50V liters of acid out of VV liters. Equating the total acid ratio to 30%30\% yields (6+0.50V)/(40+V)=0.30(6 + 0.50V) / (40 + V) = 0.30, which solves cleanly to V=30V = 30.

Step-by-Step Solution

1
Calculate the volume of pure acid initially present in Container X.
Acid in X=20% of 30=0.20×30=6 liters\text{Acid in X} = 20\% \text{ of } 30 = 0.20 \times 30 = 6 \text{ liters}.
Determining the absolute amount of solute is required to set up the mixture equation.
2
Determine the new total volume of Container X after adding pure water.
New Volume of X=30+10=40 liters\text{New Volume of X} = 30 + 10 = 40 \text{ liters}. The acid amount remains 66 liters.
Adding pure water increases the solution's total volume while leaving the acid quantity unchanged.
3
Express the amount of acid in Container Y in terms of VV.
Acid in Y=50% of V=0.50V liters\text{Acid in Y} = 50\% \text{ of } V = 0.50V \text{ liters}.
Container Y contains a 50%50\% concentration of acid across VV liters.
4
Formulate and solve the concentration equation for the final combined mixture.
6+0.50V40+V=0.30    6+0.50V=0.30(40+V)    6+0.50V=12+0.30V    0.20V=6    V=30\frac{6 + 0.50V}{40 + V} = 0.30 \implies 6 + 0.50V = 0.30(40 + V) \implies 6 + 0.50V = 12 + 0.30V \implies 0.20V = 6 \implies V = 30.
Setting total combined acid over total combined volume equal to the final concentration of 30%30\% gives the unknown volume VV.

Key Concept

Two-stage dilution and mixture concentration using weighted average equations
Estimated Time:2m 0s
Question 150Question

A publishing firm evaluated 160160 manuscript submissions based on three quality criteria: Plot Originality (PP), Character Development (CC), and Dialogue Quality (DD). Each manuscript was evaluated against all three criteria.

- 7575 manuscripts met the criteria for Plot Originality (PP).
- 8080 manuscripts met the criteria for Character Development (CC).
- 7070 manuscripts met the criteria for Dialogue Quality (DD).
- 3030 manuscripts met the criteria for both Plot Originality and Character Development (PCP \cap C).
- 2525 manuscripts met the criteria for both Character Development and Dialogue Quality (CDC \cap D).
- 2020 manuscripts met the criteria for both Plot Originality and Dialogue Quality (PDP \cap D).
- 1010 manuscripts met all three criteria (PCDP \cap C \cap D).

How many manuscripts met exactly one of the three criteria?

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Answer: 105

Answer

The number of manuscripts that met exactly one of the three criteria is 105.
To find the number of manuscripts meeting exactly one criterion, we isolate the unique regions for each set. Subtracting the overlapping intersections from the total of each set yields 35 manuscripts for Plot Originality only, 35 for Character Development only, and 35 for Dialogue Quality only. Combining these gives 105.

Step-by-Step Solution

1
Find the number of elements in the double-intersection regions only.
Exactly (P and C) = 20; Exactly (C and D) = 15; Exactly (P and D) = 10.
The given two-set intersections include manuscripts that satisfied all three criteria, so the triple intersection of 10 must be subtracted from each.
2
Calculate the count of manuscripts meeting only P, only C, and only D.
Only P = 35, Only C = 35, Only D = 35.
Subtract all shared regions (two-set-only overlaps plus the three-set overlap) from each individual total.
3
Sum the three single-criterion regions.
35 + 35 + 35 = 105.
The question asks specifically for the total number of manuscripts that met exactly one criterion.

Key Concept

3-Set Overlapping Venn Diagrams and Region Separation
Question 151Question

An athlete completes a three-leg training course consisting of cycling, running, and swimming, covering a total distance of 6060 miles in a total time of 44 hours. The athlete cycles at a constant speed of 2424 miles per hour, runs at a constant speed of 1010 miles per hour, and swims at a constant speed of 22 miles per hour. If the time spent cycling is equal to the combined time spent running and swimming, how many miles did the athlete run?

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Answer: 10

Answer

The athlete ran 10 miles.
Because cycling time equals combined running and swimming time, the 4-hour total splits evenly into 2 hours for cycling and 2 hours for the remaining legs. Cycling at 24 mph covers 48 miles, leaving 12 miles and 2 hours for running and swimming. Writing the total time equation as (d_running / 10) + ((12 - d_running) / 2) = 2 and solving yields d_running = 10 miles.

Step-by-Step Solution

1
Determine the time spent cycling
2 hours
Since the total time is 4 hours and cycling time equals the sum of running and swimming times, 2 * t_cycling = 4.
2
Calculate the distance covered while cycling
48 miles
Distance = Speed * Time = 24 mph * 2 hours = 48 miles.
3
Determine the combined distance and time for running and swimming
Combined distance = 12 miles, Combined time = 2 hours
Total distance minus cycling distance is 60 - 48 = 12 miles; total time minus cycling time is 4 - 2 = 2 hours.
4
Set up and solve the system of equations for running distance
Running distance = 10 miles
Using time = distance / speed gives (d_running / 10) + ((12 - d_running) / 2) = 2. Multiplying by 10 yields d_running + 60 - 5 * d_running = 20, leading to d_running = 10 miles.

Key Concept

Multi-leg rate, time, and distance problems using systems of linear equations.
Question 152Question

A pharmaceutical research laboratory tested 120120 chemical compounds for the presence of three distinct bio-reactivity markers: Enzyme Inhibition (EE), Protein Binding (PP), and Receptor Affinity (RR). The test results revealed the following:

- 5858 compounds exhibited Enzyme Inhibition (EE).
- 5252 compounds exhibited Protein Binding (PP).
- 4646 compounds exhibited Receptor Affinity (RR).
- 88 compounds exhibited all three markers.
- 1515 compounds exhibited none of the three markers.

How many of the tested compounds exhibited exactly two of the three markers?

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Answer: 3535

Answer

The number of compounds exhibiting exactly two markers is 3535.
The total number of compounds displaying at least one marker is 12015=105120 - 15 = 105. The sum of the single set counts is 58+52+46=15658 + 52 + 46 = 156. Using the three-set relation N(A)+N(B)+N(C)N(union)=N(exactly 2)+2N(all 3)N(A) + N(B) + N(C) - N(\text{union}) = N(\text{exactly 2}) + 2N(\text{all 3}), we get 156105=N(exactly 2)+2(8)156 - 105 = N(\text{exactly 2}) + 2(8). Simplifying gives 51=N(exactly 2)+1651 = N(\text{exactly 2}) + 16, which yields N(exactly 2)=35N(\text{exactly 2}) = 35.

Step-by-Step Solution

1
Determine the total number of compounds exhibiting at least one marker.
At least one marker=12015=105\text{At least one marker} = 120 - 15 = 105.
Subtracting the compounds with no markers from the total gives the union size N(EPR)N(E \cup P \cup R).
2
Calculate the sum of the individual set counts.
N(E)+N(P)+N(R)=58+52+46=156N(E) + N(P) + N(R) = 58 + 52 + 46 = 156.
This sum counts elements with exactly one marker once, elements with exactly two markers twice, and elements with all three markers three times.
3
Apply the 3-set region formula connecting single counts, region overlaps, and total union.
156=(Exactly 1)+2(Exactly 2)+3(All 3)156 = (\text{Exactly 1}) + 2(\text{Exactly 2}) + 3(\text{All 3}) and 105=(Exactly 1)+(Exactly 2)+(All 3)105 = (\text{Exactly 1}) + (\text{Exactly 2}) + (\text{All 3}).
Subtracting the second equation from the first yields: 156105=(Exactly 2)+2(All 3)156 - 105 = (\text{Exactly 2}) + 2(\text{All 3}).
4
Substitute the known value for all three markers to solve for compounds with exactly two markers.
51=(Exactly 2)+2(8)    51=(Exactly 2)+16    Exactly 2=3551 = (\text{Exactly 2}) + 2(8) \implies 51 = (\text{Exactly 2}) + 16 \implies \text{Exactly 2} = 35.
Solving the linear algebraic equation yields the exact number of compounds in the pairwise-only regions.

Key Concept

Three-Set Overlapping Venn Diagrams and Region Counting Equations
Estimated Time:2m 0s
Question 153Question

An automated delivery van and an electric cargo scooter both travel along a straight route from Warehouse X to Warehouse Y, which are DD miles apart.

The van travels at a constant speed of 3030 miles per hour for the first half of the total distance, and at a constant speed of 6060 miles per hour for the second half of the total distance.

The scooter travels at a constant speed of 3030 miles per hour for the first half of its total travel time, and at a constant speed of 6060 miles per hour for the second half of its total travel time.

If the scooter completes the entire journey in 1515 minutes less time than the van, what is the value of DD, in miles?

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Answer: 9090

Answer

The distance DD between Warehouse X and Warehouse Y is 9090 miles.
The option stating 9090 is correct because the van's total time is D40\frac{D}{40} hours (average speed 4040 mph across equal distances) and the scooter's total time is D45\frac{D}{45} hours (average speed 4545 mph across equal time intervals). Setting the difference D40D45\frac{D}{40} - \frac{D}{45} equal to 14\frac{1}{4} hour yields D=90D = 90 miles.

Step-by-Step Solution

1
Calculate the total travel time for the van in terms of DD.
The van travels D2\frac{D}{2} miles at 3030 mph and D2\frac{D}{2} miles at 6060 mph. Time for first half: D/230=D60\frac{D/2}{30} = \frac{D}{60} hours. Time for second half: D/260=D120\frac{D/2}{60} = \frac{D}{120} hours. Total van time Tvan=D60+D120=3D120=D40T_{\text{van}} = \frac{D}{60} + \frac{D}{120} = \frac{3D}{120} = \frac{D}{40} hours.
For equal distance legs, average speed is calculated via harmonic mean, giving total time T=Total DistanceAverage SpeedT = \frac{\text{Total Distance}}{\text{Average Speed}}.
2
Calculate the total travel time for the scooter in terms of DD.
Let TscooterT_{\text{scooter}} be the total travel time for the scooter. It spends Tscooter2\frac{T_{\text{scooter}}}{2} hours at 3030 mph and Tscooter2\frac{T_{\text{scooter}}}{2} hours at 6060 mph. Total distance D=30(Tscooter2)+60(Tscooter2)=15Tscooter+30Tscooter=45TscooterD = 30\left(\frac{T_{\text{scooter}}}{2}\right) + 60\left(\frac{T_{\text{scooter}}}{2}\right) = 15 T_{\text{scooter}} + 30 T_{\text{scooter}} = 45 T_{\text{scooter}}. Thus, Tscooter=D45T_{\text{scooter}} = \frac{D}{45} hours.
For equal time legs, average speed is the arithmetic mean of the speeds (4545 mph).
3
Set up and solve the equation using the given time difference.
1515 minutes is equal to 1560=14\frac{15}{60} = \frac{1}{4} hours. The equation is D40D45=14\frac{D}{40} - \frac{D}{45} = \frac{1}{4}. Finding a common denominator of 360360: 9D8D360=14    D360=14    D=3604=90\frac{9D - 8D}{360} = \frac{1}{4} \implies \frac{D}{360} = \frac{1}{4} \implies D = \frac{360}{4} = 90 miles.
Equating the difference between the van's travel time and the scooter's travel time to 1515 minutes (14\frac{1}{4} hour) allows solving for DD.

Key Concept

Distinction between average speed over equal distances (harmonic mean) versus average speed over equal times (arithmetic mean).
Question 154Question

At a technology firm, an evaluation was conducted for 150 software developers to test their proficiency in three programming paradigms: Object-Oriented (OO), Functional (FF), and Reactive (RR). The evaluation revealed that 85 developers are proficient in OO, 60 are proficient in FF, and 45 are proficient in RR. Furthermore, 25 developers are proficient in both OO and FF, 18 are proficient in both FF and RR, 22 are proficient in both OO and RR, and 10 developers are proficient in all three paradigms. How many of the software developers surveyed are proficient in exactly two of these three programming paradigms?

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Answer: 35

Answer

The number of software developers proficient in exactly two of the three programming paradigms is 35.
To find the number of developers proficient in exactly two paradigms, subtract the number of developers proficient in all three paradigms (1010) from each of the pairwise totals and sum the results: (2510)+(1810)+(2210)=15+8+12=35(25 - 10) + (18 - 10) + (22 - 10) = 15 + 8 + 12 = 35.

Step-by-Step Solution

1
Identify the total count for each pairwise overlap of paradigms.
The pairwise counts are OF=25|O \cap F| = 25, FR=18|F \cap R| = 18, and OR=22|O \cap R| = 22, with all three paradigms OFR=10|O \cap F \cap R| = 10.
Each given two-set intersection includes individuals who are also proficient in all three paradigms.
2
Subtract the triple intersection from each pairwise total to isolate those proficient in exactly two paradigms.
Proficient in only OO and FF: 2510=1525 - 10 = 15; proficient in only FF and RR: 1810=818 - 10 = 8; proficient in only OO and RR: 2210=1222 - 10 = 12.
The region representing 'exactly two' sets excludes the central region where all three sets overlap.
3
Sum the three distinct exclusive two-set counts.
15+8+12=3515 + 8 + 12 = 35.
Adding these mutually exclusive categories provides the total count of developers in exactly two sets.

Key Concept

Three-Set Venn Diagram Region Calculations

Alternative Method

Using a Venn diagram, enter 1010 in the central triple-intersection region. Next, fill in the three surrounding two-set-only regions by subtracting 1010 from each given pairwise total: 2510=1525 - 10 = 15, 1810=818 - 10 = 8, and 2210=1222 - 10 = 12. Adding these three region counts together gives 15+8+12=3515 + 8 + 12 = 35.
Estimated Time:1m 30s
Question 155Question

Vessel A contains 6060 liters of a beverage mixture that is 75%75\% fruit concentrate by volume, and Vessel B contains 9090 liters of a beverage mixture that is 25%25\% fruit concentrate by volume. First, 1010 liters of pure fruit concentrate are added to Vessel B. Next, xx liters of the mixture in Vessel A are removed and replaced with xx liters of pure water. If the concentration of fruit concentrate in Vessel A is now equal to the concentration of fruit concentrate in Vessel B, what is the value of xx?

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Answer: 3434

Answer

The value of xx is 3434.
To find xx, first determine the resulting concentration in Vessel B. Vessel B initially contains 0.25×90=22.50.25 \times 90 = 22.5 liters of concentrate. Adding 1010 liters of pure concentrate gives 32.532.5 liters of concentrate in 100100 liters of total mixture, which is a concentration of 32.5%32.5\%. In Vessel A, there are initially 0.75×60=450.75 \times 60 = 45 liters of concentrate. Removing xx liters of the mixture removes 0.75x0.75x liters of concentrate. When replaced with xx liters of pure water, the total volume remains 6060 liters, so the new concentration is 450.75x60\frac{45 - 0.75x}{60}. Setting 450.75x60=0.325\frac{45 - 0.75x}{60} = 0.325 gives 450.75x=19.545 - 0.75x = 19.5, which simplifies to 0.75x=25.50.75x = 25.5, yielding x=34x = 34.

Step-by-Step Solution

1
Calculate the new concentration of fruit concentrate in Vessel B after adding 1010 liters of pure concentrate.
Initial concentrate in B = 0.25×90=22.50.25 \times 90 = 22.5 liters. Adding 1010 liters of pure concentrate yields 22.5+10=32.522.5 + 10 = 32.5 liters of concentrate in a total volume of 90+10=10090 + 10 = 100 liters. Thus, the new concentration in B is 32.5100=32.5%\frac{32.5}{100} = 32.5\% (or 0.3250.325).
Adding pure concentrate increases both the solute amount and the total solution volume.
2
Express the concentration of fruit concentrate in Vessel A after removing xx liters of mixture and replacing it with xx liters of pure water.
Initial concentrate in A = 0.75×60=450.75 \times 60 = 45 liters. Removing xx liters removes 0.75x0.75x liters of concentrate. Replacing with xx liters of pure water keeps the total volume at 6060 liters. The remaining concentrate is 450.75x45 - 0.75x, so the new concentration in A is 450.75x60\frac{45 - 0.75x}{60}.
The replacement maintains constant total volume while diluting the concentrate.
3
Equate the concentrations of Vessel A and Vessel B to solve for xx.
450.75x60=0.325    450.75x=19.5    0.75x=25.5    x=25.50.75=34\frac{45 - 0.75x}{60} = 0.325 \implies 45 - 0.75x = 19.5 \implies 0.75x = 25.5 \implies x = \frac{25.5}{0.75} = 34.
Setting the two concentration expressions equal gives a single linear equation in terms of xx.

Key Concept

Dilution and multi-vessel mixture balance
Question 156Question

A management consulting firm charges clients based on three staff tiers: Junior Consultants, Senior Consultants, and Directors. Junior Consultants are billed at a rate of $120\$120 per hour, Senior Consultants at $210\$210 per hour, and Directors at $390\$390 per hour. On a strategic assignment, the ratio of hours worked by Junior Consultants to Senior Consultants to Directors was 5:3:25 : 3 : 2, respectively. What was the overall average hourly billing rate per consultant hour for this assignment?

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Answer: $201\$201

Answer

The overall average hourly billing rate per consultant hour for the assignment is $201\$201.
The weighted average is found by multiplying each tier's rate by its corresponding ratio weight, summing these products, and dividing by the sum of the ratio weights: (5×120)+(3×210)+(2×390)5+3+2=600+630+78010=201010=$201\frac{(5 \times 120) + (3 \times 210) + (2 \times 390)}{5 + 3 + 2} = \frac{600 + 630 + 780}{10} = \frac{2010}{10} = \$201.

Step-by-Step Solution

1
Determine the total weight of the ratio elements.
Sum of ratio parts = 5+3+2=105 + 3 + 2 = 10 total parts of hours.
To calculate a weighted average, the sum of all weight parts must be calculated to serve as the denominator.
2
Calculate the total revenue generated for 10 representative hours based on the ratio.
Total revenue = (5×$120)+(3×$210)+(2×$390)=$600+$630+$780=$2,010(5 \times \$120) + (3 \times \$210) + (2 \times \$390) = \$600 + \$630 + \$780 = \$2,010.
Multiplying each tier's hourly billing rate by its respective weight yields the total weighted revenue.
3
Divide the total revenue by the total number of ratio hours.
Weighted average rate = $2,01010=$201\frac{\$2,010}{10} = \$201 per hour.
Dividing total cost/revenue by total units yields the weighted average per unit.

Key Concept

Weighted Average formula: Weighted Average=(wixi)wi\text{Weighted Average} = \frac{\sum (w_i \cdot x_i)}{\sum w_i}, where wiw_i represents the weight of each component and xix_i represents the value of each component.
Estimated Time:1m 30s
Question 157Question

Two cyclists, Cyclist A and Cyclist B, depart simultaneously from Town X and Town Y, respectively, traveling toward each other along a straight 150-mile path. Cyclist A travels at a constant speed of 18 miles per hour throughout the journey. Cyclist B initially travels at a constant speed of 30 miles per hour. After traveling for 2 hours, Cyclist B encounters a steep incline and decreases speed by 40 percent, maintaining this reduced speed for the remainder of the trip. How many hours after their departure will Cyclist A and Cyclist B meet?

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Answer: 3.5

Answer

The total time elapsed from departure until Cyclist A and Cyclist B meet is 3.5 hours.
In the first 2 hours, Cyclist A covers 18×2=3618 \times 2 = 36 miles and Cyclist B covers 30×2=6030 \times 2 = 60 miles, bringing their combined distance to 96 miles and leaving 54 miles remaining. Cyclist B's speed then decreases by 40% to 18 mph (30×0.630 \times 0.6). Moving toward each other, their combined relative rate becomes 18+18=3618 + 18 = 36 mph. Dividing the remaining 54 miles by 36 mph yields 1.5 hours for the second phase. Adding the initial 2 hours gives a total time of 3.5 hours.

Step-by-Step Solution

1
Calculate cumulative distance traveled by both cyclists during the first 2 hours
Cyclist A travels 36 miles; Cyclist B travels 60 miles; Total = 96 miles
Both cyclists travel for 2 full hours at their initial constant speeds.
2
Determine the distance separating the cyclists at t = 2 hours
150 - 96 = 54 miles remaining
Subtract the combined distance covered from the total initial separation of 150 miles.
3
Calculate Cyclist B's new speed after the 40% decrease
30 * 0.60 = 18 mph
A 40% reduction means retaining 60% of the original speed of 30 mph.
4
Calculate relative speed of approach after 2 hours
18 + 18 = 36 mph
When two objects move toward each other, their relative speed is the sum of their individual speeds.
5
Compute the time to cover the remaining distance
54 / 36 = 1.5 hours
Time equals distance divided by relative speed.
6
Sum the time segments to find the total elapsed time
2 + 1.5 = 3.5 hours
The trip consists of an initial 2-hour phase plus an additional 1.5-hour phase.

Key Concept

Relative Speed and Piecewise Motion in Converging Rate Problems
Question 158Question

A pharmacist has 8080 milliliters of a topical liquid formulation that is 5%5\% active ingredient by volume. How many milliliters of pure active ingredient must the pharmacist add to this formulation so that the resulting mixture is 24%24\% active ingredient by volume?

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Answer: 20

Answer

The pharmacist must add 2020 milliliters of pure active ingredient.
Adding 2020 mL of pure active ingredient increases the amount of solute from 44 mL to 2424 mL and the total mixture volume from 8080 mL to 100100 mL, yielding a final concentration of 24100=24%\frac{24}{100} = 24\%.

Step-by-Step Solution

1
Determine initial amount of solute
Initial active ingredient volume = 80×0.05=480 \times 0.05 = 4 mL
Before adding pure active ingredient, the solution contains 5%5\% active ingredient of the total 8080 mL volume.
2
Formulate algebraic expressions for total active ingredient and total volume
Total active ingredient = 4+x4 + x mL; Total mixture volume = 80+x80 + x mL
Adding xx mL of pure active ingredient increases both the solute volume and the total mixture volume by xx.
3
Set up and solve the mixture concentration equation
4+x80+x=625    100+25x=480+6x    19x=380    x=20\frac{4 + x}{80 + x} = \frac{6}{25} \implies 100 + 25x = 480 + 6x \implies 19x = 380 \implies x = 20
Setting the solute ratio equal to the target concentration of 24%24\% yields a linear equation in xx.

Key Concept

Mixture Concentration and Algebraic Dilution/Fortification
Question 159Question

A specialty coffee roaster creates a signature espresso blend by combining Arabica coffee beans costing 18.00perkilogramwithRobustacoffeebeanscosting18.00 per kilogram with Robusta coffee beans costing 12.00 per kilogram. The final blend weighs 50 kilograms and has an overall average cost of $15.60 per kilogram. How many kilograms of Arabica beans are included in the blend?

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Answer: 30

Answer

The blend contains 30 kilograms of Arabica coffee beans.
To determine the required quantity of Arabica beans, establish the total cost equation: 18A+12(50A)=50(15.60)18A + 12(50 - A) = 50(15.60). Expanding the terms gives 18A+60012A=78018A + 600 - 12A = 780, which simplifies to 6A=1806A = 180, yielding A=30A = 30 kilograms.

Step-by-Step Solution

1
Define variables and relate component weights
Let AA represent the mass of Arabica beans in kilograms. The mass of Robusta beans is (50A)(50 - A) kilograms.
The sum of the individual component weights must equal the total blend weight of 50 kilograms.
2
Formulate the weighted average total cost equation
18A+12(50A)=15.60×50=78018A + 12(50 - A) = 15.60 \times 50 = 780
The combined monetary cost of both bean types equals the total value of the 50 kg blend.
3
Solve for the unknown variable AA
6A+600=780    6A=180    A=306A + 600 = 780 \implies 6A = 180 \implies A = 30
Simplifying the algebraic linear equation determines the exact weight of Arabica beans.

Key Concept

Weighted Averages in Applied Contexts
Question 160Question

A commercial real estate developer acquired a property for $P\$P. Over the first year, the market value of the property increased by 25%25\%. During the second year, the property's value decreased by x%x\%. In the third year, the value increased again by 20%20\% relative to its value at the end of the second year. If the final value of the property at the end of the third year was 14%14\% greater than the original acquisition price $P\$P, what is the value of xx?

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Answer: 2424

Answer

The value of xx is 24.
To find the net effect of successive percentage changes, express each period's change as a multiplier of the value from the preceding period. A 25%25\% increase corresponds to a multiplier of 1.251.25, an x%x\% decrease corresponds to a multiplier of (1x100)\left(1 - \frac{x}{100}\right), and a 20%20\% increase corresponds to a multiplier of 1.201.20. Combined, these produce a final multiplier of 1.25×1.20×(1x100)=1.50(1x100)1.25 \times 1.20 \times \left(1 - \frac{x}{100}\right) = 1.50\left(1 - \frac{x}{100}\right). Setting this equal to the net 14%14\% total increase (1.141.14) yields 1.50(1x100)=1.141.50\left(1 - \frac{x}{100}\right) = 1.14, which simplifies to 1x100=0.761 - \frac{x}{100} = 0.76, giving x=24x = 24.

Step-by-Step Solution

1
Express each sequential period's value using multiplier notation.
At Year 1 end: V1=1.25PV_1 = 1.25P. At Year 2 end: V2=1.25P(1x100)V_2 = 1.25P \left(1 - \frac{x}{100}\right). At Year 3 end: V3=1.25P(1x100)×1.20V_3 = 1.25P \left(1 - \frac{x}{100}\right) \times 1.20.
Successive percent changes compound on the intermediate values of each period, not on the original base price.
2
Equate the overall combined growth multiplier to the given total net change.
1.25×1.20×(1x100)=1.14    1.50×(1x100)=1.141.25 \times 1.20 \times \left(1 - \frac{x}{100}\right) = 1.14 \implies 1.50 \times \left(1 - \frac{x}{100}\right) = 1.14
The final value is given as 14%14\% greater than PP, which corresponds to a net multiplier of 1.141.14.
3
Solve the algebraic equation for xx.
1x100=1.141.50=0.76    x100=0.24    x=241 - \frac{x}{100} = \frac{1.14}{1.50} = 0.76 \implies \frac{x}{100} = 0.24 \implies x = 24
Subtracting 0.760.76 from 11 yields the fractional decrease of 0.240.24, which equals 24%24\%.

Key Concept

Successive Percent Change and Multipliers
Estimated Time:2m 0s
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