Tüm alıştırma soruları

387 soru

Soru 261Soru

A merchant purchased a shipment of 200 identical items for a total cost of 8,000.Themerchantmarkedupthecostpriceofeachitemby8,000. The merchant marked up the cost price of each item by P percenttoestablishitslistprice.Duringthefirstmonthofsales,60percentoftheitemsweresoldatthelistprice.Duringthesecondmonth,50percentoftheremainingitemsweresoldata20percentdiscountoffthelistprice.Theremainingitemswerethensoldduringaclearancesaleata50percentdiscountoffthelistprice.Ifthemerchantmadeanoverallnetprofitof29percentonthetotalpurchasecostoftheshipment,whatisthevalueof percent to establish its list price. During the first month of sales, 60 percent of the items were sold at the list price. During the second month, 50 percent of the remaining items were sold at a 20 percent discount off the list price. The remaining items were then sold during a clearance sale at a 50 percent discount off the list price. If the merchant made an overall net profit of 29 percent on the total purchase cost of the shipment, what is the value of P$?

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Cevap: 50

Cevap

The value of PP is 50.
The unit cost of each item is $8,000200=$40\frac{\$8,000}{200} = \$40. To make an overall net profit of 29%, total revenue must equal $8,000×1.29=$10,320\$8,000 \times 1.29 = \$10,320. Sales are divided into three tiers: 120 items sold at list price LL, 40 items sold at 0.80L0.80L, and 40 items sold at 0.50L0.50L. Summing the revenue yields 120L+32L+20L=172L120L + 32L + 20L = 172L. Solving 172L=10,320172L = 10,320 gives L=60L = 60. The percentage markup over cost price is 604040×100=50%\frac{60 - 40}{40} \times 100 = 50\%.

Adım Adım Çözüm

1
Calculate the unit cost price and required total revenue.
Unit cost C=$8,000200=$40C = \frac{\$8,000}{200} = \$40. Total target revenue R=$8,000×(1+0.29)=$10,320R = \$8,000 \times (1 + 0.29) = \$10,320.
Determining the total required revenue sets the benchmark to solve for the list price.
2
Determine the quantity of items sold at each price tier.
Month 1 sales: 0.60×200=1200.60 \times 200 = 120 items at list price LL.
Month 2 sales: 0.50×(200120)=400.50 \times (200 - 120) = 40 items at discounted price 0.80L0.80L.
Clearance sales: 20012040=40200 - 120 - 40 = 40 items at clearance price 0.50L0.50L.
Categorizing quantities by price tier enables building the total revenue expression.
3
Express total revenue in terms of list price LL and solve for LL.
Total Revenue = 120L+40(0.80L)+40(0.50L)=120L+32L+20L=172L120L + 40(0.80L) + 40(0.50L) = 120L + 32L + 20L = 172L.
Setting 172L=10,320    L=10,320172=60172L = 10,320 \implies L = \frac{10,320}{172} = 60.
Summing revenue from all tiers and equating to target revenue yields the list price per item.
4
Calculate the markup percentage PP.
P=LCC×100=604040×100=50%P = \frac{L - C}{C} \times 100 = \frac{60 - 40}{40} \times 100 = 50\%.
Percentage markup is measured relative to the initial cost price per unit.

Anahtar Kavram

Multi-tiered inventory pricing and percentage markup calculation
Tahmini Süre:2m 30s
Soru 262Soru

The scores of a student on five quizzes are 72,85,90,78,72, 85, 90, 78, and 9595. What is the positive difference between the median and the arithmetic mean of these five quiz scores?

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Cevap: 1

Cevap

The positive difference between the median and the arithmetic mean of the five quiz scores is 11.
To find the arithmetic mean, sum all five scores (72+85+90+78+95=42072 + 85 + 90 + 78 + 95 = 420) and divide by 55, yielding 8484. To find the median, list the scores in ascending order (72,78,85,90,9572, 78, 85, 90, 95); the 3rd score is 8585. The positive difference between the median (8585) and the arithmetic mean (8484) is 11.

Adım Adım Çözüm

1
Calculate the arithmetic mean of the given set of scores.
The sum of the scores is 72+78+85+90+95=42072 + 78 + 85 + 90 + 95 = 420. Dividing by 5 gives an arithmetic mean of 8484.
The mean is defined as the total sum of all values divided by the number of values.
2
Order the scores in ascending order to identify the median.
The ordered list is 72,78,85,90,9572, 78, 85, 90, 95. The middle term (the 3rd value) is 8585.
The median of a set with an odd number of elements is the middle value when arranged in numerical order.
3
Subtract the arithmetic mean from the median to find the positive difference.
8584=185 - 84 = 1.
The question asks for the positive difference between the median and the arithmetic mean.

Anahtar Kavram

Calculating and comparing the arithmetic mean and median of a finite numerical data set.
Tahmini Süre:1m 0s
Soru 263Soru

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

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Cevap: 195

Cevap

The total number of employees across all three divisions was 195.
By setting up the weighted average equation based on employee counts, we establish that the Solutions division headcount equals the Express division headcount (L=E=60L = E = 60). Incorporating the Freight division headcount (F=1.25×60=75F = 1.25 \times 60 = 75) gives a total of 60+75+60=19560 + 75 + 60 = 195 employees.

Adım Adım Çözüm

1
Relate the number of employees in the Freight and Express divisions
F=1.25EF = 1.25E
The Freight division employs 25% more people than the Express division.
2
Set up the weighted average formula and solve for the relationship between Express (EE) and Solutions (LL) headcount
L=EL = E
Setting the overall average revenue per employee to 180,000yields180,000 yields 34.5E + 24L = 40.5E + 18L ,whichsimplifiesto, which simplifies to 6L = 6E$.
3
Calculate the exact employee count of the Solutions division
L=60L = 60
Dividing the total revenue of $14,400,000\$14,400,000 by the revenue per employee of $240,000\$240,000 gives 60 employees.
4
Determine employee counts for all divisions and calculate the total
Total employees = 60+75+60=19560 + 75 + 60 = 195
Since E=L=60E = L = 60 and F=1.25(60)=75F = 1.25(60) = 75, the total employee count is 60+75+60=19560 + 75 + 60 = 195.

Anahtar Kavram

Weighted Average in Combined Sets
Soru 264Soru

A ferry travels from Port Alpha to Port Beta across a bay, a distance of 3030 miles, at a constant speed of 1515 miles per hour. On the return trip from Port Beta to Port Alpha along the exact same route, the ferry travels at a constant speed of 3030 miles per hour. What is the ferry's average speed, in miles per hour, for the entire round trip?

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Cevap: 20

Cevap

The ferry's average speed for the entire round trip is 2020 miles per hour.
Average speed is found by dividing the total distance traveled by the total time taken. The round trip consists of two 30-mile legs, making the total distance 60 miles. The outbound leg takes 2 hours (30 / 15), while the return leg takes 1 hour (30 / 30), giving a total time of 3 hours. Dividing 60 miles by 3 hours yields an average speed of 20 miles per hour.

Adım Adım Çözüm

1
Calculate outbound trip duration
2 hours
Dividing distance by rate gives time: 30/15=230 / 15 = 2 hours.
2
Calculate return trip duration
1 hour
Dividing distance by rate gives time: 30/30=130 / 30 = 1 hour.
3
Calculate total distance and total time
60 miles in 3 hours
Total distance is 30+30=6030 + 30 = 60 miles, and total time is 2+1=32 + 1 = 3 hours.
4
Calculate overall average speed
20 mph
Average speed is total distance divided by total time: 60/3=2060 / 3 = 20 mph.

Anahtar Kavram

Average speed for any multi-leg or round-trip journey is defined as Total Distance divided by Total Time, rather than the simple arithmetic mean of the speeds.
Soru 265Soru

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

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Cevap: 3600

Cevap

3600 dollars
Applying the successive percent reductions sequentially gives $5,000×0.80=$4,000\$5,000 \times 0.80 = \$4,000 at the end of Year 1, and $4,000×0.90=$3,600\$4,000 \times 0.90 = \$3,600 at the end of Year 2.

Adım Adım Çözüm

1
Calculate the value after Year 1 depreciation
$4,000
Depreciation of 20% on $5,000 leaves 80% of the original value.
2
Calculate the value after Year 2 depreciation
$3,600
Depreciation of 10% on the new base of $4,000 leaves 90% of the Year 1 value.

Anahtar Kavram

Successive Percent Change
Soru 266Soru

For how many integer values of kk does the equation x2+x+4=kx+1|x - 2| + |x + 4| = kx + 1 have no real solutions for xx?

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Cevap: 4

Cevap

4
The equation has no real solutions when the line y = kx + 1 lies completely below the graph of f(x) = |x - 2| + |x + 4|. Analyzing the three piecewise regions of f(x) shows that f(x) = 6 on [-4, 2], with linear rays extending upward outside this interval. Setting the line to remain below f(x) forces -1.25 < k <= 2. The four integers in this range are -1, 0, 1, and 2.

Adım Adım Çözüm

1
Decompose the sum of absolute values into a piecewise linear function
f(x) = -2x - 2 for x < -4; f(x) = 6 for -4 <= x < 2; f(x) = 2x + 2 for x >= 2
The critical points x = -4 and x = 2 split the real number line into three intervals where each absolute value expression maintains a constant sign.
2
Set up conditions for zero intersections with the line g(x) = kx + 1
g(x) must remain strictly below f(x) for all x
Any intersection point between y = f(x) and y = g(x) corresponds to a real solution of the equation.
3
Evaluate boundary points and slope constraints for each interval
From the middle interval and right ray: k <= 2. From the left ray: k > -1.25.
For x >= 2, the line slope k cannot exceed the ray slope of 2. For x <= -4, g(-4) < 6 requires -4k + 1 < 6, which yields k > -1.25.
4
Determine the allowable interval for k and count integer values
-1.25 < k <= 2, giving integer values k in {-1, 0, 1, 2}
The integer values strictly inside (-1.25, 2] are -1, 0, 1, and 2, making a total of 4 integer values.

Anahtar Kavram

Piecewise analysis of absolute value functions and linear line intersection conditions
Soru 267Soru

Tank X initially contains 5050 liters of a solvent solution that is 20%20\% chemical compound by volume. Tank Y initially contains 3030 liters of a solvent solution that is 70%70\% of the same chemical compound by volume. First, vv liters of the solution in Tank X are transferred to Tank Y and thoroughly mixed. Then, vv liters of the resulting mixture in Tank Y are transferred back to Tank X. If the final concentration of the chemical compound in Tank X is 32%32\% by volume, what is the value of vv?

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Cevap: 20

Cevap

The value of vv is 20.
By setting up the conservation of solute equation for Tank X, the initial 1010 liters of compound minus the 0.20v0.20v liters transferred to Tank Y, plus the return portion v21+0.20v30+vv \cdot \frac{21 + 0.20v}{30 + v} equals the final 1616 liters of compound. Solving this relation yields v=20v = 20.

Adım Adım Çözüm

1
Calculate the initial volume of pure chemical compound in each tank.
Tank X initially contains 50×0.20=1050 \times 0.20 = 10 liters of compound. Tank Y initially contains 30×0.70=2130 \times 0.70 = 21 liters of compound.
Establishing initial solute quantities is essential for building the mass balance equations.
2
Determine the concentration of Tank Y after the first transfer of vv liters from Tank X.
Tank Y contains 30+v30 + v total liters of solution and 21+0.20v21 + 0.20v liters of compound. Its concentration becomes CY=21+0.20v30+vC_Y = \frac{21 + 0.20v}{30 + v}.
The solution transferred from Tank X carries a 20% concentration into Tank Y, altering Tank Y's volume and concentration.
3
Formulate the total compound equation for Tank X after transferring vv liters back from Tank Y.
Remaining compound in X after first transfer = 100.20v10 - 0.20v. Compound returned from Y = vCY=v21+0.20v30+vv \cdot C_Y = v \cdot \frac{21 + 0.20v}{30 + v}. Final compound in X = 50×0.32=1650 \times 0.32 = 16 liters.
Tank X ends with its original total volume (50 liters) at a 32% concentration.
4
Solve the algebraic equation for vv.
(100.20v)+21v+0.20v230+v=16(10 - 0.20v) + \frac{21v + 0.20v^2}{30 + v} = 16
0.20v+21v+0.20v230+v=6-0.20v + \frac{21v + 0.20v^2}{30 + v} = 6
Multiply by (30+v)(30 + v):
0.20v(30+v)+21v+0.20v2=6(30+v)-0.20v(30 + v) + 21v + 0.20v^2 = 6(30 + v)
6v0.20v2+21v+0.20v2=180+6v-6v - 0.20v^2 + 21v + 0.20v^2 = 180 + 6v
15v=180+6v    9v=180    v=2015v = 180 + 6v \implies 9v = 180 \implies v = 20
The quadratic terms cancel out cleanly, leaving a simple linear equation.

Anahtar Kavram

Two-Stage Transfer and Replacement Mixture Balance
Soru 268Soru

A research team needs to select a subcommittee of 3 scientists from a department containing 7 scientists. How many different 3-member subcommittees can be selected?

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Cevap: 35

Cevap

35 different subcommittees can be selected.
The total number of ways to choose a committee of 3 members from a group of 7 without regard to order is given by the combination formula C(7,3)=7×6×53×2×1=35C(7,3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.

Adım Adım Çözüm

1
Determine whether the selection depends on order.
Order does not matter since all 3 members of the subcommittee have equal roles.
When order does not matter in group selection, combinations (nCrnCr) must be used rather than permutations (nPrnPr).
2
Apply the combination formula C(n,k)=n!k!(nk)!C(n,k) = \frac{n!}{k!(n-k)!} with n=7n=7 and k=3k=3.
C(7,3)=7×6×53×2×1C(7,3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1}
This counts the unique groups of 3 that can be chosen from a pool of 7.
3
Calculate the numerical result.
35
Dividing 7×6×5=2107 \times 6 \times 5 = 210 by 3×2×1=63 \times 2 \times 1 = 6 yields 35.

Anahtar Kavram

Combinations and Group Selections
Soru 269Soru

In a certain training program, Class A has 1515 participants with an average score of 8080 on a final exam, and Class B has 2525 participants with an average score of 8888 on the same exam. What is the average score for all 4040 participants combined?

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Cevap: 85

Cevap

The combined average score for all 40 participants is 85.
To find the combined average of two groups of different sizes, sum the total values of both groups (15×80+25×88=3,40015 \times 80 + 25 \times 88 = 3,400) and divide by the total number of participants (15+25=4015 + 25 = 40). This gives a weighted average of 8585.

Adım Adım Çözüm

1
Find total score of Class A
1,200 points
Multiply the number of participants in Class A by their average score.
2
Find total score of Class B
2,200 points
Multiply the number of participants in Class B by their average score.
3
Find the total combined score
3,400 points
Add the total points from Class A and Class B together.
4
Calculate the weighted average score
85
Divide the total combined score by the total number of participants (15 + 25 = 40).

Anahtar Kavram

Weighted Average of Combined Sets
Tahmini Süre:1m 0s
Soru 270Soru

A financial consulting firm audited 500500 corporate investment portfolios for exposure to three alternative asset classes: Private Equity (PP), Venture Capital (VV), and Infrastructure (II). Every audited portfolio contained at least one of the three asset classes.

The audit revealed the following:
- Exactly 270270 portfolios contained Private Equity.
- Exactly 250250 portfolios contained Venture Capital.
- Exactly 220220 portfolios contained Infrastructure.
- The ratio of the number of portfolios containing both Private Equity and Venture Capital to those containing both Venture Capital and Infrastructure to those containing both Private Equity and Infrastructure was 11:9:1011 : 9 : 10, respectively.
- The number of portfolios containing all three asset classes was 6060.

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

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Cevap: 120

Cevap

The total number of portfolios containing exactly two of the three asset classes is 120.
Using the 3-set inclusion-exclusion formula, we establish that 500=270+250+220(11k+9k+10k)+60500 = 270 + 250 + 220 - (11k + 9k + 10k) + 60, which simplifies to 30k=30030k = 300, so k=10k = 10. The sum of the pairwise intersections is 110+90+100=300110 + 90 + 100 = 300. Since each pairwise intersection includes the 60 portfolios that contain all three asset classes, the number of portfolios containing exactly two asset classes is (11060)+(9060)+(10060)=50+30+40=120(110-60) + (90-60) + (100-60) = 50 + 30 + 40 = 120.

Adım Adım Çözüm

1
Set up the 3-Set Inclusion-Exclusion Principle equation
Total=P+V+I(PV+VI+PI)+PVI\text{Total} = |P| + |V| + |I| - (|P \cap V| + |V \cap I| + |P \cap I|) + |P \cap V \cap I|
This formula accounts for all region overlaps without double-counting or triple-counting.
2
Define variables using the given ratio for pairwise intersections
PV=11k|P \cap V| = 11k, VI=9k|V \cap I| = 9k, and PI=10k|P \cap I| = 10k
Expressing the pairwise intersection sizes with a common ratio constant kk allows us to substitute them into a single-variable linear equation.
3
Substitute the known numerical values and solve for kk
500=270+250+220(11k+9k+10k)+60    500=80030k    k=10500 = 270 + 250 + 220 - (11k + 9k + 10k) + 60 \implies 500 = 800 - 30k \implies k = 10
Using algebraic simplification to solve for kk gives the scaling factor for the pairwise intersections.
4
Calculate the actual sizes of the pairwise intersections
PV=110|P \cap V| = 110, VI=90|V \cap I| = 90, and PI=100|P \cap I| = 100; Sum of pairwise intersections = 300300
Multiplying each ratio term by k=10k=10 gives the exact count of portfolios containing at least the respective pairs.
5
Calculate portfolios containing exactly two asset classes
(11060)+(9060)+(10060)=50+30+40=120(110 - 60) + (90 - 60) + (100 - 60) = 50 + 30 + 40 = 120
The triple intersection (6060) must be subtracted from each pairwise overlap to isolate the regions corresponding to portfolios holding exactly two asset classes.

Anahtar Kavram

Three-Set Overlapping Venn Diagrams and Inclusion-Exclusion Principle
Tahmini Süre:2m 0s
Soru 271Soru

Two automated document redaction pipelines, Pipeline AA and Pipeline BB, process digital legal archives. Working alone at its constant rate, Pipeline AA can redact a standard archive in 88 hours. Working alone at its constant rate, Pipeline BB can redact the same standard archive in 1212 hours. Both pipelines begin processing a standard archive together. After 33 hours of joint operation, Pipeline AA encounters a system error and stops. Pipeline BB continues working alone at its constant rate until the entire archive is redacted. How many total hours does it take from start to finish to complete the redaction of the archive?

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Cevap: 7.5

Cevap

The total time required to redact the entire archive from start to finish is 7.5 hours.
To find the total time required, first determine the combined rate of Pipeline A and Pipeline B: 18+112=524\frac{1}{8} + \frac{1}{12} = \frac{5}{24} archive per hour. In the first 3 hours of joint work, they complete 3×524=583 \times \frac{5}{24} = \frac{5}{8} of the job, leaving 158=381 - \frac{5}{8} = \frac{3}{8} of the archive remaining. Pipeline B processes the remaining 38\frac{3}{8} at its individual rate of 112\frac{1}{12} per hour, taking 3/81/12=4.5\frac{3/8}{1/12} = 4.5 hours. Adding the initial 3 hours to the 4.5 additional hours yields a total of 7.5 hours.

Adım Adım Çözüm

1
Determine the individual work rate for each pipeline.
Pipeline A's rate is 18\frac{1}{8} archive per hour; Pipeline B's rate is 112\frac{1}{12} archive per hour.
Work rate is the fraction of job completed per hour (1/time required alone1 / \text{time required alone}).
2
Calculate the combined work rate during joint operation.
Combined rate is 18+112=524\frac{1}{8} + \frac{1}{12} = \frac{5}{24} archive per hour.
When entities work concurrently, their rates are added together.
3
Determine the amount of work completed in the first 3 hours.
Work completed is 3×524=583 \times \frac{5}{24} = \frac{5}{8} of the total archive.
Work done equals rate multiplied by time.
4
Find the fraction of the archive remaining to be redacted.
Remaining work is 158=381 - \frac{5}{8} = \frac{3}{8} of the archive.
The full task represents 1 complete unit of work.
5
Calculate how long Pipeline B needs to finish the remaining work alone.
Solo time for Pipeline B is 3/81/12=4.5\frac{3/8}{1/12} = 4.5 hours.
Time equals remaining work divided by Pipeline B's individual rate.
6
Sum the joint operation time and the solo operation time.
Total time is 3+4.5=7.53 + 4.5 = 7.5 hours.
The overall duration consists of two sequential periods: 3 hours working together plus 4.5 hours for Pipeline B working alone.

Anahtar Kavram

Combined Work Rates and Multi-Stage Work Scenarios
Soru 272Soru

A logistics company operates two models of cargo trucks, Model X and Model Y, to transport freight between two distribution hubs. Model X consumes 1010 liters of fuel per 100100 kilometers driven in city traffic and 66 liters of fuel per 100100 kilometers driven on open highway. Model Y consumes 1414 liters of fuel per 100100 kilometers in city traffic and 88 liters per 100100 kilometers on open highway. On a trip between the two hubs along a route composed entirely of city traffic segments and open highway segments, Model X consumed a total of 4242 liters of fuel, and Model Y consumed a total of 5858 liters of fuel. What is the total length, in kilometers, of the open highway segments along this route?

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Cevap: 200

Cevap

The total length of the open highway segments along the route is 200 kilometers.
By converting the per-100-kilometer fuel consumption into per-kilometer rates, we obtain two linear equations in two variables: 0.10C+0.06H=420.10C + 0.06H = 42 and 0.14C+0.08H=580.14C + 0.08H = 58. Solving this system by elimination yields H=200H = 200 kilometers for the open highway distance.

Adım Adım Çözüm

1
Define variables for city traffic distance and open highway distance.
Let CC equal the distance in km of city traffic segments, and HH equal the distance in km of open highway segments.
Establishing explicit variables allows the translation of the word problem context into algebraic equations.
2
Convert consumption rates to liters per kilometer and formulate total fuel equations for both trucks.
Model X equation: 0.10C+0.06H=420.10C + 0.06H = 42; Model Y equation: 0.14C+0.08H=580.14C + 0.08H = 58.
Dividing the consumption per 100 km by 100 gives the unit rates per km for city and highway driving.
3
Simplify the system of linear equations by eliminating decimals and common factors.
First equation: 5C+3H=21005C + 3H = 2100; Second equation: 7C+4H=29007C + 4H = 2900.
Simplifying equations reduces computation errors during elimination.
4
Solve the system using elimination to isolate the highway distance variable HH.
Multiplying the equations to match coefficients of CC (35C+21H=1470035C + 21H = 14700 and 35C+20H=1450035C + 20H = 14500) and subtracting gives H=200H = 200.
Eliminating CC directly yields the target quantity HH without needing additional substitution steps.

Anahtar Kavram

Formulating and Solving Systems of Linear Equations from Multi-Condition Word Problems
Tahmini Süre:2m 0s
Soru 273Soru

A logistics company operates a fleet consisting of standard electric delivery vans and high-capacity electric delivery vans. A standard van consumes a fixed 2.52.5 kilowatt-hours (kWh) of electricity per delivery route, while a high-capacity van consumes 50%50\% more electricity per route than a standard van. On a given day, the fleet completed a total of 4040 delivery routes and consumed a total of 130130 kWh of electricity. How many delivery routes were completed by high-capacity vans?

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Cevap: 24

Cevap

The total number of delivery routes completed by high-capacity vans is 24.
By modeling the relationship using the system S+H=40S + H = 40 and 2.5S+3.75H=1302.5S + 3.75H = 130, substituting S=40HS = 40 - H gives 100+1.25H=130100 + 1.25H = 130, which resolves to H=24H = 24 high-capacity van routes.

Adım Adım Çözüm

1
Calculate the power consumption per route for a high-capacity van.
High-capacity rate = 2.5×1.5=3.752.5 \times 1.5 = 3.75 kWh per route.
High-capacity vans consume 50%50\% more energy than standard vans.
2
Formulate a system of linear equations representing total routes and total electricity used.
S+H=40S + H = 40 and 2.5S+3.75H=1302.5S + 3.75H = 130, where SS represents standard van routes and HH represents high-capacity van routes.
The sum of routes equals 40 and the combined energy consumption equals 130 kWh.
3
Solve for the target variable HH.
Substitute S=40HS = 40 - H into the energy equation: 2.5(40H)+3.75H=130    100+1.25H=130    1.25H=30    H=242.5(40 - H) + 3.75H = 130 \implies 100 + 1.25H = 130 \implies 1.25H = 30 \implies H = 24.
Substituting SS yields a single-variable linear equation for HH.

Anahtar Kavram

Algebraic Modeling of Systems of Linear Equations
Tahmini Süre:1m 30s
Soru 274Soru

Eight executive team members—3 vice presidents, 3 directors, and 2 managers—are to stand in a single line for a company photograph. If no two vice presidents can stand next to each other, and the 2 managers must stand next to each other, how many different linear arrangements of the eight team members are possible?

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Cevap: 2880

Cevap

2,880
To satisfy both constraints simultaneously, first treat the 2 managers as a single block, which has 2!=22! = 2 internal arrangements. Next, arrange the non-vice-president units—consisting of the 3 distinct directors and the 1 manager block—giving 4 units total, which can be ordered in 4!=244! = 24 ways. These 4 units create 5 distinct slot locations (at the ends and between adjacent units). To ensure no two vice presidents are adjacent, place the 3 distinct vice presidents into 3 of these 5 slots in P(5,3)=60P(5, 3) = 60 ways. Multiplying these independent decisions yields 2×24×60=2,8802 \times 24 \times 60 = 2,880.

Adım Adım Çözüm

1
Group the elements that must remain adjacent and calculate internal permutations
The 2 managers form 1 block with 2!=22! = 2 internal arrangements.
Since the 2 managers must stand next to each other, treating them as a single block ensures they are never separated.
2
Arrange all non-restricted base units in a line
The 3 directors and 1 manager block yield 4!=244! = 24 linear arrangements.
Establishing the sequence of non-vice-president units creates the fixed slots into which the vice presidents will later be inserted.
3
Calculate the available slot arrangements for the separated elements
4 base units create 5 available slots. Permuting 3 vice presidents into 5 slots yields P(5,3)=60P(5, 3) = 60 ways.
Placing at most one vice president per slot guarantees that no two vice presidents are placed adjacently.
4
Apply the Fundamental Counting Principle to determine total arrangements
Total arrangements = 2×24×60=2,8802 \times 24 \times 60 = 2,880.
The choices for internal block arrangement, base unit ordering, and slot placement are independent sequential events.

Anahtar Kavram

Linear Permutations with Simultaneous Grouping and Non-Adjacency Constraints
Soru 275Soru

A board of directors consists of 77 members. In how many different ways can a subcommittee of 33 members be chosen from the board?

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Cevap: 35

Cevap

35 different subcommittees can be chosen.
The total number of ways to choose a subcommittee of 3 members from a group of 7 members when order does not matter is given by the combination formula (73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.

Adım Adım Çözüm

1
Determine whether the selection requires permutations or combinations.
Since selecting members A, B, and C forms the exact same committee as selecting B, C, and A, the order of selection does not matter. Therefore, this is a combination problem.
Committees are unordered groups.
2
Calculate the number of combinations of 7 items taken 3 at a time using (73)=7!3!(73)!\binom{7}{3} = \frac{7!}{3!(7-3)!}.
(73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35.
Simplify the factorial expression by canceling out common terms.

Anahtar Kavram

Combinations (Selection Without Regard to Order)
Soru 276Soru

A dataset consists of 20 distinct numerical values arranged in increasing order. The lower quartile (Q1Q_1) of the dataset is equal to the average of the 5th and 6th values, and the upper quartile (Q3Q_3) is equal to the average of the 15th and 16th values. Given that the lower quartile Q1=42Q_1 = 42, the interquartile range (IQR=Q3Q1\text{IQR} = Q_3 - Q_1) is 38, and the 15th value in the dataset is 74, what is the value of the 16th term?

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Cevap: 86

Cevap

86
To find the 16th value, first determine the upper quartile Q3Q_3 using the interquartile range: Q3=Q1+IQR=42+38=80Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80. Since Q3Q_3 is the average of the 15th and 16th terms, set up the equation 74+x162=80\frac{74 + x_{16}}{2} = 80. Multiplying by 2 gives 74+x16=16074 + x_{16} = 160, so x16=86x_{16} = 86.

Adım Adım Çözüm

1
Calculate the upper quartile (Q3Q_3) using the given lower quartile (Q1Q_1) and interquartile range (IQR).
Q3=Q1+IQR=42+38=80Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80
By definition, the interquartile range is the difference between the upper and lower quartiles (IQR=Q3Q1\text{IQR} = Q_3 - Q_1).
2
Relate the upper quartile (Q3Q_3) to the 15th and 16th values of the ordered dataset.
Q3=x15+x162=80Q_3 = \frac{x_{15} + x_{16}}{2} = 80
For an ordered dataset of 20 elements, Q3Q_3 is the median of the upper half of the data (the 11th through 20th terms), which equals the arithmetic mean of the 15th and 16th terms.
3
Substitute the known value of the 15th term (x15=74x_{15} = 74) into the equation and solve for the 16th term (x16x_{16}).
\frac{74 + x_{16}}{2} = 80 \implies 74 + x_{16} = 160 \implies x_{16} = 86
Multiplying both sides by 2 gives 160, and subtracting 74 yields 86.

Anahtar Kavram

Interquartile Range and Quartile Calculations
Soru 277Soru

Set SS consists of five distinct integers: 14,22,18,9,14, 22, 18, 9, and 3131. What is the range of the numbers in Set SS?

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Cevap: 22

Cevap

The range of the numbers in Set SS is 2222.
To find the range of a set of numbers, subtract the minimum value from the maximum value in the set. For Set S={14,22,18,9,31}S = \{14, 22, 18, 9, 31\}, the maximum value is 3131 and the minimum value is 99. Subtracting the minimum from the maximum gives 319=2231 - 9 = 22.

Adım Adım Çözüm

1
Identify the maximum and minimum elements in the given set.
The maximum element is 3131 and the minimum element is 99.
The range of a dataset is defined as the difference between its greatest and least values.
2
Subtract the minimum value from the maximum value.
319=2231 - 9 = 22.
Applying the formula Range=MaximumMinimum\text{Range} = \text{Maximum} - \text{Minimum} yields the range of the set.

Anahtar Kavram

Range of a Numerical Data Set
Soru 278Soru

A dataset contains 2020 distinct test scores arranged in ascending order:

10,12,15,17,20,22,25,28,30,32,35,38,40,43,45,48,50,52,55,6010, 12, 15, 17, 20, 22, 25, 28, 30, 32, 35, 38, 40, 43, 45, 48, 50, 52, 55, 60

If the pthp\text{th} percentile of a dataset of NN values is defined as the value at position k=p100×Nk = \frac{p}{100} \times N when ordered from least to greatest, what is the 75th75\text{th} percentile of these test scores?

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Cevap: 45

Cevap

The 75th percentile of the given test scores is 45.
The 75th percentile corresponds to position k = (75/100) * 20 = 15 in the ordered list of 20 scores, which is equal to 45.

Adım Adım Çözüm

1
Calculate the position index kk for the 75th75\text{th} percentile.
k=75100×20=15k = \frac{75}{100} \times 20 = 15
The rank formula determines the 1-based index of the target percentile value in an ordered set of size N=20N = 20.
2
Locate the 15th15\text{th} score in the ordered dataset.
The 15th15\text{th} score is 4545.
Counting from the lowest score (1010 at position 1), the 15th15\text{th} score in the sequence is 4545.

Anahtar Kavram

Percentiles and Quartiles
Soru 279Soru

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

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Cevap: 37.5

Cevap

37.5%
Using the given relationship WX=2WYW_X = 2W_Y and total weight constraint WX+WY+WZ=1W_X + W_Y + W_Z = 1, Class Z's weight can be written as 13WY1 - 3W_Y. Plugging these into the overall weighted return formula yields 12(2WY)+18WY+6(13WY)=1112(2W_Y) + 18W_Y + 6(1 - 3W_Y) = 11, which solves to WY=5/24W_Y = 5/24. Substituting back gives WZ=9/24=37.5%W_Z = 9/24 = 37.5\%.

Adım Adım Çözüm

1
Define fractional weight variables for each asset class.
Let WXW_X, WYW_Y, and WZW_Z be the proportions of total capital invested in Class X, Class Y, and Class Z, such that WX+WY+WZ=1W_X + W_Y + W_Z = 1.
Establishing a standard framework for weighted average calculation.
2
Express WXW_X and WZW_Z in terms of WYW_Y.
WX=2WYW_X = 2W_Y, so 2WY+WY+WZ=1    WZ=13WY2W_Y + W_Y + W_Z = 1 \implies W_Z = 1 - 3W_Y.
Reducing the number of unknown variables to one.
3
Formulate the weighted average equation using component returns.
12WX+18WY+6WZ=11    12(2WY)+18WY+6(13WY)=1112W_X + 18W_Y + 6W_Z = 11 \implies 12(2W_Y) + 18W_Y + 6(1 - 3W_Y) = 11.
The fund's overall return equals the weighted sum of individual returns.
4
Solve the algebraic equation for WYW_Y.
24WY+18WY+618WY=11    24WY+6=11    24WY=5    WY=52424W_Y + 18W_Y + 6 - 18W_Y = 11 \implies 24W_Y + 6 = 11 \implies 24W_Y = 5 \implies W_Y = \frac{5}{24}.
Determining the exact weight of Class Y.
5
Calculate the weight percentage for Class Z.
WZ=13(524)=11524=924=38=0.375 or 37.5%W_Z = 1 - 3\left(\frac{5}{24}\right) = 1 - \frac{15}{24} = \frac{9}{24} = \frac{3}{8} = 0.375 \text{ or } 37.5\%.
Evaluating the target component weight requested by the question.

Anahtar Kavram

Weighted Average in Combined Financial Assets
Soru 280Soru

A merchant blends two types of coffee beans: Type A, which costs $12\$12 per kilogram and contains 15%15\% caffeine by weight, and Type B, which costs $18\$18 per kilogram and contains 25%25\% caffeine by weight. The merchant mixes Type A and Type B to create a 6060-kilogram batch of Blend C. If the total cost of the 6060-kilogram batch of Blend C is $936\$936, what is the percentage of caffeine by weight in Blend C?

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Cevap: 21

Cevap

The percentage of caffeine by weight in Blend C is 21%.
Solving the linear system for total mass (A+B=60A + B = 60) and total cost (12A+18B=93612A + 18B = 936) reveals that Blend C consists of 24 kg of Type A beans and 36 kg of Type B beans. Type A contributes 24×0.15=3.624 \times 0.15 = 3.6 kg of caffeine, and Type B contributes 36×0.25=9.036 \times 0.25 = 9.0 kg of caffeine. The total caffeine mass of 12.6 kg out of 60 kg total mass equals a concentration of 12.6/60=21%12.6 / 60 = 21\%.

Adım Adım Çözüm

1
Formulate linear system for bean masses based on total weight and cost
A+B=60A + B = 60 and 12A+18B=93612A + 18B = 936, solving to A=24A = 24 kg and B=36B = 36 kg
The weight of each constituent bean type must be determined before total caffeine content can be evaluated.
2
Calculate caffeine mass from each component
Type A provides 3.63.6 kg (24×0.1524 \times 0.15) and Type B provides 9.09.0 kg (36×0.2536 \times 0.25)
Solute mass is found by multiplying total component weight by its percentage concentration.
3
Sum caffeine masses to find total caffeine in the mixture
Total caffeine = 3.6+9.0=12.63.6 + 9.0 = 12.6 kg
The total amount of caffeine present in Blend C is the combined sum from both bean types.
4
Determine caffeine concentration percentage of Blend C
12.660×100=21%\frac{12.6}{60} \times 100 = 21\%
Concentration is calculated as the ratio of total solute mass to total solution mass expressed as a percentage.

Anahtar Kavram

Weighted concentration combined with multi-variable cost systems
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