Tüm alıştırma soruları

2195 soru

Soru 101Soru

An artisanal coffee roaster creates a custom blend using Arabica, Robusta, and Liberica beans in the initial ratio of 5:3:25 : 3 : 2 by weight, respectively. To adjust the flavor profile, the roaster adds 6060 kilograms of Arabica beans and 2020 kilograms of Robusta beans to the blend, while leaving the amount of Liberica beans unchanged. If the new ratio of Arabica to Robusta to Liberica beans is 4:2:14 : 2 : 1, respectively, what was the initial total weight, in kilograms, of the blend?

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Cevap: 200

Cevap

200 kilograms
Let the initial weights of Arabica, Robusta, and Liberica beans be 5x5x, 3x3x, and 2x2x kilograms, respectively. The initial total weight is 5x+3x+2x=10x5x + 3x + 2x = 10x kg. After adding 6060 kg of Arabica and 2020 kg of Robusta, the new weights are 5x+605x + 60, 3x+203x + 20, and 2x2x. Using the new ratio of Robusta to Liberica (2:12 : 1), we set up the equation 3x+202x=2\frac{3x + 20}{2x} = 2, which simplifies to 3x+20=4x3x + 20 = 4x, giving x=20x = 20. Substituting x=20x = 20 into the initial total weight expression yields 10(20)=20010(20) = 200 kg.

Adım Adım Çözüm

1
Define initial quantities using a common multiplier xx.
Arabica =5x= 5x, Robusta =3x= 3x, Liberica =2x= 2x. Initial total weight =5x+3x+2x=10x= 5x + 3x + 2x = 10x.
Expressing ratio terms in terms of a single multiplier allows algebraic modeling of changes to each component.
2
Express the new quantities after adding the specified weights.
New Arabica =5x+60= 5x + 60, New Robusta =3x+20= 3x + 20, New Liberica =2x= 2x.
Arabica and Robusta amounts increase by 60 kg and 20 kg respectively, while Liberica remains constant.
3
Set up a proportion using the new ratio of Robusta to Liberica (2:12 : 1).
3x+202x=21    3x+20=4x    x=20\frac{3x + 20}{2x} = \frac{2}{1} \implies 3x + 20 = 4x \implies x = 20.
Equating the ratio of the updated Robusta and Liberica quantities to 2:12:1 yields the multiplier xx.
4
Verify consistency with the Arabica to Liberica ratio (4:14 : 1).
5(20)+602(20)=16040=41\frac{5(20) + 60}{2(20)} = \frac{160}{40} = \frac{4}{1}, which matches the given final ratio.
Ensures that x=20x = 20 satisfies the complete 3-part ratio of 4:2:14 : 2 : 1.
5
Calculate the initial total weight.
Initial total =10x=10×20=200= 10x = 10 \times 20 = 200 kg.
Multiplying the initial sum of ratio parts (10) by x=20x = 20 gives the original total blend weight.

Anahtar Kavram

Ratio and Proportion Word Problems
Soru 102Soru

A municipal water treatment facility utilizes three primary purification units—Unit P, Unit Q, and Unit R—to process large water batches. Operating alone at its constant rate, Unit P can process a full batch in 1212 hours, Unit Q can process a full batch in 1818 hours, and Unit R can process a full batch in 3636 hours. All three units begin processing a batch together. After 33 hours, Unit P stops operating due to scheduled maintenance. Unit Q and Unit R continue operating together for an additional 22 hours, after which Unit Q is shut down. Unit R then completes the remainder of the batch working alone. How many total hours does it take from start to finish to process the complete batch?

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Cevap: 17

Cevap

The total time required to process the complete batch from start to finish is 17 hours.
To find the total time from start to finish, analyze the work completed in each of the three stages:

1. In Stage 1 (3 hours), all three units operate together. Their combined rate is 112+118+136=3+2+136=636=16\frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{3 + 2 + 1}{36} = \frac{6}{36} = \frac{1}{6} batch per hour. In 33 hours, they complete 3×16=123 \times \frac{1}{6} = \frac{1}{2} of the total batch.

2. In Stage 2 (2 hours), Unit P stops and Units Q and R continue together. Their combined rate is 118+136=2+136=336=112\frac{1}{18} + \frac{1}{36} = \frac{2 + 1}{36} = \frac{3}{36} = \frac{1}{12} batch per hour. In 22 hours, they complete 2×112=162 \times \frac{1}{12} = \frac{1}{6} of the total batch.

3. After Stage 2, the total fraction of work completed is 12+16=46=23\frac{1}{2} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3} of the batch. The remaining fraction of work is 123=131 - \frac{2}{3} = \frac{1}{3} of the batch.

4. In Stage 3, Unit R completes the remaining 13\frac{1}{3} batch alone at its rate of 136\frac{1}{36} batch per hour. The time taken by Unit R is 1/31/36=12\frac{1/3}{1/36} = 12 hours.

Adding the durations of all three stages gives 3+2+12=173 + 2 + 12 = 17 total hours.

Adım Adım Çözüm

1
Calculate individual work rates
Rate of P = 112\frac{1}{12} batch/hr, Rate of Q = 118\frac{1}{18} batch/hr, Rate of R = 136\frac{1}{36} batch/hr
Work rate is the reciprocal of the total time required to complete one entire job.
2
Calculate work done during Stage 1 (first 3 hours)
Combined rate of P, Q, R = 112+118+136=636=16\frac{1}{12} + \frac{1}{18} + \frac{1}{36} = \frac{6}{36} = \frac{1}{6} batch/hr. Work done = 3×16=123 \times \frac{1}{6} = \frac{1}{2} batch
All three units operate simultaneously for 3 hours.
3
Calculate work done during Stage 2 (next 2 hours)
Combined rate of Q and R = 118+136=336=112\frac{1}{18} + \frac{1}{36} = \frac{3}{36} = \frac{1}{12} batch/hr. Work done = 2×112=162 \times \frac{1}{12} = \frac{1}{6} batch
Unit P stops working, leaving Q and R to operate together for 2 hours.
4
Determine remaining work and time required for Stage 3
Remaining work = 1(12+16)=131 - \left(\frac{1}{2} + \frac{1}{6}\right) = \frac{1}{3} batch. Time for R alone = 1/31/36=12\frac{1/3}{1/36} = 12 hours
Unit R must complete the remaining fraction of the batch operating alone at its constant rate.
5
Calculate total elapsed time
3 hours+2 hours+12 hours=17 hours3\text{ hours} + 2\text{ hours} + 12\text{ hours} = 17\text{ hours}
Sum the durations of all three distinct operational stages.

Anahtar Kavram

Combined Work Rate in Multi-Stage Work Scenarios
Soru 103Soru

If mm and nn are non-zero real numbers such that mm and nn are the roots of the quadratic equation x2+mx+n=0x^2 + mx + n = 0, what is the value of mnm - n?

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Cevap: 3

Cevap

3
The correct answer is 3. By Vieta's formulas, for x2+mx+n=0x^2 + mx + n = 0, the sum of roots m+n=mm + n = -m gives 2m+n=02m + n = 0, and the product of roots mn=nm \cdot n = n gives m=1m = 1 since n0n \neq 0. Substituting m=1m = 1 into 2m+n=02m + n = 0 yields n=2n = -2. Thus, mn=1(2)=3m - n = 1 - (-2) = 3.

Adım Adım Çözüm

1
Apply Vieta's formulas to express the sum and product of roots in terms of coefficients.
For the quadratic equation x2+mx+n=0x^2 + mx + n = 0, the sum of roots is m+n=mm + n = -m, and the product of roots is mn=nm \cdot n = n.
By Vieta's relations for a standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0, the sum of roots is ba-\frac{b}{a} and the product is ca\frac{c}{a}.
2
Solve the product relation for mm.
Since n0n \neq 0, divide both sides of mn=nm \cdot n = n by nn to get m=1m = 1.
Because nn is specified to be non-zero, division by nn is valid and yields a unique non-zero value for mm.
3
Substitute m=1m = 1 into the sum relation to determine nn.
1+n=1    n=21 + n = -1 \implies n = -2.
Rearranging m+n=mm + n = -m gives 2m+n=02m + n = 0, so n=2(1)=2n = -2(1) = -2.
4
Calculate the target value mnm - n.
mn=1(2)=3m - n = 1 - (-2) = 3.
Subtracting 2-2 from 11 produces 1+2=31 + 2 = 3.

Anahtar Kavram

Vieta's Formulas and Root-Coefficient Relationships in Quadratic Equations
Tahmini Süre:2m 0s
Soru 104Soru

A clinical study evaluated 150150 patient records for the presence of three specific health biomarkers: Biomarker X, Biomarker Y, and Biomarker Z. The study revealed that 7070 patients had Biomarker X, 6060 patients had Biomarker Y, and 5050 patients had Biomarker Z. Exactly 3535 patients had exactly two of these biomarkers, and exactly 1010 patients had all three biomarkers. How many of the patients evaluated had none of the three biomarkers?

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Cevap: 25

Cevap

25 patients evaluated had none of the three biomarkers.
By the inclusion-exclusion principle for three sets, the total number of elements in at least one set is given by XYZ=X+Y+Z(Elements in exactly two sets)2(Elements in all three sets)|X \cup Y \cup Z| = |X| + |Y| + |Z| - (\text{Elements in exactly two sets}) - 2(\text{Elements in all three sets}). Substituting the given values yields 70+60+50352(10)=1803520=12570 + 60 + 50 - 35 - 2(10) = 180 - 35 - 20 = 125. Since there are 150150 total patients evaluated, the number of patients with none of the three biomarkers is 150125=25150 - 125 = 25.

Adım Adım Çözüm

1
Sum the individual counts for all three biomarker groups.
Sum of individual sets=70+60+50=180\text{Sum of individual sets} = 70 + 60 + 50 = 180
This sum counts individuals with 1 biomarker once, individuals with 2 biomarkers twice, and individuals with 3 biomarkers three times.
2
Apply the inclusion-exclusion formula adapted for 'exactly two' and 'all three' overlapping subsets.
At least one biomarker=180(Exactly two)2×(All three)=180352(10)=125\text{At least one biomarker} = 180 - (\text{Exactly two}) - 2 \times (\text{All three}) = 180 - 35 - 2(10) = 125
To count each person with at least one biomarker exactly once, we subtract the count of patients with exactly two biomarkers once, and the count of patients with all three biomarkers twice.
3
Subtract the number of patients with at least one biomarker from the total population of patients evaluated.
None=150125=25\text{None} = 150 - 125 = 25
The total group consists of patients with at least one biomarker plus patients with none of the three biomarkers.

Anahtar Kavram

Three-Set Overlapping Sets and Inclusion-Exclusion Principle
Tahmini Süre:1m 45s
Soru 105Soru

If 2x5=9|2x - 5| = 9, what is the sum of all possible values of xx?

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Cevap: 55

Cevap

The sum of all possible values of xx is 55.
To solve 2x5=9|2x - 5| = 9, split the equation into two linear cases: 2x5=92x - 5 = 9 and 2x5=92x - 5 = -9. Solving 2x5=92x - 5 = 9 gives 2x=142x = 14, so x=7x = 7. Solving 2x5=92x - 5 = -9 gives 2x=42x = -4, so x=2x = -2. Summing these two solutions gives 7+(2)=57 + (-2) = 5.

Adım Adım Çözüm

1
Set up the two equations based on the definition of absolute value.
2x5=92x - 5 = 9 and 2x5=92x - 5 = -9
For any expression AA and real number B0B \ge 0, A=B|A| = B implies A=BA = B or A=BA = -B.
2
Solve the first equation for xx.
2x=14    x=72x = 14 \implies x = 7
Add 55 to both sides and divide by 22.
3
Solve the second equation for xx.
2x=4    x=22x = -4 \implies x = -2
Add 55 to both sides and divide by 22.
4
Calculate the sum of all valid solutions.
7+(2)=57 + (-2) = 5
The question asks for the sum of all possible solutions for xx.

Anahtar Kavram

Solving Absolute Value Equations
Soru 106Soru
What is the product of all real solutions to the equation (x26x+10)22(x26x+10)35=0?(x^2 - 6x + 10)^2 - 2(x^2 - 6x + 10) - 35 = 0?
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Cevap: 3

Cevap

The product of all real solutions to the equation is 3.
Substituting u=x26x+10u = x^2 - 6x + 10 converts the given equation into u22u35=0u^2 - 2u - 35 = 0, which factors as (u7)(u+5)=0(u - 7)(u + 5) = 0. Substituting back yields two quadratic equations: x26x+3=0x^2 - 6x + 3 = 0 and x26x+15=0x^2 - 6x + 15 = 0. Checking the discriminants reveals that x26x+3=0x^2 - 6x + 3 = 0 has D=24>0D = 24 > 0, producing two real roots with product ca=3\frac{c}{a} = 3, whereas x26x+15=0x^2 - 6x + 15 = 0 has D=24<0D = -24 < 0, producing no real roots. Therefore, the product of all real solutions is 3.

Adım Adım Çözüm

1
Substitute a dummy variable to simplify the equation structure.
Let u=x26x+10u = x^2 - 6x + 10. The equation becomes u22u35=0u^2 - 2u - 35 = 0.
Recognizing the repeated quadratic expression allows transforming a fourth-degree equation into a standard quadratic form.
2
Factor the quadratic equation in terms of uu.
(u7)(u+5)=0    u=7 or u=5(u - 7)(u + 5) = 0 \implies u = 7 \text{ or } u = -5.
Finding the values of uu establishes the intermediate equations for xx.
3
Analyze the first case u=7u = 7 for real solutions.
x26x+3=0x^2 - 6x + 3 = 0. Discriminant D1=3612=24>0D_1 = 36 - 12 = 24 > 0. Product of real roots is 31=3\frac{3}{1} = 3.
A positive discriminant guarantees two real roots, and Vieta's formulas give their product directly.
4
Analyze the second case u=5u = -5 for real solutions.
x26x+15=0x^2 - 6x + 15 = 0. Discriminant D2=3660=24<0D_2 = 36 - 60 = -24 < 0. No real roots.
A negative discriminant indicates complex conjugate roots, which must be excluded when finding the product of real solutions.
5
Combine results to find the final product of all real solutions.
Product = 3.
Only the two roots from the first case are real, so their product is the product of all real solutions.

Anahtar Kavram

Solving disguised quadratics via algebraic substitution and testing discriminants to filter out non-real roots before applying Vieta's formulas.
Soru 107Soru

A manufacturing plant has a fixed daily setup cost of $1,500\$1,500. For a specific product, the production cost for each of the first 5050 units is $60\$60 per unit, while each additional unit produced beyond the first 5050 costs $30\$30 per unit. If the average total cost per unit produced on a given day was $45\$45, how many total units were produced on that day?

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Cevap: 200

Cevap

The total number of units produced on that day was 200.
Setting up the piecewise linear total cost expression C(n)=1,500+50(60)+(n50)30=3,000+30nC(n) = 1,500 + 50(60) + (n-50)30 = 3,000 + 30n and equating average cost 3,000+30nn\frac{3,000 + 30n}{n} to 4545 yields 15n=3,00015n = 3,000, which solves to n=200n = 200.

Adım Adım Çözüm

1
Model total daily production cost as a piecewise linear algebraic equation.
C(n)=1,500+(50×60)+(n50)×30=3,000+30nC(n) = 1,500 + (50 \times 60) + (n - 50) \times 30 = 3,000 + 30n for n>50n > 50.
Total cost combines fixed setup fees, cost of the initial 50 units, and tier-2 cost for units exceeding 50.
2
Formulate the equation for average cost per unit.
C(n)n=3,000+30nn=45\frac{C(n)}{n} = \frac{3,000 + 30n}{n} = 45
Average cost is total daily cost divided by total quantity produced, given as $45 per unit.
3
Solve the algebraic equation for n.
3,000+30n=45n    15n=3,000    n=2003,000 + 30n = 45n \implies 15n = 3,000 \implies n = 200
Isolating n yields the exact volume of units needed to satisfy the average cost target.

Anahtar Kavram

Algebraic Equation Modeling with Piecewise Cost Functions
Tahmini Süre:2m 0s
Soru 108Soru

Four distinct books—two mathematics books and two history books—are to be arranged in a single line on a shelf. If the two mathematics books must stand next to each other, in how many different linear arrangements can the four books be placed?

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Cevap: 12

Cevap

12 linear arrangements
To find the total number of linear arrangements where the two mathematics books are adjacent, treat the two mathematics books as a single combined block. This leaves 3 items to arrange (the mathematics block and the 2 individual history books), which can be arranged in 3!=63! = 6 ways. Within the block, the 2 mathematics books can be ordered in 2!=22! = 2 ways. Multiplying the independent arrangements yields 6×2=126 \times 2 = 12 total arrangements.

Adım Adım Çözüm

1
Group the adjacent items into a single unit and count total units to arrange
3 units (1 grouped math unit + 2 individual history books), which can be arranged in 3!=63! = 6 ways
Treating items that must be adjacent as a single block simplifies the linear arrangement into distinct available positions.
2
Determine internal permutations within the grouped unit
2!=22! = 2 ways to order the two mathematics books inside their block
The mathematics books can switch positions with each other within their designated block.
3
Multiply external and internal arrangements
6×2=126 \times 2 = 12 total linear arrangements
By the Fundamental Counting Principle, total arrangements equal the product of unit arrangements and internal arrangements.

Anahtar Kavram

Linear Permutations with Adjacent Restrictions (Tie-Together Method)
Soru 109Soru

In an effort to reduce operating expenses, a major maritime freight carrier replaced its entire fleet of traditional steel shipping containers with newly engineered carbon-fiber containers that are 25 percent lighter. The carrier anticipated a significant drop in fuel consumption because vessel displacement was reduced while payload volume per voyage remained identical. However, during the year following the complete container replacement, the average quantity of fuel consumed per ton-mile of freight transported across the carrier's ocean routes increased by nearly 6 percent. Which of the following, if true, most helps to resolve the apparent paradox described above?

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Cevap: Because the carbon-fiber containers were substantially lighter, vessels carrying them rode higher in the water, exposing a larger cross-section of the hull to surface winds and hydrodynamic turbulence that required significantly more engine power to maintain cruising speed.

Cevap

The apparent paradox is resolved by the option explaining that lighter containers caused vessels to ride higher in the water, thereby increasing aerodynamic and hydrodynamic drag and forcing engines to burn more fuel to maintain standard speeds.
The correct answer identifies a secondary hydrodynamic consequence of the weight reduction: because the lighter containers reduced vessel draft, more of the ship's hull was exposed above the waterline. The resulting wind and surface turbulence resistance demanded increased engine output to sustain cruising speed, neutralizing the mass benefit and driving up fuel consumption per ton-mile.

Adım Adım Çözüm

1
Identify the two seemingly contradictory facts in the stimulus.
Fact 1: Carbon-fiber containers are 25% lighter, reducing ship displacement with identical payload volume. Fact 2: Fuel consumed per ton-mile across ocean routes increased by 6% after replacing the containers.
Resolving a paradox requires finding a missing factor that allows both facts to be simultaneously true without denying either premise.
2
Evaluate the mechanism required to reconcile the premises.
The correct explanation must show how the reduction in container weight directly or indirectly caused an operational inefficiency that outweighed the expected energy savings.
A valid resolution introduces a third variable that accounts for the unexpected negative outcome resulting from the initial change.
3
Assess the correct choice against the required mechanism.
The statement regarding vessel draft explains that lighter weight caused ships to sit higher in the water, increasing exposure to wind and surface turbulence. Overcoming this extra drag required more engine power than was saved by the lower mass, explaining the 6% increase in fuel burn per ton-mile.
This logical bridge reconciles the 25% mass reduction with the 6% fuel consumption increase.

Anahtar Kavram

Resolving Paradoxes through Confounding Physical and Operational Side Effects
Soru 110Soru

A sequence of positive integers ana_n is defined by an=2n+5na_n = 2^n + 5^n for all integers n1n \ge 1. What is the remainder when a100a_{100} is divided by 77?

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Cevap: 4

Cevap

The remainder when a100a_{100} is divided by 77 is 44.
By reducing 52(mod7)5 \equiv -2 \pmod 7, we observe that 5100(2)100=2100(mod7)5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7 because the exponent 100100 is even. The powers of 2(mod7)2 \pmod 7 follow a 3-step cycle (2,4,12, 4, 1). Dividing the exponent 100100 by 33 leaves a remainder of 11, meaning 210021=2(mod7)2^{100} \equiv 2^1 = 2 \pmod 7. Adding the remainders for both terms yields 2+2=42 + 2 = 4.

Adım Adım Çözüm

1
Use modular arithmetic to simplify the base 5(mod7)5 \pmod 7.
52(mod7)5 \equiv -2 \pmod 7, which implies 5100(2)100=2100(mod7)5^{100} \equiv (-2)^{100} = 2^{100} \pmod 7 because the exponent 100100 is even.
Converting 55 to 2-2 allows both terms to be expressed using powers of 22.
2
Determine the remainder cycle of powers of 22 when divided by 77.
212(mod7)2^1 \equiv 2 \pmod 7, 224(mod7)2^2 \equiv 4 \pmod 7, and 23=81(mod7)2^3 = 8 \equiv 1 \pmod 7. The pattern repeats every 33 powers.
Finding the period of cyclicity simplifies evaluating large powers.
3
Divide the exponent 100100 by the cycle length 33.
100=3×33+1100 = 3 \times 33 + 1, leaving a remainder of 11. Thus, 210021=2(mod7)2^{100} \equiv 2^1 = 2 \pmod 7.
The remainder of the exponent modulo the cycle length determines the equivalent reduced power.
4
Combine the remainders for 21002^{100} and 51005^{100}.
a100=2100+51002+2=4(mod7)a_{100} = 2^{100} + 5^{100} \equiv 2 + 2 = 4 \pmod 7.
Adding the individual modular values gives the overall remainder.

Anahtar Kavram

Modular arithmetic cyclicity and negative congruences
Soru 111Soru

A set SS consists of consecutive integers. The arithmetic mean of all the positive integers in set SS is 18.518.5, and the arithmetic mean of all the negative integers in set SS is 12-12. How many integers are in set SS?

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Cevap: 60

Cevap

The total number of integers in set SS is 6060.
The positive integers in set SS are 1,2,,361, 2, \dots, 36, which have an arithmetic mean of 1+362=18.5\frac{1 + 36}{2} = 18.5. The negative integers in set SS are 23,22,,1-23, -22, \dots, -1, which have an arithmetic mean of 23+(1)2=12\frac{-23 + (-1)}{2} = -12. Since set SS consists of consecutive integers spanning from 23-23 to 3636, it includes 2323 negative integers, 3636 positive integers, and the integer 00. The total number of elements is 23+1+36=6023 + 1 + 36 = 60.

Adım Adım Çözüm

1
Find the largest positive integer in set SS.
The largest positive integer is 3636, meaning there are 3636 positive integers in SS.
The positive integers in SS must form a consecutive sequence starting at 11 up to some maximum integer mm. The average of consecutive integers from 11 to mm is 1+m2\frac{1 + m}{2}. Setting 1+m2=18.5\frac{1 + m}{2} = 18.5 yields 1+m=371 + m = 37, so m=36m = 36.
2
Find the smallest negative integer in set SS.
The smallest negative integer is 23-23, meaning there are 2323 negative integers in SS.
The negative integers in SS must form a consecutive sequence ending at 1-1 down to some minimum integer k-k. The average of consecutive integers from k-k to 1-1 is k+(1)2\frac{-k + (-1)}{2}. Setting k12=12\frac{-k - 1}{2} = -12 yields k1=24-k - 1 = -24, so k=23k = 23.
3
Calculate the total number of elements in set SS.
Set SS contains 6060 integers.
Because set SS contains consecutive integers ranging from negative to positive values, it must also contain 00. The total count is 23 (negative integers)+1 (the integer zero)+36 (positive integers)=6023\text{ (negative integers)} + 1\text{ (the integer zero)} + 36\text{ (positive integers)} = 60.

Anahtar Kavram

Evenly spaced set averages and classification of zero in consecutive integer sets.
Tahmini Süre:2m 0s
Soru 112Soru

An arithmetic sequence a1,a2,a3,a_1, a_2, a_3, \dots with a positive common difference dd and a geometric sequence b1,b2,b3,b_1, b_2, b_3, \dots with a positive common ratio rr both have the same positive first term (a1=b1>0a_1 = b_1 > 0). If the 3rd term of the arithmetic sequence equals the 3rd term of the geometric sequence (a3=b3a_3 = b_3), and the 7th term of the arithmetic sequence equals the 5th term of the geometric sequence (a7=b5a_7 = b_5), what is the value of rr?

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Cevap: 2\sqrt{2}

Cevap

2\sqrt{2}
Using the formulas for the nn-th terms, a3=a1+2da_3 = a_1 + 2d and b3=a1r2b_3 = a_1 r^2, giving 2d=a1(r21)2d = a_1(r^2 - 1). Substituting 6d=3a1(r21)6d = 3a_1(r^2 - 1) into a7=a1+6d=a1r4a_7 = a_1 + 6d = a_1 r^4 yields a1+3a1(r21)=a1r4a_1 + 3a_1(r^2 - 1) = a_1 r^4. Dividing by a1>0a_1 > 0 gives 1+3r23=r41 + 3r^2 - 3 = r^4, which simplifies to r43r2+2=0r^4 - 3r^2 + 2 = 0. Factoring gives (r21)(r22)=0(r^2 - 1)(r^2 - 2) = 0. Since d>0d > 0 and a1>0a_1 > 0, we must have r2>1r^2 > 1, eliminating r2=1r^2 = 1. Therefore, r2=2r^2 = 2, and since r>0r > 0, r=2r = \sqrt{2}.

Adım Adım Çözüm

1
Express sequence terms in terms of first term a1a_1, common difference dd, and common ratio rr
a3=a1+2da_3 = a_1 + 2d, a7=a1+6da_7 = a_1 + 6d, b3=a1r2b_3 = a_1 r^2, and b5=a1r4b_5 = a_1 r^4
Apply the standard formulas for the nn-th term of arithmetic (an=a1+(n1)da_n = a_1 + (n-1)d) and geometric (bn=b1rn1b_n = b_1 r^{n-1}) sequences.
2
Set up equations based on given equality of terms
Equation 1: a1+2d=a1r2    2d=a1(r21)a_1 + 2d = a_1 r^2 \implies 2d = a_1(r^2 - 1);
Equation 2: a1+6d=a1r4a_1 + 6d = a_1 r^4
Translate the given conditions a3=b3a_3 = b_3 and a7=b5a_7 = b_5 into algebraic relations.
3
Substitute 2d2d from Equation 1 into Equation 2
a1+3(2d)=a1+3a1(r21)=a1r4a_1 + 3(2d) = a_1 + 3a_1(r^2 - 1) = a_1 r^4
Express 6d6d as 3(2d)3(2d) to eliminate dd from the system.
4
Divide by a1a_1 (since a1>0a_1 > 0) and simplify to solve for rr
1+3r23=r4    r43r2+2=0    (r21)(r22)=01 + 3r^2 - 3 = r^4 \implies r^4 - 3r^2 + 2 = 0 \implies (r^2 - 1)(r^2 - 2) = 0
Reduce the equation to a quadratic in terms of r2r^2.
5
Determine the valid root for rr
r2=2    r=2r^2 = 2 \implies r = \sqrt{2} (since r>0r > 0 and d>0d > 0 implies r2>1r^2 > 1)
Since d>0d > 0 and a1>0a_1 > 0, 2d=a1(r21)>02d = a_1(r^2 - 1) > 0, which requires r2>1r^2 > 1. Thus r2=1r^2 = 1 is rejected, leaving r2=2r^2 = 2.

Anahtar Kavram

Arithmetic and Geometric Sequences Alignment
Tahmini Süre:2m 0s
Soru 113Soru

A manufacturing plant uses three assembly lines, L1L_1, L2L_2, and L3L_3, to produce three custom components, XX, YY, and ZZ.

- Producing one unit of component XX requires 2 hours on L1L_1, 1 hour on L2L_2, and 3 hours on L3L_3.
- Producing one unit of component YY requires 3 hours on L1L_1, 4 hours on L2L_2, and 2 hours on L3L_3.
- Producing one unit of component ZZ requires 1 hour on L1L_1, 2 hours on L2L_2, and 4 hours on L3L_3.

During a given week, assembly lines L1L_1, L2L_2, and L3L_3 were operated for a total of 140 hours, 165 hours, and 235 hours, respectively, with zero idle time. Assuming full capacity utilization, how many units of component ZZ were produced during that week?

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Cevap: 30

Cevap

30 units of component Z were produced.
Translating the assembly line operational hours into a 3x3 system of linear equations yields 2X+3Y+Z=1402X + 3Y + Z = 140, X+4Y+2Z=165X + 4Y + 2Z = 165, and 3X+2Y+4Z=2353X + 2Y + 4Z = 235. Eliminating XX results in two equations in YY and ZZ: 5Y+3Z=1905Y + 3Z = 190 and 5Y+Z=1305Y + Z = 130. Subtracting these equations gives 2Z=602Z = 60, so Z=30Z = 30.

Adım Adım Çözüm

1
Formulate a system of 3 linear equations representing total hours logged on each assembly line.
Line 1: 2X+3Y+Z=1402X + 3Y + Z = 140; Line 2: X+4Y+2Z=165X + 4Y + 2Z = 165; Line 3: 3X+2Y+4Z=2353X + 2Y + 4Z = 235.
Each component requires specific line processing time, and total time per line equals total available capacity.
2
Eliminate variable XX by substituting X=1654Y2ZX = 165 - 4Y - 2Z into the other two equations.
Equation A: 5Y+3Z=1905Y + 3Z = 190 and Equation B: 5Y+Z=1305Y + Z = 130.
Reducing a 3-variable system to a 2-variable system simplifies linear elimination.
3
Subtract Equation B from Equation A.
2Z = 60, so Z = 30.
Since the coefficients of YY in both reduced equations are identical (5Y5Y), subtraction directly isolates ZZ.

Anahtar Kavram

Solving Systems of Three Linear Equations via Variable Substitution and Elimination
Soru 114Soru

A tech firm observed that employees who voluntarily chose to work remotely reported higher job satisfaction than those who worked in the office. Based on this observation, management concluded that transitioning all employees to remote work will increase overall employee job satisfaction across the company. Which of the following, if true, most seriously weakens management's conclusion?

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Cevap: Employees who chose remote work were primarily those whose job duties require minimal collaboration, whereas most office-based roles require constant group coordination that suffers remotely.

Cevap

The argument is most weakened by showing that the remote employees surveyed had roles unrepresentative of the entire workforce, creating a selection bias that makes the conclusion invalid for office-based positions.
The correct answer weakens the argument by exposing a selection bias. The premise relies on data from employees who voluntarily chose remote work, likely because their specific roles suited it. If office-based roles depend heavily on real-time group coordination that deteriorates in a remote environment, forcing those employees to work remotely will likely decrease satisfaction, invalidating the conclusion that total remote work will raise overall satisfaction.

Adım Adım Çözüm

1
Identify the main conclusion and premise.
Premise: Voluntarily remote employees report higher satisfaction. Conclusion: Mandating remote work for all employees will increase overall job satisfaction.
Understanding the logic gap requires clearly isolating what the premise proves versus what the conclusion claims.
2
Evaluate the underlying assumption.
Management assumes that the satisfaction experienced by voluntary remote workers will apply equally to all employees regardless of role.
Weakening questions target unstated assumptions connecting the premise to the conclusion.
3
Select the choice that undermines this assumption.
The option showing that voluntary remote workers have fundamentally different job responsibilities than office workers demonstrates that the sample is non-representative, directly undermining the conclusion.
Pointing out selection bias breaks the causal link between the initial observation and the proposed company-wide plan.

Anahtar Kavram

Weakening Arguments: Selection Bias and Unrepresentative Samples
Tahmini Süre:1m 0s
Soru 115Soru

Set SS consists of nn consecutive even integers. The arithmetic mean of all the integers in set SS is 4545. If the difference between the largest integer and the smallest integer in set SS is 3434, what is the value of the largest integer in set SS?

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Cevap: 62

Cevap

The largest integer in set SS is 6262.
In any set of evenly spaced numbers, such as consecutive even integers, the arithmetic mean is equal to the average of the smallest term and the largest term. Since the arithmetic mean is 45, the sum of the smallest term and the largest term must be 45×2=9045 \times 2 = 90. Combining this with the given fact that the difference between the largest term and the smallest term is 34 creates a system of equations: Smallest + Largest = 90 and Largest - Smallest = 34. Adding these two equations cancels out the smallest term, resulting in 2 * Largest = 124, which gives 62 for the largest integer.

Adım Adım Çözüm

1
Express the arithmetic mean of the set in terms of the smallest element (FF) and largest element (LL).
F+L=90F + L = 90
For any evenly spaced set of numbers, the arithmetic mean is equal to the average of the first and last elements: F+L2=45\frac{F + L}{2} = 45.
2
Set up the equation for the difference between the largest and smallest elements.
LF=34L - F = 34
The question specifies that the largest integer exceeds the smallest integer by 34.
3
Solve the system of two linear equations for LL.
L=62L = 62
Adding (F+L)+(LF)=90+34(F + L) + (L - F) = 90 + 34 eliminates FF, leaving 2L=1242L = 124, which yields L=62L = 62.

Anahtar Kavram

Equivalence of arithmetic mean to the average of the first and last terms in an evenly spaced set
Soru 116Soru

Four numerical quantities KK, LL, MM, and NN are defined based on arithmetic and geometric sequences as follows. Arrange these four quantities in ascending order (from smallest to largest value):

- **Quantity KK**: The sum of the first 5 terms of an arithmetic sequence with first term a1=2a_1 = 2 and common difference d=3d = 3.
- **Quantity LL**: The 4th term of a geometric sequence with first term b1=3b_1 = 3 and common ratio r=2r = 2.
- **Quantity MM**: The sum of an infinite geometric series with first term c1=18c_1 = 18 and common ratio r=12r = \frac{1}{2}.
- **Quantity NN**: The 7th term of an arithmetic sequence with first term d1=50d_1 = 50 and common difference d=4d = -4.

Which of the following represents the correct ascending order of the four quantities?

Öğeleri doğru sıraya koymak için sürükleyin

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Cevap

The correct ascending order of the quantities is Quantity L (24) < Quantity N (26) < Quantity M (36) < Quantity K (40).
Evaluating each quantity gives Quantity L = 24, Quantity N = 26, Quantity M = 36, and Quantity K = 40. Arranging these in ascending numerical order produces the sequence Quantity L, Quantity N, Quantity M, Quantity K.

Adım Adım Çözüm

1
Calculate Quantity K
K=40K = 40
The sum of an arithmetic sequence is given by Sn=n2[2a1+(n1)d]S_n = \frac{n}{2}[2a_1 + (n-1)d]. For n=5n = 5, a1=2a_1 = 2, and d=3d = 3, S5=52[2(2)+(51)(3)]=52[4+12]=52(16)=40S_5 = \frac{5}{2}[2(2) + (5-1)(3)] = \frac{5}{2}[4 + 12] = \frac{5}{2}(16) = 40.
2
Calculate Quantity L
L=24L = 24
The nn-th term of a geometric sequence is bn=b1rn1b_n = b_1 \cdot r^{n-1}. For n=4n = 4, b1=3b_1 = 3, and r=2r = 2, b4=3241=323=38=24b_4 = 3 \cdot 2^{4-1} = 3 \cdot 2^3 = 3 \cdot 8 = 24.
3
Calculate Quantity M
M=36M = 36
The sum of an infinite geometric series with r<1|r| < 1 is S=c11rS_\infty = \frac{c_1}{1 - r}. For c1=18c_1 = 18 and r=12r = \frac{1}{2}, S=1811/2=181/2=36S_\infty = \frac{18}{1 - 1/2} = \frac{18}{1/2} = 36.
4
Calculate Quantity N
N=26N = 26
The nn-th term of an arithmetic sequence is dn=d1+(n1)dd_n = d_1 + (n-1)d. For n=7n = 7, d1=50d_1 = 50, and d=4d = -4, d7=50+(71)(4)=50+6(4)=5024=26d_7 = 50 + (7-1)(-4) = 50 + 6(-4) = 50 - 24 = 26.
5
Compare the evaluated quantities to arrange them in ascending order
24<26<36<4024 < 26 < 36 < 40, which corresponds to L<N<M<KL < N < M < K
Comparing the numeric values directly yields 24 (L)<26 (N)<36 (M)<40 (K)24 \text{ (L)} < 26 \text{ (N)} < 36 \text{ (M)} < 40 \text{ (K)}.

Anahtar Kavram

Arithmetic and Geometric Sequence Formulas
Soru 117Soru

A software analytics company offers three annual subscription plans: Standard, Professional, and Enterprise.

• A client purchasing 3 Standard, 2 Professional, and 1 Enterprise plan pays a total of 1,110.���Aclientpurchasing1Standard,4Professional,and2Enterpriseplanspaysatotalof1,110. ��� A client purchasing 1 Standard, 4 Professional, and 2 Enterprise plans pays a total of 1,620.
• A client purchasing 4 Standard, 1 Professional, and 3 Enterprise plans pays a total of $1,730.

What is the cost, in dollars, of 1 Enterprise plan?

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Cevap: 350

Cevap

The cost of 1 Enterprise plan is 350 dollars.
Setting up equations for each purchase tier gives 3x+2y+z=11103x + 2y + z = 1110, x+4y+2z=1620x + 4y + 2z = 1620, and 4x+y+3z=17304x + y + 3z = 1730. Isolating xx in the second equation gives x=16204y2zx = 1620 - 4y - 2z. Substituting xx into the first and third equations yields 2y+z=7502y + z = 750 and 3y+z=9503y + z = 950, respectively. Subtracting these reduced equations gives y=200y = 200, which upon back-substitution into 2y+z=7502y + z = 750 reveals z=350z = 350. Thus, 1 Enterprise plan costs 350 dollars.

Adım Adım Çözüm

1
Formulate the linear system of equations from the given conditions.
Let xx be the price of a Standard plan, yy the price of a Professional plan, and zz the price of an Enterprise plan:
(1) 3x+2y+z=11103x + 2y + z = 1110
(2) x+4y+2z=1620x + 4y + 2z = 1620
(3) 4x+y+3z=17304x + y + 3z = 1730
Translating the verbal conditions into algebraic equations creates a solvable system.
2
Isolate variable xx in equation (2) and substitute it into equations (1) and (3).
From (2), x=16204y2zx = 1620 - 4y - 2z.
Substituting into (1):
3(16204y2z)+2y+z=1110    486012y6z+2y+z=1110    10y+5z=3750    2y+z=7503(1620 - 4y - 2z) + 2y + z = 1110 \implies 4860 - 12y - 6z + 2y + z = 1110 \implies 10y + 5z = 3750 \implies 2y + z = 750 (Equation 4)

Substituting into (3):
4(16204y2z)+y+3z=1730    648016y8z+y+3z=1730    15y+5z=4750    3y+z=9504(1620 - 4y - 2z) + y + 3z = 1730 \implies 6480 - 16y - 8z + y + 3z = 1730 \implies 15y + 5z = 4750 \implies 3y + z = 950 (Equation 5)
Eliminating xx reduces the system to two linear equations with two variables.
3
Solve the 2x2 system of equations for yy and zz.
Subtracting Equation (4) from Equation (5):
(3y+z)(2y+z)=950750    y=200(3y + z) - (2y + z) = 950 - 750 \implies y = 200.

Substitute y=200y = 200 back into Equation (4):
2(200)+z=750    400+z=750    z=3502(200) + z = 750 \implies 400 + z = 750 \implies z = 350.
Solving the reduced system yields the exact values of yy and zz.

Anahtar Kavram

Solving Systems of Three Linear Equations via Gaussian Elimination / Variable Substitution
Soru 118Soru

A class of 20 students took a 10-point mathematics quiz. The frequency table below records the quiz scores achieved by 18 of the students:

ScoreFrequency
63
75
84
94
102

The scores of the remaining 2 students were recorded later. If the score of every student is an integer from 0 to 10, inclusive, and adding the 2 missing scores causes the median score of the entire class of 20 students to be 8 and the arithmetic mean score to be an integer, what is the score of the higher-scoring student among the 2 remaining students?

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Cevap: 10

Cevap

10
The sum of the 18 known scores is 141. For the overall mean of 20 scores to be an integer, the total sum of all 20 scores must be a multiple of 20. Since each score is at most 10, the maximum possible total sum is 141+10+10=161141 + 10 + 10 = 161. The only multiple of 20 between 141 and 161 is 160, requiring the sum of the two missing scores to be 160141=19160 - 141 = 19. The only valid integer scores bounded by 10 that sum to 19 are 9 and 10. Adding scores of 9 and 10 places the 10th and 11th ordered values at 8, giving a median of 8. Thus, the higher missing score is 10.

Adım Adım Çözüm

1
Calculate the total sum and count of the 18 known student scores.
Known count = 3+5+4+4+2=183 + 5 + 4 + 4 + 2 = 18 students. Known sum = (6×3)+(7×5)+(8×4)+(9×4)+(10×2)=18+35+32+36+20=141(6 \times 3) + (7 \times 5) + (8 \times 4) + (9 \times 4) + (10 \times 2) = 18 + 35 + 32 + 36 + 20 = 141.
Establishing baseline sum and count is essential before analyzing missing values.
2
Set up the equation for the total sum of all 20 student scores and apply the integer mean condition.
Let the missing scores be aa and bb with 0ab100 \le a \le b \le 10. The total sum for 20 students is S20=141+a+bS_{20} = 141 + a + b. The arithmetic mean is 141+a+b20\frac{141 + a + b}{20}.
Since the mean must be an integer, 141+a+b141 + a + b must be a multiple of 20.
3
Determine the required sum of the two missing scores a+ba + b.
Since 0a100 \le a \le 10 and 0b100 \le b \le 10, we have 0a+b200 \le a + b \le 20. The range for S20S_{20} is [141,161][141, 161]. The only multiple of 20 in this range is 160. Thus, 141+a+b=160    a+b=19141 + a + b = 160 \implies a + b = 19.
160 is the unique multiple of 20 reachable given score bounds.
4
Find the unique integer pair (a,b)(a, b) satisfying a+b=19a + b = 19 with a,b10a, b \le 10.
Since a10a \le 10 and b10b \le 10, the only integer solution with aba \le b is a=9a = 9 and b=10b = 10.
No other pair of integers between 0 and 10 sums to 19.
5
Verify that adding scores 9 and 10 maintains a median score of 8.
With 9 and 10 added, the frequencies are: Score 6 (3), Score 7 (5), Score 8 (4), Score 9 (5), Score 10 (3). The 10th and 11th values in order are both 8, so the median is 8+82=8\frac{8 + 8}{2} = 8.
Confirms the median constraint is fully satisfied.

Anahtar Kavram

Properties of Weighted Means and Median Constraints in Frequency Tables
Tahmini Süre:2m 0s
Soru 119Soru

The sum of the first three terms of an increasing geometric sequence of positive numbers is 2121, and the sum of the squares of these same three terms is 189189. What is the first term of the sequence?

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Cevap: 3

Cevap

The first term of the sequence is 3.
By setting up the equations a(1+r+r2)=21a(1+r+r^2) = 21 and a2(1+r2+r4)=189a^2(1+r^2+r^4) = 189, we use the identity 1+r2+r4=(1+r+r2)(1r+r2)1+r^2+r^4 = (1+r+r^2)(1-r+r^2) to find a(1r+r2)=9a(1-r+r^2) = 9. Subtracting this from the first equation gives 2ar=122ar = 12, so ar=6ar = 6. Solving for rr gives r=2r = 2 for an increasing sequence, which yields a=3a = 3.

Adım Adım Çözüm

1
Set up the algebraic equations for the sum of terms and sum of squared terms.
Let the first three terms be aa, arar, and ar2ar^2. The given conditions yield a(1+r+r2)=21a(1 + r + r^2) = 21 and a2(1+r2+r4)=189a^2(1 + r^2 + r^4) = 189.
Standard representation of a geometric sequence with first term aa and common ratio rr.
2
Factor the sum of squares expression using algebraic identities.
Note that 1+r2+r4=(1+r+r2)(1r+r2)1 + r^2 + r^4 = (1 + r + r^2)(1 - r + r^2). Thus, a2(1+r+r2)(1r+r2)=189a^2(1 + r + r^2)(1 - r + r^2) = 189.
Factoring allows substitution of the first equation into the second.
3
Substitute a(1+r+r2)=21a(1 + r + r^2) = 21 into the factored equation.
Substituting 2121 gives 21a(1r+r2)=189    a(1r+r2)=921 \cdot a(1 - r + r^2) = 189 \implies a(1 - r + r^2) = 9.
Simplifies the second-degree term expression to a system of two linear equations in terms of aa and arar.
4
Subtract the simplified equation from the initial sum equation to isolate arar.
a(1+r+r2)a(1r+r2)=219    2ar=12    ar=6a(1 + r + r^2) - a(1 - r + r^2) = 21 - 9 \implies 2ar = 12 \implies ar = 6.
Eliminating terms isolates the product of the first term and ratio, which is the second term.
5
Solve for rr and find aa.
Substitute a=6ra = \frac{6}{r} into a(1+r+r2)=21a(1 + r + r^2) = 21 to get 6r215r+6=06r^2 - 15r + 6 = 0, which factors as (2r1)(r2)=0(2r - 1)(r - 2) = 0. Since the sequence is increasing, r=2r = 2, yielding a=3a = 3.
Determines the specific parameters satisfying the increasing constraint.

Anahtar Kavram

Properties and algebraic manipulation of terms in geometric sequences and series.
Tahmini Süre:2m 0s
Soru 120Soru

A positive integer NN has exactly 1212 positive integer divisors. If NN is a multiple of 1818, and exactly 23\frac{2}{3} of the positive integer divisors of NN are even, what is the value of NN?

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Cevap: 108

Cevap

108
The integer 108 prime factorizes as 2² × 3³. It has (2+1)(3+1) = 12 total positive divisors. Its odd divisors are the 4 divisors of 3³ (namely 1, 3, 9, 27), leaving 12 - 4 = 8 even divisors. Thus, exactly 8/12 = 2/3 of its divisors are even, and 108 is divisible by 18.

Adım Adım Çözüm

1
Determine the number of even and odd divisors of N.
Number of even divisors = (2/3) × 12 = 8; Number of odd divisors = 12 - 8 = 4.
The problem states that 2/3 of the 12 total divisors are even.
2
Express N as 2^a × M, where M is odd.
Total divisors = (a + 1) × (divisors of M) = 12. Since divisors of M are the odd divisors of N, (a + 1) × 4 = 12, so a + 1 = 3 and a = 2.
Odd divisors of N come entirely from the odd part M.
3
Find M such that M is odd, divisible by 9 (since N is a multiple of 18 = 2 × 3²), and has exactly 4 divisors.
M must be of the form 3^k. Since divisors count is k + 1 = 4, k = 3, so M = 3³ = 27.
If M had two prime factors p × q, each exponent would be 1, which cannot be divisible by 3².
4
Calculate N = 2^a × M.
N = 2² × 27 = 4 × 27 = 108.
Combining the even and odd prime factor components.

Anahtar Kavram

Divisibility and Counting Divisors via Prime Factorization
Tahmini Süre:1m 30s
ÖncekiSayfa 6 / 110Sonraki
Tüm alıştırma soruları — GMAT | Examkin