Tüm alıştırma soruları

387 soru

Soru 241Soru

For a constant a>0a > 0, the quadratic equation x2ax+(2a+1)=0x^2 - ax + (2a + 1) = 0 has two real roots, rr and ss. If r2+s2=43r^2 + s^2 = 43, what is the value of aa?

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Cevap: 9

Cevap

The value of aa is 9.
According to Vieta's formulas, the sum of the roots of x2ax+(2a+1)=0x^2 - ax + (2a + 1) = 0 is r+s=ar + s = a and the product of the roots is rs=2a+1rs = 2a + 1. Using the identity r2+s2=(r+s)22rsr^2 + s^2 = (r + s)^2 - 2rs, substitute the Vieta expressions to obtain r2+s2=a22(2a+1)=a24a2r^2 + s^2 = a^2 - 2(2a + 1) = a^2 - 4a - 2. Setting this equal to 43 yields a24a2=43a^2 - 4a - 2 = 43, which simplifies to a24a45=0a^2 - 4a - 45 = 0. Factoring the quadratic gives (a9)(a+5)=0(a - 9)(a + 5) = 0, giving solutions a=9a = 9 or a=5a = -5. Because the problem specifies that a>0a > 0, aa must be 9. Checking the discriminant Δ=(9)24(1)(19)=5>0\Delta = (-9)^2 - 4(1)(19) = 5 > 0 confirms that two real roots exist.

Adım Adım Çözüm

1
Express the sum and product of the roots in terms of aa using Vieta's formulas.
r+s=ar + s = a and rs=2a+1rs = 2a + 1
For a standard quadratic equation x2+bx+c=0x^2 + bx + c = 0, the sum of the roots is b-b and the product of the roots is cc.
2
Relate r2+s2r^2 + s^2 to (r+s)(r + s) and rsrs.
r2+s2=(r+s)22rs=a22(2a+1)=a24a2r^2 + s^2 = (r + s)^2 - 2rs = a^2 - 2(2a + 1) = a^2 - 4a - 2
Expanding (r+s)2=r2+2rs+s2(r + s)^2 = r^2 + 2rs + s^2 allows expressing r2+s2r^2 + s^2 in terms of known quantities.
3
Substitute r2+s2=43r^2 + s^2 = 43 into the equation and solve for aa.
a24a2=43    a24a45=0    (a9)(a+5)=0a^2 - 4a - 2 = 43 \implies a^2 - 4a - 45 = 0 \implies (a - 9)(a + 5) = 0
Rearranging terms forms a new quadratic equation in terms of aa.
4
Apply the positivity constraint a>0a > 0 and verify that the roots are real.
Since a>0a > 0, a=9a = 9. The discriminant of the original equation is Δ=(9)24(1)(19)=5>0\Delta = (-9)^2 - 4(1)(19) = 5 > 0, confirming real roots exist.
The question specifies that aa is positive and that rr and ss are real numbers.

Anahtar Kavram

Vieta's Formulas and Quadratic Modeling
Soru 242Soru

At an intellectual property law firm, a senior partner reviewed 180180 patent applications across three technical domains: Artificial Intelligence, Biotechnology, and Clean Energy. The audit revealed that 2525 applications belonged to none of these three domains. Furthermore, 9090 applications involved Artificial Intelligence, 8585 involved Biotechnology, and 8080 involved Clean Energy. If the number of applications that belonged to exactly two of these domains was three times the number of applications that belonged to all three domains, how many patent applications belonged to exactly one domain?

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Cevap: 75

Cevap

The total number of patent applications that belonged to exactly one domain is 7575.
The total number of applications in at least one domain is 18025=155180 - 25 = 155. Let E1E_1 be the number of applications in exactly one domain, E2E_2 in exactly two domains, and xx in all three domains. The total union is E1+E2+x=155E_1 + E_2 + x = 155, while the sum of individual set sizes is A+B+C=E1+2E2+3x=90+85+80=255|A| + |B| + |C| = E_1 + 2E_2 + 3x = 90 + 85 + 80 = 255. Subtracting the union equation from the sum equation gives E2+2x=100E_2 + 2x = 100. Given E2=3xE_2 = 3x, substituting gives 5x=100    x=205x = 100 \implies x = 20, which implies E2=60E_2 = 60. Finally, subtracting E2E_2 and xx from the total union yields E1=1556020=75E_1 = 155 - 60 - 20 = 75.

Adım Adım Çözüm

1
Calculate the total number of applications in at least one domain
ABC=18025=155|A \cup B \cup C| = 180 - 25 = 155
Subtracting the applications belonging to none of the domains from the overall total yields the total number of unique applications covered by the three domains combined.
2
Formulate regional Venn diagram equations
E1+E2+x=155E_1 + E_2 + x = 155 and E1+2E2+3x=255E_1 + 2E_2 + 3x = 255
Summing individual set counts counts elements in exactly two domains twice and elements in all three domains three times.
3
Deduce the relationship between E2E_2 and xx
E2+2x=100E_2 + 2x = 100
Subtracting the equation for total union from the sum of individual sets isolates the overcounted regions.
4
Use the given proportion E2=3xE_2 = 3x to solve for xx and E2E_2
x=20x = 20 and E2=60E_2 = 60
Substituting E2=3xE_2 = 3x into E2+2x=100E_2 + 2x = 100 gives 5x=1005x = 100, so x=20x = 20 and E2=60E_2 = 60.
5
Find the number of applications belonging to exactly one domain (E1E_1)
E1=75E_1 = 75
Subtracting the count of applications in exactly two domains (6060) and all three domains (2020) from the total union (155155) gives 1556020=75155 - 60 - 20 = 75.

Anahtar Kavram

Three-Set Venn Diagram Region Partitioning
Soru 243Soru

A wholesaler purchases a desk for $250\$250 and marks up the cost price by 20%20\% to establish the retail price. If the desk is sold at the retail price, what is the profit, in dollars, earned on the sale?

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Cevap: 50

Cevap

The profit earned on the sale of the desk is $50\$50.
The profit is calculated by finding 20%20\% of the cost price of $250\$250, which is 0.20×250=500.20 \times 250 = 50.

Adım Adım Çözüm

1
Identify the cost price and markup rate from the problem statement.
Cost price =$250= \$250; Markup rate =20%= 20\%.
Markup is calculated directly as a percentage of the original cost price.
2
Multiply the cost price by the markup percentage to find the profit.
Profit=0.20×250=50\text{Profit} = 0.20 \times 250 = 50.
Taking 20%20\% of the $250\$250 purchase price gives the exact dollar markup/profit.

Anahtar Kavram

Dollar profit is calculated as the markup percentage multiplied by the cost price.
Soru 244Soru

A chemical storage vessel contains three industrial solvents—Solvent A, Solvent B, and Solvent C—blended in the initial ratio of 2:5:72 : 5 : 7, respectively. After 2424 liters of Solvent A and 1212 liters of Solvent B are added to the vessel, the ratio of Solvent A to Solvent B becomes 4:74 : 7. What was the initial volume, in liters, of Solvent C in the vessel?

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Cevap: 140

Cevap

The initial volume of Solvent C in the vessel was 140 liters.
Defining the initial amounts of Solvents A, B, and C as 2x2x, 5x5x, and 7x7x, respectively, the addition of 2424 liters of Solvent A and 1212 liters of Solvent B yields the ratio equation 2x+245x+12=47\frac{2x + 24}{5x + 12} = \frac{4}{7}. Cross-multiplying gives 14x+168=20x+4814x + 168 = 20x + 48, which simplifies to 6x=1206x = 120 and x=20x = 20. The initial volume of Solvent C is therefore 7×20=1407 \times 20 = 140 liters.

Adım Adım Çözüm

1
Express the initial quantities of the three solvents in terms of a common ratio multiplier xx.
Solvent A = 2x2x, Solvent B = 5x5x, Solvent C = 7x7x.
A ratio of 2:5:72 : 5 : 7 implies the actual quantities are proportional to 2, 5, and 7 by a factor xx.
2
Set up an equation representing the modified ratio of Solvent A to Solvent B.
2x+245x+12=47\frac{2x + 24}{5x + 12} = \frac{4}{7}
Adding 24 liters to Solvent A yields 2x+242x + 24 liters, and adding 12 liters to Solvent B yields 5x+125x + 12 liters.
3
Solve the algebraic equation for xx.
7(2x+24)=4(5x+12)    14x+168=20x+48    6x=120    x=207(2x + 24) = 4(5x + 12) \implies 14x + 168 = 20x + 48 \implies 6x = 120 \implies x = 20.
Cross-multiplication converts the ratio comparison into a linear equation.
4
Compute the initial volume of Solvent C.
Initial volume of Solvent C = 7x=7(20)=1407x = 7(20) = 140 liters.
Substitute x=20x = 20 into the expression 7x7x for the original amount of Solvent C.

Anahtar Kavram

Ratio scaling and algebraic modeling of partial additions
Soru 245Soru

A chemist prepares a mixture by combining 2020 liters of a 10%10\% saline solution with 3030 liters of a 20%20\% saline solution. What is the saline concentration, as a percentage, of the resulting mixture?

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Cevap: 16

Cevap

The concentration of the resulting mixture is 16%16\%.
The total amount of pure saline in the mixture is 88 liters out of a total solution volume of 5050 liters. The overall concentration is 850×100%=16%\frac{8}{50} \times 100\% = 16\%.

Adım Adım Çözüm

1
Calculate the total amount of pure saline solute contributed by both solutions.
20 L×0.10+30 L×0.20=2+6=8 liters20 \text{ L} \times 0.10 + 30 \text{ L} \times 0.20 = 2 + 6 = 8 \text{ liters}
The total solute is the sum of the solute amounts in each component solution.
2
Calculate the total volume of the combined mixture.
20+30=50 liters20 + 30 = 50 \text{ liters}
The combined volume is the sum of the individual volumes.
3
Compute the weighted average concentration of the mixture.
850=0.16=16%\frac{8}{50} = 0.16 = 16\%
The weighted average concentration equals total solute divided by total solution volume.

Anahtar Kavram

Weighted Average in Mixture Problems
Soru 246Soru

For all real numbers xx and yy, the custom binary operator Δ\Delta is defined by xΔy=x25xy+4y2x \Delta y = x^2 - 5xy + 4y^2. The function ff is defined by f(x)=xΔ1f(x) = x \Delta 1. What is the positive integer value of kk such that f(f(k))=0f(f(k)) = 0?

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Cevap: 5

Cevap

The positive integer value of kk is 55.
Substituting y=1y = 1 into xΔy=x25xy+4y2x \Delta y = x^2 - 5xy + 4y^2 gives f(x)=x25x+4f(x) = x^2 - 5x + 4. For f(f(k))=0f(f(k)) = 0, the outer function evaluation requires f(k)f(k) to be a root of f(x)=0f(x) = 0. Solving x25x+4=0x^2 - 5x + 4 = 0 yields roots 11 and 44. Setting f(k)=4f(k) = 4 yields k25k=0k^2 - 5k = 0, which has roots k=0k = 0 and k=5k = 5. Since kk must be a positive integer, k=5k = 5. Setting f(k)=1f(k) = 1 yields irrational values, so 55 is the unique solution.

Adım Adım Çözüm

1
Substitute y=1y = 1 into the operator definition to find f(x)f(x).
f(x)=x25x+4f(x) = x^2 - 5x + 4
f(x)=xΔ1=x25x(1)+4(1)2f(x) = x \Delta 1 = x^2 - 5x(1) + 4(1)^2.
2
Set f(u)=0f(u) = 0 for u=f(k)u = f(k) and solve for uu.
u=1u = 1 or u=4u = 4
Factoring u25u+4=0u^2 - 5u + 4 = 0 yields (u1)(u4)=0(u - 1)(u - 4) = 0.
3
Solve f(k)=4f(k) = 4 for kk.
k=0k = 0 or k=5k = 5
k25k+4=4    k25k=0    k(k5)=0k^2 - 5k + 4 = 4 \implies k^2 - 5k = 0 \implies k(k - 5) = 0.
4
Solve f(k)=1f(k) = 1 for kk and check for integer solutions.
k=5±132k = \frac{5 \pm \sqrt{13}}{2} (irrational roots)
k25k+4=1    k25k+3=0k^2 - 5k + 4 = 1 \implies k^2 - 5k + 3 = 0.
5
Select the positive integer solution.
k=5k = 5
The value k=0k = 0 is not positive, and the roots from f(k)=1f(k) = 1 are not integers.

Anahtar Kavram

Custom operator evaluation, composite/nested functions, and quadratic root analysis.
Soru 247Soru

A set of 55 integers has an arithmetic mean of 1212. If four of the integers are 88, 1010, 1414, and 1515, what is the value of the fifth integer?

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Cevap: 13

Cevap

The value of the fifth integer is 1313.
The total sum of a set of numbers is given by the formula Sum=Mean×n\text{Sum} = \text{Mean} \times n. For 55 numbers with a mean of 1212, the total sum is 5×12=605 \times 12 = 60. The sum of the four provided numbers is 8+10+14+15=478 + 10 + 14 + 15 = 47. Subtracting 4747 from 6060 yields 1313, which is the value of the fifth integer.

Adım Adım Çözüm

1
Find the total sum of the 5 integers
The total sum is 6060
The sum of a set of numbers equals the arithmetic mean multiplied by the total count of numbers (12×5=6012 \times 5 = 60).
2
Sum the four given integers
The sum of the four integers is 4747
Adding the given numbers: 8+10+14+15=478 + 10 + 14 + 15 = 47.
3
Subtract the sum of the known integers from the total sum
The fifth integer is 1313
Subtracting 4747 from 6060 gives 6047=1360 - 47 = 13.

Anahtar Kavram

Arithmetic Mean
Tahmini Süre:45s
Soru 248Soru

Let SS be the set of all real numbers xx that satisfy the inequality x26x0x^2 - 6x \leq 0. How many integer values of kk are there such that the equation x4x+2=k||x - 4| - |x + 2|| = k has at least one solution xSx \in S?

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Cevap: 7

Cevap

The correct answer is 7.
Solving the quadratic inequality x26x0x^2 - 6x \leq 0 gives the domain S=[0,6]S = [0, 6]. Over this closed interval, the function g(x)=x4x+2g(x) = ||x - 4| - |x + 2|| is continuous and attains its minimum value of 0 at x=1x = 1 and its maximum value of 6 at x=4x = 4 (and throughout [4,6][4, 6]). By the Intermediate Value Theorem, g(x)g(x) takes on all real values in the interval [0,6][0, 6]. The integer values of kk for which g(x)=kg(x) = k has a solution in SS are 0,1,2,3,4,5,0, 1, 2, 3, 4, 5, and 66, making a total of 7 integers.

Adım Adım Çözüm

1
Solve the quadratic inequality to define the set SS.
x26x0    x(x6)0    0x6x^2 - 6x \leq 0 \iff x(x - 6) \leq 0 \iff 0 \leq x \leq 6. Thus, S=[0,6]S = [0, 6].
The solution set of x(x6)0x(x-6) \leq 0 lies between the roots x=0x = 0 and x=6x = 6 inclusive.
2
Analyze the inner function f(x)=x4x+2f(x) = |x - 4| - |x + 2| for x[0,6]x \in [0, 6].
Critical points of absolute values occur at x=2x = -2 and x=4x = 4. Within [0,6][0, 6], we split at x=4x = 4.
The signs of (x4)(x - 4) and (x+2)(x + 2) determine how the absolute value bars simplify.
3
Evaluate g(x)=f(x)g(x) = |f(x)| on the sub-interval [0,4][0, 4].
For 0x40 \leq x \leq 4: x4=4x|x - 4| = 4 - x and x+2=x+2|x + 2| = x + 2. Thus, f(x)=22xf(x) = 2 - 2x and g(x)=22xg(x) = |2 - 2x|.
Evaluating g(x)g(x) at key points gives g(0)=2g(0) = 2, g(1)=0g(1) = 0, and g(4)=6g(4) = 6. By continuity, g(x)g(x) covers all values in [0,6][0, 6] on this interval.
4
Evaluate g(x)g(x) on the sub-interval [4,6][4, 6].
For 4x64 \leq x \leq 6: x4=x4|x - 4| = x - 4 and x+2=x+2|x + 2| = x + 2. Thus, f(x)=6f(x) = -6 and g(x)=6=6g(x) = |-6| = 6.
g(x)g(x) remains constant at 6 for x[4,6]x \in [4, 6].
5
Determine the range of g(x)g(x) on SS and count the integer values of kk.
The range of g(x)g(x) for x[0,6]x \in [0, 6] is [0,6][0, 6]. The integer values in this interval are 0,1,2,3,4,5,60, 1, 2, 3, 4, 5, 6.
The equation g(x)=kg(x) = k has a solution in SS if and only if kk lies in the range of g(x)g(x) over SS. There are 60+1=76 - 0 + 1 = 7 such integers.

Anahtar Kavram

Absolute value functions case evaluation and finding the range over a restricted domain.
Soru 249Soru

An integer nn is chosen at random from the set of all positive integers less than or equal to 120120. What is the probability that nn is a multiple of either 33 or 55, but not a multiple of 1515?

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Cevap: 0.4

Cevap

The probability that the selected integer is a multiple of either 3 or 5, but not 15, is 0.4 (or 2/5).
The total number of integers from 1 to 120 is 120. Multiples of 3 up to 120 total 40, multiples of 5 total 24, and multiples of 15 total 8. Integers that are multiples of 3 but not 15 number 40 - 8 = 32. Integers that are multiples of 5 but not 15 number 24 - 8 = 16. The total number of favorable outcomes is 32 + 16 = 48. Thus, the probability is 48/120 = 2/5 = 0.4.

Adım Adım Çözüm

1
Count total possible outcomes in the sample space.
The total number of integers from 1 to 120 is 120.
Each integer in the set {1, 2, ..., 120} is equally likely to be selected.
2
Count the number of multiples of 3, 5, and 15 within the range.
Multiples of 3: 40; Multiples of 5: 24; Multiples of 15: 8.
Since 120 is divisible by 3, 5, and 15, the count of multiples of k up to 120 is 120/k.
3
Calculate the number of integers that are multiples of 3 or 5, but not 15.
Number of favorable outcomes = (Multiples of 3 only) + (Multiples of 5 only) = (40 - 8) + (24 - 8) = 32 + 16 = 48.
Multiples of 15 are common multiples of both 3 and 5 and must be excluded completely according to the condition 'not a multiple of 15'.
4
Compute the single-event probability.
Probability = 48 / 120 = 2 / 5 = 0.4.
Probability of a single event is defined as the ratio of favorable outcomes to total possible outcomes.

Anahtar Kavram

Basic Single-Event Probability with Set Restrictions
Soru 250Soru

A quality control inspector reviews a shipment containing a total of NN customized components, of which exactly 3 are defective. The inspector randomly selects 3 components from the shipment one by one without replacement. If the probability that at least one of the selected components is defective is equal to 3135\frac{31}{35}, what is the value of NN?

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Cevap: 7

Cevap

The total number of components in the shipment, NN, is 7.
To find NN, apply the complementary probability formula P(at least 1 defective)=1P(0 defective)P(\text{at least 1 defective}) = 1 - P(\text{0 defective}). Given that P(at least 1 defective)=3135P(\text{at least 1 defective}) = \frac{31}{35}, the probability of drawing zero defective components is 13135=4351 - \frac{31}{35} = \frac{4}{35}. Out of NN total components, N3N-3 are non-defective. Drawing 3 non-defective components without replacement gives P(0 defective)=(N3)(N4)(N5)N(N1)(N2)P(\text{0 defective}) = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}. Setting this equal to 435\frac{4}{35} and testing integer values starting at N=6N=6 yields N=7N=7, since 4×3×27×6×5=24210=435\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}.

Adım Adım Çözüm

1
Calculate the probability of the complementary event (selecting no defective components).
P(0 defective)=13135=435P(\text{0 defective}) = 1 - \frac{31}{35} = \frac{4}{35}.
Calculating 'at least one' directly requires summing three separate cases (1 defective, 2 defective, 3 defective), whereas using the complement P(at least 1)=1P(none)P(\text{at least 1}) = 1 - P(\text{none}) requires evaluating only one scenario.
2
Formulate the algebraic expression for picking 3 non-defective components without replacement.
P(0 defective)=(N33)(N3)=(N3)(N4)(N5)N(N1)(N2)P(\text{0 defective}) = \frac{\binom{N-3}{3}}{\binom{N}{3}} = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}.
There are N3N-3 non-defective components out of NN total components, and 3 are selected without replacement.
3
Equate the algebraic probability to the known complementary probability and solve for NN.
\frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)} = \frac{4}{35} \implies N = 7.
Testing N=7N = 7 yields 4×3×27×6×5=24210=435\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}. The function is strictly increasing for N6N \ge 6, making N=7N = 7 the unique integer solution.

Anahtar Kavram

Complementary Probability and Dependent Sampling (Without Replacement)
Soru 251Soru

If 3x+4=19|3x + 4| = 19, what is the positive value of xx?

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Cevap: 5

Cevap

The positive value of xx is 5.
Solving the equation 3x+4=19|3x + 4| = 19 yields two cases: 3x+4=193x + 4 = 19, giving x=5x = 5, and 3x+4=193x + 4 = -19, giving x=233x = -\frac{23}{3}. The positive value among these solutions is 5.

Adım Adım Çözüm

1
Split the absolute value equation into two linear equations.
3x+4=193x + 4 = 19 or 3x+4=193x + 4 = -19
The equation a=b|a| = b for b0b \ge 0 implies a=ba = b or a=ba = -b.
2
Solve for xx in both cases.
x=5x = 5 or x=233x = -\frac{23}{3}
Subtract 4 from both sides and divide by 3.
3
Select the value satisfying the problem constraint.
5
The problem asks specifically for the positive value of xx.

Anahtar Kavram

Solving Linear Absolute Value Equations
Soru 252Soru

A municipal water treatment facility uses three pumps—Pump PP, Pump QQ, and Pump RR—to fill a main reservoir. Working alone at their respective constant rates, Pump PP can fill the reservoir in 20 hours, Pump QQ in 30 hours, and Pump RR in 40 hours.

All three pumps begin filling the empty reservoir simultaneously. After 4 hours, Pump PP shuts off. Pumps QQ and RR continue operating together for another 8 hours before Pump QQ is also shut off. Pump RR continues to run alone until the reservoir is completely filled.

How many total hours does it take to fill the reservoir from the start of the process until it is completely filled?

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Cevap: 16

Cevap

The total time required to fill the reservoir from start to finish is 16 hours.
Converting the individual completion times into work rates per hour (1/20, 1/30, and 1/40 of the reservoir per hour) allows us to determine the combined output per phase. In the first 4 hours, all three pumps fill 52/120 of the reservoir. In the next 8 hours, Pumps Q and R fill 56/120 of the reservoir. This leaves 12/120 (or 1/10) of the job remaining. Pump R, working at a rate of 1/40 per hour, takes 4 hours to complete the final 1/10. Adding all three durations (4 + 8 + 4) gives a total of 16 hours.

Adım Adım Çözüm

1
Determine the individual hourly rates of work for each pump.
Pump P completes 120\frac{1}{20} of the job per hour, Pump Q completes 130\frac{1}{30} of the job per hour, and Pump R completes 140\frac{1}{40} of the job per hour.
Work rate is the reciprocal of the total time required to complete one whole task.
2
Calculate the work completed during Stage 1 (4 hours with all 3 pumps working).
Combined rate = 120+130+140=6+4+3120=13120\frac{1}{20} + \frac{1}{30} + \frac{1}{40} = \frac{6 + 4 + 3}{120} = \frac{13}{120}. Work done = 4×13120=521204 \times \frac{13}{120} = \frac{52}{120}.
Work done equals combined rate multiplied by time spent working together.
3
Calculate the work completed during Stage 2 (8 hours with Pumps Q and R working).
Combined rate of Q and R = 130+140=4+3120=7120\frac{1}{30} + \frac{1}{40} = \frac{4 + 3}{120} = \frac{7}{120}. Work done = 8×7120=561208 \times \frac{7}{120} = \frac{56}{120}.
Only Pumps Q and R are active during this second period.
4
Calculate the remaining fraction of work after Stage 1 and Stage 2.
Total work done so far = 52120+56120=108120=910\frac{52}{120} + \frac{56}{120} = \frac{108}{120} = \frac{9}{10}. Remaining work = 1910=1101 - \frac{9}{10} = \frac{1}{10}.
Subtracting completed work from the whole (1) leaves the remaining work to be completed.
5
Calculate the time required for Pump R to finish the remaining work alone.
Time = 1/101/40=4010=4\frac{1/10}{1/40} = \frac{40}{10} = 4 hours.
Time equals remaining work divided by the individual rate of Pump R.
6
Sum the time spent across all stages.
Total time = 4 hours (Stage 1)+8 hours (Stage 2)+4 hours (Stage 3)=164 \text{ hours (Stage 1)} + 8 \text{ hours (Stage 2)} + 4 \text{ hours (Stage 3)} = 16 hours.
The total elapsed time is the sum of the durations of each distinct phase.

Anahtar Kavram

Work Rate and Combined Work in Multi-Stage Processes

Alternatif Yöntem

Assume a convenient total reservoir volume equal to the least common multiple of the times: 120 units. Pump P produces 6 units/hr, Pump Q produces 4 units/hr, and Pump R produces 3 units/hr. Stage 1 (4 hrs): 4 × (6 + 4 + 3) = 52 units. Stage 2 (8 hrs): 8 × (4 + 3) = 56 units. Total filled = 108 units. Remaining = 12 units. Stage 3 (Pump R alone): 12 / 3 = 4 hrs. Total time = 4 + 8 + 4 = 16 hours.
Tahmini Süre:2m 0s
Soru 253Soru

Train X and Train Y depart simultaneously from Station A and Station B, respectively, heading toward each other along a straight, 360-mile track. Train X travels at a constant speed of 40 miles per hour for the first 2 hours, after which its speed increases by 50 percent. Train Y travels at a constant speed of 50 miles per hour for the first hour, stops for 30 minutes due to a signal delay, and then resumes its journey at a constant speed of 70 miles per hour. How many hours after their simultaneous departure will the two trains meet?

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Cevap: 3.5

Cevap

3.5 hours
Evaluating distance piecewise up to t=2t = 2 hours shows Train X covers 80 miles and Train Y covers 85 miles, leaving 195 miles between them. After t=2t = 2 hours, their relative rate of convergence is 60+70=13060 + 70 = 130 mph. Dividing 195 miles by 130 mph yields 1.5 additional hours, bringing the total time since departure to 3.5 hours.

Adım Adım Çözüm

1
Determine distance traveled by Train X in the first 2 hours and its new speed.
Train X distance = 40 mph×2 hours=80 miles40 \text{ mph} \times 2 \text{ hours} = 80 \text{ miles}. New speed = 40×(1+0.50)=60 mph40 \times (1 + 0.50) = 60 \text{ mph}.
Train X increases its speed by 50% after the 2-hour mark.
2
Determine distance traveled by Train Y during the first 2 hours.
In hour 1 (t=0t=0 to t=1t=1): 50 mph×1 h=50 miles50 \text{ mph} \times 1 \text{ h} = 50 \text{ miles}. During delay (t=1t=1 to t=1.5t=1.5): 0 miles0 \text{ miles}. From t=1.5t=1.5 to t=2.0t=2.0: 70 mph×0.5 h=35 miles70 \text{ mph} \times 0.5 \text{ h} = 35 \text{ miles}. Total distance = 50+0+35=85 miles50 + 0 + 35 = 85 \text{ miles}.
Accounting for Train Y's initial leg, 30-minute delay, and new speed up to t=2t = 2 hours.
3
Find the remaining distance separating the two trains at t=2t = 2 hours.
Total distance covered by both trains = 80+85=165 miles80 + 85 = 165 \text{ miles}. Remaining distance = 360165=195 miles360 - 165 = 195 \text{ miles}.
Subtracting total distance covered by both trains from the track length of 360 miles.
4
Calculate the time required to cover the remaining distance using relative speed.
Relative closing speed for t>2t > 2 is 60+70=130 mph60 + 70 = 130 \text{ mph}. Additional time = 195130=1.5 hours\frac{195}{130} = 1.5 \text{ hours}.
When two objects move toward each other, their rates add together to find the closing rate.
5
Calculate total elapsed time from departure.
Total time = 2.0+1.5=3.5 hours2.0 + 1.5 = 3.5 \text{ hours}.
Combining the initial 2-hour evaluation window with the additional 1.5 hours required to meet.

Anahtar Kavram

Relative speed and piecewise distance calculations for converging vehicles with variable speeds and delays.
Tahmini Süre:3m 0s
Soru 254Soru

The table below categorizes 100 analysts at a consulting firm by their department and experience level:

DepartmentJunior (1–3 yrs)Senior (4–7 yrs)Lead (8+ yrs)Total
Technology14161040
Analytics1218535
Operations911525
Total354520100

If one analyst is selected at random from this group, what is the probability that the selected analyst works in the Analytics department or has Lead experience, but NOT both?

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Cevap: 0.45

Cevap

0.45
To find the probability of selecting an analyst who is either in the Analytics department or at the Lead level, but not both, we count the analysts in Analytics who are not Leads (12+18=3012 + 18 = 30) and the analysts at the Lead level who are not in Analytics (10+5=1510 + 5 = 15). The total number of favorable outcomes is 30+15=4530 + 15 = 45. Dividing by the total pool of 100100 analysts yields a probability of 45100=0.45\frac{45}{100} = 0.45.

Adım Adım Çözüm

1
Determine the total size of the sample space.
Total analysts N=100N = 100.
Basic probability requires dividing favorable outcomes by total possible outcomes.
2
Calculate the number of analysts satisfying 'Analytics, but NOT Lead'.
Junior Analytics (1212) + Senior Analytics (1818) = 3030.
Excludes the 55 Lead analysts in the Analytics department.
3
Calculate the number of analysts satisfying 'Lead, but NOT Analytics'.
Technology Lead (1010) + Operations Lead (55) = 1515.
Excludes the 55 Lead analysts in the Analytics department.
4
Sum the non-overlapping favorable counts and compute probability.
Favorable outcomes =30+15=45= 30 + 15 = 45; Probability =45100=0.45= \frac{45}{100} = 0.45.
Probability of a single event is defined as Favorable OutcomesTotal Outcomes\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}.

Anahtar Kavram

Basic Single-Event Probability from Two-Way Tabular Data with Mutually Exclusive Set Conditions
Soru 255Soru

The roots of the quadratic polynomial P(x)=x2+bx+cP(x) = x^2 + bx + c are r1r_1 and r2r_2, where bb and cc are constants. If r1+2r_1 + 2 and r2+2r_2 + 2 are the roots of the quadratic equation x210x+21=0x^2 - 10x + 21 = 0, what is the value of cc?

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Cevap: 5

Cevap

The value of cc is 5.
Factoring x210x+21=0x^2 - 10x + 21 = 0 yields roots 33 and 77. Because these roots represent r1+2r_1 + 2 and r2+2r_2 + 2, subtracting 22 from each root gives the original roots r1=1r_1 = 1 and r2=5r_2 = 5. In the monic polynomial P(x)=x2+bx+cP(x) = x^2 + bx + c, the constant coefficient cc equals the product of the roots r1×r2=1×5=5r_1 \times r_2 = 1 \times 5 = 5.

Adım Adım Çözüm

1
Solve for the roots of the given equation x210x+21=0x^2 - 10x + 21 = 0
The roots are 33 and 77.
Factoring the quadratic expression yields (x3)(x7)=0(x - 3)(x - 7) = 0, so x=3x = 3 or x=7x = 7.
2
Determine the roots r1r_1 and r2r_2 of P(x)P(x)
r1=1r_1 = 1 and r2=5r_2 = 5.
Since the roots of the second equation are shifted by +2+2, we set r1+2=3    r1=1r_1 + 2 = 3 \implies r_1 = 1 and r2+2=7    r2=5r_2 + 2 = 7 \implies r_2 = 5.
3
Calculate the constant term cc
c=5c = 5.
By Vieta's formulas, for any monic quadratic polynomial x2+bx+cx^2 + bx + c, the constant term cc is the product of the roots r1r2=1×5=5r_1 r_2 = 1 \times 5 = 5.

Anahtar Kavram

Root transformation of quadratic equations and relationship between roots and coefficients via Vieta's formulas.
Soru 256Soru

What is the numerical value of 283464\frac{2^8 \cdot 3^4}{6^4}?

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Cevap: 16

Cevap

16
Rewriting the base 66 as 232 \cdot 3 gives 64=24346^4 = 2^4 \cdot 3^4. Substituting this into the original expression yields 28342434\frac{2^8 \cdot 3^4}{2^4 \cdot 3^4}. Cancelling 343^4 leaves 2824=284=24=16\frac{2^8}{2^4} = 2^{8-4} = 2^4 = 16.

Adım Adım Çözüm

1
Rewrite composite bases into prime factors
64=(23)4=24346^4 = (2 \cdot 3)^4 = 2^4 \cdot 3^4
Applying the power of a product rule (ab)n=anbn(ab)^n = a^n b^n allows base matching.
2
Simplify the quotient using exponent subtraction
28342434=284344=2430=161=16\frac{2^8 \cdot 3^4}{2^4 \cdot 3^4} = 2^{8-4} \cdot 3^{4-4} = 2^4 \cdot 3^0 = 16 \cdot 1 = 16
Applying the quotient rule aman=amn\frac{a^m}{a^n} = a^{m-n} for powers with equal bases.

Anahtar Kavram

Exponent rules: Power of a Product (ab)n=anbn(ab)^n = a^n b^n and Quotient of Powers aman=amn\frac{a^m}{a^n} = a^{m-n}
Soru 257Soru

A commuter drives from home to work at a constant speed of 4545 miles per hour and arrives 1010 minutes late. On the following day, driving along the exact same route at a constant speed of 6060 miles per hour, the commuter arrives 55 minutes early. What is the distance, in miles, from the commuter's home to work?

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Cevap: 45

Cevap

The distance from the commuter's home to work is 45 miles.
Let dd be the distance in miles between home and work. At 4545 miles per hour, the time taken is d45\frac{d}{45} hours. At 6060 miles per hour, the time taken is d60\frac{d}{60} hours. The difference between being 1010 minutes late and 55 minutes early is 1515 minutes, or 1560=14\frac{15}{60} = \frac{1}{4} hour. Setting up the equation d45d60=14\frac{d}{45} - \frac{d}{60} = \frac{1}{4} and finding a common denominator of 180180 gives 4d3d180=14\frac{4d - 3d}{180} = \frac{1}{4}, which simplifies to d180=14\frac{d}{180} = \frac{1}{4}. Multiplying both sides by 180180 yields d=45d = 45 miles.

Adım Adım Çözüm

1
Calculate the difference in travel time between the two trips in hours.
The time difference is 10 minutes late - (-5 minutes early) = 15 minutes = 1/4 hour.
Since speeds are given in miles per hour, time must be expressed in hours.
2
Set up an equation relating the two travel times using the formula time = distance / speed.
d/45 - d/60 = 1/4, where d represents the distance in miles.
The slower speed (45 mph) takes 1/4 hour longer than the faster speed (60 mph).
3
Solve the equation for the distance variable d.
(4d - 3d)/180 = 1/4 => d/180 = 1/4 => d = 45.
Find a common denominator for 45 and 60, which is 180, and solve for d.

Anahtar Kavram

Rate, Time, and Distance Problems
Soru 258Soru

At a commercial bank, an audit of 300300 business loan applications revealed that each application underwent at least one of three specialized risk evaluations: Credit Risk, Market Risk, and Operational Risk. Exactly 180180 applications underwent Credit Risk evaluation, 150150 underwent Market Risk evaluation, and 135135 underwent Operational Risk evaluation. If exactly 9595 applications underwent exactly two of the three risk evaluations, how many loan applications underwent all three risk evaluations?

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Cevap: 35

Cevap

35 loan applications underwent all three risk evaluations.
The total number of applications is 300300. The sum of individual evaluations is 180+150+135=465180 + 150 + 135 = 465. Using the Venn diagram region formula, Sum of individual setsTotal=Exactly 2+2×(Exactly 3)\text{Sum of individual sets} - \text{Total} = \text{Exactly } 2 + 2 \times (\text{Exactly } 3). Substituting the known values gives 465300=95+2x465 - 300 = 95 + 2x, which simplifies to 165=95+2x165 = 95 + 2x. Solving for xx gives 2x=702x = 70, so x=35x = 35.

Adım Adım Çözüm

1
State the fundamental overlapping set region equation for 3 sets.
\text{Total} = \text{Exactly } 1 + \text{Exactly } 2 + \text{Exactly } 3 + \text{Neither}
Every application falls into exactly one of these non-overlapping region categories.
2
Calculate the sum of the individual set counts.
180 + 150 + 135 = 465
Summing individual set counts counts items in 1 set once, 2 sets twice, and 3 sets three times.
3
Formulate the algebraic identity linking individual sums to region totals.
\text{Sum of Individual Sets} = \text{Exactly } 1 + 2(\text{Exactly } 2) + 3(\text{Exactly } 3)
This accounts for the multiple counting of overlapping regions.
4
Subtract the Total equation from the Sum equation to eliminate the 'Exactly 1' term.
465 - 300 = \text{Exactly } 2 + 2(\text{Exactly } 3) \Rightarrow 165 = 95 + 2(\text{Exactly } 3)
Subtracting (Exactly 1+Exactly 2+Exactly 3)(\text{Exactly } 1 + \text{Exactly } 2 + \text{Exactly } 3) from (Exactly 1+2Exactly 2+3Exactly 3)(\text{Exactly } 1 + 2\cdot\text{Exactly } 2 + 3\cdot\text{Exactly } 3) leaves 1Exactly 2+2Exactly 31\cdot\text{Exactly } 2 + 2\cdot\text{Exactly } 3.
5
Solve the linear equation for the 'Exactly 3' intersection.
2(\text{Exactly } 3) = 165 - 95 = 70 \Rightarrow \text{Exactly } 3 = 35
Dividing the remaining difference of 70 by 2 gives the number of applications in all three sets.

Anahtar Kavram

Three-Set Venn Diagram Region Decomposition
Soru 259Soru

For all real numbers xx and yy, the custom binary operator \odot is defined by xy=x2yxy2x \odot y = x^2 y - x y^2. The function ff is defined by f(x)=x3f(x) = x \odot 3. What is the value of f(f(2))f(f(2))?

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Cevap: 162

Cevap

The value of f(f(2))f(f(2)) is 162.
To evaluate the nested function f(f(2))f(f(2)), first determine f(2)f(2) using the definition f(x)=x3f(x) = x \odot 3. Applying the definition of the custom operator xy=x2yxy2x \odot y = x^2 y - x y^2 with x=2x = 2 and y=3y = 3 yields 23=(2)2(3)(2)(3)2=1218=62 \odot 3 = (2)^2(3) - (2)(3)^2 = 12 - 18 = -6. Then, evaluate f(6)=63=(6)2(3)(6)(3)2=108(54)=162f(-6) = -6 \odot 3 = (-6)^2(3) - (-6)(3)^2 = 108 - (-54) = 162.

Adım Adım Çözüm

1
Evaluate the inner function expression f(2)
f(2) = 2 \odot 3 = (2)^2(3) - (2)(3)^2 = 12 - 18 = -6
By definition, f(x) = x \odot 3. Substituting x = 2 gives 2 \odot 3, which applies the custom operator definition x^2 y - x y^2.
2
Evaluate the outer function expression f(f(2)) = f(-6)
f(-6) = -6 \odot 3 = (-6)^2(3) - (-6)(3)^2 = 36(3) - (-54) = 108 + 54 = 162
Substitute the inner result -6 into the function f(x).

Anahtar Kavram

Custom Binary Operators and Nested Function Evaluation
Soru 260Soru

An agricultural facility uses two automated loading conveyors, Conveyor AA and Conveyor BB, to fill a grain storage silo. Working alone at its constant rate, Conveyor AA can fill the silo in 10 hours. Working alone at its constant rate, Conveyor BB can fill the silo in 15 hours. Conveyor AA begins filling the empty silo alone. After 4 hours, Conveyor BB is turned on, and both conveyors work together at their respective constant rates until the silo is completely full. How many total hours does it take, from the moment Conveyor AA starts, to completely fill the silo?

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Cevap: 7.6

Cevap

The total time required to fill the silo is 7.6 hours.
Conveyor A fills 1/10 of the silo per hour and works alone for 4 hours, completing 2/5 of the total job. The remaining 3/5 of the job is completed by both conveyors working together at a combined rate of 1/10 + 1/15 = 1/6 silo per hour. Dividing 3/5 by 1/6 gives 3.6 hours for the second stage. Adding the initial 4 hours yields a total time of 7.6 hours.

Adım Adım Çözüm

1
Determine the individual hourly rates of Conveyor A and Conveyor B.
Conveyor A rate = 1/101/10 silo per hour; Conveyor B rate = 1/151/15 silo per hour.
Work rate is the reciprocal of the time required to complete the total job.
2
Calculate the fraction of the silo filled by Conveyor A working alone for 4 hours.
Work completed = 4×110=254 \times \frac{1}{10} = \frac{2}{5} of the silo.
Work done equals rate multiplied by time.
3
Determine the remaining fraction of work to be completed.
Remaining work = 125=351 - \frac{2}{5} = \frac{3}{5} of the silo.
Subtract the completed fraction from the total job (1).
4
Calculate the combined work rate when both conveyors operate together.
Combined rate = 110+115=3+230=530=16\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6} silo per hour.
When working together, individual rates add up.
5
Determine the time required for both conveyors to finish the remaining work.
Time = 3/51/6=185=3.6\frac{3/5}{1/6} = \frac{18}{5} = 3.6 hours.
Divide remaining work by the combined work rate.
6
Sum the time spent in both stages to find the total time.
Total time = 4+3.6=7.64 + 3.6 = 7.6 hours.
The total duration includes 4 hours of Conveyor A operating alone plus 3.6 hours of combined operation.

Anahtar Kavram

Work Rate and Combined Work
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