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231 questions

Question 161Question

A city planning board needs to form a 66-member advisory task force selected from a pool of 55 architects and 55 civil engineers. The task force must include at least 22 architects and at least 22 civil engineers. However, two specific architects, Architect X and Architect Y, cannot both serve on the task force together. How many different 66-member task forces can be formed satisfying these conditions?

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Answer: 135

Answer

The total number of different valid 6-member task forces that can be formed is 135.
To solve this problem, we apply the addition principle over mutually exclusive cases of committee composition, followed by complementary counting to enforce the exclusion restriction. First, we identify the valid breakdown of architects and engineers for a 6-member team requiring at least 2 of each profession: (4 architects, 2 engineers), (3 architects, 3 engineers), and (2 architects, 4 engineers). Calculating the combinations for each breakdown yields 50, 100, and 50 ways respectively, totaling 200 unconstrained team options. Next, we determine how many of these teams include both Architect X and Architect Y. Pre-assigning both architects reduces the remaining available architects to 3. The invalid cases for each breakdown are 30, 30, and 5 respectively, totaling 65 invalid configurations. Subtracting the 65 invalid teams from the 200 total unconstrained teams yields 135 valid task forces.

Step-by-Step Solution

1
Identify the allowed group breakdowns under the restriction of at least 2 architects and at least 2 engineers.
The valid (architect, engineer) count pairs for a 6-member task force are (4, 2), (3, 3), and (2, 4).
Choosing 5 architects would leave only 1 engineer, violating the minimum requirement of 2 engineers, and vice versa.
2
Compute the total combinations without the exclusion restriction.
Total unconstrained combinations = 200.
(54)(52)+(53)(53)+(52)(54)=(5×10)+(10×10)+(10×5)=50+100+50=200\binom{5}{4}\binom{5}{2} + \binom{5}{3}\binom{5}{3} + \binom{5}{2}\binom{5}{4} = (5 \times 10) + (10 \times 10) + (10 \times 5) = 50 + 100 + 50 = 200.
3
Calculate the number of task forces that violate the restriction by including both Architect X and Architect Y.
Total invalid combinations = 65.
If Architect X and Architect Y are both included, selecting remaining architects from the other 3 yields: (32)(52)+(31)(53)+(30)(54)=(3×10)+(3×10)+(1×5)=30+30+5=65\binom{3}{2}\binom{5}{2} + \binom{3}{1}\binom{5}{3} + \binom{3}{0}\binom{5}{4} = (3 \times 10) + (3 \times 10) + (1 \times 5) = 30 + 30 + 5 = 65.
4
Subtract the invalid combinations from the total unconstrained combinations.
200 - 65 = 135.
Using the complementary counting principle provides the exact number of valid combinations where Architect X and Architect Y do not serve together.

Key Concept

Combinations with multiple category constraints and complementary counting for exclusion rules.
Question 162Question

A museum curator is arranging 66 distinct paintings in a single row along a gallery wall. If 22 specific paintings must not be placed adjacent to each other, how many different arrangements of the 66 paintings are possible?

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Answer: 480

Answer

480
To find the number of valid arrangements where two specific paintings are not adjacent, use complementary counting. First, compute the total number of ways to arrange 6 distinct paintings without restrictions, which is 6!=7206! = 720. Next, calculate the number of arrangements where the two specific paintings are placed adjacent to one another by treating them as a single block. There are 5 units in total to arrange (the pair block plus the remaining 4 individual paintings), which gives 5!=1205! = 120 ways. Since the two specific paintings can be arranged in 2!=22! = 2 ways inside their block, the total number of adjacent arrangements is 120×2=240120 \times 2 = 240. Finally, subtract the adjacent arrangements from the total arrangements: 720240=480720 - 240 = 480.

Step-by-Step Solution

1
Calculate the total number of ways to arrange all 6 paintings in a row without any restrictions.
6! = 720
There are 6 distinct items to arrange in 6 sequential positions.
2
Calculate the number of arrangements where the 2 specific paintings are adjacent (placed next to each other).
5! × 2! = 120 × 2 = 240
Treat the 2 specific paintings as a single block unit. This leaves 5 items to arrange (the block + 4 individual paintings), which can be ordered in 5! ways. Within the block, the 2 paintings can be ordered in 2! ways.
3
Subtract the number of adjacent arrangements from the total unrestricted arrangements.
720 - 240 = 480
Complementary counting dictates that valid non-adjacent arrangements equal total possible arrangements minus adjacent arrangements.

Key Concept

Permutations with Adjacency Restrictions (Complementary Counting)
Question 163Question

In rhombus ABCDABCD, the side length is 1010 and the length of diagonal BDBD is 1212. Line segment APAP is drawn perpendicular to side BCBC, with point PP lying on segment BCBC. What is the length of segment APAP?

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Answer: 9.6

Answer

9.6
The diagonals of rhombus ABCDABCD intersect perpendicularly at OO and bisect each other. Given BD=12BD = 12, half of the diagonal is BO=6BO = 6. Right triangle AOBAOB has hypotenuse AB=10AB = 10 and leg BO=6BO = 6, so by the Pythagorean theorem, leg AO=10262=8AO = \sqrt{10^2 - 6^2} = 8. Thus, diagonal AC=16AC = 16. The area of rhombus ABCDABCD is 12×AC×BD=12×16×12=96\frac{1}{2} \times AC \times BD = \frac{1}{2} \times 16 \times 12 = 96. The area is also equal to base×height=BC×AP=10×AP\text{base} \times \text{height} = BC \times AP = 10 \times AP. Setting 10×AP=9610 \times AP = 96 gives AP=9.6AP = 9.6.

Step-by-Step Solution

1
Find half the length of diagonal BD.
Segment BO = 6.
The diagonals of a rhombus bisect each other at right angles.
2
Apply the Pythagorean theorem to right triangle AOB to determine half of diagonal AC.
AO = 8, so diagonal AC = 16.
Triangle AOB has hypotenuse 10 and leg 6, forming a 6-8-10 Pythagorean triple.
3
Calculate the total area of rhombus ABCD from its diagonal lengths.
Area = 96.
The area of a rhombus equals half the product of its two diagonals.
4
Use the alternative area formula (base × height) to solve for altitude AP.
AP = 9.6.
Base BC = 10 and height AP give Area = 10 × AP = 96.

Key Concept

Properties of rhombus diagonals, Pythagorean theorem, and dual area formulas for quadrilaterals
Question 164Question

For all positive real numbers aa and bb, the custom operation \diamondsuit is defined by ab=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}. The function ff is defined for all x>0x > 0 by f(x)=x4f(x) = x \diamondsuit 4. If f(x)=2.5f(x) = 2.5, what is the value of xx that is greater than 44?

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Answer: 8

Answer

The value of xx greater than 44 is 88.
Applying the custom operator gives f(x)=x2+164xf(x) = \frac{x^2 + 16}{4x}. Setting this equal to 2.52.5 yields x2+164x=52\frac{x^2 + 16}{4x} = \frac{5}{2}, which simplifies to x210x+16=0x^2 - 10x + 16 = 0. The roots are x=2x = 2 and x=8x = 8. Since xx must be greater than 44, the only valid answer is 88.

Step-by-Step Solution

1
Substitute a=xa = x and b=4b = 4 into the custom operation definition ab=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}.
f(x)=x2+164xf(x) = \frac{x^2 + 16}{4x}
This establishes the explicit algebraic rule for the function f(x)f(x).
2
Set f(x)f(x) equal to 2.52.5 and clear the fraction.
x2+164x=2.5    x2+16=10x\frac{x^2 + 16}{4x} = 2.5 \implies x^2 + 16 = 10x
Multiplying both sides by 4x4x converts the rational equation into a standard polynomial equation.
3
Rearrange into standard quadratic form and solve by factoring.
x210x+16=0    (x2)(x8)=0    x=2x^2 - 10x + 16 = 0 \implies (x - 2)(x - 8) = 0 \implies x = 2 or x=8x = 8
Factoring determines all potential positive real solutions for xx.
4
Select the solution satisfying the constraint x>4x > 4.
x=8x = 8
The question explicitly specifies that xx must be greater than 44, eliminating x=2x = 2.

Key Concept

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.
Question 165Question

A short-pulse laser emits a single pulse lasting 8.4×1098.4 \times 10^{-9} seconds. A high-speed optical sensor completes one measurement cycle every 1.4×10111.4 \times 10^{-11} seconds. How many measurement cycles does the sensor complete during the duration of a single laser pulse?

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Answer: 600

Answer

600
Dividing the pulse duration (8.4×1098.4 \times 10^{-9} seconds) by the sensor cycle time (1.4×10111.4 \times 10^{-11} seconds) yields 8.41.4×109(11)=6×102=600\frac{8.4}{1.4} \times 10^{-9 - (-11)} = 6 \times 10^2 = 600 complete cycles.

Step-by-Step Solution

1
Set up the division expression for the total number of cycles.
\frac{8.4 \times 10^{-9}\text{ seconds}}{1.4 \times 10^{-11}\text{ seconds}}
To find how many cycle intervals fit within the total pulse duration.
2
Divide the decimal coefficients.
8.41.4=6\frac{8.4}{1.4} = 6
Separating the numerical coefficients from the powers of ten.
3
Apply exponent rules to divide powers of 10.
10^{-9 - (-11)} = 10^{-9 + 11} = 10^2 = 100
Dividing powers with the same base requires subtracting the denominator exponent from the numerator exponent.
4
Combine results to find total cycles.
6×100=6006 \times 100 = 600
Multiplying coefficient quotient by the simplified power of ten.

Key Concept

Division of Numbers in Scientific Notation and Exponent Rules
Question 166Question

An automated risk-management system uses three independent algorithms—Algorithm X, Algorithm Y, and Algorithm Z—to detect fraudulent transactions. The probability that Algorithm X detects a given fraudulent transaction is 35\frac{3}{5}, the probability that Algorithm Y detects it is 23\frac{2}{3}, and the probability that Algorithm Z detects it is 34\frac{3}{4}. If a fraudulent transaction occurs, what is the probability that it will be detected by at least two of these three algorithms?

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Answer: 0.75

Answer

The probability that the transaction is detected by at least two of the three algorithms is 0.75 (or 3/4).
Because the algorithms operate independently, the event 'at least two algorithms detect the transaction' consists of four mutually exclusive outcomes: exactly X and Y detect (probability 6/60 = 0.10), exactly X and Z detect (probability 9/60 = 0.15), exactly Y and Z detect (probability 12/60 = 0.20), and all three detect (probability 18/60 = 0.30). Summing these four probabilities gives 0.10 + 0.15 + 0.20 + 0.30 = 0.75.

Step-by-Step Solution

1
Determine the complementary probabilities of non-detection for each algorithm.
P(X does not detect) = 2/5, P(Y does not detect) = 1/3, and P(Z does not detect) = 1/4.
The probability of an event's complement is 1 minus the probability of the event.
2
Calculate the probability for each scenario where exactly two algorithms detect the transaction.
P(X and Y only) = 6/60, P(X and Z only) = 9/60, P(Y and Z only) = 12/60.
Since the algorithms operate independently, joint probabilities are calculated by multiplying individual probabilities.
3
Calculate the probability that all three algorithms detect the transaction.
P(X and Y and Z) = 18/60.
Multiplying the individual detection probabilities of all three independent algorithms.
4
Sum the probabilities of all qualifying mutually exclusive outcomes.
(6/60) + (9/60) + (12/60) + (18/60) = 45/60 = 0.75.
The events representing different combinations of detections are mutually exclusive, so their probabilities add directly.

Key Concept

Independent Events and Addition Rule for Mutually Exclusive Outcomes
Question 167Question

A university department tracked the number of research articles published by its 8 faculty members over a five-year period. The numbers of publications for 7 of the faculty members were 4,7,9,12,15,18,4, 7, 9, 12, 15, 18, and 2323. If the arithmetic mean of the number of publications for all 8 faculty members is equal to 1.251.25 times their median, what is the number of publications for the 8th faculty member?

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Answer: 47

Answer

47
The sum of the 7 known values is 8888, making the total sum 88+x88 + x and the arithmetic mean 88+x8\frac{88 + x}{8}. Assuming x15x \ge 15, the 4th and 5th numbers when sorted are 1212 and 1515, yielding a median of 13.513.5. Setting the mean equal to 1.25×13.5=16.8751.25 \times 13.5 = 16.875 gives 88+x8=16.875\frac{88 + x}{8} = 16.875, which simplifies to 88+x=13588 + x = 135 and yields x=47x = 47.

Step-by-Step Solution

1
Calculate the sum of the 7 known data points and express the mean in terms of the unknown 8th value xx.
Sum of 7 known values = 4+7+9+12+15+18+23=884 + 7 + 9 + 12 + 15 + 18 + 23 = 88. Total mean = 88+x8\frac{88 + x}{8}.
The arithmetic mean of nn values is the sum of all values divided by nn.
2
Analyze the position of xx in sorted order to determine the median.
Assuming x15x \ge 15, the 4th and 5th values in ascending order are 1212 and 1515, giving a median of 12+152=13.5\frac{12 + 15}{2} = 13.5.
For an even number of data points (88), the median is the average of the 4th and 5th terms in sorted order.
3
Formulate and solve the equation linking the mean and median.
88+x8=1.25×13.5=16.875    88+x=135    x=47\frac{88 + x}{8} = 1.25 \times 13.5 = 16.875 \implies 88 + x = 135 \implies x = 47.
The problem specifies that the mean is equal to 1.251.25 times the median.

Key Concept

Calculating mean and median of a dataset containing an unknown value.
Estimated Time:1m 30s
Question 168Question

A laboratory technician has 1212 liters of a solution containing 15%15\% salt by weight. The technician wants to increase the salt concentration to 25%25\% by adding a second solution that contains 40%40\% salt by weight. How many liters of the 40%40\% solution must be added?

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Answer: 8

Answer

8 liters of the 40% solution must be added.
The total amount of salt contributed by the two solutions is 0.15(12)+0.40x=1.8+0.40x0.15(12) + 0.40x = 1.8 + 0.40x. The final mixture volume is (12+x)(12 + x) liters, and its concentration must be 25%25\%. Equating the total salt to 25%25\% of total volume yields 1.8+0.40x=0.25(12+x)1.8 + 0.40x = 0.25(12 + x). Expanding gives 1.8+0.40x=3.0+0.25x1.8 + 0.40x = 3.0 + 0.25x, which simplifies to 0.15x=1.20.15x = 1.2, giving x=8x = 8 liters.

Step-by-Step Solution

1
Define the unknown variable and calculate the initial solute quantity.
Let xx be the volume of the 40%40\% solution added (in liters). Salt in initial solution = 0.15×12=1.80.15 \times 12 = 1.8 liters.
Establishing the mass balance of the salt solute is necessary to construct the algebraic equation.
2
Set up the algebraic concentration equation.
1.8+0.40x=0.25(12+x)1.8 + 0.40x = 0.25(12 + x)
The combined salt from both solutions must equal 25%25\% of the total combined liquid volume (12+x)(12 + x) liters.
3
Solve the linear equation for xx.
1.8+0.40x=3.0+0.25x    0.15x=1.2    x=81.8 + 0.40x = 3.0 + 0.25x \implies 0.15x = 1.2 \implies x = 8
Isolating xx yields the exact number of liters required.

Key Concept

Algebraic mixture problems using mass balance equations.
Estimated Time:1m 30s
Question 169Question

In the geometric configuration formed by adjacent triangles ABCABC and ACDACD sharing segment ACAC, ABC=90\angle ABC = 90^\circ and ACD=90\angle ACD = 90^\circ. The lengths of the sides of triangle ABCABC are AB=9AB = 9 and BC=12BC = 12. If ADC=30\angle ADC = 30^\circ, what is the length of segment ADAD?

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Answer: 30

Answer

30
Applying the Pythagorean theorem to right triangle ABC gives AC = sqrt(9^2 + 12^2) = 15. In right triangle ACD with angle ADC = 30 degrees, side AC is opposite the 30-degree angle. By the 30-60-90 triangle side ratio (1 : sqrt(3) : 2), the hypotenuse AD is equal to twice the short leg AC, giving AD = 2 * 15 = 30.

Step-by-Step Solution

1
Calculate hypotenuse AC of right triangle ABC using the Pythagorean theorem.
AC = 15
In right triangle ABC with legs 9 and 12, AC^2 = 9^2 + 12^2 = 225, so AC = 15.
2
Identify the side relationships in special right triangle ACD.
AC is the shorter leg opposite the 30-degree angle ADC, and AD is the hypotenuse.
Since angle ACD is 90 degrees and angle ADC is 30 degrees, triangle ACD is a 30-60-90 right triangle.
3
Compute the hypotenuse AD from short leg AC.
AD = 30
In any 30-60-90 right triangle, the hypotenuse is twice the length of the leg opposite the 30-degree angle.

Key Concept

Pythagorean Theorem and Special Right Triangles
Estimated Time:1m 30s
Question 170Question

In the xyxy-plane, line L1L_1 is defined by the equation y=3x4y = 3x - 4. Line L2L_2 is perpendicular to line L1L_1 and passes through the point (6,2)(6, 2). What is the xx-intercept of line L2L_2?

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Answer: 12

Answer

12
Line L1L_1 has a slope of 33. A line perpendicular to L1L_1 must have a slope equal to the negative reciprocal of 33, which is 13-\frac{1}{3}. Substituting the point (6,2)(6, 2) into the point-slope form gives y2=13(x6)y - 2 = -\frac{1}{3}(x - 6), which simplifies to y=13x+4y = -\frac{1}{3}x + 4. Setting y=0y = 0 yields 0=13x+40 = -\frac{1}{3}x + 4, giving x=12x = 12 as the xx-intercept.

Step-by-Step Solution

1
Find the slope of line L1L_1
Slope m1=3m_1 = 3
The equation y=3x4y = 3x - 4 is in slope-intercept form y=mx+by = mx + b, where m=3m = 3.
2
Determine the slope of perpendicular line L2L_2
Slope m2=13m_2 = -\frac{1}{3}
Perpendicular lines have slopes that are negative reciprocals of each other (m1m2=1m_1 \cdot m_2 = -1).
3
Find the equation of line L2L_2
y=13x+4y = -\frac{1}{3}x + 4
Substitute point (6,2)(6, 2) into point-slope formula y2=13(x6)y - 2 = -\frac{1}{3}(x - 6).
4
Calculate the xx-intercept of line L2L_2
x=12x = 12
Set y=0y = 0 in the line equation and solve for xx: 0=13x+4    x=120 = -\frac{1}{3}x + 4 \implies x = 12.

Key Concept

Perpendicular lines have negative reciprocal slopes (m1m2=1m_1 \cdot m_2 = -1). The xx-intercept is the point where y=0y = 0.
Question 171Question

A water reservoir initially contains 450450 liters of water. Water drains out of the reservoir at a constant rate of rr liters per hour, while an inlet pipe supplies water at a constant rate of 1818 liters per hour. If the reservoir contains 390390 liters of water after 66 hours, what is the value of rr?

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Answer: 28

Answer

The value of rr is 2828.
The reservoir starts with 450450 liters. Over 66 hours, water enters at 1818 liters/hour and leaves at rr liters/hour, giving a net volume equation of 450+6(18r)=390450 + 6(18 - r) = 390. Simplifying yields 5586r=390558 - 6r = 390, which subtracts to 6r=168-6r = -168, giving r=28r = 28.

Step-by-Step Solution

1
Set up the linear equation representing the net change in water volume over time.
450+6(18r)=390450 + 6(18 - r) = 390
The final volume equals the initial volume plus the net water added (inflow rate minus outflow rate multiplied by hours).
2
Expand and simplify the linear expression.
5586r=390558 - 6r = 390
Distribute 66 across (18r)(18 - r) to obtain 1086r108 - 6r, then add to 450450.
3
Isolate the variable term 6r-6r.
6r=168-6r = -168
Subtract 558558 from both sides of the equation.
4
Solve for the rate rr.
r=28r = 28
Divide both sides by 6-6.

Key Concept

Formulating and solving a linear equation in one variable from a rate problem context.
Estimated Time:1m 30s
Question 172Question

If N=24×33×52N = 2^4 \times 3^3 \times 5^2, how many positive integer factors of NN are divisible by 66 but not divisible by 1212?

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Answer: 9

Answer

The number of positive integer factors of NN divisible by 6 but not by 12 is 9.
Any factor of N=24×33×52N = 2^4 \times 3^3 \times 5^2 is of the form 2a×3b×5c2^a \times 3^b \times 5^c. For the factor to be divisible by 66, we must have a1a \ge 1 and b1b \ge 1. For it to not be divisible by 12=22×3112 = 2^2 \times 3^1, we must have a<2a < 2. Therefore, aa must equal 11. The possible choices for aa are 11 value (11), for bb are 33 values (1,2,31, 2, 3), and for cc are 33 values (0,1,20, 1, 2). Multiplying these options gives 1×3×3=91 \times 3 \times 3 = 9.

Step-by-Step Solution

1
Express the prime factorization structure of a factor of NN.
Any factor of NN takes the form 2a×3b×5c2^a \times 3^b \times 5^c with bounds 0a40 \le a \le 4, 0b30 \le b \le 3, and 0c20 \le c \le 2.
Divisors of a number are formed by taking prime factors with exponents between zero and their maximum powers in the original number.
2
Determine exponent constraints for divisibility by 6 and non-divisibility by 12.
Divisibility by 66 requires a1a \ge 1 and b1b \ge 1. Non-divisibility by 1212 requires a<2a < 2. Thus, a=1a = 1 exactly.
A factor must contain at least one factor of 2 and one factor of 3 to be a multiple of 6, but containing two or more factors of 2 makes it a multiple of 12.
3
Count combinations of choices for the exponents.
Exponent aa has 11 option (a=1a = 1), bb has 33 options (b{1,2,3}b \in \{1, 2, 3\}), and cc has 33 options (c{0,1,2}c \in \{0, 1, 2\}). Total =1×3×3=9= 1 \times 3 \times 3 = 9.
Applying the fundamental counting principle by multiplying the number of choices for independent prime factor exponents.

Key Concept

Counting Divisors with Prime Factorization and Divisibility Constraints
Question 173Question

Let nn be an integer such that 15n15-15 \le n \le 15. How many integer values of nn satisfy both of the following conditions?

1. (1)n2+n+1<0(-1)^{n^2 + n + 1} < 0
2. (n3)(n+4)2<0(n - 3)(n + 4)^2 < 0

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Answer: 17

Answer

17
Condition 1 is satisfied by every integer because n2+n=n(n+1)n^2 + n = n(n+1) is always even (as the product of two consecutive integers), which makes n2+n+1n^2 + n + 1 always odd and (1)odd=1<0(-1)^{\text{odd}} = -1 < 0. Condition 2 requires (n3)(n+4)2<0(n - 3)(n + 4)^2 < 0. Since (n+4)2>0(n+4)^2 > 0 for all n4n \neq -4, this reduces to n<3n < 3 while excluding n=4n = -4. Within the range 15n15-15 \le n \le 15, there are 18 integers strictly less than 3, and removing n=4n = -4 leaves 17 valid values.

Step-by-Step Solution

1
Determine the parity of the exponent n2+n+1n^2 + n + 1
n2+n+1n^2 + n + 1 is always odd for any integer nn, making (1)n2+n+1=1<0(-1)^{n^2 + n + 1} = -1 < 0 unconditionally true.
The product of consecutive integers n(n+1)n(n+1) is always even, so adding 1 results in an odd number.
2
Solve the sign inequality (n3)(n+4)2<0(n - 3)(n + 4)^2 < 0
n<3n < 3 with n4n \neq -4.
A squared expression is strictly positive except when its base is zero. At n=4n = -4, the product becomes 0, violating the strict inequality.
3
Count the integer solutions in the range 15n15-15 \le n \le 15
17 integers satisfy both conditions.
There are 18 integers less than 3 in the interval [15,15][-15, 15], and excluding n=4n = -4 gives 181=1718 - 1 = 17.

Key Concept

Parity of consecutive integer products and sign rules for squared terms in inequalities
Estimated Time:2m 0s
Question 174Question

A car travels a total distance of 120120 miles. For the first 6060 miles, the car travels at an average speed of 3030 miles per hour, and for the remaining 6060 miles, the car travels at an average speed of 6060 miles per hour. What is the average speed of the car, in miles per hour, for the entire 120120-mile trip?

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Answer: 40

Answer

The average speed of the car for the entire trip is 4040 miles per hour.
The overall average speed is determined by dividing total distance (120120 miles) by total time (33 hours), yielding 4040 miles per hour.

Step-by-Step Solution

1
Calculate time taken for the first segment
Time = 22 hours
Using Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}, 6030=2\frac{60}{30} = 2 hours.
2
Calculate time taken for the second segment
Time = 11 hour
Using Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}, 6060=1\frac{60}{60} = 1 hour.
3
Determine total distance and total time
Total distance = 120120 miles, Total time = 33 hours
Add distances (60+60=12060 + 60 = 120) and times (2+1=32 + 1 = 3).
4
Calculate overall average speed
Average speed = 4040 miles per hour
Divide total distance by total time: 1203=40\frac{120}{3} = 40.

Key Concept

Average Speed and Rates
Estimated Time:1m 0s
Question 175Question

A sequence of numbers begins with t1=4t_1 = 4. For all integers n2n \geq 2, each term is defined by tn=3tn12t_n = 3t_{n-1} - 2. What is the value of t5t_5?

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Answer: 244

Answer

The 5th term of the sequence, t5t_5, is equal to 244.
Applying the recursive relation tn=3tn12t_n = 3t_{n-1} - 2 step-by-step starting from t1=4t_1 = 4 gives t2=10t_2 = 10, t3=28t_3 = 28, t4=82t_4 = 82, and finally t5=244t_5 = 244.

Step-by-Step Solution

1
Find the second term, t2t_2, using t1=4t_1 = 4.
t2=3(4)2=122=10t_2 = 3(4) - 2 = 12 - 2 = 10
Substitute n=2n = 2 into the recursive rule tn=3tn12t_n = 3t_{n-1} - 2.
2
Find the third term, t3t_3, using t2=10t_2 = 10.
t3=3(10)2=302=28t_3 = 3(10) - 2 = 30 - 2 = 28
Substitute n=3n = 3 into the recursive rule.
3
Find the fourth term, t4t_4, using t3=28t_3 = 28.
t4=3(28)2=842=82t_4 = 3(28) - 2 = 84 - 2 = 82
Substitute n=4n = 4 into the recursive rule.
4
Find the fifth term, t5t_5, using t4=82t_4 = 82.
t5=3(82)2=2462=244t_5 = 3(82) - 2 = 246 - 2 = 244
Substitute n=5n = 5 into the recursive rule.

Key Concept

Evaluating terms in a recursively defined sequence
Question 176Question

At the beginning of a quarter, the cost of raw material for a manufacturing plant was xx dollars per ton. During the first month, the cost per ton increased by 15%15\%. During the second month, the cost per ton decreased by 20%20\% relative to the first month's price. If the cost per ton at the end of the second month was $552\$552, what was the initial cost xx, in dollars?

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Answer: 600

Answer

600
A 15%15\% increase followed by a 20%20\% decrease yields a combined multiplier of (1+0.15)(10.20)=1.15×0.80=0.92(1 + 0.15)(1 - 0.20) = 1.15 \times 0.80 = 0.92. Setting 0.92x=5520.92x = 552 gives x=5520.92=600x = \frac{552}{0.92} = 600.

Step-by-Step Solution

1
Calculate the price multiplier for the first month
Price after month 1 is 1.15x1.15x
A 15%15\% increase scales the initial value by (1+0.15)=1.15(1 + 0.15) = 1.15.
2
Calculate the price multiplier for the second month
Price after month 2 is 1.15x×0.80=0.92x1.15x \times 0.80 = 0.92x
A 20%20\% decrease applies to the new base price, scaling it by (10.20)=0.80(1 - 0.20) = 0.80.
3
Solve for the initial price xx
x=600x = 600
Divide the given final price of $552\$552 by the overall multiplier 0.920.92.

Key Concept

Successive Percent Changes and Base Shifts
Estimated Time:1m 30s
Question 177Question

A plumbing service charges a flat diagnostic fee of $45\$45 plus $65\$65 for each hour of repair work. If the total bill for a repair job was $305\$305, how many hours of repair work were performed?

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Answer: 4

Answer

The total number of hours of repair work performed was 4.
To determine the number of repair hours, represent the scenario with the linear equation 45+65h=30545 + 65h = 305, where hh is the number of hours. Subtracting 45 from both sides yields 65h=26065h = 260. Dividing 260 by 65 gives h=4h = 4.

Step-by-Step Solution

1
Formulate the linear equation from the word problem context.
45+65h=30545 + 65h = 305
The total charge consists of a one-time fixed fee of 45plusavariablefeeof45 plus a variable fee of 65 per hour hh.
2
Isolate the variable term by subtracting 45 from both sides of the equation.
65h=26065h = 260
Subtracting the constant fee isolates the total cost incurred from hourly work.
3
Divide both sides by the coefficient of the variable to solve for hh.
h=4h = 4
Dividing the total labor charge (260)bythehourlyrate(260) by the hourly rate ( 65) yields the number of hours worked.

Key Concept

Linear Equations in One Variable
Question 178Question

On a real number line, point AA has coordinate 5-5 and point BB has coordinate 1111. Point CC has coordinate xx such that the distance between CC and the midpoint of line segment ABAB is equal to 13\frac{1}{3} of the distance between CC and point BB. What is the maximum possible value of xx?

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Answer: 5

Answer

The maximum possible value of xx is 55.
The midpoint of A(5)A(-5) and B(11)B(11) is M=3M = 3. Setting up the distance equation x3=13x11|x - 3| = \frac{1}{3}|x - 11| leads to 3x3=x113|x - 3| = |x - 11|. Evaluating the two cases 3(x3)=x113(x - 3) = x - 11 and 3(x3)=(x11)3(x - 3) = -(x - 11) yields solutions x=1x = -1 and x=5x = 5. The maximum possible value among these is 5.

Step-by-Step Solution

1
Calculate the coordinate of the midpoint of line segment ABAB.
Midpoint coordinate M=3M = 3.
The midpoint of two coordinates aa and bb on a number line is given by a+b2=5+112=3\frac{a + b}{2} = \frac{-5 + 11}{2} = 3.
2
Formulate the distance equation using absolute values.
3x3=x113|x - 3| = |x - 11|.
The distance between xx and 33 is x3|x - 3| and the distance between xx and 1111 is x11|x - 11|. The given condition is x3=13x11|x - 3| = \frac{1}{3}|x - 11|.
3
Solve the absolute value equation for all possible values of xx.
x=1x = -1 and x=5x = 5.
Splitting 3(x3)=x113(x - 3) = x - 11 yields x=1x = -1, and splitting 3(x3)=(x11)3(x - 3) = -(x - 11) yields x=5x = 5.
4
Select the maximum value among all valid solutions.
55
Comparing x=1x = -1 and x=5x = 5, the maximum value is 55.

Key Concept

Distance on a number line using absolute value and midpoint formula
Estimated Time:2m 0s
Question 179Question

For how many integer values of xx does the inequality 2x754||2x - 7| - 5| \le 4 hold true?

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Answer: 10

Answer

10
To solve 2x754||2x - 7| - 5| \le 4, break the outer absolute value into 42x754-4 \le |2x - 7| - 5 \le 4. Adding 5 to all parts yields 12x791 \le |2x - 7| \le 9. This produces two simultaneous conditions: 2x79|2x - 7| \le 9, which gives 1x8-1 \le x \le 8, and 2x71|2x - 7| \ge 1, which gives x3x \le 3 or x4x \ge 4. Intersecting these solution sets yields the real intervals [1,3][-1, 3] and [4,8][4, 8]. The integers contained within these intervals are 1,0,1,2,3-1, 0, 1, 2, 3 and 4,5,6,7,84, 5, 6, 7, 8, giving a total of 10 valid integer values.

Step-by-Step Solution

1
Unpack the outer absolute value inequality
42x754-4 \le |2x - 7| - 5 \le 4
An inequality of the form uk|u| \le k for k0k \ge 0 is equivalent to kuk-k \le u \le k.
2
Isolate the inner absolute value term
12x791 \le |2x - 7| \le 9
Adding 5 across the compound inequality isolates the term 2x7|2x - 7|.
3
Solve the upper bound inequality 2x79|2x - 7| \le 9
1x8-1 \le x \le 8
92x79    22x16    1x8-9 \le 2x - 7 \le 9 \implies -2 \le 2x \le 16 \implies -1 \le x \le 8.
4
Solve the lower bound inequality 2x71|2x - 7| \ge 1
x3x \le 3 or x4x \ge 4
The inequality u1|u| \ge 1 splits into u1u \ge 1 or u1u \le -1, yielding 2x71    x42x - 7 \ge 1 \implies x \ge 4 or 2x71    x32x - 7 \le -1 \implies x \le 3.
5
Determine the overlapping interval and count integer solutions
10 integer solutions
The intersection of [1,8][-1, 8] with ((,3][4,))((-\infty, 3] \cup [4, \infty)) is [1,3][4,8][-1, 3] \cup [4, 8]. The integers in this domain are 1,0,1,2,3,4,5,6,7,8-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, which equals 10 integer values.

Key Concept

Solving nested absolute value inequalities on the real number line
Question 180Question

At a public university, the ratio of the number of undergraduate students to the number of graduate students at the beginning of an academic year was 44 to 11. By the end of the academic year, the number of undergraduate students had increased by 15%15\%, and the total student population (undergraduate and graduate combined) had increased by 18%18\%. By what percent did the number of graduate students increase?

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Answer: 30

Answer

30%
Representing the initial populations by their ratio components (44 parts undergraduate, 11 part graduate, 55 parts total), an overall growth of 18%18\% across 55 total parts yields 5×0.18=0.905 \times 0.18 = 0.90 total parts of growth. The undergraduate population grew by 15%15\%, contributing 4×0.15=0.604 \times 0.15 = 0.60 parts of growth. The remaining 0.900.60=0.300.90 - 0.60 = 0.30 parts of growth must come from the graduate student component. Since the initial graduate population was 11 part, a growth of 0.300.30 parts corresponds to a 30%30\% increase.

Step-by-Step Solution

1
Set up initial component values using the ratio 4:14:1
Undergraduates = 400400, Graduates = 100100, Total = 500500
Choosing convenient numbers compatible with the 4:14:1 ratio simplifies percentage calculations.
2
Calculate the updated undergraduate count after a 15%15\% increase
New Undergraduates = 460460
An increase of 15%15\% means the new value is 115%115\% of the original 400400.
3
Calculate the updated total count after an 18%18\% increase
New Total = 590590
An overall increase of 18%18\% means the new total is 118%118\% of the original 500500.
4
Subtract the new undergraduate count from the new total count to find the new graduate count
New Graduates = 130130
The total population is the sum of undergraduate and graduate populations.
5
Calculate the percent change for the graduate population
30%30\%
The increase of 3030 graduate students over the initial 100100 graduate students represents a 30%30\% increase.

Key Concept

Weighted percentage change across subgroup components
Estimated Time:2m 0s
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