Arithmetic

306 soru

Soru 241Soru

A cyclist completes a journey consisting of three distinct segments: an uphill segment, a flat segment, and a downhill segment. The ratio of the distances of the uphill, flat, and downhill segments is 2:3:52 : 3 : 5, respectively. The cyclist's average speed on the flat segment is twice her average speed on the uphill segment, and her average speed on the downhill segment is three times her average speed on the uphill segment. If the cyclist's overall average speed for the entire journey is 3030 miles per hour, what is her average speed, in miles per hour, on the flat segment?

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Cevap: 31

Cevap

31
The correct average speed on the flat segment is 31 miles per hour. Setting up segment distances as 2x2x, 3x3x, and 5x5x (total distance 10x10x) and segment speeds as vv, 2v2v, and 3v3v, the segment times are t1=2xvt_1 = \frac{2x}{v}, t2=3x2vt_2 = \frac{3x}{2v}, and t3=5x3vt_3 = \frac{5x}{3v}. The total travel time is T=31x6vT = \frac{31x}{6v}. Dividing total distance 10x10x by total time TT yields an overall average speed of 60v31=30\frac{60v}{31} = 30. Solving for vv gives v=15.5v = 15.5 miles per hour. Thus, the average speed on the flat segment is 2v=312v = 31 miles per hour.

Adım Adım Çözüm

1
Define segment distances using ratio multipliers.
Distances are d1=2xd_1 = 2x, d2=3xd_2 = 3x, and d3=5xd_3 = 5x, giving total distance D=10xD = 10x.
The distances of the three segments are in the ratio 2:3:52 : 3 : 5.
2
Express segment speeds relative to the uphill speed vv.
Uphill speed is vv, flat speed is 2v2v, and downhill speed is 3v3v.
The problem states flat speed is twice uphill speed, and downhill speed is three times uphill speed.
3
Calculate the time spent on each segment.
t1=2xvt_1 = \frac{2x}{v}, t2=3x2vt_2 = \frac{3x}{2v}, and t3=5x3vt_3 = \frac{5x}{3v}.
Time equals distance divided by speed (t=dvt = \frac{d}{v}).
4
Calculate total travel time by summing individual segment times.
T=xv(2+32+53)=31x6vT = \frac{x}{v} \left(2 + \frac{3}{2} + \frac{5}{3}\right) = \frac{31x}{6v}.
Combining fractions with a common denominator of 6 gives 12+9+106=316\frac{12 + 9 + 10}{6} = \frac{31}{6}.
5
Relate total distance and total time to the overall average speed.
Average Speed=Total DistanceTotal Time=10x31x6v=60v31=30\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{10x}{\frac{31x}{6v}} = \frac{60v}{31} = 30.
Overall average speed is defined as total distance divided by total time.
6
Solve for vv and calculate the flat segment speed 2v2v.
v=15.5v = 15.5 mph, so flat segment speed =2(15.5)=31= 2(15.5) = 31 mph.
Solving 60v31=30\frac{60v}{31} = 30 yields v=15.5v = 15.5, making 2v=312v = 31.

Anahtar Kavram

Weighted Average Speed and Multi-Segment Distance-Rate-Time Ratios
Soru 242Soru

A sequence of positive real numbers a1,a2,a3,a_1, a_2, a_3, \dots is defined by a1=3a_1 = 3 and an+1=an+2ana_{n+1} = a_n + \frac{2}{a_n} for all integers n1n \ge 1. Which of the following statements must be true? Select all that apply.

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Cevap: The sequence a1,a2,a3,a_1, a_2, a_3, \dots is strictly increasing.; The term a50a_{50} is strictly greater than 1414.; The sequence of consecutive term differences dn=an+1and_n = a_{n+1} - a_n is strictly decreasing for n1n \ge 1.

Cevap

The correct statements are that the sequence is strictly increasing, the term a50a_{50} is strictly greater than 14, and the sequence of consecutive term differences dn=an+1and_n = a_{n+1} - a_n is strictly decreasing.
The statement asserting strict monotonicity is correct because an+1an=2an>0a_{n+1} - a_n = \frac{2}{a_n} > 0 for positive terms. The statement regarding a50>14a_{50} > 14 is correct because a502>32+4(49)=205>196a_{50}^2 > 3^2 + 4(49) = 205 > 196. The statement concerning consecutive differences is correct because 2an\frac{2}{a_n} strictly decreases as ana_n increases.

Adım Adım Çözüm

1
Analyze monotonicity of the sequence
an+1an=2an>0a_{n+1} - a_n = \frac{2}{a_n} > 0 for all n1n \ge 1
Since a1=3>0a_1 = 3 > 0, all terms remain positive, making each term strictly larger than the previous.
2
Analyze the sequence of term differences
dn=2and_n = \frac{2}{a_n} decreases as ana_n increases
Since ana_n grows strictly monotonically, its reciprocal strictly decreases, so dn+1<dnd_{n+1} < d_n.
3
Derive a lower bound for a50a_{50} using quadratic expansion
a502>205    a50>14a_{50}^2 > 205 \implies a_{50} > 14
Expanding ak+12=ak2+4+4ak2>ak2+4a_{k+1}^2 = a_k^2 + 4 + \frac{4}{a_k^2} > a_k^2 + 4 and summing from k=1k=1 to 4949 gives a502>32+4(49)=205>196=142a_{50}^2 > 3^2 + 4(49) = 205 > 196 = 14^2.
4
Derive an upper bound for a50a_{50} to test the rounding statement
a50<14.45a_{50} < 14.45, so rounding to the nearest integer yields 14
Using 4ak2<44k+5\frac{4}{a_k^2} < \frac{4}{4k+5}, the sum of error terms is bounded above by 04944x+5dx=ln(40.2)3.7\int_0^{49} \frac{4}{4x+5} dx = \ln(40.2) \approx 3.7. Thus a502<208.7<14.52a_{50}^2 < 208.7 < 14.5^2, so a50a_{50} rounds to 14, not 15.
5
Estimate a100a_{100} to test the magnitude bound
a100<409.420.23<25a_{100} < \sqrt{409.4} \approx 20.23 < 25
Summing the squared recurrence up to 100 terms gives a1002<405+ln(80.2)409.4a_{100}^2 < 405 + \ln(80.2) \approx 409.4, showing a100a_{100} cannot exceed 25.

Anahtar Kavram

Estimation of non-linear recursive sequences using squared bounds and integral comparison.
Tahmini Süre:3m 0s
Soru 243Soru

Two deep-space radio signals have power measurements of S1=4.5×107S_1 = 4.5 \times 10^{-7} watts and S2=3.5×106S_2 = 3.5 \times 10^{-6} watts. A combined signal power is defined as S3=S1+S2S_3 = S_1 + S_2. When S3S_3 is written in scientific notation as a×10na \times 10^n, where 1a<101 \leq a < 10 and nn is an integer, what is the value of a+na + n?

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Cevap: 2.05-2.05

Cevap

The value of a+na + n is 2.05-2.05.
To add numbers in scientific notation, first rewrite them with identical powers of 10. Expressing 4.5×1074.5 \times 10^{-7} as 0.45×1060.45 \times 10^{-6} allows direct addition with 3.5×1063.5 \times 10^{-6}, resulting in 3.95×1063.95 \times 10^{-6}. Here a=3.95a = 3.95 (which satisfies 1a<101 \leq a < 10) and n=6n = -6. Adding a+na + n yields 3.95+(6)=2.053.95 + (-6) = -2.05.

Adım Adım Çözüm

1
Express both quantities with a common power of 10 to allow addition.
S1=4.5×107=0.45×106S_1 = 4.5 \times 10^{-7} = 0.45 \times 10^{-6} watts.
Before adding numbers in scientific notation, their powers of 10 must match.
2
Add the coefficients while maintaining the common power of 10.
S3=(0.45+3.5)×106=3.95×106S_3 = (0.45 + 3.5) \times 10^{-6} = 3.95 \times 10^{-6} watts.
Distributive property allows adding coefficients once exponents match.
3
Verify proper scientific notation form a×10na \times 10^n where 1a<101 \leq a < 10.
a=3.95a = 3.95 and n=6n = -6.
The coefficient 3.953.95 satisfies 13.95<101 \leq 3.95 < 10, so no further decimal shift is needed.
4
Calculate the requested sum a+na + n.
3.95+(6)=2.053.95 + (-6) = -2.05.
Adding the coefficient 3.953.95 to the negative exponent 6-6 gives 2.05-2.05.

Anahtar Kavram

Scientific Notation Addition and Place Value Alignment
Soru 244Soru

If xx and yy are positive integers such that 3x+24y3x4y+1=11,5203^{x+2} \cdot 4^y - 3^x \cdot 4^{y+1} = 11,520, what is the value of x+yx + y?

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Cevap: 6

Cevap

6
Factoring 3x4y3^x \cdot 4^y from the expression 3x+24y3x4y+13^{x+2} \cdot 4^y - 3^x \cdot 4^{y+1} yields 3x4y(3241)=53x4y3^x \cdot 4^y (3^2 - 4^1) = 5 \cdot 3^x \cdot 4^y. Setting this equal to 11,520 and dividing by 5 gives 3x4y=2,3043^x \cdot 4^y = 2,304. Prime factorization of 2,304 gives 32443^2 \cdot 4^4, so x=2x = 2 and y=4y = 4. The sum x+yx + y is equal to 6.

Adım Adım Çözüm

1
Factor out the greatest common exponential factor 3x4y3^x \cdot 4^y from the left side of the equation.
3x4y(3241)=11,5203^x \cdot 4^y (3^2 - 4^1) = 11,520
By exponent rules, 3x+2=3x323^{x+2} = 3^x \cdot 3^2 and 4y+1=4y414^{y+1} = 4^y \cdot 4^1.
2
Evaluate the constant factor inside the parentheses.
3241=94=53^2 - 4^1 = 9 - 4 = 5, so 53x4y=11,5205 \cdot 3^x \cdot 4^y = 11,520
Simplifying numerical exponents.
3
Divide both sides of the equation by 5.
3x4y=2,3043^x \cdot 4^y = 2,304
Isolating the variable exponential terms.
4
Determine the prime factorization of 2,304 into powers of 3 and 4.
2,304=9256=32442,304 = 9 \cdot 256 = 3^2 \cdot 4^4, which implies x=2x = 2 and y=4y = 4
Unique factorization for integer bases.
5
Calculate the sum x+yx + y.
x+y=2+4=6x + y = 2 + 4 = 6
Answering the explicit prompt.

Anahtar Kavram

Factoring Exponential Expressions and Unique Factorization
Soru 245Soru

For two positive integers mm and nn, the greatest common divisor is gcd(m,n)=15\gcd(m, n) = 15 and the least common multiple is lcm(m,n)=900\text{lcm}(m, n) = 900. If m=75m = 75, what is the total number of positive divisors of nn?

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Cevap: 18

Cevap

18
The correct answer is 18. First, use the relation mn=gcd(m,n)lcm(m,n)m \cdot n = \gcd(m, n) \cdot \text{lcm}(m, n) to find n=1590075=180n = \frac{15 \cdot 900}{75} = 180. Next, write 180 in prime factorized form: 180=223251180 = 2^2 \cdot 3^2 \cdot 5^1. The total number of positive divisors is found by adding 1 to each exponent and multiplying the results: (2+1)(2+1)(1+1)=332=18(2 + 1)(2 + 1)(1 + 1) = 3 \cdot 3 \cdot 2 = 18.

Adım Adım Çözüm

1
Calculate the value of nn using the fundamental product identity for GCD and LCM.
n=gcd(m,n)lcm(m,n)m=1590075=180n = \frac{\gcd(m, n) \cdot \text{lcm}(m, n)}{m} = \frac{15 \cdot 900}{75} = 180.
For any two positive integers, the product of the integers equals the product of their GCD and LCM.
2
Find the prime factorization of 180180.
180=223251180 = 2^2 \cdot 3^2 \cdot 5^1.
Breaking down 180 into prime powers allows determination of the total count of positive divisors.
3
Apply the divisor counting formula by adding 1 to each prime exponent and multiplying.
(2+1)(2+1)(1+1)=332=18(2+1)(2+1)(1+1) = 3 \cdot 3 \cdot 2 = 18.
If an integer has prime factorization p1e1p2e2pkekp_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}, the number of positive divisors is (e1+1)(e2+1)(ek+1)(e_1+1)(e_2+1)\cdots(e_k+1).

Anahtar Kavram

GCD and LCM fundamental identity (ab=gcd(a,b)lcm(a,b)a \cdot b = \gcd(a,b) \cdot \text{lcm}(a,b)) combined with the prime factorization divisor counting formula.
Tahmini Süre:1m 30s
Soru 246Soru

In Year 1, a software company had a total of 400 subscribers divided between a Basic plan and a Premium plan, with 70%70\% of the subscribers enrolled in the Basic plan. In Year 2, the number of Basic plan subscribers decreased by 20%20\%, while the total number of subscribers across both plans increased by 10%10\%. What was the percentage increase in the number of Premium plan subscribers from Year 1 to Year 2?

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Cevap: 80%80\%

Cevap

The percentage increase in Premium plan subscribers from Year 1 to Year 2 was 80%80\%.
In Year 1, 70%70\% of the 400 subscribers were on the Basic plan (0.70×400=2800.70 \times 400 = 280), which leaves 120 subscribers on the Premium plan. In Year 2, the total subscriber count increased by 10%10\% to 440 (400×1.10=440400 \times 1.10 = 440), while Basic subscribers decreased by 20%20\% to 224 (280×0.80=224280 \times 0.80 = 224). Subtracting 224 Basic subscribers from the total 440 gives 216 Premium subscribers in Year 2. The increase in Premium subscribers is 216120=96216 - 120 = 96. To find the percentage increase, divide the increase of 96 by the original Premium subscriber count of 120, yielding 96120=0.80\frac{96}{120} = 0.80, or 80%80\%.

Adım Adım Çözüm

1
Calculate the number of Basic and Premium subscribers in Year 1
Basic subscribers = 0.70×400=2800.70 \times 400 = 280; Premium subscribers = 400280=120400 - 280 = 120
Determining the starting baseline values for each category is necessary to evaluate subsequent changes.
2
Calculate the total subscribers and Basic subscribers in Year 2
Total subscribers in Year 2 = 400×(1+0.10)=440400 \times (1 + 0.10) = 440; Basic subscribers in Year 2 = 280×(10.20)=224280 \times (1 - 0.20) = 224
Apply the given percentage changes to find the updated group totals in Year 2.
3
Find the number of Premium subscribers in Year 2 and the absolute increase
Premium subscribers in Year 2 = 440224=216440 - 224 = 216; Absolute increase = 216120=96216 - 120 = 96
Subtracting the Basic subscriber count from the total count yields the Premium subscriber count for Year 2.
4
Compute the percentage increase for Premium subscribers
Percentage increase = 96120×100%=80%\frac{96}{120} \times 100\% = 80\%
Divide the absolute increase by the initial Year 1 Premium subscriber count (the correct base value) and convert to a percentage.

Anahtar Kavram

Successive percentage changes and base shifting in multi-part totals
Tahmini Süre:1m 30s
Soru 247Soru

If xx and yy are real numbers such that x32|x - 3| \le 2 and y+14|y + 1| \le 4, which of the following could be the value of xy|x - y|? Select all such values.

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Cevap: 00; 55; 1010

Cevap

The possible values of xy|x - y| are 0, 5, and 10.
Solving the inequalities yields 1x51 \le x \le 5 and 5y3-5 \le y \le 3. The expression xy|x - y| represents the distance between xx and yy on the number line. Because the intervals overlap between 1 and 3, xx and yy can be equal, making the minimum distance 0. The maximum distance occurs at the extreme points x=5x = 5 and y=5y = -5, giving a distance of 5(5)=10|5 - (-5)| = 10. Therefore, any value from 0 to 10 inclusive is possible, making 0, 5, and 10 valid values.

Adım Adım Çözüm

1
Solve the absolute value inequality for xx.
2x32    1x5-2 \le x - 3 \le 2 \implies 1 \le x \le 5
Unwrapping x32|x - 3| \le 2 gives the bounds for xx on the real number line.
2
Solve the absolute value inequality for yy.
4y+14    5y3-4 \le y + 1 \le 4 \implies -5 \le y \le 3
Unwrapping y+14|y + 1| \le 4 gives the bounds for yy on the real number line.
3
Determine the minimum and maximum possible values for xy|x - y|.
Minimum value is 0 (since the intervals [1,5][1, 5] and [5,3][-5, 3] overlap at [1,3][1, 3]). Maximum value is 5(5)=10|5 - (-5)| = 10. Thus, 0xy100 \le |x - y| \le 10.
The absolute value xy|x - y| represents the distance between xx and yy on the number line, which can take any real value from 0 to 10.
4
Evaluate the given choices against the range [0,10][0, 10].
The values 0, 5, and 10 fall within [0,10][0, 10], whereas 2-2 is impossible for absolute values and 1212 exceeds the maximum bound.
Determines which specific options are valid outcomes for xy|x - y|.

Anahtar Kavram

Absolute value as distance on the real number line and range of differences between bounded real variables
Tahmini Süre:1m 30s
Soru 248Soru

The positive integer NN is divisible by 6060. If the greatest common divisor of NN and 500500 is 100100, which of the following statements MUST be true? Select all such statements.

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Cevap: NN is a multiple of 300300.; The prime factorization of NN contains at least two factors of 22.; The greatest common divisor of NN and 250250 is 5050.

Cevap

The statements asserting that NN is a multiple of 300300, that the prime factorization of NN contains at least two factors of 22, and that the greatest common divisor of NN and 250250 is 5050 must all be true.
Analyzing the prime factorizations: 60=22315160 = 2^2 \cdot 3^1 \cdot 5^1 and 500=2253500 = 2^2 \cdot 5^3. Since NN is divisible by 6060, the prime factorization of NN must have exponents a2a \ge 2 for prime 22, b1b \ge 1 for prime 33, and c1c \ge 1 for prime 55. Furthermore, gcd(N,500)=100=2252\gcd(N, 500) = 100 = 2^2 \cdot 5^2. The GCD rule requires taking the minimum exponent for each prime factor: min(a,2)=2    a2\min(a, 2) = 2 \implies a \ge 2, and min(c,3)=2    c=2\min(c, 3) = 2 \implies c = 2. Therefore, N=2a3b52N = 2^{a} \cdot 3^{b} \cdot 5^{2} where a2a \ge 2 and b1b \ge 1. This means NN is divisible by 223152=3002^2 \cdot 3^1 \cdot 5^2 = 300, contains at least two prime factors of 22, and has gcd(N,250)=gcd(2a3b52,2153)=2152=50\gcd(N, 250) = \gcd(2^a \cdot 3^b \cdot 5^2, 2^1 \cdot 5^3) = 2^1 \cdot 5^2 = 50.

Adım Adım Çözüm

1
Express 6060 and 500500 in their prime factorizations.
60=22315160 = 2^2 \cdot 3^1 \cdot 5^1 and 500=2253500 = 2^2 \cdot 5^3.
Prime factorization allows us to analyze divisibility and GCD conditions in terms of prime exponents.
2
Apply the condition that NN is divisible by 6060.
The exponent of 22 in NN is at least 22, the exponent of 33 is at least 11, and the exponent of 55 is at least 11.
For NN to be divisible by an integer, NN must contain at least as many of each prime factor as that integer.
3
Apply the GCD condition gcd(N,500)=100=2252\gcd(N, 500) = 100 = 2^2 \cdot 5^2.
The exponent of 55 in NN must be exactly 22.
Since 500500 has 535^3 and the GCD has 525^2, taking the minimum exponent of 55 between NN and 500500 yields 22. Thus, NN has 525^2 and not 535^3 or higher.
4
Evaluate each statement against the established exponent bounds for NN.
NN has prime factor powers 2231522^{\ge 2} \cdot 3^{\ge 1} \cdot 5^2. This guarantees NN is a multiple of 223152=3002^2 \cdot 3^1 \cdot 5^2 = 300, contains at least two factors of 22, and yields gcd(N,250)=2152=50\gcd(N, 250) = 2^1 \cdot 5^2 = 50.
Comparing the prime factor requirements of each choice confirms which statements MUST be true.

Anahtar Kavram

Prime factorization rules for divisibility and greatest common divisor (GCD)
Soru 249Soru

In Year 1, a retail company's total revenue was generated by two divisions: Division X, which accounted for 40%40\% of total revenue, and Division Y, which accounted for the remaining 60%60\%. From Year 1 to Year 2, revenue from Division X increased by 20%20\%, while revenue from Division Y decreased by 10%10\%. From Year 2 to Year 3, revenue from Division X decreased by 10%10\%, while revenue from Division Y increased by 20%20\%. Which of the following statements must be true? Select all such statements.

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Cevap: Total company revenue in Year 3 was 8%8\% greater than total company revenue in Year 1.; From Year 1 to Year 3, revenue from Division X increased by 8%8\%.; From Year 1 to Year 3, revenue from Division Y increased by 8%8\%.

Cevap

The correct statements are: total company revenue in Year 3 was 8% greater than in Year 1; revenue from Division X increased by 8% from Year 1 to Year 3; and revenue from Division Y increased by 8% from Year 1 to Year 3.
The statements confirming an 8% overall increase for total company revenue, Division X revenue, and Division Y revenue from Year 1 to Year 3 are all correct. A 20% increase followed by a 10% decrease (and vice versa) multiplies the starting amount by 1.20×0.90=1.081.20 \times 0.90 = 1.08, which represents a net 8% increase on each respective division's initial base.

Adım Adım Çözüm

1
Assign a variable to initial total revenue and calculate division revenues for Year 1.
Let total revenue in Year 1 be RR. Revenue from Division X is 0.40R0.40 R and revenue from Division Y is 0.60R0.60 R.
Establishing initial quantities provides the base for computing percentage changes across subsequent years.
2
Calculate division revenues and total revenue for Year 2.
Division X revenue in Year 2 = 0.40R×1.20=0.48R0.40 R \times 1.20 = 0.48 R. Division Y revenue in Year 2 = 0.60R×0.90=0.54R0.60 R \times 0.90 = 0.54 R. Total Year 2 revenue = 0.48R+0.54R=1.02R0.48 R + 0.54 R = 1.02 R.
Applying the respective 20% increase and 10% decrease to Year 1 base values yields Year 2 amounts.
3
Calculate division revenues and total revenue for Year 3.
Division X revenue in Year 3 = 0.48R×0.90=0.432R0.48 R \times 0.90 = 0.432 R. Division Y revenue in Year 3 = 0.54R×1.20=0.648R0.54 R \times 1.20 = 0.648 R. Total Year 3 revenue = 0.432R+0.648R=1.08R0.432 R + 0.648 R = 1.08 R.
Applying the respective 10% decrease and 20% increase to Year 2 values yields Year 3 amounts.
4
Evaluate the percent changes from Year 1 to Year 3 for each division and the total company.
Division X change: 0.432R0.40R0.40R=8%\frac{0.432 R - 0.40 R}{0.40 R} = 8\%. Division Y change: 0.648R0.60R0.60R=8%\frac{0.648 R - 0.60 R}{0.60 R} = 8\%. Total revenue change: 1.08R1.00R1.00R=8%\frac{1.08 R - 1.00 R}{1.00 R} = 8\%. Also, Division X proportion in Year 2 was 0.48R1.02R47.06%\frac{0.48 R}{1.02 R} \approx 47.06\%.
Determines which of the given statements hold true based on accurate relative changes.

Anahtar Kavram

Weighted and Successive Percent Change
Tahmini Süre:2m 0s
Soru 250Soru

Let pp, qq, and rr be non-zero integers satisfying the following three conditions:

I. p3qr2<0p^3 q r^2 < 0
II. (p)q<0(-p)^q < 0
III. (1)p+r=1(-1)^{p + r} = 1

Which of the following expressions MUST be negative?

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Cevap: qprq \cdot p^r

Cevap

The expression qprq \cdot p^r MUST be negative.
Condition II establishes that p>0p > 0 and qq is odd. Condition I establishes that q<0q < 0. Since p>0p > 0, raising pp to any integer exponent rr results in a strictly positive value (pr>0p^r > 0). Multiplying this positive value by negative qq guarantees a negative outcome regardless of the value or sign of rr.

Adım Adım Çözüm

1
Analyze Condition II: (p)q<0(-p)^q < 0
Base p<0    p>0-p < 0 \implies p > 0, and exponent qq must be an odd integer.
A negative number raised to an integer power is negative if and only if the exponent is odd.
2
Analyze Condition I: p3qr2<0p^3 q r^2 < 0
q<0q < 0 (negative odd integer).
Since p>0p > 0, p3>0p^3 > 0. Also r0    r2>0r \neq 0 \implies r^2 > 0. For the overall product p3qr2p^3 q r^2 to be negative, qq must be negative.
3
Analyze Condition III: (1)p+r=1(-1)^{p + r} = 1
p+rp + r is an even integer, so pp and rr have the same parity.
(1)k=1(-1)^k = 1 requires kk to be an even integer.
4
Evaluate the sign of qprq \cdot p^r
qpr<0q \cdot p^r < 0 for all valid values.
Since p>0p > 0, any integer power pr>0p^r > 0. Multiplying positive prp^r by negative qq yields a negative result.

Anahtar Kavram

Even-odd exponent rules and negative base sign determination
Soru 251Soru

On the real number line, the distance between xx and 33 is equal to twice the distance between xx and 9-9. What is the sum of all possible real values of xx?

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Cevap: 26-26

Cevap

The sum of all possible real values of xx is 26-26.
The distance between xx and aa on the real number line is expressed as xa|x - a|. Thus, the condition translates to x3=2x(9)|x - 3| = 2|x - (-9)|, which simplifies to x3=2x+9|x - 3| = 2|x + 9|. Setting up the two algebraic cases gives x3=2(x+9)x - 3 = 2(x + 9), leading to x=21x = -21, and x3=2(x+9)x - 3 = -2(x + 9), leading to x=5x = -5. Adding these two solutions yields (21)+(5)=26(-21) + (-5) = -26.

Adım Adım Çözüm

1
Translate the geometric statement on the number line into an absolute value equation.
The distance between xx and 33 is x3|x - 3|, and the distance between xx and 9-9 is x(9)=x+9|x - (-9)| = |x + 9|. Therefore, x3=2x+9|x - 3| = 2|x + 9|.
Distance between two points aa and bb on the number line is given by ab|a - b|.
2
Solve Case 1 where the expressions inside the absolute values have the same sign.
x3=2(x+9)    x3=2x+18    x=21x - 3 = 2(x + 9) \implies x - 3 = 2x + 18 \implies x = -21.
Removing absolute values with identical signs gives a linear equation in xx.
3
Solve Case 2 where the expressions inside the absolute values have opposite signs.
x3=2(x+9)    x3=2x18    3x=15    x=5x - 3 = -2(x + 9) \implies x - 3 = -2x - 18 \implies 3x = -15 \implies x = -5.
Removing absolute values with opposite signs accounts for the second possible geometric location.
4
Sum the valid real solutions.
(21)+(5)=26(-21) + (-5) = -26.
The question asks for the sum of all possible real values of xx.

Anahtar Kavram

Absolute Value as Distance on the Real Number Line
Soru 252Soru

Three positive integers xx, yy, and zz satisfy gcd(x,y)=12\gcd(x, y) = 12, gcd(y,z)=18\gcd(y, z) = 18, and gcd(x,z)=30\gcd(x, z) = 30, where gcd(a,b)\gcd(a, b) denotes the greatest common divisor of aa and bb. What is the minimum possible value of the least common multiple lcm(x,y,z)\text{lcm}(x, y, z)?

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Cevap: 180

Cevap

180
The correct answer is 180. Expressing the pairwise GCDs in prime factorized form shows that gcd(x,y)=2231\gcd(x, y) = 2^2 \cdot 3^1, gcd(y,z)=2132\gcd(y, z) = 2^1 \cdot 3^2, and gcd(x,z)=213151\gcd(x, z) = 2^1 \cdot 3^1 \cdot 5^1. To satisfy these simultaneously while minimizing the overall LCM, the maximum power of 2 across the numbers must be 222^2, the maximum power of 3 must be 323^2, and the maximum power of 5 must be 515^1. Thus, lcm(x,y,z)=223251=180\text{lcm}(x, y, z) = 2^2 \cdot 3^2 \cdot 5^1 = 180.

Adım Adım Çözüm

1
Express the given GCD values in their prime factorized forms.
gcd(x,y)=12=223150\gcd(x, y) = 12 = 2^2 \cdot 3^1 \cdot 5^0, gcd(y,z)=18=213250\gcd(y, z) = 18 = 2^1 \cdot 3^2 \cdot 5^0, and gcd(x,z)=30=213151\gcd(x, z) = 30 = 2^1 \cdot 3^1 \cdot 5^1.
Analyzing prime factor exponents determines the minimum required power of each prime in xx, yy, and zz.
2
Determine the minimum necessary exponents for prime 2 in xx, yy, and zz.
Since gcd(x,y)\gcd(x, y) has 222^2, both xx and yy must contain at least 222^2. Thus, the maximum exponent of 2 across the integers is at least 2.
The least common multiple takes the maximum exponent for each prime factor among all three numbers.
3
Determine the minimum necessary exponents for prime 3 in xx, yy, and zz.
Since gcd(y,z)\gcd(y, z) has 323^2, both yy and zz must contain at least 323^2. Thus, the maximum exponent of 3 across the integers is at least 2.
To satisfy gcd(y,z)=18\gcd(y, z) = 18, the prime factor 3 must appear with an exponent of at least 2 in both yy and zz.
4
Determine the minimum necessary exponents for prime 5 in xx, yy, and zz.
Since gcd(x,z)\gcd(x, z) has 515^1, both xx and zz must contain at least 515^1. Thus, the maximum exponent of 5 across the integers is at least 1.
To satisfy gcd(x,z)=30\gcd(x, z) = 30, the prime factor 5 must appear with an exponent of at least 1 in both xx and zz.
5
Calculate the minimum least common multiple lcm(x,y,z)\text{lcm}(x, y, z).
lcm(x,y,z)=223251=495=180\text{lcm}(x, y, z) = 2^2 \cdot 3^2 \cdot 5^1 = 4 \cdot 9 \cdot 5 = 180. (A valid assignment is x=60x = 60, y=36y = 36, z=90z = 90).
Multiplying the minimum maximum prime factor powers yields the smallest possible value for the LCM.

Anahtar Kavram

Prime Factor Exponents in Pairwise GCD and LCM Relations
Tahmini Süre:2m 0s
Soru 253Soru

In January, a coffee roasting facility produced two specialty blends: Roast A and Roast B. Roast A sold for $10\$10 per pound and Roast B sold for $30\$30 per pound. A distributor purchased a total of 100100 pounds of coffee consisting of these two blends at an average price of $18\$18 per pound. In February, the price of Roast A increased by 20%20\%, while the price of Roast B decreased by 4%4\%. If the distributor purchased the exact same quantities of Roast A and Roast B in February as in January, by what percent did the total cost of the order increase?

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Cevap: 4%4\%

Cevap

The total cost of the order increased by 4%4\%.
The choice stating 4%4\% is correct because the total cost in January is $1,800\$1,800 (6060 lbs of Roast A at $10\$10 and 4040 lbs of Roast B at $30\$30). In February, the price of Roast A rises to $12\$12 and Roast B drops to $28.80\$28.80, bringing the total cost to $1,872\$1,872. The change of $72\$72 represents a 4%4\% increase relative to the baseline January cost of $1,800\$1,800.

Adım Adım Çözüm

1
Determine the quantities of Roast A and Roast B purchased in January.
Roast A: 6060 lbs, Roast B: 4040 lbs.
Let aa be the pounds of Roast A and bb be the pounds of Roast B. We are given a+b=100a + b = 100 and 10a+30b=18×100=180010a + 30b = 18 \times 100 = 1800. Substituting b=100ab = 100 - a gives 10a+30(100a)=1800    20a=1200    a=6010a + 30(100 - a) = 1800 \implies -20a = -1200 \implies a = 60 and b=40b = 40.
2
Calculate the total cost of the order in January.
Total January Cost = $1,800\$1,800.
Total cost = (60 lbs×$10/lb)+(40 lbs×$30/lb)=$600+$1,200=$1,800(60 \text{ lbs} \times \$10/\text{lb}) + (40 \text{ lbs} \times \$30/\text{lb}) = \$600 + \$1,200 = \$1,800.
3
Calculate the new prices and total cost in February.
Total February Cost = $1,872\$1,872.
New price of Roast A = $10×(1+0.20)=$12/lb\$10 \times (1 + 0.20) = \$12/\text{lb}. New price of Roast B = $30×(10.04)=$28.80/lb\$30 \times (1 - 0.04) = \$28.80/\text{lb}. Total February Cost = (60×$12)+(40×$28.80)=$720+$1,152=$1,872(60 \times \$12) + (40 \times \$28.80) = \$720 + \$1,152 = \$1,872.
4
Calculate the percentage increase in total cost from January to February.
Percent Increase = 4%4\%.
Percent Change = February CostJanuary CostJanuary Cost×100%=$1,872$1,800$1,800×100%=$72$1,800×100%=4%\frac{\text{February Cost} - \text{January Cost}}{\text{January Cost}} \times 100\% = \frac{\$1,872 - \$1,800}{\$1,800} \times 100\% = \frac{\$72}{\$1,800} \times 100\% = 4\%.

Anahtar Kavram

Weighted Percent Change and Systems of Equations
Tahmini Süre:1m 30s
Soru 254Soru

On a number line, point AA is located at 1010 and point BB is located at 22. If point PP, with coordinate x>0x > 0, is three times as far from point AA as it is from point BB, what is the value of xx?

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Cevap: 4

Cevap

4
The distance from P(x)P(x) to A(10)A(10) is x10|x - 10| and to B(2)B(2) is x2|x - 2|. Setting x10=3x2|x - 10| = 3|x - 2| gives two equations: x10=3x6x - 10 = 3x - 6, which yields x=2x = -2, and x10=3x+6x - 10 = -3x + 6, which yields x=4x = 4. Since x>0x > 0, the correct value is 4.

Adım Adım Çözüm

1
Formulate the distance relationship using absolute value notation.
x10=3x2|x - 10| = 3|x - 2|
The distance between two points uu and vv on the real number line is given by uv|u - v|.
2
Split the absolute value equation into two linear equations representing possible cases.
x10=3(x2)x - 10 = 3(x - 2) or x10=3(x2)x - 10 = -3(x - 2)
The equality a=b|a| = |b| implies a=ba = b or a=ba = -b.
3
Solve each case algebraically.
Case 1 gives x10=3x6    2x=4    x=2x - 10 = 3x - 6 \implies 2x = -4 \implies x = -2. Case 2 gives x10=3x+6    4x=16    x=4x - 10 = -3x + 6 \implies 4x = 16 \implies x = 4.
Standard linear equation solving.
4
Apply the given domain condition x>0x > 0.
x=4x = 4
The solution x=2x = -2 is negative and therefore violates the condition x>0x > 0.

Anahtar Kavram

Distance on a number line represented by absolute value equations
Soru 255Soru

A commercial coffee roasting facility operates three roasters: XX, YY, and ZZ.

- Roaster XX processes coffee beans at a constant rate of 60 kg/hr60\text{ kg/hr}, producing a blend with an Arabica-to-Robusta ratio of 3:23:2 by weight.
- Roaster YY processes coffee beans at a constant rate of 90 kg/hr90\text{ kg/hr}, producing a blend with an Arabica-to-Robusta ratio of 2:12:1 by weight.
- Roaster ZZ processes coffee beans at a constant rate of 150 kg/hr150\text{ kg/hr}, producing a blend with an Arabica-to-Robusta ratio of 1:41:4 by weight.

All three roasters operate simultaneously for 44 hours to complete a production order. Which of the following statements regarding the production output over this 44-hour period must be true? Select all such statements.

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Cevap: The total weight of Arabica beans processed across all three roasters is 504 kg504\text{ kg}.; Arabica beans account for exactly 42%42\% of the total weight of coffee beans processed.; The overall ratio of total Arabica weight to total Robusta weight produced is 21:2921:29.

Cevap

The statements confirming that total Arabica weight is 504 kg504\text{ kg}, Arabica accounts for 42%42\% of total weight, and the overall ratio of Arabica to Robusta is 21:2921:29 are all correct.
The total production from Roaster XX is 240 kg240\text{ kg} (144 kg144\text{ kg} Arabica, 96 kg96\text{ kg} Robusta), from Roaster YY is 360 kg360\text{ kg} (240 kg240\text{ kg} Arabica, 120 kg120\text{ kg} Robusta), and from Roaster ZZ is 600 kg600\text{ kg} (120 kg120\text{ kg} Arabica, 480 kg480\text{ kg} Robusta). The total Arabica weight is 144+240+120=504 kg144 + 240 + 120 = 504\text{ kg}, which constitutes 5041,200=42%\frac{504}{1,200} = 42\% of the 1,200 kg1,200\text{ kg} total output. The remaining 696 kg696\text{ kg} is Robusta, giving an Arabica to Robusta ratio of 504:696=21:29504:696 = 21:29.

Adım Adım Çözüm

1
Calculate total output and Arabica quantity for each roaster over 4 hours
Roaster XX: 240 kg240\text{ kg} total, 35×240=144 kg\frac{3}{5} \times 240 = 144\text{ kg} Arabica. Roaster YY: 360 kg360\text{ kg} total, 23×360=240 kg\frac{2}{3} \times 360 = 240\text{ kg} Arabica. Roaster ZZ: 600 kg600\text{ kg} total, 15×600=120 kg\frac{1}{5} \times 600 = 120\text{ kg} Arabica.
Converting hourly rates to total batch weights over 44 hours and applying part-to-whole fractions determined by each ratio.
2
Aggregate total Arabica and Robusta weights across all roasters
Total Arabica = 144+240+120=504 kg144 + 240 + 120 = 504\text{ kg}. Total coffee = 240+360+600=1,200 kg240 + 360 + 600 = 1,200\text{ kg}. Total Robusta = 1,200504=696 kg1,200 - 504 = 696\text{ kg}.
Adding individual component weights to get total aggregate weights.
3
Evaluate percentage and ratio statements
Arabica percentage = 5041,200×100%=42%\frac{504}{1,200} \times 100\% = 42\%. Arabica to Robusta ratio = 504696=2129\frac{504}{696} = \frac{21}{29}.
Verifying overall proportions and simplifying the fraction by dividing by 2424.

Anahtar Kavram

Multi-stage weighted component ratios and rate conversion
Soru 256Soru

If aa and bb are nonzero real numbers such that a2b3<0a^2 b^3 < 0 and a2b4=ab2\sqrt{a^2 b^4} = -a b^2, which of the following statements must be true? Select all such statements.

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Cevap: a+b<0a + b < 0; a3b>0\frac{a^3}{b} > 0

Cevap

The statements that must be true are the inequality asserting that the sum of the variables is negative (a+b<0a + b < 0) and the inequality asserting that the quotient of the cubed variable and the second variable is positive (a3b>0\frac{a^3}{b} > 0).
Analyzing the given constraints reveals that both variables are negative. From a2b3<0a^2 b^3 < 0, since a2>0a^2 > 0, we must have b3<0b^3 < 0, so b<0b < 0. Next, from a2b4=ab2=ab2\sqrt{a^2 b^4} = |a| b^2 = -a b^2, dividing by b2>0b^2 > 0 gives a=a|a| = -a, which implies a<0a < 0. Thus, a<0a < 0 and b<0b < 0. The statement asserting a+b<0a + b < 0 is true because the sum of two negative numbers is negative. The statement asserting a3b>0\frac{a^3}{b} > 0 is true because a3<0a^3 < 0 and b<0b < 0, and dividing two negative numbers yields a positive quotient.

Adım Adım Çözüm

1
Determine the sign of bb using the given inequality a2b3<0a^2 b^3 < 0.
b<0b < 0
Since aa is a nonzero real number, a2>0a^2 > 0. For the product a2b3a^2 b^3 to be negative, b3b^3 must be negative, which implies b<0b < 0.
2
Determine the sign of aa using the identity a2b4=ab2\sqrt{a^2 b^4} = -a b^2.
a<0a < 0
Simplify the radical: a2b4=a2(b2)2=ab2\sqrt{a^2 b^4} = \sqrt{a^2} \cdot \sqrt{(b^2)^2} = |a| b^2. Equating this to ab2-a b^2 gives ab2=ab2|a| b^2 = -a b^2. Since b0b \neq 0, b2>0b^2 > 0, so dividing by b2b^2 yields a=a|a| = -a. For a nonzero real number, a=a|a| = -a implies a<0a < 0.
3
Evaluate statement a+b<0a + b < 0.
True
The sum of two negative numbers (a<0a < 0 and b<0b < 0) is always negative.
4
Evaluate statement a3b>0\frac{a^3}{b} > 0.
True
Since a<0a < 0, a3<0a^3 < 0. Dividing the negative quantity a3a^3 by the negative quantity bb yields a positive result.
5
Evaluate statement a4b2=a2b\sqrt{a^4 b^2} = a^2 b.
False
a4b2=a4b2=a2b\sqrt{a^4 b^2} = \sqrt{a^4}\sqrt{b^2} = a^2 |b|. Since b<0b < 0, b=b|b| = -b, so a4b2=a2b\sqrt{a^4 b^2} = -a^2 b.
6
Evaluate statement (a)3b2<0(-a)^3 b^2 < 0.
False
Since a<0a < 0, a>0-a > 0, making (a)3>0(-a)^3 > 0. Since b0b \neq 0, b2>0b^2 > 0. The product of two positive numbers is positive, so (a)3b2>0(-a)^3 b^2 > 0.
7
Evaluate statement (a+b)2=a+b\sqrt{(a + b)^2} = a + b.
False
x2=x\sqrt{x^2} = |x| for any real xx. Since a+b<0a + b < 0, (a+b)2=a+b=(a+b)a+b\sqrt{(a + b)^2} = |a + b| = -(a + b) \neq a + b.

Anahtar Kavram

Properties of even exponents, odd exponents, and principal square roots of negative variable terms.
Soru 257Soru

The prime factorizations of two positive integers AA and BB are given by A=2a×35×5bA = 2^a \times 3^5 \times 5^b and B=24×3c×72B = 2^4 \times 3^c \times 7^2, where aa, bb, and cc are positive integers. If the greatest common divisor of AA and BB is gcd(A,B)=22×33\gcd(A, B) = 2^2 \times 3^3, and their least common multiple is \text{lcm}(A,B)=24×35×53×72(A, B) = 2^4 \times 3^5 \times 5^3 \times 7^2, what is the value of a+b+ca + b + c?

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Cevap: 8

Cevap

The correct answer is 8.
For any two positive integers expressed in prime factorized form, the greatest common divisor contains each prime factor raised to the minimum of its exponents in the two numbers, while the least common multiple contains each prime factor raised to the maximum of its exponents. For prime factor 2, the GCD has exponent 2, so min(a, 4) = 2, giving a = 2. For prime factor 3, the GCD has exponent 3, so min(5, c) = 3, giving c = 3. For prime factor 5, the LCM has exponent 3, so max(b, 0) = 3, giving b = 3. Adding these values together yields a + b + c = 2 + 3 + 3 = 8.

Adım Adım Çözüm

1
Analyze the prime factor 22
min(a, 4) = 2, so a = 2
The greatest common divisor takes the minimum exponent for each prime factor shared between A and B.
2
Analyze the prime factor 33
min(5, c) = 3, so c = 3
The exponent of 3 in the GCD is 3, which must equal the smaller of the two exponents 5 and c.
3
Analyze the prime factor 55
max(b, 0) = 3, so b = 3
The least common multiple takes the maximum exponent for each prime factor present in either number.
4
Calculate the sum a+b+ca + b + c
2 + 3 + 3 = 8
Add the solved values of the three unknown prime exponents.

Anahtar Kavram

Relating prime factor exponents to GCD (minimum exponent) and LCM (maximum exponent)
Soru 258Soru
If xx is a positive integer such that
4x+4x+4x+4x2x+2x=512\frac{4^x + 4^x + 4^x + 4^x}{2^x + 2^x} = 512
what is the value of xx?
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Cevap: 8

Cevap

The value of xx is 8.
Combining four terms of 4x4^x yields 44x=4x+1=22x+24 \cdot 4^x = 4^{x+1} = 2^{2x+2}. Combining two terms of 2x2^x yields 22x=2x+12 \cdot 2^x = 2^{x+1}. Dividing the numerator by the denominator gives 22x+2(x+1)=2x+12^{2x+2 - (x+1)} = 2^{x+1}. Since 512=29512 = 2^9, setting 2x+1=292^{x+1} = 2^9 gives x+1=9x + 1 = 9, which leads directly to x=8x = 8.

Adım Adım Çözüm

1
Simplify the numerator by combining identical added terms.
The numerator 4x+4x+4x+4x4^x + 4^x + 4^x + 4^x equals 44x4 \cdot 4^x, which simplifies to 4x+14^{x+1}.
Adding four identical quantities is equivalent to multiplying that quantity by 4.
2
Simplify the denominator by combining identical added terms.
The denominator 2x+2x2^x + 2^x equals 22x2 \cdot 2^x, which simplifies to 2x+12^{x+1}.
Adding two identical quantities is equivalent to multiplying that quantity by 2.
3
Convert the numerator to base 2 and simplify the fraction.
Since 4x+1=(22)x+1=22x+24^{x+1} = (2^2)^{x+1} = 2^{2x+2}, the fraction becomes 22x+22x+1=2(2x+2)(x+1)=2x+1\frac{2^{2x+2}}{2^{x+1}} = 2^{(2x+2)-(x+1)} = 2^{x+1}.
Converting all terms to a common base allows using the exponent quotient rule am/an=amna^m / a^n = a^{m-n}.
4
Solve for xx by equating the simplified power to 512.
Setting 2x+1=512=292^{x+1} = 512 = 2^9 yields x+1=9x + 1 = 9, so x=8x = 8.
When exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Anahtar Kavram

Combining repeated addition of exponential terms and converting powers to a common base using exponent laws (aman=am+na^m \cdot a^n = a^{m+n} and aman=amn\frac{a^m}{a^n} = a^{m-n}).
Soru 259Soru

A textile mill uses two weaving looms, Loom PP and Loom QQ, to produce a fabric blend composed of silk, wool, and cotton.

- Loom PP produces silk, wool, and cotton in the ratio 2:3:52 : 3 : 5 by weight, operating at a constant output rate of 120 kg per hour120\text{ kg per hour}.
- Loom QQ produces silk, wool, and cotton in the ratio 1:4:31 : 4 : 3 by weight, operating at a constant output rate of 160 kg per hour160\text{ kg per hour}.

If both looms operate simultaneously for 5 hours, what is the ratio of the total weight of wool produced to the total weight of cotton produced in the combined output?

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Cevap: 29:3029 : 30

Cevap

The ratio of the total weight of wool produced to the total weight of cotton produced is 29:3029 : 30.
To find the overall ratio of wool to cotton, determine the actual mass of each material produced per hour by each machine. Loom PP outputs 310×120=36 kg\frac{3}{10} \times 120 = 36\text{ kg} of wool and 510×120=60 kg\frac{5}{10} \times 120 = 60\text{ kg} of cotton per hour. Loom QQ outputs 48×160=80 kg\frac{4}{8} \times 160 = 80\text{ kg} of wool and 38×160=60 kg\frac{3}{8} \times 160 = 60\text{ kg} of cotton per hour. Combining both looms yields 116 kg116\text{ kg} of wool and 120 kg120\text{ kg} of cotton per hour. Over 5 hours, the combined output is 580 kg580\text{ kg} of wool and 600 kg600\text{ kg} of cotton. The ratio of total wool to total cotton is 580:600580 : 600, which simplifies to 29:3029 : 30.

Adım Adım Çözüm

1
Calculate the hourly production rates of wool and cotton for Loom PP.
Loom PP produces fabric at 120 kg/hr120\text{ kg/hr} with ratio parts 2+3+5=102 + 3 + 5 = 10.
- Wool rate from P=310×120=36 kg/hrP = \frac{3}{10} \times 120 = 36\text{ kg/hr}
- Cotton rate from P=510×120=60 kg/hrP = \frac{5}{10} \times 120 = 60\text{ kg/hr}
Converting the ratio into component rates by multiplying each component's fraction by Loom PP's total hourly output.
2
Calculate the hourly production rates of wool and cotton for Loom QQ.
Loom QQ produces fabric at 160 kg/hr160\text{ kg/hr} with ratio parts 1+4+3=81 + 4 + 3 = 8.
- Wool rate from Q=48×160=80 kg/hrQ = \frac{4}{8} \times 160 = 80\text{ kg/hr}
- Cotton rate from Q=38×160=60 kg/hrQ = \frac{3}{8} \times 160 = 60\text{ kg/hr}
Converting the ratio into component rates by multiplying each component's fraction by Loom QQ's total hourly output.
3
Determine total quantities produced over the 5-hour period.
Total wool =(36+80)×5=116×5=580 kg= (36 + 80) \times 5 = 116 \times 5 = 580\text{ kg}.
Total cotton =(60+60)×5=120×5=600 kg= (60 + 60) \times 5 = 120 \times 5 = 600\text{ kg}.
Combining the hourly outputs from both machines and multiplying by the total duration of 5 hours.
4
Compute and simplify the ratio of total wool to total cotton.
\text{Ratio} = \frac{580}{600} = \frac{29}{30},whichisexpressedas, which is expressed as 29 : 30$.
Dividing both terms of the ratio by their greatest common divisor, 20.

Anahtar Kavram

Combining weighted rates across multiple ratio-based sub-components
Tahmini Süre:2m 0s
Soru 260Soru

Let n=2a3b5cn = 2^a \cdot 3^b \cdot 5^c be a positive integer, where aa, bb, and cc are non-negative integers. If the greatest common divisor of nn and 360360 is 4545, and the least common multiple of nn and 9090 is 450450, what is the value of a+b+ca + b + c?

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Cevap: 4

Cevap

The value of a+b+ca + b + c is 44.
Prime factorizing the given values yields 360=233251360 = 2^3 \cdot 3^2 \cdot 5^1, 45=20325145 = 2^0 \cdot 3^2 \cdot 5^1, 90=21325190 = 2^1 \cdot 3^2 \cdot 5^1, and 450=213252450 = 2^1 \cdot 3^2 \cdot 5^2. Because gcd(n,360)=45\gcd(n, 360) = 45, taking the minimum exponent of 2 implies min(a,3)=0\min(a, 3) = 0, so a=0a = 0. Taking the minimum exponent of 3 implies b2b \ge 2, and for 5 implies c1c \ge 1. Next, using lcm(n,90)=450\text{lcm}(n, 90) = 450, taking the maximum exponent of 3 gives max(b,2)=2\max(b, 2) = 2, which forces b=2b = 2. Taking the maximum exponent of 5 gives max(c,1)=2\max(c, 1) = 2, which forces c=2c = 2. Therefore, a+b+c=0+2+2=4a + b + c = 0 + 2 + 2 = 4.

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1
Express all given integers in their prime factorizations.
360=233251360 = 2^3 \cdot 3^2 \cdot 5^1, 45=3251=20325145 = 3^2 \cdot 5^1 = 2^0 \cdot 3^2 \cdot 5^1, 90=21325190 = 2^1 \cdot 3^2 \cdot 5^1, and 450=213252450 = 2^1 \cdot 3^2 \cdot 5^2.
Prime factorizations allow determination of exponents using exponent rules for GCD and LCM.
2
Apply the GCD condition gcd(n,360)=45\gcd(n, 360) = 45.
min(a,3)=0    a=0\min(a, 3) = 0 \implies a = 0, min(b,2)=2    b2\min(b, 2) = 2 \implies b \ge 2, and min(c,1)=1    c1\min(c, 1) = 1 \implies c \ge 1.
The greatest common divisor takes the minimum exponent for each prime factor shared between the numbers.
3
Apply the LCM condition lcm(n,90)=450\text{lcm}(n, 90) = 450 using a=0a = 0.
max(b,2)=2    b2\max(b, 2) = 2 \implies b \le 2 (so b=2b = 2), and max(c,1)=2    c=2\max(c, 1) = 2 \implies c = 2.
The least common multiple takes the maximum exponent for each prime factor.
4
Calculate the sum a+b+ca + b + c.
a+b+c=0+2+2=4a + b + c = 0 + 2 + 2 = 4.
Adding the individual prime factor exponents yields the requested sum.

Anahtar Kavram

Relating prime factor exponents to GCD (minimum powers) and LCM (maximum powers)
Tahmini Süre:1m 30s
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