Arithmetic

306 soru

Soru 101Soru

An asset management firm divides its total portfolio into three distinct funds: Fund X, Fund Y, and Fund Z. Initially, 13\frac{1}{3} of the total portfolio value is in Fund X, and 25\frac{2}{5} is in Fund Y, with the remaining fraction in Fund Z. Over the course of one year, the value of Fund X increases by 15\frac{1}{5}, the value of Fund Y decreases by 14\frac{1}{4}, and the value of Fund Z increases by 12\frac{1}{2}. At the end of the year, what fraction of the portfolio's total value is contained in Fund Z?

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Cevap: 411\frac{4}{11}

Cevap

The fraction of the portfolio's total final value contained in Fund Z is 411\frac{4}{11}.
Fund Z grows to 25\frac{2}{5} of the original portfolio value, while the overall portfolio value increases to 1110\frac{11}{10} of the original total. Comparing Fund Z's final value to the new total yields 2/511/10=411\frac{2/5}{11/10} = \frac{4}{11}.

Adım Adım Çözüm

1
Calculate the initial fraction of the total portfolio allocated to Fund Z.
The initial allocation for Fund Z is 1(13+25)=11115=4151 - \left(\frac{1}{3} + \frac{2}{5}\right) = 1 - \frac{11}{15} = \frac{4}{15}.
The sum of all initial fund allocations must equal 1.
2
Determine the final value of each fund relative to the initial total portfolio value VV.
Fund X: 13V×(1+15)=25V\frac{1}{3}V \times \left(1 + \frac{1}{5}\right) = \frac{2}{5}V. Fund Y: 25V×(114)=310V\frac{2}{5}V \times \left(1 - \frac{1}{4}\right) = \frac{3}{10}V. Fund Z: 415V×(1+12)=25V\frac{4}{15}V \times \left(1 + \frac{1}{2}\right) = \frac{2}{5}V.
Multiply each initial fund share by its respective growth or reduction factor.
3
Calculate the new total portfolio value relative to the initial value VV.
Total final value = 25V+310V+25V=410V+310V+410V=1110V\frac{2}{5}V + \frac{3}{10}V + \frac{2}{5}V = \frac{4}{10}V + \frac{3}{10}V + \frac{4}{10}V = \frac{11}{10}V.
Sum the final values of all three individual funds.
4
Compute the final fraction of the portfolio contained in Fund Z.
Fraction = 25V1110V=25×1011=411\frac{\frac{2}{5}V}{\frac{11}{10}V} = \frac{2}{5} \times \frac{10}{11} = \frac{4}{11}.
Divide the final value of Fund Z by the new total portfolio value.

Anahtar Kavram

Rational number operations and shifting base values in multi-step fraction problems
Tahmini Süre:2m 0s
Soru 102Soru

If kk is a positive integer such that kk is divisible by 1515 and k2k^2 is divisible by 360360, what is the minimum possible number of positive divisors of kk?

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Cevap: 12

Cevap

The minimum possible number of positive divisors of kk is 12.
The minimum possible number of positive divisors occurs when kk has the smallest possible prime exponents satisfying all divisibility conditions. Since 15=3×515 = 3 \times 5 divides kk, kk must have prime factors 313^1 and 515^1. For k2k^2 to be divisible by 360=23×32×51360 = 2^3 \times 3^2 \times 5^1, k2k^2 must have at least 232^3, which requires kk to have at least 22=42^2 = 4. Thus, the minimal prime factorization of kk is k=22×31×51=60k = 2^2 \times 3^1 \times 5^1 = 60. Using the divisor count formula (e1+1)(e2+1)(e3+1)(e_1 + 1)(e_2 + 1)(e_3 + 1), we get (2+1)(1+1)(1+1)=12(2+1)(1+1)(1+1) = 12 positive divisors.

Adım Adım Çözüm

1
Find the prime factorizations of 15 and 360.
15=31×5115 = 3^1 \times 5^1 and 360=23×32×51360 = 2^3 \times 3^2 \times 5^1.
Expressing numbers in terms of prime factors reveals the required prime exponents for divisibility.
2
Determine the minimum exponents of the prime factors required for kk.
Since 1515 divides kk, kk must contain at least 313^1 and 515^1. For k2k^2 to be divisible by 360=23×32×51360 = 2^3 \times 3^2 \times 5^1, k2k^2 must contain at least 232^3, which means kk must contain at least 222^2. The required factors 323^2 and 515^1 in k2k^2 are automatically provided since (31)2=32(3^1)^2 = 3^2 and (51)2=52(5^1)^2 = 5^2. Thus, the minimal kk is 22×31×51=602^2 \times 3^1 \times 5^1 = 60.
Taking the minimum required exponent for each prime factor minimizes the total number of divisors of kk.
3
Calculate the number of positive divisors of the minimal kk.
The number of divisors for k=22×31×51k = 2^2 \times 3^1 \times 5^1 is (2+1)(1+1)(1+1)=3×2×2=12(2 + 1)(1 + 1)(1 + 1) = 3 \times 2 \times 2 = 12.
The divisor count formula adds 1 to each prime exponent and multiplies the results.

Anahtar Kavram

Properties of Integer Divisibility and Divisor Counting
Tahmini Süre:1m 30s
Soru 103Soru

What is the value of 4.8×1031.2×107\frac{4.8 \times 10^{-3}}{1.2 \times 10^{-7}} expressed in scientific notation?

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Cevap: 4.0×1044.0 \times 10^4

Cevap

4.0×1044.0 \times 10^4
Dividing 4.8×1034.8 \times 10^{-3} by 1.2×1071.2 \times 10^{-7} requires dividing the coefficient 4.84.8 by 1.21.2 to obtain 4.04.0, and subtracting the exponent of the denominator from that of the numerator: 3(7)=4-3 - (-7) = 4. This yields 4.0×1044.0 \times 10^4.

Adım Adım Çözüm

1
Divide the numerical coefficients.
4.8÷1.2=4.04.8 \div 1.2 = 4.0
When dividing numbers in scientific notation, divide the coefficients independently from the powers of ten.
2
Apply exponent rules to divide the powers of ten.
103÷107=103(7)=10410^{-3} \div 10^{-7} = 10^{-3 - (-7)} = 10^4
Subtract the exponent of the denominator from the exponent of the numerator.
3
Combine the coefficient and power of ten into scientific notation.
4.0×1044.0 \times 10^4
The coefficient 4.04.0 is between 11 and 1010, so the expression is in proper standard scientific notation form.

Anahtar Kavram

Division of numbers in scientific notation
Soru 104Soru

The price of an item was increased by 25%25\% and then the new price was decreased by 20%20\%. Which of the following statements must be true regarding the final price of the item relative to its initial price? Select all such statements.

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Cevap: The final price of the item is equal to its initial price.; The overall net percent change in the price of the item is 0%0\%.; The ratio of the final price to the initial price is 11 to 11.

Cevap

The statements confirming that the final price equals the initial price, that the overall net percent change is 0%, and that the ratio of the final price to initial price is 1 to 1 are all correct.
Let PP be the original price. After a 25%25\% increase, the price becomes 1.25P1.25P. Decreasing this intermediate price by 20%20\% yields 1.25P×0.80=1.00P1.25P \times 0.80 = 1.00P. Because the final price is identical to the initial price, the final price is equal to the initial price, the net percentage change is 0%0\%, and the ratio of final to initial price is 1:11:1.

Adım Adım Çözüm

1
Represent the initial price with a variable
Let the initial price of the item be PP.
Using a variable makes tracking successive percent changes straightforward.
2
Apply the 25%25\% increase
Increased price =P×(1+0.25)=1.25P= P \times (1 + 0.25) = 1.25P.
A 25%25\% increase corresponds to multiplying the base by 1.251.25.
3
Apply the 20%20\% decrease to the new base
Final price =1.25P×(10.20)=1.25P×0.80=1.00P= 1.25P \times (1 - 0.20) = 1.25P \times 0.80 = 1.00P.
The 20%20\% decrease applies to the updated base value of 1.25P1.25P.
4
Evaluate each statement based on the final price 1.00P1.00P
The final price equals the initial price (PP), the net percent change is 0%0\%, and the ratio of final to initial price is 1:11:1.
All three true statements directly reflect that the final price is equal to the initial price.

Anahtar Kavram

Successive Percentage Changes and Base Shift
Tahmini Süre:45s
Soru 105Soru

If xx is an odd integer and yy is an even integer, which of the following expressions must result in an even integer? Select all that apply.

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Cevap: xy+yxy + y; (x+1)(y+1)(x + 1)(y + 1)

Cevap

The expressions that must be even are xy+yxy + y and (x+1)(y+1)(x + 1)(y + 1).
The expression xy+yxy + y is guaranteed to be even because xyxy is even (product of odd and even) and adding another even integer yy produces an even sum. The expression (x+1)(y+1)(x + 1)(y + 1) is also guaranteed to be even because adding 1 to an odd integer xx creates an even integer x+1x + 1, and multiplying an even number by any integer always yields an even product.

Adım Adım Çözüm

1
Analyze fundamental parity rules for addition and multiplication of integers.
Recall that odd×even=even\text{odd} \times \text{even} = \text{even}, odd×odd=odd\text{odd} \times \text{odd} = \text{odd}, even+even=even\text{even} + \text{even} = \text{even}, and odd+even=odd\text{odd} + \text{even} = \text{odd}.
Parity rules govern the even/odd behavior of combined expressions.
2
Evaluate the expression xy+yxy + y.
Since xx is odd and yy is even, xyxy is even. Then even+y(even)=even\text{even} + y\,(\text{even}) = \text{even}.
Adding two even terms always results in an even number.
3
Evaluate the expression (x+1)(y+1)(x + 1)(y + 1).
Since xx is odd, x+1x + 1 is even. The product of an even integer and any integer (y+1y + 1) is always even.
An even factor guarantees an even product.

Anahtar Kavram

Even and Odd Parity Rules under Arithmetic Operations
Soru 106Soru

A laboratory container holds a solution composed solely of alcohol, acid, and water. Initially, alcohol accounts for 38\frac{3}{8} of the solution's total volume, and acid accounts for 14\frac{1}{4} of the total volume. In a two-step process, a chemist first removes 13\frac{1}{3} of the alcohol present and 12\frac{1}{2} of the acid present, with no water removed. Next, the chemist adds pure water until water accounts for 35\frac{3}{5} of the solution's new total volume. What fraction of the final solution's total volume is alcohol?

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Cevap: 415\frac{4}{15}

Cevap

The fraction of the final solution's total volume that is alcohol is 415\frac{4}{15}.
The correct fraction 415\frac{4}{15} is derived by calculating the remaining alcohol volume as 14\frac{1}{4} of the initial total volume VV, and determining the final total volume Vfinal=1516VV_{\text{final}} = \frac{15}{16}V using the constant non-water volume of 38V\frac{3}{8}V. Dividing 14V\frac{1}{4}V by 1516V\frac{15}{16}V yields 415\frac{4}{15}.

Adım Adım Çözüm

1
Determine initial volume fractions for all three components.
Let VV be the initial total volume. Alcohol is 38V\frac{3}{8}V, acid is 14V=28V\frac{1}{4}V = \frac{2}{8}V, and water is 1(38+28)=38V1 - (\frac{3}{8} + \frac{2}{8}) = \frac{3}{8}V.
Establishing the initial baseline volumes allows accurate tracking through the multi-step changes.
2
Calculate component volumes after the removal step.
Remaining alcohol = 38V×(113)=14V\frac{3}{8}V \times (1 - \frac{1}{3}) = \frac{1}{4}V. Remaining acid = 14V×(112)=18V\frac{1}{4}V \times (1 - \frac{1}{2}) = \frac{1}{8}V. Water remains 38V\frac{3}{8}V.
Removing specified fractions of individual components changes their absolute volumes.
3
Calculate non-water volume and final total volume after water is added.
Total non-water volume = 14V+18V=38V\frac{1}{4}V + \frac{1}{8}V = \frac{3}{8}V. Since water becomes 35\frac{3}{5} of the final volume VfinalV_{\text{final}}, non-water is 135=251 - \frac{3}{5} = \frac{2}{5} of VfinalV_{\text{final}}. Thus, 25Vfinal=38V    Vfinal=52×38V=1516V\frac{2}{5} V_{\text{final}} = \frac{3}{8}V \implies V_{\text{final}} = \frac{5}{2} \times \frac{3}{8}V = \frac{15}{16}V.
Adding only water keeps the non-water volume constant, providing a fixed reference point to find the new total volume.
4
Compute the final fraction of alcohol in the solution.
\text{Alcohol Fraction} = \frac{\text{Alcohol Volume}}{V_{\text{final}}} = \frac{\frac{1}{4}V}{\frac{15}{16}V} = \frac{1}{4} \times \frac{16}{15} = \frac{4}{15}.
The required quantity is the part-to-whole ratio of remaining alcohol to the final total volume.

Anahtar Kavram

Multi-step fraction operations involving component-wise removals and fixed non-water volumes in liquid mixtures.
Tahmini Süre:2m 30s
Soru 107Soru

What is the largest two-digit positive integer nn such that when nn is divided by 44, the remainder is 33, and when nn is divided by 55, the remainder is 22?

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Cevap: 87

Cevap

The largest two-digit positive integer satisfying both remainder conditions is 87.
Any integer satisfying both remainder requirements must be of the form n=20m+7n = 20m + 7 for an integer mm. Testing values for mm shows that m=4m = 4 yields n=87n = 87, which is the largest two-digit integer fitting this rule.

Adım Adım Çözüm

1
Set up modular arithmetic equations for the given remainder conditions.
n3(mod4)n \equiv 3 \pmod 4 and n2(mod5)n \equiv 2 \pmod 5
Translate the verbal description of remainders into mathematical congruence relations.
2
Combine the congruence relations to find the general form of nn.
n=20m+7n = 20m + 7 for non-negative integers mm
The least common multiple of 44 and 55 is 2020, meaning solutions repeat every 2020 units.
3
Find the maximum integer mm that produces a two-digit integer.
For m=4m = 4, n=87n = 87. For m=5m = 5, n=107n = 107.
Two-digit integers are strictly less than 100100.

Anahtar Kavram

Simultaneous Remainders and Divisibility Cycles
Tahmini Süre:1m 30s
Soru 108Soru

If nn is a negative odd integer, which of the following expressions MUST be a positive even integer?

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Cevap: n2+1n^2 + 1

Cevap

The expression n2+1n^2 + 1 must be a positive even integer.
Squaring any negative odd integer nn produces a positive odd integer (n21n^2 \ge 1). Adding 11 to a positive odd integer always yields a positive even integer (n2+12n^2 + 1 \ge 2).

Adım Adım Çözüm

1
Analyze the parity and sign of n2n^2
Since nn is negative and odd, any even power of nn (such as n2n^2) produces a positive odd integer.
Negative times negative yields positive, and odd times odd yields odd.
2
Analyze the effect of adding 11 to n2n^2
n2+1n^2 + 1 is the sum of a positive odd integer and 11, which results in a positive even integer.
Adding 11 to any odd integer results in an even integer, and adding 11 to a positive integer maintains positivity.

Anahtar Kavram

Parity and sign rules for exponents and addition
Soru 109Soru

A water storage tank is initially filled to 45\frac{4}{5} of its total capacity with water. First, 14\frac{1}{4} of the water in the tank is drained for irrigation. Next, an amount of water equal to 13\frac{1}{3} of the remaining water in the tank is added back. Finally, 38\frac{3}{8} of the water currently in the tank is removed for domestic use. Which of the following statements about the volume of water in the tank must be true? Select all such statements.

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Cevap: After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations.; The final volume of water in the tank is equal to 58\frac{5}{8} of the initial volume of water.; The total volume of water removed from the tank during the first and third operations combined is equal to 12\frac{1}{2} of the total capacity of the tank.

Cevap

The correct statements are: (1) After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations; (2) The final volume of water in the tank is equal to 5/8 of the initial volume of water; and (3) The total volume of water removed from the tank during the first and third operations combined is equal to 1/2 of the total capacity of the tank.
The statement regarding the volume after the second operation is correct because adding 1/3 of the remaining 3/5 capacity adds 1/5 capacity, returning the total volume to 4/5 capacity. The statement comparing the final volume to the initial volume is correct because (1/2) divided by (4/5) equals 5/8. The statement regarding total water removed in the first and third operations is correct because 1/5 capacity plus 3/10 capacity equals 1/2 capacity.

Adım Adım Çözüm

1
Define total tank capacity as TT and find initial water volume.
Initial volume V0=45TV_0 = \frac{4}{5}T.
The problem states the tank is initially 45\frac{4}{5} full.
2
Calculate water remaining and removed after Operation 1.
Removed = 14×45T=15T\frac{1}{4} \times \frac{4}{5}T = \frac{1}{5}T. Remaining V1=45T15T=35TV_1 = \frac{4}{5}T - \frac{1}{5}T = \frac{3}{5}T.
Draining 14\frac{1}{4} of existing water leaves 34\frac{3}{4} of the existing water.
3
Calculate water remaining after Operation 2.
Added = 13×35T=15T\frac{1}{3} \times \frac{3}{5}T = \frac{1}{5}T. New volume V2=35T+15T=45TV_2 = \frac{3}{5}T + \frac{1}{5}T = \frac{4}{5}T.
Adding 13\frac{1}{3} of the remaining volume increases it by a factor of 1+13=431 + \frac{1}{3} = \frac{4}{3}.
4
Calculate water remaining and removed after Operation 3.
Removed = 38×45T=310T\frac{3}{8} \times \frac{4}{5}T = \frac{3}{10}T. Final volume V3=45T×(138)=45T×58=12TV_3 = \frac{4}{5}T \times \left(1 - \frac{3}{8}\right) = \frac{4}{5}T \times \frac{5}{8} = \frac{1}{2}T.
Removing 38\frac{3}{8} of current water leaves 58\frac{5}{8} of that volume.
5
Evaluate each given statement against calculated values.
V2=V0=45TV_2 = V_0 = \frac{4}{5}T (True). V3V0=12T45T=58\frac{V_3}{V_0} = \frac{\frac{1}{2}T}{\frac{4}{5}T} = \frac{5}{8} (True). Total removed = 15T+310T=12T\frac{1}{5}T + \frac{3}{10}T = \frac{1}{2}T (True).
Direct comparison with calculated step outcomes confirms these three statements are valid.

Anahtar Kavram

Sequential Fraction Multiplication and Part-to-Whole Relationships
Tahmini Süre:1m 45s
Soru 110Soru

An investment fund had an initial value of $4000\$4{}000. During the first year, the value of the fund increased by 15%15\%. During the second year, the new value of the fund increased by 10%10\%. What was the total value of the fund at the end of the second year?

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Cevap: $5060\$5{}060

Cevap

The total value of the fund at the end of the second year was $5060\$5{}060.
To find the final value after successive percent increases, compute the growth sequentially. After Year 1, the fund grows to $4000×1.15=$4600\$4{}000 \times 1.15 = \$4{}600. In Year 2, the 10%10\% increase applies to the new amount of $4600\$4{}600, resulting in $4600×1.10=$5060\$4{}600 \times 1.10 = \$5{}060. Thus, $5060\$5{}060 is the correct final value.

Adım Adım Çözüm

1
Calculate the fund's value at the end of the first year.
$4000×(1+0.15)=$4000×1.15=$4600\$4{}000 \times (1 + 0.15) = \$4{}000 \times 1.15 = \$4{}600
A 15%15\% increase on the initial $4000\$4{}000 base increases the value to 115%115\% of its original amount.
2
Calculate the fund's value at the end of the second year using the updated base value.
$4600×(1+0.10)=$4600×1.10=$5060\$4{}600 \times (1 + 0.10) = \$4{}600 \times 1.10 = \$5{}060
The second year's 10%10\% increase applies to the new base value of $4600\$4{}600, not the initial $4000\$4{}000.

Anahtar Kavram

Successive Percent Changes and Base Shift
Soru 111Soru

In a manufacturing facility, three automated machines—Machine A, Machine B, and Machine C—produce components at constant individual rates. Machine A working alone completes 310\frac{3}{10} of a standard daily order in 33 hours. Machine B working alone completes 25\frac{2}{5} of the order in 66 hours. Machine C working alone completes 14\frac{1}{4} of the order in 55 hours.

Initially, all three machines work together for 33 hours. At the end of 33 hours, Machine A malfunctions and stops, while Machines B and C continue working together without interruption until the daily order is finished. How many total hours does it take from the start to complete the entire daily order?

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Cevap: 6

Cevap

The total time required from the start to complete the entire daily order is 6 hours.
Each machine's rate per hour is found by dividing the given fraction by hours: Machine A is 110\frac{1}{10}, B is 115\frac{1}{15}, and C is 120\frac{1}{20}. Summing these gives an initial combined rate of 1360\frac{13}{60} per hour. In the first 3 hours, the three machines complete 3×1360=13203 \times \frac{13}{60} = \frac{13}{20} of the order, leaving 11320=7201 - \frac{13}{20} = \frac{7}{20} of the order remaining. When Machine A stops, the remaining combined rate of B and C is 115+120=760\frac{1}{15} + \frac{1}{20} = \frac{7}{60} per hour. Dividing the remaining 720\frac{7}{20} by 760\frac{7}{60} yields 3 additional hours. Adding the initial 3 hours gives a total duration of 6 hours.

Adım Adım Çözüm

1
Determine individual hourly work rates for each machine.
Machine A rate = 110\frac{1}{10} order/hr, Machine B rate = 115\frac{1}{15} order/hr, Machine C rate = 120\frac{1}{20} order/hr.
Divide the fraction of the job completed by the duration in hours to find the unit rate for each machine.
2
Calculate the combined rate of all three machines and the fraction completed in the first 3 hours.
Combined rate = 1360\frac{13}{60} order/hr; Work completed in 3 hours = 1320\frac{13}{20}, leaving 720\frac{7}{20} of the order unfinished.
Sum the three rates using a common denominator of 60, then multiply by 3 hours to find the total work done during the first stage.
3
Calculate the combined rate of Machines B and C, and find the additional time required to complete the remaining fraction.
Combined rate of B and C = 760\frac{7}{60} order/hr; Additional time needed = 3 hours.
Divide the remaining fraction 720\frac{7}{20} by the combined rate 760\frac{7}{60} to determine the remaining hours needed.
4
Add the initial duration and additional duration to obtain the total time.
Total time = 3+3=63 + 3 = 6 hours.
The question asks for the total elapsed time from the start of the job.

Anahtar Kavram

Adding rational fractions with different denominators to solve multi-stage work and rate problems.
Soru 112Soru

A technology company allocates its quarterly research budget among three key projects: Project Alpha, Project Beta, and Project Gamma. Initially, Project Alpha receives 25\frac{2}{5} of the total budget. Of the remaining budget, Project Beta is allocated 47\frac{4}{7}, and Project Gamma receives the rest. During a mid-quarter review, 16\frac{1}{6} of Project Alpha's allocated budget and 14\frac{1}{4} of Project Beta's allocated budget are transferred to Project Gamma. After these transfers, Project Gamma's final budget is what fraction of the total initial quarterly research budget?

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Cevap: 43105\frac{43}{105}

Cevap

43105\frac{43}{105}
Project Alpha receives 25\frac{2}{5} of the total budget, leaving 35\frac{3}{5}. Project Beta gets 47\frac{4}{7} of 35\frac{3}{5}, which is 1235\frac{12}{35}. Project Gamma's initial share is 351235=935\frac{3}{5} - \frac{12}{35} = \frac{9}{35}. Transferring 16\frac{1}{6} of Alpha's share contributes 115\frac{1}{15}, and transferring 14\frac{1}{4} of Beta's share contributes 335\frac{3}{35}. Adding these to Gamma's initial share yields 935+335+115=1235+115=36+7105=43105\frac{9}{35} + \frac{3}{35} + \frac{1}{15} = \frac{12}{35} + \frac{1}{15} = \frac{36 + 7}{105} = \frac{43}{105}.

Adım Adım Çözüm

1
Determine Project Alpha's initial allocation and the remaining budget fraction.
Project Alpha gets 25\frac{2}{5} of the total budget BB. The remaining fraction is 125=35B1 - \frac{2}{5} = \frac{3}{5} B.
Subtractions from the whole yield the remaining unallocated portion.
2
Calculate the initial allocations for Project Beta and Project Gamma.
Project Beta receives 47×35B=1235B\frac{4}{7} \times \frac{3}{5} B = \frac{12}{35} B. Project Gamma receives the rest of the remaining budget: 35B1235B=2135B1235B=935B\frac{3}{5} B - \frac{12}{35} B = \frac{21}{35} B - \frac{12}{35} B = \frac{9}{35} B.
Project Beta's share is a fraction of the remaining budget, not the total budget.
3
Calculate the transferred amounts from Project Alpha and Project Beta to Project Gamma.
Transfer from Alpha = 16×25B=115B\frac{1}{6} \times \frac{2}{5} B = \frac{1}{15} B. Transfer from Beta = 14×1235B=335B\frac{1}{4} \times \frac{12}{35} B = \frac{3}{35} B.
Transfers are fractional parts of each project's individual initial allocation.
4
Sum Project Gamma's initial share and the two transferred amounts.
Gamma's final fraction = 935+115+335=1235+115=36105+7105=43105\frac{9}{35} + \frac{1}{15} + \frac{3}{35} = \frac{12}{35} + \frac{1}{15} = \frac{36}{105} + \frac{7}{105} = \frac{43}{105}.
Finding the least common denominator (105105) allows exact addition of the rational numbers.

Anahtar Kavram

Multi-step sequential operations with fractions and rational numbers
Soru 113Soru

A technology company's annual revenue increased by 40%40\% from 2021 to 2022, and then decreased by 15%15\% from 2022 to 2023. If the company's annual revenue in 2023 was $476,000\$476,000, what was its annual revenue in 2021?

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Cevap: $400,000\$400,000

Cevap

$400,000\$400,000
The correct answer is $400,000\$400,000. Expressing the multi-step percentage change relative to the initial 2021 revenue RR gives an intermediate 2022 value of 1.40R1.40R and a final 2023 value of 1.40R×0.85=1.19R1.40R \times 0.85 = 1.19R. Solving 1.19R=476,0001.19R = 476,000 yields R=400,000R = 400,000.

Adım Adım Çözüm

1
Define the unknown variable and express successive percentage changes algebraically.
Let RR be the 2021 annual revenue. The 2022 revenue is R×(1+0.40)=1.40RR \times (1 + 0.40) = 1.40R. The 2023 revenue is 1.40R×(10.15)=1.40R×0.851.40R \times (1 - 0.15) = 1.40R \times 0.85.
Sequential percentage adjustments apply to the newly resulting base values rather than the initial value.
2
Calculate the net multiplier for the two-year change.
1.40×0.85=1.191.40 \times 0.85 = 1.19. Thus, the 2023 revenue equals 1.19R1.19R.
Multiplying the growth factor (1.401.40) by the decay factor (0.850.85) gives the overall percentage relationship between 2021 and 2023.
3
Solve for the initial revenue RR using the given 2023 revenue.
1.19R=476,000    R=476,0001.19=400,0001.19R = 476,000 \implies R = \frac{476,000}{1.19} = 400,000.
Dividing the final amount by the combined multiplier determines the original starting revenue.

Anahtar Kavram

Successive Percent Change and Base Shift
Soru 114Soru

Let nn be a positive integer that is a factor of 180180. If nn is divisible by 66 but is not divisible by 44, what is the maximum possible number of positive divisors of nn?

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Cevap: 12

Cevap

12
The prime factorization of 180180 is 22×32×512^2 \times 3^2 \times 5^1. Any factor nn of 180180 is of the form 2a×3b×5c2^a \times 3^b \times 5^c. For nn to be divisible by 66, we must have a1a \ge 1 and b1b \ge 1. For nn not to be divisible by 44, we must have a<2a < 2, which forces a=1a = 1. To maximize the total number of divisors (a+1)(b+1)(c+1)(a+1)(b+1)(c+1), we select the largest possible values for bb and cc, which are b=2b = 2 and c=1c = 1. This yields n=21×32×51=90n = 2^1 \times 3^2 \times 5^1 = 90, giving (1+1)(2+1)(1+1)=12(1+1)(2+1)(1+1) = 12 positive divisors.

Adım Adım Çözüm

1
Find the prime factorization of 180.
180=22×32×51180 = 2^2 \times 3^2 \times 5^1
Any positive factor nn of 180180 must have the form n=2a×3b×5cn = 2^a \times 3^b \times 5^c, where 0a20 \le a \le 2, 0b20 \le b \le 2, and 0c10 \le c \le 1.
2
Apply the divisibility conditions to determine the possible values of the exponent aa.
a=1a = 1
Since nn is divisible by 6=2×36 = 2 \times 3, a1a \ge 1 and b1b \ge 1. Since nn is not divisible by 4=224 = 2^2, a<2a < 2. Thus, aa must equal 11.
3
Maximize the divisor count formula (a+1)(b+1)(c+1)(a+1)(b+1)(c+1) using the available ranges for bb and cc.
b=2b = 2 and c=1c = 1
To maximize the number of positive divisors, choose the maximum allowable values for bb (b=2b=2) and cc (c=1c=1).
4
Calculate the maximum number of positive divisors.
(1+1)(2+1)(1+1)=2×3×2=12(1+1)(2+1)(1+1) = 2 \times 3 \times 2 = 12
For n=21×32×51=90n = 2^1 \times 3^2 \times 5^1 = 90, the total number of positive divisors is 1212.

Anahtar Kavram

Divisor count formula and prime factor constraints
Tahmini Süre:1m 30s
Soru 115Soru

A charity foundation allocates its annual grant budget among three sectors: Healthcare, Education, and Environmental Conservation. First, 25\frac{2}{5} of the total annual budget is allocated to Healthcare. Next, 59\frac{5}{9} of the remaining budget is allocated to Education. The final remainder of $120,000\$120,000 is allocated to Environmental Conservation. What was the foundation's total annual grant budget?

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Cevap: $450,000\$450,000

Cevap

$450,000\$450,000
The option specifying $450,000\$450,000 is correct because allocating 25\frac{2}{5} leaves 35\frac{3}{5} of the budget. Taking 59\frac{5}{9} of 35\frac{3}{5} gives 13\frac{1}{3} of the total. The combined allocations leave 1(25+13)=4151 - (\frac{2}{5} + \frac{1}{3}) = \frac{4}{15} of the total budget, which equals $120,000\$120,000. Dividing $120,000\$120,000 by 415\frac{4}{15} yields $450,000\$450,000.

Adım Adım Çözüm

1
Determine the fraction of the total budget remaining after the Healthcare allocation.
Since Healthcare receives 25\frac{2}{5} of the total budget, the remaining fraction is 125=351 - \frac{2}{5} = \frac{3}{5}.
Sequential fraction operations require updating the remaining whole after each step.
2
Calculate the fraction of the total budget allocated to Education.
Education receives 59\frac{5}{9} of the remaining 35\frac{3}{5}, which is 59×35=1545=13\frac{5}{9} \times \frac{3}{5} = \frac{15}{45} = \frac{1}{3} of the total budget.
To find a fraction of a remaining fraction, multiply the two fractions together.
3
Find the final remaining fraction allocated to Environmental Conservation.
Subtract both allocations from the whole: 1(25+13)=11115=4151 - \left(\frac{2}{5} + \frac{1}{3}\right) = 1 - \frac{11}{15} = \frac{4}{15}.
The final dollar amount corresponds to the fraction of the total budget left over.
4
Solve for the total annual grant budget.
415×Total=$120,000    Total=$120,000×154=$450,000\frac{4}{15} \times \text{Total} = \$120,000 \implies \text{Total} = \$120,000 \times \frac{15}{4} = \$450,000.
Dividing the part by its corresponding fractional proportion yields the total whole.

Anahtar Kavram

Sequential Fraction Operations and Finding the Whole from a Fractional Part
Soru 116Soru

Let mm and nn be positive integers such that 722m=n372^2 \cdot m = n^3. Which of the following statements MUST be true? Select all that apply.

Geçerli olan tümünü seçin

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Cevap: mm is divisible by 99; nn is divisible by 3636; The product mnm \cdot n is divisible by 108108

Cevap

The statements that mm is divisible by 99, nn is divisible by 3636, and the product mnm \cdot n is divisible by 108108 MUST be true.
Analyzing prime factorizations shows that n3=2634mn^3 = 2^6 \cdot 3^4 \cdot m. For the right-hand side to be a perfect cube, the power of 33 must be elevated to at least 66, requiring mm to contain 32=93^2 = 9 as a factor. Consequently, n3n^3 contains at least 2636=3632^6 \cdot 3^6 = 36^3, forcing nn to be a multiple of 3636. Finally, multiplying m=9k3m = 9k^3 by n=36kn = 36k yields 324k4324k^4, which is a multiple of 108108 for all integer values of kk.

Adım Adım Çözüm

1
Express 72272^2 in terms of its prime factorization.
72=2332    722=(2332)2=263472 = 2^3 \cdot 3^2 \implies 72^2 = (2^3 \cdot 3^2)^2 = 2^6 \cdot 3^4
Prime factorization allows us to analyze the exponents required for n3n^3 to be a perfect cube.
2
Determine the minimum prime factor requirements for mm and nn.
2634m=n3    2^6 \cdot 3^4 \cdot m = n^3 \implies exponent of 22 in mm must be 0\ge 0 (a multiple of 3), exponent of 33 in mm must be 2\ge 2 (since 4+2=64 + 2 = 6 is a multiple of 3).
Every prime factor in a perfect cube must have an exponent divisible by 3.
3
Establish general algebraic expressions for mm and nn.
m=9k3m = 9k^3 and n=36kn = 36k for any positive integer kk.
This captures all possible integer solutions for mm and nn.
4
Test each statement using the general forms m=9k3m = 9k^3 and n=36kn = 36k.
mm is divisible by 99 (9k39k^3 is a multiple of 99); nn is divisible by 3636 (36k36k is a multiple of 3636); mn=324k4=108(3k4)m \cdot n = 324k^4 = 108(3k^4), which is divisible by 108108. Counterexamples show mm need not be a square (m=72m=72) and nn need not be divisible by 5454 (n=36n=36).
Verifies which properties hold universally versus which can fail.

Anahtar Kavram

Prime Factor Exponents in Perfect Powers and Divisibility Rules
Soru 117Soru

An employee's hourly wage increased from $20.00\$20.00 to $25.00\$25.00. What was the percent increase in the employee's hourly wage?

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Cevap: 25

Cevap

The percent increase in the employee's hourly wage is 25%25\%.
The percent increase is calculated using the formula Percent Increase=New ValueOriginal ValueOriginal Value×100%\text{Percent Increase} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%. Substituting the given values gives 252020×100%=520×100%=25%\frac{25 - 20}{20} \times 100\% = \frac{5}{20} \times 100\% = 25\%.

Adım Adım Çözüm

1
Calculate the absolute increase in hourly wage
Increase = $25.00$20.00=$5.00\$25.00 - \$20.00 = \$5.00
Percent change is calculated relative to the absolute change from the initial value.
2
Calculate the percent change using the original wage as the base
Percent Increase=$5.00$20.00×100%=25%\text{Percent Increase} = \frac{\$5.00}{\$20.00} \times 100\% = 25\%
The formula for percent change is Amount of ChangeOriginal Amount×100%\frac{\text{Amount of Change}}{\text{Original Amount}} \times 100\%.

Anahtar Kavram

Percent Change Formula
Soru 118Soru

If mm is an even integer and nn is an odd integer, what is the value of (1)m+(1)n(-1)^m + (-1)^n?

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Cevap: 0

Cevap

0
Since mm is an even integer, (1)m=1(-1)^m = 1. Since nn is an odd integer, (1)n=1(-1)^n = -1. Adding 11 and 1-1 yields 00.

Adım Adım Çözüm

1
Evaluate (1)m(-1)^m for an even integer mm
(1)m=1(-1)^m = 1
Raising 1-1 to an even integer power always results in 11.
2
Evaluate (1)n(-1)^n for an odd integer nn
(1)n=1(-1)^n = -1
Raising 1-1 to an odd integer power always results in 1-1.
3
Sum the two evaluated expressions
1+(1)=01 + (-1) = 0
Combining 11 and 1-1 equals 00.

Anahtar Kavram

Exponent sign rules for even and odd powers
Soru 119Soru

On the real number line, the point PP corresponds to the real number xx. If x3=7|x - 3| = 7, which of the following is a possible value of x+5x + 5?

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Cevap: 11

Cevap

The correct answer is 11.
Solving x3=7|x - 3| = 7 produces two possible values for xx: x=10x = 10 and x=4x = -4. Substituting x=4x = -4 into x+5x + 5 yields 4+5=1-4 + 5 = 1, which is listed among the choices.

Adım Adım Çözüm

1
Set up the two linear equations defined by the absolute value equation x3=7|x - 3| = 7.
The two cases are x3=7x - 3 = 7 and x3=7x - 3 = -7.
By definition of absolute value, k=7|k| = 7 means k=7k = 7 or k=7k = -7.
2
Solve each case for xx.
From x3=7x - 3 = 7, x=10x = 10. From x3=7x - 3 = -7, x=4x = -4.
Adding 33 to both sides of each equation isolates xx.
3
Evaluate x+5x + 5 for both solutions of xx.
If x=10x = 10, x+5=15x + 5 = 15. If x=4x = -4, x+5=1x + 5 = 1.
Substitute each possible value of xx into the expression x+5x + 5.

Anahtar Kavram

Absolute Value as Distance on the Number Line
Tahmini Süre:45s
Soru 120Soru

In a community library, 415\frac{4}{15} of the total book collection consists of fiction books, 25\frac{2}{5} consists of non-fiction books, and the remaining 240 books are children's books. If 38\frac{3}{8} of the fiction books are hardcovers and the rest are softcovers, how many softcover fiction books are in the library?

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Cevap: 120

Cevap

The total number of softcover fiction books in the library is 120.
To find the number of softcover fiction books, first calculate the total number of books in the library. The combined fraction of fiction and non-fiction books is 415+25=415+615=1015=23\frac{4}{15} + \frac{2}{5} = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. Therefore, children's books represent the remaining 123=131 - \frac{2}{3} = \frac{1}{3} of the collection. Since 13\frac{1}{3} of the total equals 240 books, the total collection is 240×3=720240 \times 3 = 720 books. Next, compute the total number of fiction books: 415×720=192\frac{4}{15} \times 720 = 192. Finally, since 38\frac{3}{8} of fiction books are hardcovers, the remaining fraction of softcovers is 138=581 - \frac{3}{8} = \frac{5}{8}. Multiplying 58×192\frac{5}{8} \times 192 gives 120 softcover fiction books.

Adım Adım Çözüm

1
Find the combined fraction of fiction and non-fiction books, then determine the fraction of children's books.
Combined fraction = 415+615=1015=23\frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. Children's books fraction = 123=131 - \frac{2}{3} = \frac{1}{3}.
The sum of all non-overlapping fractions comprising the entire collection must equal 1.
2
Calculate the total number of books in the library.
Total collection = 240÷13=720240 \div \frac{1}{3} = 720 books.
The 240 children's books represent exactly 13\frac{1}{3} of the whole collection.
3
Determine the total count of fiction books.
Fiction books = 415×720=192\frac{4}{15} \times 720 = 192 books.
Fiction books make up 415\frac{4}{15} of the total 720 books.
4
Calculate the number of softcover fiction books.
Softcover fiction books = (138)×192=58×192=120\left(1 - \frac{3}{8}\right) \times 192 = \frac{5}{8} \times 192 = 120 books.
If 38\frac{3}{8} of the fiction subset are hardcovers, the remaining 58\frac{5}{8} of that subset are softcovers.

Anahtar Kavram

Solving multi-step rational number word problems by combining fractions and determining fractional parts of a subset.
Tahmini Süre:1m 30s
ÖncekiSayfa 6 / 16Sonraki
Arithmetic Alıştırma Soruları — GRE General Test — Sayfa 6 | Examkin