Tüm alıştırma soruları

2131 soru

Soru 1381Soru →

Printer X prints at a constant rate of 6060 pages per minute, and Printer Y prints at a constant rate of 4040 pages per minute. If both printers operate simultaneously at their respective constant rates, which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

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Cevap: Working together, the two printers print 100100 pages in 11 minute.; Printer X prints 60%60\% of the total number of pages produced when both printers operate for the same duration.; Working together, the two printers require 55 minutes to print 500500 pages.

Cevap

The correct statements are that working together the printers print 100 pages in 1 minute, Printer X prints 60% of the total pages, and working together they require 5 minutes to print 500 pages.
The combined rate is 60+40=10060 + 40 = 100 pages per minute. In 11 minute, 100100 pages are printed. In 55 minutes, 5×100=5005 \times 100 = 500 pages are printed. Out of every 100100 pages, Printer X produces 6060, which corresponds to 60%60\%.

Adım Adım Çözüm

1
Calculate the combined rate of both printers
Combined rate = 60+40=10060 + 40 = 100 pages per minute
When two entities work together simultaneously, their work rates add directly.
2
Evaluate the proportion of total output printed by Printer X
Proportion = 6060+40=60100=60%\frac{60}{60 + 40} = \frac{60}{100} = 60\%
The fraction of work done by one printer is its rate divided by the total combined rate.
3
Determine the time ratio between Printer Y and Printer X for a fixed job
Time ratio Y : X = 1/401/60=6040=3:2\frac{1/40}{1/60} = \frac{60}{40} = 3:2
Time required for a fixed task is inversely proportional to the work rate.
4
Calculate time needed for 500 pages working together
Time = 500 pages100 pages/min=5 minutes\frac{500\text{ pages}}{100\text{ pages/min}} = 5\text{ minutes}
Time equals total work divided by combined rate.

Anahtar Kavram

Additive work rates and inverse relationship between rate and time in ratio problems
Tahmini Süre:1m 0s
Soru 1382Soru →

Fill in the blank in the sentence below with the word that best completes the passage based on the structural contrast clues.

Aşağıdaki boşlukları doldurun

Paradoxically, despite the author's longstanding reputation for writing prose that alienated lay readers, her final manuscript was widely celebrated for its remarkable lucidity and broad accessibility.
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Cevap

abstruse (or semantically equivalent contrast terms such as recondite, esoteric, arcane, or opaque)
The sentence relies on dual contrast signals ('Paradoxically' and 'despite') to establish a reversal between the author's previous writing style and her final manuscript. Since the final manuscript is praised for its 'lucidity and broad accessibility,' the correct word for the blank must describe prose that is obscure or difficult to comprehend. The word 'abstruse' (or synonyms like 'recondite' or 'esoteric') fits both the contrast requirement and the contextual clue that her earlier work 'alienated lay readers.'

Adım Adım Çözüm

1
Identify structural transition signals in the sentence
The words 'Paradoxically' and 'despite' act as reversal and contrast pivots signaling an opposing relationship between the target blank and the description in the second half of the sentence.
Contrast pivots dictate that the missing word must carry a meaning opposite to the subsequent outcome described.
2
Analyze contextual clues and target valence
The manuscript was celebrated for 'lucidity and broad accessibility' and previously 'alienated lay readers.'
The blank describes prose that is difficult for general readers to comprehend.
3
Determine the required meaning for the blank
The blank requires a term meaning obscure, dense, or hard to understand, such as 'abstruse'.
'Abstruse' directly contrasts with 'lucidity' and logically accounts for alienating lay readers.

Anahtar Kavram

Identifying structural reversal signals (such as 'paradoxically' and 'despite') to determine semantic opposition in Text Completion questions.
Soru 1383Soru →

Passage:

In late eighteenth-century France, agronomic advocate Antoine-Augustin Parmentier sought to elevate the potato from a despised livestock feed to a staple crop for human consumption. Following his imprisonment in Prussia during the Seven Years' War—where he observed prisoners surviving solely on potato tubers—Parmentier realized the crop's caloric efficiency and resilience against famine. Prior to his campaign, French peasant communities widely believed that consuming potatoes transmitted leprosy, a superstition reinforced by municipal edicts prohibiting their cultivation in several northern provinces. To dismantle this institutional resistance, Parmentier employed unconventional publicity tactics rather than relying exclusively on academic treatises. In 1787, he planted a high-yield potato plot on royal land at Sablons, surrounding the perimeter with heavily armed guards during the day to convey an aura of immense value. Crucially, Parmentier instructed the guards to accept bribes and abandon their posts at night, allowing local farmers to intentionally 'steal' the tubers and cultivate them in private plots. This psychological strategy successfully transformed public perception, accelerating widespread domestic cultivation prior to the harvest failures of 1789.

According to the passage, French peasant resistance to potato cultivation prior to Parmentier's campaign was partly attributable to which of the following?

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Cevap: A widespread superstitious belief that consuming the tubers transmitted leprosy

Cevap

A widespread superstitious belief that consuming the tubers transmitted leprosy
The passage explicitly states that prior to Parmentier's campaign, 'French peasant communities widely believed that consuming potatoes transmitted leprosy.' The correct choice accurately retrieves and paraphrases this explicitly stated fact.

Adım Adım Çözüm

1
Analyze the prompt to determine the target detail
The prompt asks for a reason cited in the passage for French peasant resistance to potato cultivation prior to Parmentier's campaign.
Explicit detail retrieval requires matching the question's target criteria directly to the text.
2
Scan the passage for keywords related to peasant belief and resistance prior to the campaign
Located the sentence: 'Prior to his campaign, French peasant communities widely believed that consuming potatoes transmitted leprosy, a superstition reinforced by municipal edicts prohibiting their cultivation in several northern provinces.'
This section specifically details the cause of peasant resistance.
3
Evaluate the options against the explicit text statement
The option identifying a widespread superstitious belief that consuming tubers transmitted leprosy directly matches the statement in the text.
The correct answer must accurately paraphrase facts explicitly stated in the passage without adding outside assumptions.

Anahtar Kavram

Explicit Detail Retrieval
Tahmini Süre:1m 30s
Soru 1384Soru →

A circle has a radius of 66. A central angle of 60∘60^\circ intercepts an arc on the circle. What is the perimeter of the sector defined by this central angle?

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Cevap: 2π+122\pi + 12

Cevap

2π+122\pi + 12
The sector's perimeter includes the curved arc length and the two straight radii bounding it. The central angle of 60∘60^\circ represents 60360=16\frac{60}{360} = \frac{1}{6} of the full circle. The arc length is 16×2π(6)=2π\frac{1}{6} \times 2\pi(6) = 2\pi. Adding the two radii of length 66 gives 2π+6+6=2π+122\pi + 6 + 6 = 2\pi + 12.

Adım Adım Çözüm

1
Calculate the arc length of the sector
Arc length = 2π2\pi
The arc length formula is Arc Length=θ360∘×2πr\text{Arc Length} = \frac{\theta}{360^\circ} \times 2\pi r. Substituting θ=60∘\theta = 60^\circ and r=6r = 6 yields 60360×12π=16×12π=2π\frac{60}{360} \times 12\pi = \frac{1}{6} \times 12\pi = 2\pi.
2
Calculate the perimeter of the sector by adding the arc length to the two radii
Perimeter = 2π+122\pi + 12
The perimeter of a sector consists of the arc length plus two radii (2r2r). Thus, Perimeter=2π+2(6)=2π+12\text{Perimeter} = 2\pi + 2(6) = 2\pi + 12.

Anahtar Kavram

Perimeter of a sector equals arc length plus twice the radius (L+2rL + 2r).
Tahmini Süre:45s
Soru 1385Soru →

If x>0x > 0 and x12+x−12=3x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3, what is the value of x2+x−2x^2 + x^{-2}?

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Cevap: 47

Cevap

47
Squaring both sides of x12+x−12=3x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3 gives x+2+x−1=9x + 2 + x^{-1} = 9, which simplifies to x+x−1=7x + x^{-1} = 7. Squaring both sides of x+x−1=7x + x^{-1} = 7 gives x2+2+x−2=49x^2 + 2 + x^{-2} = 49, which yields x2+x−2=47x^2 + x^{-2} = 47.

Adım Adım Çözüm

1
Square both sides of the given equation x12+x−12=3x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 3.
(x12+x−12)2=32  ⟹  x+2(x12)(x−12)+x−1=9(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^2 = 3^2 \implies x + 2(x^{\frac{1}{2}})(x^{-\frac{1}{2}}) + x^{-1} = 9
Applying the binomial expansion identity (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2.
2
Simplify the middle term and solve for x+x−1x + x^{-1}.
x+2(1)+x−1=9  ⟹  x+x−1=7x + 2(1) + x^{-1} = 9 \implies x + x^{-1} = 7
Since x12⋅x−12=x0=1x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = x^0 = 1, subtracting 2 from both sides isolates x+x−1x + x^{-1}.
3
Square both sides of x+x−1=7x + x^{-1} = 7.
(x+x−1)2=72  ⟹  x2+2(x)(x−1)+x−2=49(x + x^{-1})^2 = 7^2 \implies x^2 + 2(x)(x^{-1}) + x^{-2} = 49
Squaring x+x−1x + x^{-1} generates the terms x2x^2 and x−2x^{-2}.
4
Simplify the middle term and solve for x2+x−2x^2 + x^{-2}.
x2+2+x−2=49  ⟹  x2+x−2=47x^2 + 2 + x^{-2} = 49 \implies x^2 + x^{-2} = 47
Subtracting 2 from both sides isolates the desired expression x2+x−2x^2 + x^{-2}.

Anahtar Kavram

Algebraic Exponents and Binomial Expansion
Soru 1386Soru →

A dataset consists of 15 numerical values with a mean of 50 and a standard deviation of 8. If a constant value of 5 is added to every number in the dataset, what is the standard deviation of the resulting dataset?

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Cevap: 8

Cevap

The standard deviation of the resulting dataset is 8.
Adding a constant to every data point in a set shifts the central location (the mean and median) by that constant, but leaves the distances between points unchanged. Because standard deviation depends solely on the distances between each point and the mean, the standard deviation remains 8.

Adım Adım Çözüm

1
Recall the effect of adding a constant to data points on measures of dispersion.
Measures of dispersion (range, IQR, standard deviation) describe the spread or scatter of data relative to its center.
When a constant kk is added to every value xix_i, the new mean becomes mean+k\text{mean} + k.
2
Evaluate the distance of each transformed data point from the new mean.
(xi+k)−(mean+k)=xi−mean(x_i + k) - (\text{mean} + k) = x_i - \text{mean}.
The deviation of each point from the mean remains identical to its original deviation.
3
Determine the standard deviation of the new dataset.
The standard deviation is unchanged and remains 8.
Because all individual deviations from the mean are preserved, the average distance from the mean does not change.

Anahtar Kavram

Invariance of standard deviation under addition of a constant
Soru 1387Soru →

If xx is an integer that satisfies both ∣2x−5∣≤9|2x - 5| \le 9 and 3−x<53 - x < 5, how many possible values of xx exist?

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Cevap: 9

Cevap

There are 9 possible integer values for x.
Solving ∣2x−5∣≤9|2x - 5| \le 9 gives −2≤x≤7-2 \le x \le 7. Solving 3−x<53 - x < 5 gives x>−2x > -2. Taking the intersection yields −2<x≤7-2 < x \le 7. The integers in this interval are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, amounting to 9 values in total.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣2x−5∣≤9|2x - 5| \le 9
−2≤x≤7-2 \le x \le 7
Unfold ∣2x−5∣≤9|2x - 5| \le 9 as −9≤2x−5≤9-9 \le 2x - 5 \le 9, add 5 to obtain −4≤2x≤14-4 \le 2x \le 14, and divide by 2.
2
Solve the linear inequality 3−x<53 - x < 5
x>−2x > -2
Subtract 3 to get −x<2-x < 2, then divide by −1-1 and reverse the inequality symbol.
3
Determine the intersection of both inequalities
−2<x≤7-2 < x \le 7
Combine −2≤x≤7-2 \le x \le 7 and x>−2x > -2 on the real number line.
4
Count the integer values within the intersection −2<x≤7-2 < x \le 7
9 integer values
The valid integers are −1,0,1,2,3,4,5,6,7-1, 0, 1, 2, 3, 4, 5, 6, 7, giving a total count of 7−(−1)+1=97 - (-1) + 1 = 9.

Anahtar Kavram

Linear Inequalities and Absolute Value Bounds
Tahmini Süre:1m 30s
Soru 1388Soru →

If xx is a real number that satisfies the equation x+6x−9+x−6x−9=10\sqrt{x + 6\sqrt{x - 9}} + \sqrt{x - 6\sqrt{x - 9}} = 10, what is the value of xx?

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Cevap: 34

Cevap

34
Using the substitution u=x−9≥0u = \sqrt{x - 9} \ge 0, we have x=u2+9x = u^2 + 9. The expressions under the square roots become x+6x−9=u2+6u+9=(u+3)2x + 6\sqrt{x - 9} = u^2 + 6u + 9 = (u + 3)^2 and x−6x−9=u2−6u+9=(u−3)2x - 6\sqrt{x - 9} = u^2 - 6u + 9 = (u - 3)^2. Taking square roots gives (u+3)2+(u−3)2=(u+3)+∣u−3∣=10\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10. For u≥3u \ge 3, this simplifies to (u+3)+(u−3)=10  ⟹  2u=10  ⟹  u=5(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5. Finally, substituting u=5u = 5 back yields x=52+9=34x = 5^2 + 9 = 34.

Adım Adım Çözüm

1
Define a variable substitution to simplify the nested radical structure.
Let u=x−9u = \sqrt{x - 9} where u≥0u \ge 0. Squaring both sides gives u2=x−9u^2 = x - 9, so x=u2+9x = u^2 + 9.
This substitution allows the expressions inside the outer square roots to be rewritten as polynomials in terms of uu.
2
Rewrite the expressions under each square root as perfect square trinomials.
x+6x−9=(u2+9)+6u=(u+3)2x + 6\sqrt{x - 9} = (u^2 + 9) + 6u = (u + 3)^2 and x−6x−9=(u2+9)−6u=(u−3)2x - 6\sqrt{x - 9} = (u^2 + 9) - 6u = (u - 3)^2.
Expressing terms as perfect squares allows the outer radicals to be simplified.
3
Simplify the square root expressions using absolute values.
(u+3)2+(u−3)2=(u+3)+∣u−3∣=10\sqrt{(u + 3)^2} + \sqrt{(u - 3)^2} = (u + 3) + |u - 3| = 10.
For any real number aa, a2=∣a∣\sqrt{a^2} = |a|. Since u≥0u \ge 0, u+3>0u + 3 > 0, so ∣u+3∣=u+3|u + 3| = u + 3.
4
Solve the absolute value equation across valid domain intervals.
If u≥3u \ge 3, ∣u−3∣=u−3|u - 3| = u - 3, giving (u+3)+(u−3)=10  ⟹  2u=10  ⟹  u=5(u + 3) + (u - 3) = 10 \implies 2u = 10 \implies u = 5. If 0≤u<30 \le u < 3, ∣u−3∣=3−u|u - 3| = 3 - u, giving (u+3)+(3−u)=6≠10(u + 3) + (3 - u) = 6 \neq 10 (no solution). Thus, u=5u = 5.
Splitting into cases based on the definition of absolute value isolates the valid root.
5
Substitute u=5u = 5 back into the expression for xx.
x=52+9=25+9=34x = 5^2 + 9 = 25 + 9 = 34.
Converting from uu back to xx provides the solution to the original equation.

Anahtar Kavram

Simplifying nested radicals by completing the square under the radical sign and applying the identity a2=∣a∣\sqrt{a^2} = |a|.
Soru 1389Soru →

A company allocated a total budget of BB dollars for a project. In the first phase of the project, 25\frac{2}{5} of the total budget plus $3,000\$3,000 was spent. In the second phase, 13\frac{1}{3} of the remaining budget after the first phase was spent. If the unspent amount after both phases is $14,000\$14,000, what was the total initial budget BB?

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Cevap: $40,000\$40,000

Cevap

$40,000\$40,000
The correct answer of $40,000\$40,000 is obtained by systematically tracking the remaining budget after each phase. After Phase 1, the remaining amount is B−(25B+3,000)=35B−3,000B - (\frac{2}{5}B + 3,000) = \frac{3}{5}B - 3,000. Spending 13\frac{1}{3} of this balance in Phase 2 leaves 23\frac{2}{3} of it unspent: 23(35B−3,000)=25B−2,000\frac{2}{3}(\frac{3}{5}B - 3,000) = \frac{2}{5}B - 2,000. Setting this equal to the final unspent amount of $14,000\$14,000 gives 25B=16,000\frac{2}{5}B = 16,000, which solves to B=40,000B = 40,000.

Adım Adım Çözüm

1
Express the remaining budget after the first phase in terms of BB.
Amount spent in Phase 1 = 25B+3,000\frac{2}{5}B + 3,000. Remaining after Phase 1 = B−(25B+3,000)=35B−3,000B - \left(\frac{2}{5}B + 3,000\right) = \frac{3}{5}B - 3,000.
Subtracting the first phase expenses from the initial total budget BB determines the balance available for the second phase.
2
Express the unspent budget after the second phase.
Since 13\frac{1}{3} of the remaining budget was spent in Phase 2, 1−13=231 - \frac{1}{3} = \frac{2}{3} of that remaining budget is left. Remaining after Phase 2 = 23(35B−3,000)\frac{2}{3}\left(\frac{3}{5}B - 3,000\right).
Taking 23\frac{2}{3} of the Phase 1 remainder directly gives the final unspent amount.
3
Expand and simplify the algebraic equation setting the unspent amount equal to $14,000\$14,000.
\frac{2}{3}\left(\frac{3}{5}B - 3,000\right) = 14,000 \implies \frac{2}{5}B - 2,000 = 14,000.
Distributing 23\frac{2}{3} across both terms inside the parentheses clears the fraction product.
4
Solve the linear equation for BB.
\frac{2}{5}B = 16,000 \implies B = 16,000 \times \frac{5}{2} = 40,000.
Adding 2,0002,000 to both sides and multiplying by the reciprocal 52\frac{5}{2} yields the total budget BB.

Anahtar Kavram

Formulating and solving multi-step linear equations in one variable with fractional quantities and consecutive remaining balances.
Tahmini Süre:2m 0s
Soru 1390Soru →

If xx and yy are positive integers satisfying 3x−3y=7023^x - 3^y = 702 and x+y=3\sqrt{x + y} = 3, what is the value of x2−y2x^2 - y^2?

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Cevap: 2727

Cevap

The correct value of x2−y2x^2 - y^2 is 27.
Squaring x+y=3\sqrt{x + y} = 3 gives x+y=9x + y = 9. Factoring 3x−3y=7023^x - 3^y = 702 gives 3y(3x−y−1)=7023^y(3^{x-y} - 1) = 702. Since 702=33×26702 = 3^3 \times 26 and (3x−y−1)(3^{x-y} - 1) is coprime to 3, we deduce 3y=33  ⟹  y=33^y = 3^3 \implies y = 3, and 3x−3−1=26  ⟹  3x−3=27  ⟹  x=63^{x-3} - 1 = 26 \implies 3^{x-3} = 27 \implies x = 6. Substituting x=6x = 6 and y=3y = 3 into x2−y2x^2 - y^2 gives 36−9=2736 - 9 = 27.

Adım Adım Çözüm

1
Eliminate the radical from the given linear equation.
Squaring both sides of x+y=3\sqrt{x + y} = 3 gives x+y=9x + y = 9.
Squaring both sides removes the square root operator to establish a linear relationship between xx and yy.
2
Factor out the common exponential term 3y3^y from 3x−3y=7023^x - 3^y = 702.
3y(3x−y−1)=7023^y(3^{x-y} - 1) = 702.
Since xx and yy are positive integers and 702>0702 > 0, it must be true that x>yx > y, allowing factoring by exponent rules 3x=3y⋅3x−y3^x = 3^y \cdot 3^{x-y}.
3
Find the prime factorization of 702 and match the power of 3.
702=27×26=33×26702 = 27 \times 26 = 3^3 \times 26, so 3y(3x−y−1)=33×263^y(3^{x-y} - 1) = 3^3 \times 26.
The factor (3x−y−1)(3^{x-y} - 1) is not divisible by 3 because 3x−y3^{x-y} is a multiple of 3 for x>yx > y. Therefore, all powers of 3 in 702 must belong to 3y3^y.
4
Solve for the values of yy and xx.
y=3y = 3 and 3x−3−1=26  ⟹  3x−3=27=33  ⟹  x−3=3  ⟹  x=63^{x-3} - 1 = 26 \implies 3^{x-3} = 27 = 3^3 \implies x - 3 = 3 \implies x = 6.
Equating prime component bases yields y=3y = 3 and x=6x = 6, which satisfies x+y=6+3=9x + y = 6 + 3 = 9.
5
Calculate the target expression x2−y2x^2 - y^2.
x2−y2=62−32=36−9=27x^2 - y^2 = 6^2 - 3^2 = 36 - 9 = 27.
Substituting x=6x = 6 and y=3y = 3 into x2−y2x^2 - y^2 yields 27 (or using (x+y)(x−y)=9×3=27(x+y)(x-y) = 9 \times 3 = 27).

Anahtar Kavram

Factoring exponential expressions using prime factorization, radical simplification, and difference of squares.
Soru 1391Soru →

A dataset of 2020 numerical measurements has a range of 3030 and a standard deviation of 6.46.4. If every measurement in the dataset is multiplied by −2-2 and then increased by 55 to construct a new dataset, what are the range and standard deviation of the new dataset, respectively?

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Cevap: Range is 6060 and standard deviation is 12.812.8

Cevap

The range of the new dataset is 6060 and the standard deviation is 12.812.8.
Under a linear transformation Y=aX+bY = aX + b, all measures of dispersion (range, interquartile range, standard deviation) are multiplied by ∣a∣|a| and are unaffected by the additive constant bb. Here, a=−2a = -2 and b=5b = 5. The absolute multiplier is ∣−2∣=2|-2| = 2. Therefore, the new range is 2×30=602 \times 30 = 60 and the new standard deviation is 2×6.4=12.82 \times 6.4 = 12.8.

Adım Adım Çözüm

1
Analyze the general transformation formula for measures of dispersion under linear transformations of the form Y=aX+bY = aX + b.
For any constant multiplier aa and constant shift bb, the new range is ∣a∣×RangeX|a| \times \text{Range}_X and the new standard deviation is ∣a∣×σX|a| \times \sigma_X.
Measures of dispersion reflect spread; multiplying each data value by aa scales the spread by ∣a∣|a|, while adding a constant shift bb translates all values equally without expanding or contracting their relative distance.
2
Calculate the new range using a=−2a = -2 and b=5b = 5.
\text{New Range} = |-2| \times 30 = 2 \times 30 = 60.
The range scales by the absolute value of the multiplier ∣−2∣=2|-2| = 2, while the additive constant 55 has no impact on the spread.
3
Calculate the new standard deviation using a=−2a = -2 and b=5b = 5.
\text{New Standard Deviation} = |-2| \times 6.4 = 2 \times 6.4 = 12.8.
The standard deviation scales by ∣−2∣=2|-2| = 2, and the shift of +5+5 does not change the dispersion around the mean.

Anahtar Kavram

Effect of Linear Transformations on Measures of Dispersion
Tahmini Süre:1m 30s
Soru 1392Soru →

How many integer values of xx satisfy the compound absolute value inequality 1≤∣∣x−4∣−3∣≤51 \le ||x - 4| - 3| \le 5?

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Cevap: 15

Cevap

There are 15 integer values of xx that satisfy the given compound inequality.
Solving 1≤∣∣x−4∣−3∣≤51 \le ||x - 4| - 3| \le 5 requires breaking the nested absolute value into its boundary constraints. The upper bound ∥x−4∣−3∣≤5\|x - 4| - 3| \le 5 restricts xx to [−4,12][-4, 12]. The lower bound ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1 requires either ∣x−4∣≥4|x - 4| \ge 4 (giving x≤0x \le 0 or x≥8x \ge 8) or ∣x−4∣≤2|x - 4| \le 2 (giving 2≤x≤62 \le x \le 6). Taking the intersection produces three distinct inclusive integer intervals: [−4,0][-4, 0], [2,6][2, 6], and [8,12][8, 12]. Each interval contains 5 integers, yielding a total of 15 integer solutions.

Adım Adım Çözüm

1
Decompose the double inequality into two separate absolute value inequalities: ∣∣x−4∣−3∣≤5||x - 4| - 3| \le 5 and ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1.
Two simultaneous inequalities to solve for xx.
A double inequality a≤∣u∣≤ba \le |u| \le b requires satisfying both ∣u∣≤b|u| \le b and ∣u∣≥a|u| \ge a.
2
Solve the upper bound inequality ∣∣x−4∣−3∣≤5||x - 4| - 3| \le 5.
−5≤∣x−4∣−3≤5  ⟹  −2≤∣x−4∣≤8-5 \le |x - 4| - 3 \le 5 \implies -2 \le |x - 4| \le 8. Since ∣x−4∣≥0≥−2|x - 4| \ge 0 \ge -2 is always true, this simplifies to ∣x−4∣≤8  ⟹  −4≤x≤12|x - 4| \le 8 \implies -4 \le x \le 12.
Absolute value is non-negative, so the lower bound of −2-2 imposes no extra constraint.
3
Solve the lower bound inequality ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1.
This splits into two cases: ∣x−4∣−3≥1|x - 4| - 3 \ge 1 OR ∣x−4∣−3≤−1|x - 4| - 3 \le -1.
Case A: ∣x−4∣≥4  ⟹  x−4≥4|x - 4| \ge 4 \implies x - 4 \ge 4 or x−4≤−4  ⟹  x≥8x - 4 \le -4 \implies x \ge 8 or x≤0x \le 0.
Case B: ∣x−4∣≤2  ⟹  −2≤x−4≤2  ⟹  2≤x≤6|x - 4| \le 2 \implies -2 \le x - 4 \le 2 \implies 2 \le x \le 6.
The absolute value inequality ∣u∣≥1|u| \ge 1 holds when u≥1u \ge 1 or u≤−1u \le -1.
4
Intersect the solution set from the upper bound [−4,12][-4, 12] with the solution set from the lower bound (−∞,0]∪[2,6]∪[8,∞)(-\infty, 0] \cup [2, 6] \cup [8, \infty).
The valid solution set is x∈[−4,0]∪[2,6]∪[8,12]x \in [-4, 0] \cup [2, 6] \cup [8, 12].
Both conditions must hold simultaneously.
5
Count the integer values in each of the three valid intervals.
Interval [−4,0][-4, 0] has 5 integers: {−4,−3,−2,−1,0}\{-4, -3, -2, -1, 0\}.
Interval [2,6][2, 6] has 5 integers: {2,3,4,5,6}\{2, 3, 4, 5, 6\}.
Interval [8,12][8, 12] has 5 integers: {8,9,10,11,12}\{8, 9, 10, 11, 12\}.
Total integer solutions = 5+5+5=155 + 5 + 5 = 15.
The number of integers in an inclusive integer range [a,b][a, b] is b−a+1b - a + 1.

Anahtar Kavram

Linear Inequalities and Absolute Value
Soru 1393Soru →

A company's annual budget of $84,000\$84,000 is split among three departments: Research, Marketing, and Operations. The Marketing department receives 23\frac{2}{3} as much funding as the Research department. The Operations department receives $6,000\$6,000 more than half of the combined funding of the Research and Marketing departments. What is the amount, in dollars, allocated to the Research department?

Cevabı ve açıklamayı göster

Cevap: 31,20031,200

Cevap

31,20031,200
Defining Research funding as xx, Marketing funding becomes 23x\frac{2}{3}x, and Operations funding becomes 12(x+23x)+6000=56x+6000\frac{1}{2}(x + \frac{2}{3}x) + 6000 = \frac{5}{6}x + 6000. Summing all three department allocations yields x+23x+56x+6000=84,000x + \frac{2}{3}x + \frac{5}{6}x + 6000 = 84,000. Combining the variable terms gives 52x+6000=84,000\frac{5}{2}x + 6000 = 84,000, which simplifies to 52x=78,000\frac{5}{2}x = 78,000 and yields x=31,200x = 31,200.

Adım Adım Çözüm

1
Define the unknown variable for the target quantity.
Let xx represent the dollar amount allocated to the Research department.
The problem asks specifically for the Research department allocation.
2
Express the allocations of Marketing and Operations in terms of xx.
Marketing =23x= \frac{2}{3}x. Combined Research and Marketing =x+23x=53x= x + \frac{2}{3}x = \frac{5}{3}x. Operations =12(53x)+6,000=56x+6,000= \frac{1}{2}\left(\frac{5}{3}x\right) + 6,000 = \frac{5}{6}x + 6,000.
Translating word problem relationships into algebraic expressions.
3
Set up the single-variable linear equation for the total budget.
x+23x+(56x+6,000)=84,000x + \frac{2}{3}x + \left(\frac{5}{6}x + 6,000\right) = 84,000
The sum of allocations across all three departments must equal the total budget of $84,000\$84,000.
4
Combine like terms using a common denominator.
66x+46x+56x+6,000=84,000  ⟹  156x+6,000=84,000  ⟹  52x+6,000=84,000\frac{6}{6}x + \frac{4}{6}x + \frac{5}{6}x + 6,000 = 84,000 \implies \frac{15}{6}x + 6,000 = 84,000 \implies \frac{5}{2}x + 6,000 = 84,000
Simplifying fractional coefficients by finding the common denominator 6.
5
Isolate xx to solve the linear equation.
52x=78,000  ⟹  5x=156,000  ⟹  x=31,200\frac{5}{2}x = 78,000 \implies 5x = 156,000 \implies x = 31,200
Subtracting 6,0006,000 from both sides and multiplying by 25\frac{2}{5}.

Anahtar Kavram

Linear Equations in One Variable
Soru 1394Soru →

If xx is a real number such that ∣4−3x∣≤13|4 - 3x| \leq 13, and yy is an integer such that −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3, what is the least possible integer value of x2−yx^2 - y?

Cevabı ve açıklamayı göster

Cevap: −7-7

Cevap

The least possible integer value of x2−yx^2 - y is −7-7.
To find the minimum value of x2−yx^2 - y, we must minimize x2x^2 and maximize yy. The absolute value inequality ∣4−3x∣≤13|4 - 3x| \leq 13 simplifies to −3≤x≤173-3 \leq x \leq \frac{17}{3}, which contains 00, so the minimum of x2x^2 is 00. The inequality −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3 simplifies to −4≤y<8-4 \leq y < 8. Since yy is an integer, its maximum value is 77. Therefore, the minimum possible value of x2−yx^2 - y is 0−7=−70 - 7 = -7.

Adım Adım Çözüm

1
Solve the absolute value inequality ∣4−3x∣≤13|4 - 3x| \leq 13 for xx.
−13≤4−3x≤13  ⟹  −17≤−3x≤9  ⟹  173≥x≥−3  ⟹  −3≤x≤173-13 \leq 4 - 3x \leq 13 \implies -17 \leq -3x \leq 9 \implies \frac{17}{3} \geq x \geq -3 \implies -3 \leq x \leq \frac{17}{3}.
An absolute value inequality ∣u∣≤k|u| \leq k expands to −k≤u≤k-k \leq u \leq k. Dividing by a negative number flips the inequality signs.
2
Determine the minimum possible value of x2x^2.
Minimum x2=0x^2 = 0.
Since xx can take any real value in the interval [−3,173][-3, \frac{17}{3}], which contains 00, the minimum square of any real number in this interval is 00 at x=0x = 0.
3
Solve the double inequality −5<1−2y3≤3-5 < \frac{1 - 2y}{3} \leq 3 for yy.
−15<1−2y≤9  ⟹  −16<−2y≤8  ⟹  8>y≥−4  ⟹  −4≤y<8-15 < 1 - 2y \leq 9 \implies -16 < -2y \leq 8 \implies 8 > y \geq -4 \implies -4 \leq y < 8.
Multiplying by 33 preserves inequalities, subtracting 11 preserves inequalities, and dividing by −2-2 reverses all inequality signs.
4
Find the maximum integer value of yy.
Maximum integer y=7y = 7.
The solution set for yy is the half-open interval [−4,8)[-4, 8). Since yy is constrained to be an integer, the largest integer strictly less than 88 is 77.
5
Minimize the expression x2−yx^2 - y.
Minimum (x2−y)=Minimum (x2)−Maximum (y)=0−7=−7\text{Minimum } (x^2 - y) = \text{Minimum } (x^2) - \text{Maximum } (y) = 0 - 7 = -7.
To minimize a difference A−BA - B, one must minimize the minuend AA and maximize the subtrahend BB.

Anahtar Kavram

Linear Inequalities and Absolute Value
Tahmini Süre:2m 0s
Soru 1395Soru →

A circle has a radius of 1010 units. A sector within this circle is defined by a central angle of 72∘72^\circ. Which of the following statements about this sector are true? Select all such statements.

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Cevabı ve açıklamayı göster

Cevap: The arc length of the sector is 4π4\pi units.; The area of the sector is 20π20\pi square units.; The ratio of the sector's area to the circle's total area is 11 to 55.

Cevap

The statements confirming an arc length of 4π4\pi units, a sector area of 20π20\pi square units, and a sector-to-total area ratio of 11 to 55 are correct.
The central angle fraction is 72∘360∘=15\frac{72^\circ}{360^\circ} = \frac{1}{5}. Multiplying the full circumference 20π20\pi by 15\frac{1}{5} gives an arc length of 4π4\pi. Multiplying the total area 100π100\pi by 15\frac{1}{5} gives a sector area of 20π20\pi. The ratio of sector area to total area is also equal to 15\frac{1}{5}.

Adım Adım Çözüm

1
Find the central angle fraction of the circle
The fraction is 72∘360∘=15\frac{72^\circ}{360^\circ} = \frac{1}{5}
Arc length and sector area are proportional to the central angle relative to a full 360∘360^\circ turn.
2
Calculate the arc length of the sector
Arc Length =15×2π(10)=4π= \frac{1}{5} \times 2\pi(10) = 4\pi units
The arc length is the circle's circumference multiplied by the central angle fraction.
3
Calculate the area of the sector
Sector Area =15×π(10)2=20π= \frac{1}{5} \times \pi(10)^2 = 20\pi square units
The sector area is the circle's total area multiplied by the central angle fraction.
4
Determine the area ratio
Ratio =Sector AreaTotal Area=20π100π=15= \frac{\text{Sector Area}}{\text{Total Area}} = \frac{20\pi}{100\pi} = \frac{1}{5}
The ratio of the sector area to the total area is identical to the central angle fraction.

Anahtar Kavram

Arc Length and Sector Area Formulas
Tahmini Süre:1m 0s
Soru 1396Soru →

For all real numbers x>0x > 0, which of the following expressions are equivalent to (x3/2⋅x3x1/6)2\left(\frac{x^{3/2} \cdot \sqrt[3]{x}}{x^{1/6}}\right)^2? Select all such expressions.

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Cevabı ve açıklamayı göster

Cevap: x103\sqrt[3]{x^{10}}; x3x3x^3 \sqrt[3]{x}; (x53)2\left(\sqrt[3]{x^5}\right)^2

Cevap

The expressions equivalent to the given expression are x103\sqrt[3]{x^{10}}, x3x3x^3 \sqrt[3]{x}, and (x53)2\left(\sqrt[3]{x^5}\right)^2.
Simplifying the given expression by converting all radical forms to fractional exponents yields x10/3x^{10/3}. The three expressions x103\sqrt[3]{x^{10}}, x3x3x^3 \sqrt[3]{x}, and (x53)2\left(\sqrt[3]{x^5}\right)^2 each rewrite to x10/3x^{10/3} when evaluated using standard power laws.

Adım Adım Çözüm

1
Convert radical expressions to fractional exponent form inside the parentheses.
x3=x1/3\sqrt[3]{x} = x^{1/3}, so the numerator becomes x3/2⋅x1/3x^{3/2} \cdot x^{1/3}.
Converting all terms to exponent notation allows applying standard exponent addition and subtraction rules.
2
Simplify the numerator by adding exponents.
x3/2+1/3=x9/6+2/6=x11/6x^{3/2 + 1/3} = x^{9/6 + 2/6} = x^{11/6}.
When multiplying exponential terms with the same base, add their exponents using a common denominator.
3
Divide by the denominator by subtracting exponents.
x11/6x1/6=x11/6−1/6=x10/6=x5/3\frac{x^{11/6}}{x^{1/6}} = x^{11/6 - 1/6} = x^{10/6} = x^{5/3}.
When dividing exponential terms with the same base, subtract the denominator's exponent from the numerator's exponent.
4
Apply the outer exponent of 2.
(x5/3)2=x(5/3)⋅2=x10/3\left(x^{5/3}\right)^2 = x^{(5/3) \cdot 2} = x^{10/3}.
When raising a power to another power, multiply the inner and outer exponents.
5
Compare x10/3x^{10/3} to each option.
x103=x10/3\sqrt[3]{x^{10}} = x^{10/3}, x3x3=x3+1/3=x10/3x^3 \sqrt[3]{x} = x^{3 + 1/3} = x^{10/3}, and (x53)2=(x5/3)2=x10/3\left(\sqrt[3]{x^5}\right)^2 = (x^{5/3})^2 = x^{10/3} are all equivalent.
Matching each candidate expression in fractional exponent form identifies all equivalent choices.

Anahtar Kavram

Simplification of algebraic expressions using rules of fractional exponents and radicals
Tahmini Süre:1m 30s
Soru 1397Soru →

Three straight lines, RR, SS, and TT, all intersect at a single point PP. Line RR is perpendicular to line SS. Line TT intersects line RR such that one of the acute angles formed between line RR and line TT measures 35∘35^\circ. Which of the following degree measures represent angles formed between any two of the intersecting lines at point PP? Select all that apply.

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Cevabı ve açıklamayı göster

Cevap: 55∘55^\circ; 90∘90^\circ; 145∘145^\circ

Cevap

The degree measures 55∘55^\circ, 90∘90^\circ, and 145∘145^\circ are all valid angle measures formed by the intersecting lines at point PP.
The intersecting lines form several distinct angle measures at point PP. The right angle between lines RR and SS measures 90∘90^\circ. The acute angle between lines TT and SS is complementary to the 35∘35^\circ angle, giving 90∘−35∘=55∘90^\circ - 35^\circ = 55^\circ. The obtuse angle between lines TT and RR along straight line RR is supplementary to 35∘35^\circ, giving 180∘−35∘=145∘180^\circ - 35^\circ = 145^\circ. Therefore, the options stating 55∘55^\circ, 90∘90^\circ, and 145∘145^\circ are all correct.

Adım Adım Çözüm

1
Identify the angle measure between perpendicular lines R and S.
Since line R⊥SR \perp S, the angle between them is 90∘90^\circ.
Perpendicular lines intersect at right angles (90∘90^\circ).
2
Calculate the acute angle between line T and line S.
Angle between TT and SS = 90∘−35∘=55∘90^\circ - 35^\circ = 55^\circ.
The given 35∘35^\circ angle between RR and TT and the adjacent angle between TT and SS form the 90∘90^\circ right angle between RR and SS.
3
Determine the supplementary obtuse angles formed by line T with line R and line S.
Supplementary angle to 35∘35^\circ is 180∘−35∘=145∘180^\circ - 35^\circ = 145^\circ. Supplementary angle to 55∘55^\circ is 180∘−55∘=125∘180^\circ - 55^\circ = 125^\circ.
Adjacent angles along a straight line sum to 180∘180^\circ.
4
Compare calculated angle measures with the options.
The valid angle measures are 35∘35^\circ, 55∘55^\circ, 90∘90^\circ, 125∘125^\circ, and 145∘145^\circ. Thus 55∘55^\circ, 90∘90^\circ, and 145∘145^\circ are correct.
Matching calculated angle measures with the choices provided.

Anahtar Kavram

Perpendicular line relationships and supplementary angle properties of intersecting lines.
Tahmini Süre:1m 0s
Soru 1398Soru →

If −3∣x−2∣+4>−11-3|x - 2| + 4 > -11, which of the following inequalities represents all possible real values of xx?

Cevabı ve açıklamayı göster

Cevap: −3<x<7-3 < x < 7

Cevap

−3<x<7-3 < x < 7
Subtracting 44 from both sides gives −3∣x−2∣>−15-3|x - 2| > -15. Dividing by −3-3 and reversing the inequality sign results in ∣x−2∣<5|x - 2| < 5. Converting to the compound inequality −5<x−2<5-5 < x - 2 < 5 and adding 22 to each part yields the correct interval −3<x<7-3 < x < 7.

Adım Adım Çözüm

1
Subtract 4 from both sides of the inequality
−3∣x−2∣>−15-3|x - 2| > -15
Isolate the absolute value term on the left side.
2
Divide both sides by −3-3 and reverse the inequality sign
∣x−2∣<5|x - 2| < 5
Dividing an inequality by a negative number reverses the direction of the inequality symbol.
3
Express the absolute value inequality as a compound inequality
−5<x−2<5-5 < x - 2 < 5
An inequality of the form ∣u∣<c|u| < c (where c>0c > 0) is equivalent to −c<u<c-c < u < c.
4
Add 2 to all three parts of the compound inequality
−3<x<7-3 < x < 7
Isolate xx to find the complete range of solution values.

Anahtar Kavram

Linear Inequalities and Absolute Value
Soru 1399Soru →

A circle has a circumference of 20π20\pi. A sector of this circle is defined by a central angle that intercepts an arc of length 5π5\pi. What is the area of this sector?

Cevabı ve açıklamayı göster

Cevap: 25π25\pi

Cevap

The area of the sector is 25π25\pi.
First, the radius is found using 2πr=20π2\pi r = 20\pi, which gives r=10r = 10. The total area of the circle is π(10)2=100π\pi (10)^2 = 100\pi. The fraction of the circle defined by the sector is 5π20π=14\frac{5\pi}{20\pi} = \frac{1}{4}. Multiplying the total area by this fraction gives 14×100π=25π\frac{1}{4} \times 100\pi = 25\pi.

Adım Adım Çözüm

1
Find the radius of the circle from the circumference.
Since C=2πr=20πC = 2\pi r = 20\pi, solving for rr gives r=10r = 10.
The radius is required to calculate the total circle area.
2
Determine the fraction of the circle that the sector represents.
\text{Fraction} = \frac{\text{Arc Length}}{\text{Circumference}} = \frac{5\pi}{20\pi} = \frac{1}{4}.
The ratio of arc length to circumference gives the proportion of the total circle occupied by the sector.
3
Calculate the total area of the circle.
A = \pi r^2 = \pi (10)^2 = 100\pi.
The area of a circle with radius 10 is 100π100\pi.
4
Multiply the total area by the sector fraction.
\text{Sector Area} = \frac{1}{4} \times 100\pi = 25\pi.
Applying the proportional fraction yields the sector area.

Anahtar Kavram

The ratio of a sector's arc length to the full circumference is equal to the ratio of the sector's area to the full circle's area.
Soru 1400Soru →

An original dataset XX consists of 2020 distinct positive numerical values with standard deviation σ\sigma, range RR, and interquartile range IQRIQR. A new dataset YY is created by transforming each data value xix_i in XX according to the rule yi=−3xi+5y_i = -3x_i + 5. Which of the following statements about the dispersion metrics of dataset YY compared to dataset XX must be true? Select all such statements.

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Cevabı ve açıklamayı göster

Cevap: The standard deviation of dataset YY is 3σ3\sigma.; The interquartile range of dataset YY is 3×IQR3 \times IQR.

Cevap

The statements asserting that the standard deviation of dataset Y is 3 times the standard deviation of dataset X, and that the interquartile range of dataset Y is 3 times the interquartile range of dataset X, are both correct.
For any data set undergoing a linear transformation y=ax+by = ax + b, all measures of dispersion (including standard deviation, range, and interquartile range) are scaled by the absolute value of the multiplier ∣a∣|a| and remain completely unaffected by the additive constant bb. Since a=−3a = -3, ∣a∣=∣−3∣=3|a| = |-3| = 3. Therefore, both the standard deviation and the interquartile range scale by a factor of 3.

Adım Adım Çözüm

1
Recall the general rule for linear transformation of measures of dispersion.
For any linear transformation y=ax+by = ax + b, measures of dispersion (Range, IQR, Standard Deviation) scale by ∣a∣|a| and are unaffected by bb.
Measures of dispersion evaluate the spread/distance between points, which shifts unchanged when a constant is added, and scales non-negatively when multiplied by a constant.
2
Identify the values of the multiplicative constant aa and additive constant bb.
a=−3a = -3 and b=5b = 5, giving ∣a∣=∣−3∣=3|a| = |-3| = 3.
The transformation equation is yi=−3xi+5y_i = -3x_i + 5.
3
Apply the scaling factor ∣a∣=3|a| = 3 to each measure of dispersion.
Standard deviation of YY is 3σ3\sigma, IQR of YY is 3×IQR3 \times IQR, and Range of YY is 3R3R.
All measures of dispersion scale by a factor of 3 regardless of the negative sign of the multiplier or the addition of 5.

Anahtar Kavram

Linear Transformations on Measures of Dispersion
ÖncekiSayfa 70 / 107Sonraki
Tüm alıştırma soruları — GRE General Test | Examkin