Algebra

356 soru

Soru 261Soru

A theater sells student tickets for $20\$20 each and adult tickets for $35\$35 each. For a specific performance, the theater sold a total of 150150 tickets and collected total revenue of RR dollars. If at least 4040 student tickets were sold and at most 9090 adult tickets were sold, which of the following values could be the total revenue RR? Select all such values.

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Cevap: $3450\$3{}450; $4200\$4{}200

Cevap

The possible values for the total revenue RR are $3450\$3{}450 and $4200\$4{}200.
The linear revenue model is R=525015sR = 5250 - 15s. Considering both conditions (s40s \ge 40 and a=150s90    s60a = 150 - s \le 90 \implies s \ge 60), the valid range for student tickets is 60s15060 \le s \le 150. This restricts the possible revenue RR to multiples of $15\$15 between $3000\$3{}000 and $4350\$4{}350. Both $3450\$3{}450 and $4200\$4{}200 fall within this valid range and correspond to integer ticket quantities (s=120,a=30s = 120, a = 30 and s=70,a=80s = 70, a = 80, respectively).

Adım Adım Çözüm

1
Set up equations for the total number of tickets and revenue.
Let ss be the number of student tickets and aa be the number of adult tickets. Then s+a=150s + a = 150, so a=150sa = 150 - s. Total revenue R=20s+35a=20s+35(150s)=525015sR = 20s + 35a = 20s + 35(150 - s) = 5250 - 15s.
Expressing revenue in terms of a single variable ss simplifies finding the domain and range.
2
Determine the constraints on the variable ss.
We are given s40s \ge 40 and a90a \le 90. Substituting a=150s90a = 150 - s \le 90 yields s60s \ge 60. Combining constraints gives 60s15060 \le s \le 150.
The number of adult tickets being at most 9090 forces the number of student tickets to be at least 6060.
3
Calculate the upper and lower bounds for the revenue RR.
Maximum revenue occurs when s=60s = 60: Rmax=525015(60)=$4350R_{\text{max}} = 5250 - 15(60) = \$4{}350. Minimum revenue occurs when s=150s = 150: Rmin=525015(150)=$3000R_{\text{min}} = 5250 - 15(150) = \$3{}000.
Since R=525015sR = 5250 - 15s is a decreasing linear function of ss, the maximum revenue occurs at the minimum valid value of ss and vice versa.
4
Evaluate the given choices against the range and divisibility requirements.
RR must be an integer multiple of 1515 subtracted from 52505250, meaning RR must be between $3000\$3{}000 and $4350\$4{}350 inclusive, and (5250R)(5250 - R) must be divisible by 1515. $3450\$3{}450 (where s=120s = 120) and $4200\$4{}200 (where s=70s = 70) are both valid. $2850\$2{}850 is below the minimum bound, $4400\$4{}400 does not yield an integer value for ss, and $4650\$4{}650 violates the adult ticket upper bound.
Only options meeting both inequality constraints and integer ticket requirements are valid.

Anahtar Kavram

Linear modeling of word problems under linear system constraints and inequalities
Tahmini Süre:1m 30s
Soru 262Soru

For a real constant kk, the quadratic equation x22kx+(k2k6)=0x^2 - 2kx + (k^2 - k - 6) = 0 has two distinct real roots rr and ss such that r<0<sr < 0 < s and r<s|r| < |s|. Which of the following inequalities expresses all possible values of kk?

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Cevap: 0<k<30 < k < 3

Cevap

The inequality expressing all possible values of kk is 0<k<30 < k < 3.
The condition that one root is negative and one root is positive (r<0<sr < 0 < s) requires the product of the roots rs=k2k6rs = k^2 - k - 6 to be negative, which resolves to 2<k<3-2 < k < 3. Furthermore, since the positive root ss has a larger absolute magnitude than the negative root rr (r<s|r| < |s|), the sum of the roots r+s=2kr + s = 2k must be positive, requiring k>0k > 0. Taking the intersection of 2<k<3-2 < k < 3 and k>0k > 0 yields 0<k<30 < k < 3.

Adım Adım Çözüm

1
Apply Vieta's formulas to express the sum and product of the roots in terms of kk.
For x22kx+(k2k6)=0x^2 - 2kx + (k^2 - k - 6) = 0, the sum of roots is r+s=2kr + s = 2k and the product of roots is rs=k2k6rs = k^2 - k - 6.
Vieta's formulas directly relate the coefficients of a quadratic equation to the sum and product of its roots.
2
Analyze the condition r<0<sr < 0 < s.
Since one root is negative and the other is positive, their product must be negative: rs=k2k6<0rs = k^2 - k - 6 < 0. Factoring gives (k3)(k+2)<0(k - 3)(k + 2) < 0, which yields 2<k<3-2 < k < 3.
A positive number multiplied by a negative number produces a negative product.
3
Analyze the condition r<s|r| < |s|.
Since r<0r < 0, r=r|r| = -r. Since s>0s > 0, s=s|s| = s. The inequality r<s|r| < |s| becomes r<s-r < s, which simplifies to r+s>0r + s > 0. Substituting r+s=2kr + s = 2k gives 2k>02k > 0, or k>0k > 0.
The positive root having greater magnitude than the absolute value of the negative root means the sum of the roots must be positive.
4
Combine the conditions to find the valid range for kk.
Combining 2<k<3-2 < k < 3 and k>0k > 0 yields the intersection 0<k<30 < k < 3. (The discriminant condition Δ=4(k+6)>0    k>6\Delta = 4(k + 6) > 0 \implies k > -6 is satisfied for all k(0,3)k \in (0, 3)).
The parameter kk must satisfy both root sign constraints simultaneously.

Anahtar Kavram

Using Vieta's formulas and root magnitude conditions to solve quadratic parameter inequality problems.
Soru 263Soru

A clean energy technology company manufactures two models of solar panels: Model X and Model Y. Model X produces 150150 kilowatt-hours (kWh) of electricity per day and costs $400\$400 to manufacture, while Model Y produces 200200 kWh of electricity per day and costs $550\$550 to manufacture. A solar energy project purchased a total of 5050 panels for a total manufacturing cost of $24,500\$24,500. What is the total daily electricity production, in kilowatt-hours, of all 5050 panels combined?

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Cevap: 9,0009,000

Cevap

The total daily electricity production of all 5050 panels combined is 9,0009,000 kWh.
The correct option is 9,0009,000 kWh. Defining xx as the number of Model X panels and yy as the number of Model Y panels gives the equations x+y=50x + y = 50 and 400x+550y=24,500400x + 550y = 24,500. Substituting y=50xy = 50 - x yields 400x+550(50x)=24,500400x + 550(50 - x) = 24,500, which simplifies to 150x=3,000-150x = -3,000, so x=20x = 20 and y=30y = 30. Computing the total output yields 20(150)+30(200)=3,000+6,000=9,00020(150) + 30(200) = 3,000 + 6,000 = 9,000 kWh.

Adım Adım Çözüm

1
Define variables for the quantities of each panel model.
Let xx represent the number of Model X panels and yy represent the number of Model Y panels.
Establishing explicit variables allows formulating a system of linear equations from the word problem.
2
Set up equations for total panel quantity and total manufacturing cost.
System equations: x+y=50x + y = 50 and 400x+550y=24,500400x + 550y = 24,500.
The total number of panels is 5050, and the combined manufacturing cost equals $24,500\$24,500.
3
Solve the system using substitution.
Substitute y=50xy = 50 - x into the cost equation: 400x+550(50x)=24,500    400x+27,500550x=24,500    150x=3,000    x=20400x + 550(50 - x) = 24,500 \implies 400x + 27,500 - 550x = 24,500 \implies -150x = -3,000 \implies x = 20. Consequently, y=5020=30y = 50 - 20 = 30.
Determining x=20x = 20 and y=30y = 30 gives the exact number of Model X and Model Y panels purchased.
4
Calculate the total daily electricity production.
Total production =20(150)+30(200)=3,000+6,000=9,000= 20(150) + 30(200) = 3,000 + 6,000 = 9,000 kWh.
Multiply the quantity of each panel model by its respective daily output rate and sum the products.

Anahtar Kavram

Formulating and solving a system of two linear equations from a real-world scenario to find unknown quantities and evaluate a secondary combination function.
Tahmini Süre:1m 30s
Soru 264Soru

For all real numbers xx and yy, the custom operation \odot is defined by xy=xyyxx \odot y = x|y| - y|x|. Which of the following statements must be true for all real numbers xx and yy? Select all such statements.

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Cevap: xy=0x \odot y = 0 whenever xx and yy have the same sign; xy=(yx)x \odot y = -(y \odot x); If x>0x > 0 and y<0y < 0, then xy>0x \odot y > 0

Cevap

The correct statements are the statement asserting xy=0x \odot y = 0 when xx and yy have the same sign, the statement asserting anti-commutativity xy=(yx)x \odot y = -(y \odot x), and the statement asserting xy>0x \odot y > 0 when x>0x > 0 and y<0y < 0.
The operation xy=xyyxx \odot y = x|y| - y|x| produces 0 whenever xx and yy share the same sign because terms evaluate to identical quantities. Swapping variables negates the expression, establishing anti-commutativity. When xx is positive and yy is negative, xyx \odot y simplifies to 2xy-2xy, which is strictly greater than 0 since xy<0xy < 0.

Adım Adım Çözüm

1
Analyze the first statement regarding same-sign inputs
If x>0x > 0 and y>0y > 0, x=x|x|=x and y=y|y|=y, so xy=xyyx=0x \odot y = xy - yx = 0. If x<0x < 0 and y<0y < 0, x=x|x|=-x and y=y|y|=-y, so xy=x(y)y(x)=xy+xy=0x \odot y = x(-y) - y(-x) = -xy + xy = 0. Thus, xy=0x \odot y = 0 when xx and yy have the same sign.
Verifying the definition under both positive and negative cases of identical sign.
2
Analyze the second statement regarding operand order reversal
yx=yxxy=(xyyx)=(xy)y \odot x = y|x| - x|y| = -(x|y| - y|x|) = -(x \odot y), which holds universally for all real numbers.
Testing anti-commutativity by algebraic substitution into the custom operation.
3
Analyze the third statement for opposite signs (x>0x > 0 and y<0y < 0)
Since x>0x > 0, x=x|x|=x. Since y<0y < 0, y=y|y|=-y. Substituting yields x(y)y(x)=xyxy=2xyx(-y) - y(x) = -xy - xy = -2xy. Because x>0x > 0 and y<0y < 0, the product xyxy is negative, making 2xy-2xy strictly positive.
Determining the overall algebraic sign of the expression when variables have opposite signs.
4
Counter-test the remaining statements to verify incorrectness
For x(x)x \odot (-x) with x=1x=1: 1(1)=11(1)1=1(1)=201 \odot (-1) = 1|-1| - (-1)|1| = 1 - (-1) = 2 \neq 0. For associativity with x=2,y=1,z=1x=2, y=-1, z=-1: (21)1=41=8(2 \odot -1) \odot -1 = 4 \odot -1 = 8, but 2(11)=20=02 \odot (-1 \odot -1) = 2 \odot 0 = 0.
Demonstrating specific counterexamples for false generalizations.

Anahtar Kavram

Custom Binary Operations and Absolute Value Properties
Tahmini Süre:1m 30s
Soru 265Soru

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

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Cevap: 8

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.
Soru 266Soru

A courier service calculates the shipping cost for a package based on its weight ww, in pounds. For packages weighing up to 2020 pounds, the cost is a flat fee of $12.00\$12.00 plus $2.50\$2.50 per pound. For packages weighing more than 2020 pounds, the cost is a flat fee of $20.00\$20.00 plus $2.00\$2.00 per pound. If the total shipping cost for a package was strictly greater than $45.00\$45.00 and at most $70.00\$70.00, which of the following could be the weight of the package, in pounds? Select all such weights.

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Cevap: 1515 pounds; 2020 pounds; 2424 pounds

Cevap

The valid weights of the package are 15 pounds, 20 pounds, and 24 pounds.
The weight of the package must result in a total cost C(w)C(w) such that $45.00<C(w)$70.00\$45.00 < C(w) \le \$70.00. Evaluating the options:
- For 15 pounds: 12+2.50(15)=$49.5012 + 2.50(15) = \$49.50, which falls within the range.
- For 20 pounds: 12+2.50(20)=$62.0012 + 2.50(20) = \$62.00, which falls within the range.
- For 24 pounds: 20+2.00(24)=$68.0020 + 2.00(24) = \$68.00, which falls within the range.

Adım Adım Çözüm

1
Formulate the cost function C(w)C(w) as a piecewise model.
C(w)=12+2.50wC(w) = 12 + 2.50w for 0<w200 < w \le 20, and C(w)=20+2.00wC(w) = 20 + 2.00w for w>20w > 20.
The cost structure changes depending on whether the weight exceeds 2020 pounds.
2
Set up and solve the inequality 45<C(w)7045 < C(w) \le 70 for packages up to 2020 pounds.
45<12+2.50w70    33<2.50w58    13.2<w23.245 < 12 + 2.50w \le 70 \implies 33 < 2.50w \le 58 \implies 13.2 < w \le 23.2. Restricting to w20w \le 20 gives 13.2<w2013.2 < w \le 20.
This determines the valid weight interval for packages billed under the first tier.
3
Set up and solve the inequality 45<C(w)7045 < C(w) \le 70 for packages over 2020 pounds.
45<20+2.00w70    25<2.00w50    12.5<w2545 < 20 + 2.00w \le 70 \implies 25 < 2.00w \le 50 \implies 12.5 < w \le 25. Restricting to w>20w > 20 gives 20<w2520 < w \le 25.
This determines the valid weight interval for packages billed under the second tier.
4
Combine the valid intervals and evaluate each option.
The overall valid weight interval is 13.2<w2513.2 < w \le 25. Testing the values: 12 pounds is outside the range; 15 pounds, 20 pounds, and 24 pounds are within the range; 26 pounds is outside the range.
Any weight strictly greater than 13.213.2 pounds and up to 2525 pounds produces a cost between $45.00\$45.00 and $70.00\$70.00.

Anahtar Kavram

Algebraic Modeling of Piecewise Functions and Compound Inequalities
Soru 267Soru

A logistics company operates two delivery vans, Van A and Van B. Van A consumes fuel at a rate of 11 gallon for every 1515 miles driven, and Van B consumes fuel at a rate of 11 gallon for every 2525 miles driven. On a specific trip, the two vans were driven a total combined distance of 450450 miles and together consumed exactly 2222 gallons of fuel. What was the total distance, in miles, driven by Van A?

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Cevap: 150 miles

Cevap

The total distance driven by Van A was 150 miles.
The correct answer is 150 miles. By setting up the total fuel consumption equation as dA15+450dA25=22\frac{d_A}{15} + \frac{450 - d_A}{25} = 22, multiplying by the least common multiple 7575 gives 5dA+13503dA=16505d_A + 1350 - 3d_A = 1650, which simplifies to 2dA=3002d_A = 300 and dA=150d_A = 150 miles.

Adım Adım Çözüm

1
Define variables for the distance traveled by each van.
Let dAd_A be the distance driven by Van A. Then the distance driven by Van B is 450dA450 - d_A.
The total combined distance for both vans is 450450 miles.
2
Express the total fuel consumed in terms of dAd_A.
\frac{d_A}{15} + \frac{450 - d_A}{25} = 22
Fuel used equals distance divided by miles per gallon for each vehicle.
3
Clear denominators by multiplying the entire equation by the common denominator 75.
5d_A + 3(450 - d_A) = 1650
Simplifies fractions to solve the linear algebraic equation easily.
4
Solve for dAd_A.
5d_A + 1350 - 3d_A = 1650 \implies 2d_A = 300 \implies d_A = 150
Isolates the target variable dAd_A to find the distance driven by Van A.

Anahtar Kavram

Linear Equations and Rate-Distance Modeling
Soru 268Soru

In the xyxy-plane, line kk has an xx-intercept of (8,0)(8, 0) and a yy-intercept of (0,6)(0, 6). Line pp is perpendicular to line kk and intersects line kk at its yy-intercept. What is the xx-intercept of line pp?

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Cevap: 92-\frac{9}{2}

Cevap

92-\frac{9}{2}
The line kk passes through (8,0)(8, 0) and (0,6)(0, 6), giving a slope of mk=6008=34m_k = \frac{6 - 0}{0 - 8} = -\frac{3}{4}. A line perpendicular to kk must have a slope equal to the negative reciprocal of 34-\frac{3}{4}, which is 43\frac{4}{3}. Because line pp intersects line kk at (0,6)(0, 6), its yy-intercept is also 66, making its equation y=43x+6y = \frac{4}{3}x + 6. Setting y=0y = 0 gives 0=43x+60 = \frac{4}{3}x + 6, which solves to x=92x = -\frac{9}{2}.

Adım Adım Çözüm

1
Calculate the slope of line kk
Slope of line kk is mk=6008=68=34m_k = \frac{6 - 0}{0 - 8} = -\frac{6}{8} = -\frac{3}{4}
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} using given intercepts (8,0)(8, 0) and (0,6)(0, 6).
2
Determine the slope of line pp
Slope of line pp is mp=1mk=43m_p = -\frac{1}{m_k} = \frac{4}{3}
Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.
3
Write the equation of line pp
Equation of line pp is y=43x+6y = \frac{4}{3}x + 6
Line pp intersects line kk at its yy-intercept (0,6)(0, 6), so line pp has a yy-intercept of 66.
4
Find the xx-intercept of line pp
x=92x = -\frac{9}{2} (or 4.5-4.5)
Set y=0y = 0 in the line equation: 0=43x+6    43x=6    x=634=920 = \frac{4}{3}x + 6 \implies \frac{4}{3}x = -6 \implies x = -6 \cdot \frac{3}{4} = -\frac{9}{2}.

Anahtar Kavram

Perpendicular Slopes and Line Intercepts
Tahmini Süre:1m 30s
Soru 269Soru

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

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Cevap: 8

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Algebraic mixture problems using mass balance equations.
Tahmini Süre:1m 30s
Soru 270Soru

In the xyxy-plane, line LL has a slope of 34-\frac{3}{4} and passes through the point (4,1)(4, 1). Which of the following statements about line LL must be true? Select all such statements.

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Cevap: Line LL passes through the point (4,7)(-4, 7).; Line LL is perpendicular to the line defined by 4x3y=124x - 3y = 12.; Line LL is parallel to the line defined by 3x+4y=253x + 4y = 25.

Cevap

The true statements are that line L passes through (-4, 7), is perpendicular to 4x - 3y = 12, and is parallel to 3x + 4y = 25.
The correct statements are those identifying that (4,7)(-4, 7) satisfies y=34x+4y = -\frac{3}{4}x + 4, that 4x3y=124x - 3y = 12 has a perpendicular slope of 43\frac{4}{3}, and that 3x+4y=253x + 4y = 25 has an identical parallel slope of 34-\frac{3}{4}.

Adım Adım Çözüm

1
Find the equation of line L using point-slope form.
y1=34(x4)    y=34x+4y - 1 = -\frac{3}{4}(x - 4) \implies y = -\frac{3}{4}x + 4, or in standard form 3x+4y=163x + 4y = 16.
Establishing the linear equation allows systematic verification of points, intercepts, and slope properties.
2
Verify point (4,7)(-4, 7) on line L.
3(4)+4(7)=12+28=163(-4) + 4(7) = -12 + 28 = 16, which satisfies the equation.
Checking if the point coordinates satisfy 3x+4y=163x + 4y = 16 confirms it lies on the line.
3
Determine quadrant coverage.
The yy-intercept is (0,4)(0, 4) and the xx-intercept is (163,0)\left(\frac{16}{3}, 0\right). For x<0x < 0, y>4y > 4 (Quadrant II); for 0<x<1630 < x < \frac{16}{3}, y>0y > 0 (Quadrant I); for x>163x > \frac{16}{3}, y<0y < 0 (Quadrant IV).
Line L passes through Quadrants I, II, and IV only, so it does not enter Quadrant III.
4
Analyze slope relationships for perpendicularity and parallelism.
Line 4x3y=124x - 3y = 12 has slope 43\frac{4}{3} (negative reciprocal of 34-\frac{3}{4}). Line 3x+4y=253x + 4y = 25 has slope 34-\frac{3}{4} (identical slope).
Perpendicular lines have slopes that multiply to 1-1, while parallel lines have equal slopes and different intercepts.

Anahtar Kavram

Linear line equations, parallel/perpendicular slopes, and point/intercept evaluations in coordinate geometry.
Tahmini Süre:1m 30s
Soru 271Soru

A municipal water treatment facility uses two intake pipes, Pipe XX and Pipe YY, to fill a main reservoir. Operating alone at its constant rate, Pipe XX can fill the empty reservoir in 1010 hours. Operating alone at its constant rate, Pipe YY can fill the empty reservoir in 1515 hours. Pipe XX is turned on first and operates alone for 33 hours. Then, Pipe YY is also turned on, and both pipes operate together until the reservoir is completely full. What is the total number of hours Pipe XX operates from the moment it is turned on until the reservoir is completely filled?

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Cevap: 7.27.2

Cevap

The total number of hours Pipe X operates is 7.27.2 hours.
The correct answer is 7.27.2 hours. Pipe X completes 3/103/10 of the reservoir in 3 hours, leaving 7/107/10 of the reservoir to be filled. The combined rate of Pipe X and Pipe Y is 1/10+1/15=1/61/10 + 1/15 = 1/6 per hour. The joint time required to fill the remaining 7/107/10 is (7/10)/(1/6)=4.2(7/10) / (1/6) = 4.2 hours. Adding the initial 3 hours Pipe X worked alone yields a total of 3+4.2=7.23 + 4.2 = 7.2 hours.

Adım Adım Çözüm

1
Determine the individual hourly work rates of Pipe X and Pipe Y.
Rate of Pipe X = 110\frac{1}{10} reservoir per hour; Rate of Pipe Y = 115\frac{1}{15} reservoir per hour.
The rate is the reciprocal of the total time required to complete the job individually.
2
Calculate the fraction of the reservoir filled by Pipe X during its initial 3-hour solo operation.
Work done in first 3 hours = 3×110=3103 \times \frac{1}{10} = \frac{3}{10} of the reservoir.
Multiplying Pipe X's rate by its solo operating time gives the completed portion of the work.
3
Find the remaining fraction of the reservoir that needs to be filled.
Remaining work = 1310=7101 - \frac{3}{10} = \frac{7}{10} of the reservoir.
Subtracting the completed portion from the whole (1) leaves the uncompleted portion.
4
Determine the combined hourly rate when both pipes operate simultaneously.
Combined rate = 110+115=330+230=530=16\frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} of the reservoir per hour.
Simultaneous operation rates are additive.
5
Calculate the time tt during which both pipes operate together to finish the remaining work.
t=71016=710×6=4210=4.2t = \frac{\frac{7}{10}}{\frac{1}{6}} = \frac{7}{10} \times 6 = \frac{42}{10} = 4.2 hours.
Dividing the remaining work by the combined rate yields the joint operation time.
6
Add Pipe X's solo time to the joint operation time to get Pipe X's total operating time.
Total time = 3+4.2=7.23 + 4.2 = 7.2 hours.
Pipe X was active during both the initial solo period and the joint operating period.

Anahtar Kavram

Combined Rate and Staggered Work Modeling
Tahmini Süre:2m 0s
Soru 272Soru

In the xyxy-plane, line kk is defined by the equation 2xy=82x - y = 8. Line mm is perpendicular to line kk and passes through the points (1,9)(-1, 9) and (t,5)(t, 5). What is the value of tt?

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Cevap: 77

Cevap

The value of tt is 77.
The line kk has equation y=2x8y = 2x - 8, giving a slope of 22. A line perpendicular to it must have a slope of 12-\frac{1}{2}. Using the slope formula for line mm passing through (1,9)(-1, 9) and (t,5)(t, 5) yields 59t(1)=12\frac{5 - 9}{t - (-1)} = -\frac{1}{2}, which simplifies to 4t+1=12\frac{-4}{t + 1} = -\frac{1}{2}, leading directly to t=7t = 7.

Adım Adım Çözüm

1
Find the slope of line kk.
Converting 2xy=82x - y = 8 to slope-intercept form gives y=2x8y = 2x - 8, so the slope of line kk is mk=2m_k = 2.
The slope-intercept form y=mx+by = mx + b directly identifies the slope mm of a line.
2
Determine the slope of line mm.
Since line mm is perpendicular to line kk, its slope is mm=1mk=12m_m = -\frac{1}{m_k} = -\frac{1}{2}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Set up the slope formula for line mm using the given points and solve for tt.
59t(1)=12    4t+1=12    4t+1=12    t+1=8    t=7\frac{5 - 9}{t - (-1)} = -\frac{1}{2} \implies \frac{-4}{t + 1} = -\frac{1}{2} \implies \frac{4}{t + 1} = \frac{1}{2} \implies t + 1 = 8 \implies t = 7.
The slope between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by y2y1x2x1\frac{y_2 - y_1}{x_2 - x_1}.

Anahtar Kavram

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.
Tahmini Süre:1m 30s
Soru 273Soru

An investor allocated a total principal of $10,000\$10,000 between two accounts, Account X and Account Y. Account X pays simple annual interest at a rate of 6%6\%, while Account Y pays simple annual interest at a rate of 8%8\%. Let xx represent the amount of money, in dollars, invested in Account X, and let dd represent the total annual interest earned, in dollars, from both accounts after one year. Which of the following statements must be true? Select all that apply.

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Cevap: The total annual interest dd satisfies 600d800600 \le d \le 800.; The amount invested in Account X can be represented as x=800d0.02x = \frac{800 - d}{0.02}.; If equal amounts were invested in both accounts, the total annual interest earned is $700\$700.

Cevap

The true statements are that the total annual interest dd satisfies 600d800600 \le d \le 800, the amount in Account X is given by x=800d0.02x = \frac{800 - d}{0.02}, and equal investment in both accounts yields $700\$700 in total interest.
The total interest equation d=8000.02xd = 800 - 0.02x dictates all valid relationships. Because 0x10,0000 \le x \le 10,000, the bounds for dd are strictly between 600600 and 800800. Rearranging the equation yields x=800d0.02x = \frac{800 - d}{0.02}, which correctly models the Account X investment. Substituting x=5,000x = 5,000 yields d=700d = 700, confirming the equal-allocation scenario.

Adım Adım Çözüm

1
Formulate the total interest model as a linear equation in terms of xx.
d=0.06x+0.08(10,000x)=8000.02xd = 0.06x + 0.08(10,000 - x) = 800 - 0.02x
The interest earned from Account X is 0.06x0.06x and the interest from Account Y is 0.08(10,000x)0.08(10,000 - x).
2
Determine the range of possible interest values for 0x10,0000 \le x \le 10,000.
When x=10,000x = 10,000, d=600d = 600. When x=0x = 0, d=800d = 800. Thus 600d800600 \le d \le 800.
The minimum and maximum interest values occur at the extreme allocation bounds.
3
Rearrange the interest equation to express xx in terms of dd.
0.02x=800d    x=800d0.020.02x = 800 - d \implies x = \frac{800 - d}{0.02}
Isolating xx provides a formula for calculating the Account X principal directly from the total interest.
4
Evaluate the specific numerical cases given in the options.
Equal investment (x=5,000x = 5,000) gives d=800100=700d = 800 - 100 = 700. For d=750d = 750, x=2,500x = 2,500 (Account Y = 7,5007,500). For d=680d = 680, x=6,000x = 6,000 (Account Y = 4,0004,000, ratio 3:23:2).
Substituting specific values tests the validity of each conditional statement.

Anahtar Kavram

Linear Modeling and Algebraic Rate Allocation
Soru 274Soru

If 3x+y=143x + y = 14 and x2y=7x - 2y = -7, what is the value of x+yx + y?

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Cevap: 8

Cevap

8
Solving the linear system yields x=3x = 3 and y=5y = 5. Adding these values together gives 3+5=83 + 5 = 8, which is the correct answer.

Adım Adım Çözüm

1
Express yy in terms of xx using the first equation
y=143xy = 14 - 3x
Isolating one variable allows for direct substitution into the second equation.
2
Substitute y=143xy = 14 - 3x into the second equation x2y=7x - 2y = -7
x2(143x)=7    x28+6x=7    7x=21    x=3x - 2(14 - 3x) = -7 \implies x - 28 + 6x = -7 \implies 7x = 21 \implies x = 3
Solving the single-variable equation gives the exact value for xx.
3
Calculate yy using x=3x = 3
y=143(3)=5y = 14 - 3(3) = 5
Substituting x=3x = 3 back into y=143xy = 14 - 3x gives the value for yy.
4
Compute the sum x+yx + y
x+y=3+5=8x + y = 3 + 5 = 8
The question specifically asks for the sum of both variables.

Anahtar Kavram

Solving a 2x2 System of Linear Equations by Substitution or Elimination
Tahmini Süre:45s
Soru 275Soru

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

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Cevap: 12

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Perpendicular lines have negative reciprocal slopes (m1m2=1m_1 \cdot m_2 = -1). The xx-intercept is the point where y=0y = 0.
Soru 276Soru

In the xyxy-plane, line kk passes through the points (2,5)(-2, 5) and (4,7)(4, -7). Which of the following statements about line kk must be true? Select all that apply.

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Cevap: Line kk has a slope of 2-2.; Line kk passes through the point (3,5)(3, -5).; Line kk does not pass through Quadrant III.

Cevap

The correct statements are: line kk has a slope of 2-2, line kk passes through the point (3,5)(3, -5), and line kk does not pass through Quadrant III.
The slope of line kk is m=754(2)=2m = \frac{-7 - 5}{4 - (-2)} = -2. The line equation is y=2x+1y = -2x + 1. Substituting x=3x = 3 yields y=5y = -5, confirming (3,5)(3, -5) is on the line. For all negative xx-values (x<0x < 0), y=2x+1y = -2x + 1 remains strictly positive (y>1y > 1), so no point on the line has both negative coordinates; hence line kk never passes through Quadrant III.

Adım Adım Çözüm

1
Calculate the slope of line kk
slope m=754(2)=126=2m = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2
The slope formula is m=ΔyΔx=y2y1x2x1m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}.
2
Determine the equation of line kk
y=2x+1y = -2x + 1
Using point-slope form yy1=m(xx1)y - y_1 = m(x - x_1) with point (2,5)(-2, 5) gives y5=2(x+2)    y=2x+1y - 5 = -2(x + 2) \implies y = -2x + 1.
3
Test point (3,5)(3, -5) and calculate the xx-intercept
Point (3,5)(3, -5) lies on the line since 2(3)+1=5-2(3) + 1 = -5. Setting y=0y = 0 gives 0=2x+1    x=120 = -2x + 1 \implies x = \frac{1}{2}.
A point lies on a line if its coordinates satisfy the line equation. The xx-intercept occurs where y=0y = 0.
4
Analyze quadrant passage and perpendicular slope
For x<0x < 0, y=2x+1>0y = -2x + 1 > 0, so points in Quadrant III (x<0,y<0x < 0, y < 0) are never reached. Perpendicular slope is 12=12-\frac{1}{-2} = \frac{1}{2}.
Quadrant III requires both coordinates to be negative. Perpendicular lines have negative reciprocal slopes.

Anahtar Kavram

Coordinate Geometry of Lines: Slope, Line Equations, Intercepts, Quadrant Passage, and Perpendicular Slopes
Soru 277Soru

A technician charges a one-time fixed diagnostic fee of $35\$35 plus $25\$25 for each hour of repair work. If the total bill for a repair job was $160\$160, how many hours of repair work were performed?

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Cevap: 55

Cevap

5 hours
Let hh represent the number of hours of repair work. The total charge is given by the sum of the fixed fee ($35\$35) and the hourly rate times hours worked (25h25h). Setting up the linear equation 35+25h=16035 + 25h = 160 and isolating hh gives 25h=12525h = 125, which simplifies to h=5h = 5. Thus, 5 hours of work were performed.

Adım Adım Çözüm

1
Define the variable and set up the linear equation
35+25h=16035 + 25h = 160, where hh is the number of repair hours.
The total cost is the sum of the fixed diagnostic fee and the hourly charge multiplied by hours worked.
2
Subtract the fixed fee from both sides of the equation
25h=16035    25h=12525h = 160 - 35 \implies 25h = 125
Isolate the variable term containing hh.
3
Divide both sides by the hourly rate coefficient
h=12525=5h = \frac{125}{25} = 5
Solve for hh.

Anahtar Kavram

Formulating and solving linear equations in one variable from real-world scenarios
Soru 278Soru
If xx satisfies the linear equation
3(x4)42x+16=x+235\frac{3(x - 4)}{4} - \frac{2x + 1}{6} = \frac{x + 2}{3} - 5
what is the value of 2x+52x + 5?
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Cevap: 23-23

Cevap

The value of 2x+52x + 5 is 23-23.
Clearing denominators by multiplying the entire equation by 1212 gives 9(x4)2(2x+1)=4(x+2)609(x - 4) - 2(2x + 1) = 4(x + 2) - 60. Expanding both sides yields 9x364x2=4x+8609x - 36 - 4x - 2 = 4x + 8 - 60, which simplifies to 5x38=4x525x - 38 = 4x - 52. Isolating xx gives x=14x = -14. Substituting x=14x = -14 into the target expression 2x+52x + 5 yields 2(14)+5=232(-14) + 5 = -23.

Adım Adım Çözüm

1
Clear the denominators by multiplying every term on both sides of the equation by the least common multiple of 4,6,4, 6, and 33, which is 1212.
123(x4)4122x+16=12x+2312512 \cdot \frac{3(x - 4)}{4} - 12 \cdot \frac{2x + 1}{6} = 12 \cdot \frac{x + 2}{3} - 12 \cdot 5, leading to 9(x4)2(2x+1)=4(x+2)609(x - 4) - 2(2x + 1) = 4(x + 2) - 60.
Clearing denominators simplifies the multi-step fractional equation into an integer linear equation.
2
Expand all grouping symbols and combine like terms on both sides.
9x364x2=4x+8609x - 36 - 4x - 2 = 4x + 8 - 60, which simplifies to 5x38=4x525x - 38 = 4x - 52.
Distributing terms carefully ensures proper sign distribution, especially for negative signs across parentheses.
3
Isolate the variable xx on one side of the equation.
5x4x=52+385x - 4x = -52 + 38, giving x=14x = -14.
Subtracting 4x4x and adding 3838 isolates xx.
4
Evaluate the target expression 2x+52x + 5 using the solved value x=14x = -14.
2(14)+5=28+5=232(-14) + 5 = -28 + 5 = -23.
The question asks for the value of 2x+52x + 5, not xx.

Anahtar Kavram

Solving linear equations in one variable with fractional coefficients and evaluating algebraic expressions
Tahmini Süre:1m 30s
Soru 279Soru

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

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Cevap: 28

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Formulating and solving a linear equation in one variable from a rate problem context.
Tahmini Süre:1m 30s
Soru 280Soru
If xx satisfies the linear equation 4(x2)15=2x+34(x - 2) - 15 = 2x + 3 which of the following statements must be true? Select all that apply.

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Cevap: xx is a prime number; 2x52x - 5 is equal to 2121; x+7x + 7 is a multiple of 55

Cevap

The correct statements are that xx is a prime number, 2x52x - 5 is equal to 2121, and x+7x + 7 is a multiple of 55.
Solving the equation 4(x2)15=2x+34(x - 2) - 15 = 2x + 3 yields x=13x = 13. Evaluating the choices with x=13x = 13 confirms that 1313 is a prime number, 2(13)5=212(13) - 5 = 21, and 13+7=2013 + 7 = 20 (a multiple of 55).

Adım Adım Çözüm

1
Expand the left side of the linear equation by distributing the constant term.
4x815=2x+34x - 8 - 15 = 2x + 3
Distribution removes parentheses so like terms can be combined.
2
Combine constant terms on the left side of the equation.
4x23=2x+34x - 23 = 2x + 3
Simplifying 815-8 - 15 yields 23-23.
3
Isolate variable terms on the left side and constant terms on the right side by subtracting 2x2x and adding 2323 to both sides.
2x=262x = 26
Subtracting 2x2x gives 2x2x on the left, and adding 2323 gives 2626 on the right.
4
Divide both sides of the equation by 22 to solve for xx.
x=13x = 13
Dividing 2626 by 22 determines the unique value of xx.
5
Test each statement using x=13x = 13.
1313 is prime (True); 2(13)5=212(13) - 5 = 21 (True); 13+7=2013 + 7 = 20 which is a multiple of 55 (True); 1313 is even (False); 13+12=76\frac{13 + 1}{2} = 7 \neq 6 (False).
Direct evaluation identifies which statements hold true.

Anahtar Kavram

Solving linear equations in one variable and evaluating numerical properties of the solution.
Tahmini Süre:1m 0s
ÖncekiSayfa 14 / 18Sonraki
Algebra Alıştırma Soruları — GRE General Test — Sayfa 14 | Examkin