Algebra and Functions

215 soru

Soru 181Soru

An event catering service charges a fixed setup fee plus a constant hourly rate for event staff. For a 4-hour event, the total charge is 680.Fora7houreventwiththesamestaffrequirements,thetotalchargeis680. For a 7-hour event with the same staff requirements, the total charge is 1,070. What is the fixed setup fee, in dollars, charged by the catering service?

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Cevap: 160

Cevap

The fixed setup fee charged by the catering service is 160 dollars.
Modeling the total cost as C=S+rtC = S + rt, where SS is the fixed setup fee and rr is the hourly rate, yields the system of equations S+4r=680S + 4r = 680 and S+7r=1070S + 7r = 1070. Subtracting the first equation from the second yields 3r=3903r = 390, so r=130r = 130. Substituting r=130r = 130 back into the first equation yields S+520=680S + 520 = 680, which gives S=160S = 160.

Adım Adım Çözüm

1
Define the algebraic model for total cost.
Let SS be the fixed setup fee in dollars and rr be the hourly rate in dollars per hour. The total cost CC for tt hours is modeled by C=S+rtC = S + rt.
The problem presents a linear relationship between time and total cost, consisting of a fixed component and a variable component.
2
Construct a system of linear equations.
S+4r=680S + 4r = 680 and S+7r=1070S + 7r = 1070.
Substitute the two given combinations of time (t=4t = 4 and t=7t = 7) and total cost (C=680C = 680 and C=1070C = 1070) into the algebraic model.
3
Solve for the variable hourly rate rr.
(S+7r)(S+4r)=1070680    3r=390    r=130(S + 7r) - (S + 4r) = 1070 - 680 \implies 3r = 390 \implies r = 130.
Subtracting the first equation from the second eliminates the fixed setup fee variable SS.
4
Calculate the fixed setup fee SS.
S+4(130)=680    S+520=680    S=160S + 4(130) = 680 \implies S + 520 = 680 \implies S = 160.
Substitute r=130r = 130 back into the first equation to find the value of SS.

Anahtar Kavram

Linear Equation Modeling and Systems of Equations
Soru 182Soru

If xx is a real number that satisfies the equation 52x=3x10|5 - 2x| = 3x - 10, what is the value of x4+2x|x - 4| + 2x?

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Cevap: 11

Cevap

11
Solving the equation 52x=3x10|5 - 2x| = 3x - 10 yields two potential values: x=3x = 3 and x=5x = 5. Substituting x=3x = 3 into the original equation yields 1=1|-1| = -1, which is impossible because an absolute value cannot be negative. Therefore, x=3x = 3 is an extraneous root. Substituting x=5x = 5 yields 5=5|-5| = 5, which is true. Evaluating the requested expression x4+2x|x - 4| + 2x at x=5x = 5 gives 54+2(5)=1+10=11|5 - 4| + 2(5) = 1 + 10 = 11.

Adım Adım Çözüm

1
Set up the two algebraic cases for the absolute value equation 52x=3x10|5 - 2x| = 3x - 10.
Case 1: 52x=3x105 - 2x = 3x - 10 when 52x05 - 2x \ge 0 (x2.5x \le 2.5).
Case 2: (52x)=3x10-(5 - 2x) = 3x - 10 when 52x<05 - 2x < 0 (x>2.5x > 2.5).
An absolute value expression u|u| equals uu when u0u \ge 0 and u-u when u<0u < 0.
2
Solve Case 1 algebraically.
52x=3x10    15=5x    x=35 - 2x = 3x - 10 \implies 15 = 5x \implies x = 3.
Isolate the variable xx on one side of the equation.
3
Solve Case 2 algebraically.
2x - 5 = 3x - 10 \implies 5 = x \implies x = 5$.
Simplify and isolate xx.
4
Check both candidate solutions in the original equation to filter out extraneous roots.
For x=3x = 3: 52(3)=3(3)10    1=1    1=1|5 - 2(3)| = 3(3) - 10 \implies |-1| = -1 \implies 1 = -1 (False, extraneous).
For x=5x = 5: 52(5)=3(5)10    5=5    5=5|5 - 2(5)| = 3(5) - 10 \implies |-5| = 5 \implies 5 = 5 (True, valid).
Since the right-hand side 3x103x - 10 must be non-negative for the absolute value to hold, candidate solutions must be tested in the original equation.
5
Substitute the valid solution x=5x = 5 into the targeted expression x4+2x|x - 4| + 2x.
54+2(5)=1+10=1+10=11|5 - 4| + 2(5) = |1| + 10 = 1 + 10 = 11.
Evaluate the target expression using the single real value x=5x = 5 that satisfies the given linear absolute value equation.

Anahtar Kavram

Solving absolute value linear equations and identifying extraneous solutions
Tahmini Süre:2m 0s
Soru 183Soru

If yy is a real number satisfying the inequality 3y6y+2<1\frac{|3y - 6|}{y + 2} < 1, which of the following represents the complete set of all possible values of yy?

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Cevap: y<2y < -2 or 1<y<41 < y < 4

Cevap

y<2y < -2 or 1<y<41 < y < 4
To evaluate 3y6y+2<1\frac{|3y - 6|}{y + 2} < 1, analyze the sign of the denominator y+2y + 2. If y<2y < -2, the denominator is negative, so multiplying both sides by y+2y + 2 reverses the inequality to 3y6>y+2|3y - 6| > y + 2. Because absolute values are always non-negative and y+2y + 2 is negative for y<2y < -2, this inequality holds for all y<2y < -2. If y>2y > -2, the denominator is positive, yielding 3y6<y+2|3y - 6| < y + 2, which expands to y2<3y6<y+2-y - 2 < 3y - 6 < y + 2. Solving the left inequality gives y>1y > 1, and solving the right gives y<4y < 4, producing 1<y<41 < y < 4. Combining both valid cases gives y<2y < -2 or 1<y<41 < y < 4.

Adım Adım Çözüm

1
Determine the domain restriction and split into cases based on the denominator's sign.
The expression is undefined when y=2y = -2. We analyze Case 1 (y>2y > -2) and Case 2 (y<2y < -2).
Multiplying an inequality by an algebraic expression requires knowing its sign to preserve or reverse the inequality direction.
2
Solve Case 1 where y>2y > -2 (positive denominator).
Multiplying by y+2y + 2 gives 3y6<y+2|3y - 6| < y + 2, which expands to (y+2)<3y6<y+2-(y + 2) < 3y - 6 < y + 2. Solving y2<3y6-y - 2 < 3y - 6 gives 4<4y    y>14 < 4y \implies y > 1. Solving 3y6<y+23y - 6 < y + 2 gives 2y<8    y<42y < 8 \implies y < 4. Combining gives 1<y<41 < y < 4.
Since y+2>0y + 2 > 0, multiplying preserves the inequality sign.
3
Solve Case 2 where y<2y < -2 (negative denominator).
Multiplying by y+2y + 2 flips the inequality sign to 3y6>y+2|3y - 6| > y + 2. Since 3y60|3y - 6| \ge 0 for all real yy and y+2<0y + 2 < 0 when y<2y < -2, a non-negative number is always strictly greater than a negative number. Thus, all y<2y < -2 are valid solutions.
Any non-negative real value is strictly greater than any negative value.
4
Combine the valid intervals from both cases.
y<2y < -2 or 1<y<41 < y < 4.
The complete solution set is the union of solutions from Case 1 and Case 2.

Anahtar Kavram

Solving Rational Absolute Value Inequalities via Denominator Sign Case Analysis
Tahmini Süre:2m 0s
Soru 184Soru

An artisan workshop produces custom wooden chairs and tables. Each chair requires 33 hours of carving and 22 hours of finishing, while each table requires 55 hours of carving and 44 hours of finishing. If the workshop logged a total of 110110 hours of carving and 8484 hours of finishing last week, how many tables were produced?

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Cevap: 16

Cevap

The workshop produced 16 tables.
By defining cc as the number of chairs and tt as the number of tables, we can set up two linear equations representing total hours: 3c+5t=1103c + 5t = 110 for carving and 2c+4t=842c + 4t = 84 for finishing. Multiplying the second equation by 1.51.5 yields 3c+6t=1263c + 6t = 126. Subtracting 3c+5t=1103c + 5t = 110 from 3c+6t=1263c + 6t = 126 leaves t=16t = 16. Thus, the workshop produced 16 tables.

Adım Adım Çözüm

1
Define variables and translate the word problem into a system of linear equations.
Let cc be the number of chairs and tt be the number of tables. Carving equation: 3c+5t=1103c + 5t = 110. Finishing equation: 2c+4t=842c + 4t = 84.
The total hours for each activity equal the sum of hours spent on chairs and tables.
2
Eliminate variable cc to solve for tt.
Multiply the finishing equation by 1.51.5 to get 3c+6t=1263c + 6t = 126. Subtract the carving equation (3c+5t=1103c + 5t = 110) from this equation: (3c+6t)(3c+5t)=126110    t=16(3c + 6t) - (3c + 5t) = 126 - 110 \implies t = 16.
Aligning the coefficient of cc in both equations allows direct elimination of cc to isolate tt.
3
Verify the solution by calculating cc and checking both original equations.
Substitute t=16t = 16 into 2c+4(16)=84    2c+64=84    2c=20    c=102c + 4(16) = 84 \implies 2c + 64 = 84 \implies 2c = 20 \implies c = 10. Check carving: 3(10)+5(16)=30+80=1103(10) + 5(16) = 30 + 80 = 110.
Ensures that t=16t = 16 and c=10c = 10 satisfy both resource constraints without calculation errors.

Anahtar Kavram

Setting up and solving a system of two linear equations in two variables

İpuçları

1
Set up two separate linear equations: one for total carving hours and one for total finishing hours.
2
Let cc be the number of chairs and tt be the number of tables. Your system is 3c+5t=1103c + 5t = 110 and 2c+4t=842c + 4t = 84.
3
Multiply 2c+4t=842c + 4t = 84 by 1.51.5 to get 3c+6t=1263c + 6t = 126, then subtract 3c+5t=1103c + 5t = 110 to find tt directly.

Daha Fazla Pratik

Try solving a similar problem where the total revenue and total unit count are given to practice standard linear system modeling.

Alternatif Yöntem

Divide the finishing equation 2c+4t=842c + 4t = 84 by 22 to get c+2t=42    c=422tc + 2t = 42 \implies c = 42 - 2t. Substitute this into the carving equation: 3(422t)+5t=110    1266t+5t=110    t=16    t=163(42 - 2t) + 5t = 110 \implies 126 - 6t + 5t = 110 \implies -t = -16 \implies t = 16.
Tahmini Süre:1m 30s
Soru 185Soru

If xx is a positive real number such that x12+x12=4x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4, what is the value of x32+x32x^{\frac{3}{2}} + x^{-\frac{3}{2}}?

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Cevap: 52

Cevap

52
Cubing both sides of x12+x12=4x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4 yields (x12+x12)3=64(x^{\frac{1}{2}} + x^{-\frac{1}{2}})^3 = 64. By the identity (a+b)3=a3+b3+3ab(a+b)(a+b)^3 = a^3 + b^3 + 3ab(a+b), the left side expands to (x32+x32)+3(x12x12)(x12+x12)(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 3(x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}})(x^{\frac{1}{2}} + x^{-\frac{1}{2}}). Substituting x12x12=1x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = 1 and x12+x12=4x^{\frac{1}{2}} + x^{-\frac{1}{2}} = 4 gives (x32+x32)+12=64(x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 = 64. Subtracting 12 yields 52.

Adım Adım Çözüm

1
Set up the cubic identity for the sum of fractional exponents.
Let a=x12a = x^{\frac{1}{2}} and b=x12b = x^{-\frac{1}{2}}. Then a+b=4a + b = 4 and ab=x12x12=1ab = x^{\frac{1}{2}} \cdot x^{-\frac{1}{2}} = 1.
Recognizing that x32=a3x^{\frac{3}{2}} = a^3 and x32=b3x^{-\frac{3}{2}} = b^3 allows the use of binomial expansion.
2
Cube both sides of the given equation a+b=4a + b = 4.
(a+b)3=a3+b3+3ab(a+b)    43=(x32+x32)+3(1)(4)(a + b)^3 = a^3 + b^3 + 3ab(a + b) \implies 4^3 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 3(1)(4).
Expanding the cube retains the desired expression a3+b3a^3 + b^3 alongside simpler terms.
3
Simplify and solve for x32+x32x^{\frac{3}{2}} + x^{-\frac{3}{2}}.
64 = (x^{\frac{3}{2}} + x^{-\frac{3}{2}}) + 12 \implies x^{\frac{3}{2}} + x^{-\frac{3}{2}} = 64 - 12 = 52.
Subtracting 12 from 64 gives the exact value of the expression.

Anahtar Kavram

Algebraic identities with fractional exponents
Tahmini Süre:2m 0s
Soru 186Soru

If xx is an integer such that x3+x+512|x - 3| + |x + 5| \le 12, how many possible values of xx exist?

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Cevap: 13

Cevap

13
Evaluating the inequality x3+x+512|x - 3| + |x + 5| \le 12 across the three intervals defined by critical points x=5x = -5 and x=3x = 3 yields the continuous real solution set [7,5][-7, 5]. Counting all integers in this inclusive range gives 5(7)+1=135 - (-7) + 1 = 13 possible values.

Adım Adım Çözüm

1
Identify the critical points of the absolute value expressions.
The critical points where the expressions inside the absolute values change sign are x=3x = 3 and x=5x = -5.
Setting x3=0x - 3 = 0 gives x=3x = 3, and setting x+5=0x + 5 = 0 gives x=5x = -5.
2
Solve the inequality for the region x<5x < -5.
(x3)(x+5)12    2x212    2x14    x7-(x - 3) - (x + 5) \le 12 \implies -2x - 2 \le 12 \implies -2x \le 14 \implies x \ge -7. Thus, 7x<5-7 \le x < -5.
When x<5x < -5, both x3<0x - 3 < 0 and x+5<0x + 5 < 0, so x3=(x3)|x - 3| = -(x - 3) and x+5=(x+5)|x + 5| = -(x + 5).
3
Solve the inequality for the region 5x3-5 \le x \le 3.
(x3)+(x+5)12    812-(x - 3) + (x + 5) \le 12 \implies 8 \le 12, which is universally true for all xx in this interval.
When 5x3-5 \le x \le 3, x30x - 3 \le 0 and x+50x + 5 \ge 0, so x3=(x3)|x - 3| = -(x - 3) and x+5=x+5|x + 5| = x + 5.
4
Solve the inequality for the region x>3x > 3.
(x3)+(x+5)12    2x+212    2x10    x5(x - 3) + (x + 5) \le 12 \implies 2x + 2 \le 12 \implies 2x \le 10 \implies x \le 5. Thus, 3<x53 < x \le 5.
When x>3x > 3, both x3>0x - 3 > 0 and x+5>0x + 5 > 0.
5
Combine the valid intervals and count the integer solutions.
The full solution range is [7,5][-7, 5]. The number of integer values is 5(7)+1=135 - (-7) + 1 = 13.
The number of integers in an inclusive range [a,b][a, b] is given by ba+1b - a + 1.

Anahtar Kavram

Solving multi-term absolute value inequalities by dividing the domain at critical points into distinct cases.
Soru 187Soru

For what value of the constant kk does the system of linear equations below have no solution?

(k2)x+3y=64x+(k+2)y=12\begin{aligned} (k - 2)x + 3y &= 6 \\ 4x + (k + 2)y &= 12 \end{aligned}
Cevabı ve açıklamayı göster

Cevap: 4-4

Cevap

The constant value k=4k = -4 results in a system with no solution.
For a 2×22 \times 2 linear system a1x+b1y=c1a_1 x + b_1 y = c_1 and a2x+b2y=c2a_2 x + b_2 y = c_2 to have no solution, the equations must have proportional variable coefficients but non-proportional constant terms: a1a2=b1b2c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}. Solving k24=3k+2\frac{k-2}{4} = \frac{3}{k+2} gives k24=12k^2 - 4 = 12, so k=±4k = \pm 4. Substituting k=4k = -4 yields 6x+3y=6-6x + 3y = 6 and 4x2y=124x - 2y = 12, which simplify to 2x+y=2-2x + y = 2 and 2x+y=6-2x + y = -6. Since the constants 22 and 6-6 differ, the lines are parallel and distinct, meaning the system has no solution.

Adım Adım Çözüm

1
Set up the condition for parallel lines (equal slopes) by equating the ratio of the coefficients of xx and yy.
\frac{k - 2}{4} = \frac{3}{k + 2}
A system of two linear equations has either zero solutions (parallel, non-intersecting lines) or infinitely many solutions (coincident lines) when the slopes are equal.
2
Cross-multiply and solve the quadratic equation for kk.
(k - 2)(k + 2) = 12 \implies k^2 - 4 = 12 \implies k^2 = 16 \implies k = 4 \text{ or } k = -4
Finding all values of kk where the coefficient matrix determinant is zero.
3
Test k=4k = 4 in the original system.
(4 - 2)x + 3y = 6 \implies 2x + 3y = 6 \quad \text{and} \quad 4x + (4 + 2)y = 12 \implies 4x + 6y = 12
Dividing 4x+6y=124x + 6y = 12 by 22 yields 2x+3y=62x + 3y = 6, which is identical to the first equation. Thus, k=4k = 4 yields infinitely many solutions.
4
Test k=4k = -4 in the original system.
(-4 - 2)x + 3y = 6 \implies -6x + 3y = 6 \quad \text{and} \quad 4x + (-4 + 2)y = 12 \implies 4x - 2y = 12
Simplifying both equations gives 2x+y=2-2x + y = 2 and 2x+y=6-2x + y = -6. The lines have identical slopes but different constants, so they are parallel and distinct, producing no solution.

Anahtar Kavram

System Solvability and Linear Consistency
Tahmini Süre:2m 0s
Soru 188Soru

A beverage producer creates two liquid mixtures, Mixture AA and Mixture BB. Mixture AA consists of 30%30\% fruit concentrate by volume, and Mixture BB consists of 70%70\% fruit concentrate by volume. A lab technician combines xx liters of Mixture AA with yy liters of Mixture BB to prepare an 8080-liter batch that contains 45%45\% fruit concentrate by volume. What is the value of xx?

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Cevap: 50

Cevap

The value of xx is 50.
To find xx, we construct two linear equations based on total liquid volume and total fruit concentrate. The total volume equation is x+y=80x + y = 80, which gives y=80xy = 80 - x. The concentrate equation is 0.30x+0.70y=0.45(80)=360.30x + 0.70y = 0.45(80) = 36. Substituting y=80xy = 80 - x yields 0.30x+0.70(80x)=360.30x + 0.70(80 - x) = 36. Expanding gives 0.30x+560.70x=360.30x + 56 - 0.70x = 36, so 0.40x=20-0.40x = -20, which results in x=50x = 50.

Adım Adım Çözüm

1
Formulate a system of two linear equations representing total volume and total concentrate volume.
System equations: x+y=80x + y = 80 and 0.30x+0.70y=360.30x + 0.70y = 36.
The sum of the component volumes equals the total mixture volume, and the sum of the pure concentrate from each component equals the total concentrate in the final mixture.
2
Substitute y=80xy = 80 - x into the concentrate equation to eliminate yy.
0.30x+0.70(80x)=360.30x + 0.70(80 - x) = 36.
Substituting one variable reduces the system to a single linear equation in one variable.
3
Simplify the single-variable linear equation and solve for xx.
0.30x+560.70x=36    0.40x=20    x=500.30x + 56 - 0.70x = 36 \implies -0.40x = -20 \implies x = 50.
Combining like terms isolates the variable xx.

Anahtar Kavram

Solving systems of two linear equations formed by weighted mixture word problems.
Soru 189Soru

For all positive real numbers aa and bb such that b=4ab = 4a, if ab=baa^b = b^a, then the value of aa is 2\sqrt{2}.

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Cevap: False

Cevap

False. The statement is false because the exact value of aa is 43\sqrt[3]{4}, which is not equal to 2\sqrt{2}.
The statement is false because simplifying a4a=(4a)aa^{4a} = (4a)^a leads directly to a3=4a^3 = 4, giving a=43=22/3a = \sqrt[3]{4} = 2^{2/3}. The claimed value 2=21/2\sqrt{2} = 2^{1/2} fails to satisfy the original equation, making the assertion mathematically false.

Adım Adım Çözüm

1
Substitute b=4ab = 4a into the given exponent equation ab=baa^b = b^a.
a4a=(4a)aa^{4a} = (4a)^a
Eliminate variable bb to express the equation solely in terms of aa.
2
Raise both sides of the equation to the power of 1a\frac{1}{a}.
(a4a)1/a=((4a)a)1/a    a4=4a(a^{4a})^{1/a} = ((4a)^a)^{1/a} \implies a^4 = 4a
Apply power of a power exponent rule (xm)n=xmn(x^m)^n = x^{mn} to simplify the exponents.
3
Divide both sides by aa (since a>0a > 0) and solve for aa.
a4a=4aa    a3=4    a=43=22/3\frac{a^4}{a} = \frac{4a}{a} \implies a^3 = 4 \implies a = \sqrt[3]{4} = 2^{2/3}
Isolate aa using standard division and radical extraction rules.
4
Compare the calculated value of aa with 2\sqrt{2}.
a=22/31.587a = 2^{2/3} \approx 1.587, whereas 2=21/21.414\sqrt{2} = 2^{1/2} \approx 1.414. Thus a2a \neq \sqrt{2}.
Determine the truth value of the claimed conclusion.

Anahtar Kavram

Solving variable exponent equations of the form ab=baa^b = b^a using exponent power rules and root extraction.
Soru 190Soru

What is the sum of all real solutions to the equation x210=3x|x^2 - 10| = 3x?

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Cevap: 7

Cevap

The sum of all real solutions to the equation is 7.
To solve x210=3x|x^2 - 10| = 3x, note that 3x03x \ge 0 (so x0x \ge 0). Splitting into two algebraic cases gives x23x10=0x^2 - 3x - 10 = 0 (yielding x=5x = 5 and extraneous x=2x = -2) and x2+3x10=0x^2 + 3x - 10 = 0 (yielding x=2x = 2 and extraneous x=5x = -5). The valid real solutions are x=5x = 5 and x=2x = 2, whose sum is 7.

Adım Adım Çözüm

1
Determine domain restriction based on the absolute value definition.
x0x \ge 0
Because an absolute value expression cannot be negative, x210=3x|x^2 - 10| = 3x requires 3x03x \ge 0, which means x0x \ge 0.
2
Solve Case 1: x210=3xx^2 - 10 = 3x.
x=5x = 5
Rearranging yields x23x10=0x^2 - 3x - 10 = 0, which factors as (x5)(x+2)=0(x - 5)(x + 2) = 0. The roots are x=5x = 5 and x=2x = -2. Reject x=2x = -2 because x0x \ge 0.
3
Solve Case 2: (x210)=3x-(x^2 - 10) = 3x.
x=2x = 2
Rearranging yields x2+3x10=0x^2 + 3x - 10 = 0, which factors as (x+5)(x2)=0(x + 5)(x - 2) = 0. The roots are x=2x = 2 and x=5x = -5. Reject x=5x = -5 because x0x \ge 0.
4
Sum all valid real solutions.
7
The valid real solutions are x=5x = 5 and x=2x = 2. Their sum is 5+2=75 + 2 = 7.

Anahtar Kavram

Solving absolute value equations with a variable expression on one side requires verifying non-negativity constraints to eliminate extraneous roots.
Tahmini Süre:1m 30s
Soru 191Soru

A biotechnology laboratory formulates three custom reagent mixtures—Solution XX, Solution YY, and Solution ZZ—using three chemical compounds: Alpha, Beta, and Gamma.

- Solution XX contains 2 mL2\text{ mL} of Alpha, 3 mL3\text{ mL} of Beta, and 1 mL1\text{ mL} of Gamma, and costs $13.00\$13.00.
- Solution YY contains 1 mL1\text{ mL} of Alpha, 2 mL2\text{ mL} of Beta, and 4 mL4\text{ mL} of Gamma, and costs $11.00\$11.00.
- Solution ZZ contains 3 mL3\text{ mL} of Alpha, 1 mL1\text{ mL} of Beta, and 2 mL2\text{ mL} of Gamma, and costs $13.00\$13.00.

What is the cost of a mixture containing 5 mL5\text{ mL} of Alpha, 4 mL4\text{ mL} of Beta, and 3 mL3\text{ mL} of Gamma?

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Cevap: $26.00\$26.00

Cevap

$26.00\$26.00
The target quantity consists of 5 mL5\text{ mL} of Alpha, 4 mL4\text{ mL} of Beta, and 3 mL3\text{ mL} of Gamma. By inspecting the given system, adding Solution XX (2 mL2\text{ mL} Alpha, 3 mL3\text{ mL} Beta, 1 mL1\text{ mL} Gamma) and Solution ZZ (3 mL3\text{ mL} Alpha, 1 mL1\text{ mL} Beta, 2 mL2\text{ mL} Gamma) directly yields 5 mL5\text{ mL} Alpha, 4 mL4\text{ mL} Beta, and 3 mL3\text{ mL} Gamma. Therefore, the required cost is simply the sum of the costs of Solution XX and Solution ZZ: $13.00+$13.00=$26.00\$13.00 + \$13.00 = \$26.00.

Adım Adım Çözüm

1
Set up the linear system of equations representing the cost of each solution.
Let aa, bb, and gg be the cost per mL of Alpha, Beta, and Gamma, respectively.
Equation 1: 2a+3b+g=132a + 3b + g = 13
Equation 2: a+2b+4g=11a + 2b + 4g = 11
Equation 3: 3a+b+2g=133a + b + 2g = 13
Translate the given word problem into algebraic equations representing system relationships.
2
Identify the requested quantity and evaluate whether it can be formed as a linear combination of the given equations.
Target expression: 5a+4b+3g5a + 4b + 3g
Recognizing linear combinations avoids solving for individual variable values when not required.
3
Add Equation 1 and Equation 3.
(2a+3b+g)+(3a+b+2g)=13+13    5a+4b+3g=26(2a + 3b + g) + (3a + b + 2g) = 13 + 13 \implies 5a + 4b + 3g = 26
The sum of coefficients for Alpha (2+3=52+3=5), Beta (3+1=43+1=4), and Gamma (1+2=31+2=3) exactly matches the target mixture.

Anahtar Kavram

Solving systems of linear equations using linear combinations without full variable elimination
Tahmini Süre:2m 0s
Soru 192Soru

If xx is a real number satisfying the equation 152x=2x3\sqrt{15 - 2x} = 2x - 3, what is the sum of all real values of xx that satisfy this equation?

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Cevap: 33

Cevap

The sum of all real values of xx satisfying the equation is 33.
Squaring both sides of 152x=2x3\sqrt{15 - 2x} = 2x - 3 gives 152x=4x212x+915 - 2x = 4x^2 - 12x + 9, which simplifies to 2x25x3=02x^2 - 5x - 3 = 0. Factoring yields candidate roots x=3x = 3 and x=1/2x = -1/2. Substituting x=3x = 3 into the original equation gives 9=3\sqrt{9} = 3, which is true. Substituting x=1/2x = -1/2 yields 16=4\sqrt{16} = -4, which is false because principal radicals cannot evaluate to negative values. Therefore, x=3x = 3 is the sole valid solution, making the sum equal to 33.

Adım Adım Çözüm

1
Isolate the radical and state domain constraints.
The principal square root 152x\sqrt{15 - 2x} must be non-negative, requiring 152x0    x7.515 - 2x \ge 0 \implies x \le 7.5, and 2x30    x1.52x - 3 \ge 0 \implies x \ge 1.5.
Radical expressions produce non-negative principal square roots.
2
Square both sides of the equation.
152x=(2x3)2    152x=4x212x+915 - 2x = (2x - 3)^2 \implies 15 - 2x = 4x^2 - 12x + 9.
Eliminate the radical to form a polynomial equation.
3
Rearrange into standard quadratic form and solve for xx.
4x210x6=0    2x25x3=0    (2x+1)(x3)=04x^2 - 10x - 6 = 0 \implies 2x^2 - 5x - 3 = 0 \implies (2x + 1)(x - 3) = 0, giving candidate solutions x=3x = 3 and x=1/2x = -1/2.
Solve the quadratic equation using factoring.
4
Test candidate solutions in the original equation to eliminate extraneous roots.
For x=3x = 3: 152(3)=9=3\sqrt{15 - 2(3)} = \sqrt{9} = 3 and 2(3)3=32(3) - 3 = 3 (Valid). For x=1/2x = -1/2: 152(1/2)=16=4\sqrt{15 - 2(-1/2)} = \sqrt{16} = 4, but 2(1/2)3=42(-1/2) - 3 = -4 (Extraneous).
Squaring an equation can introduce extraneous roots.

Anahtar Kavram

Solving Radical Equations and Identifying Extraneous Solutions
Soru 193Soru

An arithmetic sequence a1,a2,a3,a_1, a_2, a_3, \dots has a first term a1=4a_1 = 4 and a common difference d=3d = 3. A geometric sequence b1,b2,b3,b_1, b_2, b_3, \dots has a first term b1=2b_1 = 2 and a common ratio r=2r = 2. If the kk-th term of the arithmetic sequence and the mm-th term of the geometric sequence are both equal to 6464, what is the value of k+mk + m?

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Cevap: 27

Cevap

The value of k+mk + m is 27.
For the arithmetic sequence, the kk-th term is ak=a1+(k1)da_k = a_1 + (k-1)d. Setting 4+3(k1)=644 + 3(k-1) = 64 gives 3(k1)=603(k-1) = 60, so k1=20k-1 = 20 and k=21k = 21. For the geometric sequence, the mm-th term is bm=b1rm1b_m = b_1 r^{m-1}. Setting 22m1=642 \cdot 2^{m-1} = 64 gives 2m=642^m = 64, which implies m=6m = 6. Adding the two values gives k+m=21+6=27k + m = 21 + 6 = 27.

Adım Adım Çözüm

1
Determine the term position kk in the arithmetic sequence.
k=21k = 21
Using ak=a1+(k1)da_k = a_1 + (k-1)d, set 4+3(k1)=64    3(k1)=60    k1=20    k=214 + 3(k-1) = 64 \implies 3(k-1) = 60 \implies k - 1 = 20 \implies k = 21.
2
Determine the term position mm in the geometric sequence.
m=6m = 6
Using bm=b1rm1b_m = b_1 \cdot r^{m-1}, set 22m1=64    2m=64    m=62 \cdot 2^{m-1} = 64 \implies 2^m = 64 \implies m = 6.
3
Compute the sum of the two position indices kk and mm.
2727
k+m=21+6=27k + m = 21 + 6 = 27.

Anahtar Kavram

Calculating term indices in arithmetic and geometric sequences using general term formulas
Tahmini Süre:1m 30s
Soru 194Soru

A web hosting company charges each enterprise client a one-time fixed setup fee plus a constant monthly maintenance fee per server. A client operating 44 servers pays a total of $1,100\$1,100 for the setup fee and the first 66 months of server maintenance. A client operating 99 servers pays a total of $1,850\$1,850 for the setup fee and the first 66 months of server maintenance. What is the one-time fixed setup fee, in dollars?

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Cevap: 500

Cevap

500
The fixed setup fee is $500\$500. Letting SS represent the fixed setup fee and MM represent the 6-month maintenance fee per server, the two given scenarios yield S+4M=1100S + 4M = 1100 and S+9M=1850S + 9M = 1850. Subtracting the first equation from the second gives 5M=7505M = 750, which simplifies to M=150M = 150. Substituting M=150M = 150 into S+4(150)=1100S + 4(150) = 1100 gives S+600=1100S + 600 = 1100, so S=500S = 500.

Adım Adım Çözüm

1
Define variables for the unknown fixed cost and per-server cost, and construct the system of linear equations.
Let SS be the fixed setup fee in dollars and MM be the 6-month maintenance fee per server in dollars. The equations are S+4M=1100S + 4M = 1100 and S+9M=1850S + 9M = 1850.
Modeling the situational relationships as a linear system allows isolated solution of each unknown.
2
Subtract the two linear equations to eliminate the fixed fee SS and solve for MM.
5M=750    M=1505M = 750 \implies M = 150.
Since the coefficient of SS is 1 in both equations, elimination by subtraction directly isolates MM.
3
Substitute the value of MM back into the first equation to solve for SS.
S+4(150)=1100    S+600=1100    S=500S + 4(150) = 1100 \implies S + 600 = 1100 \implies S = 500.
Replacing MM with 150 yields a linear equation in one variable for the setup fee.

Anahtar Kavram

Linear Equations in One and Two Variables
Tahmini Süre:1m 30s
Soru 195Soru

How many integer values of xx satisfy the inequality x25x6|x^2 - 5x| \le 6?

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Cevap: 8

Cevap

The correct answer is 8.
To solve x25x6|x^2 - 5x| \le 6, express it as 6x25x6-6 \le x^2 - 5x \le 6. Solving x25x60x^2 - 5x - 6 \le 0 gives [1,6][-1, 6], while solving x25x+60x^2 - 5x + 6 \ge 0 gives (,2][3,)(-\infty, 2] \cup [3, \infty). Taking their intersection yields the solution set [1,2][3,6][-1, 2] \cup [3, 6]. The integers contained in this set are 1,0,1,2,3,4,5,6-1, 0, 1, 2, 3, 4, 5, 6, which gives 8 distinct integer values.

Adım Adım Çözüm

1
Rewrite the absolute value inequality
6x25x6-6 \le x^2 - 5x \le 6
For any real expression AA and constant k0k \ge 0, Ak|A| \le k is equivalent to kAk-k \le A \le k.
2
Solve the upper bound condition x25x6x^2 - 5x \le 6
1x6-1 \le x \le 6
Subtract 6 from both sides to obtain x25x60x^2 - 5x - 6 \le 0. Factoring gives (x6)(x+1)0(x - 6)(x + 1) \le 0.
3
Solve the lower bound condition x25x6x^2 - 5x \ge -6
x2x \le 2 or x3x \ge 3
Add 6 to both sides to obtain x25x+60x^2 - 5x + 6 \ge 0. Factoring gives (x2)(x3)0(x - 2)(x - 3) \ge 0.
4
Combine the solution sets
[1,2][3,6][-1, 2] \cup [3, 6]
The intersection of [1,6][-1, 6] with (,2][3,)(-\infty, 2] \cup [3, \infty) is the set of intervals [1,2][-1, 2] and [3,6][3, 6].
5
Count all integer solutions in the combined set
8 integer values
The integer values in [1,2][-1, 2] are 1,0,1,2-1, 0, 1, 2 (4 integers), and in [3,6][3, 6] are 3,4,5,63, 4, 5, 6 (4 integers), totaling 4+4=84 + 4 = 8 integers.

Anahtar Kavram

Solving quadratic absolute value inequalities using compound inequality decomposition.
Soru 196Soru

A specialized coffee roastery produces three custom blends—Roast Alpha, Roast Beta, and Roast Gamma—using three varieties of single-origin beans: Grade A, Grade B, and Grade C.

- One batch of Roast Alpha requires 3 kg of Grade A, 1 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of 64.OnebatchofRoastBetarequires1kgofGradeA,4kgofGradeB,and2kgofGradeCbeans,andhasatotalrawmaterialcostof64. - One batch of Roast Beta requires 1 kg of Grade A, 4 kg of Grade B, and 2 kg of Grade C beans, and has a total raw material cost of 64.
- One batch of Roast Gamma requires 2 kg of Grade A, 2 kg of Grade B, and 5 kg of Grade C beans, and has a total raw material cost of $90.

What is the cost, in dollars, of 1 kg of Grade A beans?

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Cevap: 12

Cevap

The cost of 1 kg of Grade A beans is 12 dollars.
Setting up the 3-variable linear system 3x+y+2z=643x + y + 2z = 64, x+4y+2z=64x + 4y + 2z = 64, and 2x+2y+5z=902x + 2y + 5z = 90 allows us to eliminate zz by subtracting the second equation from the first, yielding 2x3y=02x - 3y = 0, or x=1.5yx = 1.5y. Substituting this relationship back into the system leads to y=8y = 8 and x=12x = 12. Thus, 1 kg of Grade A beans costs 12 dollars.

Adım Adım Çözüm

1
Formulate linear equations representing the total cost of each coffee blend batch.
Let xx be the cost per kg of Grade A beans, yy be the cost per kg of Grade B beans, and zz be the cost per kg of Grade C beans:
(1)3x+y+2z=64(2)x+4y+2z=64(3)2x+2y+5z=90\begin{aligned} (1)\quad 3x + y + 2z &= 64 \\ (2)\quad x + 4y + 2z &= 64 \\ (3)\quad 2x + 2y + 5z &= 90 \end{aligned}
Translating the word problem into a system of 3 linear equations with 3 variables.
2
Eliminate variable zz by subtracting Equation (2) from Equation (1).
(3x+y+2z)(x+4y+2z)=6464    2x3y=0    x=1.5y(3x + y + 2z) - (x + 4y + 2z) = 64 - 64 \implies 2x - 3y = 0 \implies x = 1.5y
Since both equations (1) and (2) contain the term +2z+2z, subtracting them removes zz directly and provides a simple relation between xx and yy.
3
Substitute x=1.5yx = 1.5y into Equation (1) and Equation (3) to obtain a system in terms of yy and zz.
From Equation (1):
3(1.5y)+y+2z=64    5.5y+2z=64    11y+4z=128(4)3(1.5y) + y + 2z = 64 \implies 5.5y + 2z = 64 \implies 11y + 4z = 128 \quad (4)
From Equation (3):
2(1.5y)+2y+5z=90    5y+5z=90    y+z=18    z=18y2(1.5y) + 2y + 5z = 90 \implies 5y + 5z = 90 \implies y + z = 18 \implies z = 18 - y
Reducing the system from 3 variables down to 2 variables.
4
Substitute z=18yz = 18 - y into Equation (4) to solve for yy, and subsequently calculate xx.
11y+4(18y)=128    7y+72=128    7y=56    y=811y + 4(18 - y) = 128 \implies 7y + 72 = 128 \implies 7y = 56 \implies y = 8
Using x=1.5yx = 1.5y:
x=1.5×8=12x = 1.5 \times 8 = 12
Solving the single-variable linear equation for yy, then substituting back to find the required cost xx for Grade A beans.

Anahtar Kavram

Solving a 3-Variable System of Linear Equations via Variable Elimination
Soru 197Soru

In an arithmetic sequence, the sum of the first 44 terms is 2828 and the sum of the first 88 terms is 8888. What is the 10th10\text{th} term of this sequence?

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Cevap: 2222

Cevap

The 10th term of the sequence is 22.
Using the arithmetic series sum formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d), the conditions yield two linear equations: 2a1+3d=142a_1 + 3d = 14 and 2a1+7d=222a_1 + 7d = 22. Subtracting these equations gives 4d=84d = 8, so d=2d = 2. Substituting d=2d = 2 back into 2a1+3(2)=142a_1 + 3(2) = 14 yields a1=4a_1 = 4. The 10th term is then calculated as a1+9d=4+9(2)=22a_1 + 9d = 4 + 9(2) = 22.

Adım Adım Çözüm

1
Express the given sums using the arithmetic series sum formula Sn=n2(2a1+(n1)d)S_n = \frac{n}{2}(2a_1 + (n-1)d).
For S4=28S_4 = 28: 42(2a1+3d)=28    2a1+3d=14\frac{4}{2}(2a_1 + 3d) = 28 \implies 2a_1 + 3d = 14.
For S8=88S_8 = 88: 82(2a1+7d)=88    2a1+7d=22\frac{8}{2}(2a_1 + 7d) = 88 \implies 2a_1 + 7d = 22.
Setting up linear equations in terms of the first term a1a_1 and common difference dd allows us to solve for both sequence parameters.
2
Subtract the first equation from the second equation to solve for dd.
(2a1+7d)(2a1+3d)=2214    4d=8    d=2(2a_1 + 7d) - (2a_1 + 3d) = 22 - 14 \implies 4d = 8 \implies d = 2.
Eliminating a1a_1 isolates the common difference dd.
3
Substitute d=2d = 2 back into the first equation to solve for a1a_1.
2a1+3(2)=14    2a1+6=14    2a1=8    a1=42a_1 + 3(2) = 14 \implies 2a_1 + 6 = 14 \implies 2a_1 = 8 \implies a_1 = 4.
Finding a1a_1 completes the essential parameters of the sequence.
4
Calculate the 10th term using the formula an=a1+(n1)da_n = a_1 + (n-1)d.
a10=4+(101)(2)=4+18=22a_{10} = 4 + (10 - 1)(2) = 4 + 18 = 22.
Evaluating the formula at n=10n = 10 provides the target term.

Anahtar Kavram

Arithmetic sequence term and series sum formulas.
Tahmini Süre:2m 0s
Soru 198Soru

An event design company offers three distinct decorative bundles—Bundle X, Bundle Y, and Bundle Z—for corporate gala setups.

• Bundle X contains 3 floral arrangements, 2 LED uplights, and 1 table runner.
• Bundle Y contains 1 floral arrangement, 3 LED uplights, and 2 table runners.
• Bundle Z contains 2 floral arrangements, 1 LED uplight, and 3 table runners.

To decorate a venue, a coordinator orders a combination of these bundles containing a total of 26 floral arrangements, 23 LED uplights, and 23 table runners. If every bundle ordered is used in its entirety, what is the total number of bundles ordered by the coordinator?

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Cevap: 12

Cevap

The total number of bundles ordered by the coordinator is 12.
Summing all three equations gives 6x+6y+6z=726x + 6y + 6z = 72. Factoring out 66 yields 6(x+y+z)=726(x + y + z) = 72, so dividing by 66 gives x+y+z=12x + y + z = 12. Alternatively, solving the system explicitly gives x=5x = 5, y=4y = 4, and z=3z = 3, whose sum is 5+4+3=125 + 4 + 3 = 12.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations based on item counts.
Let xx, yy, and zz represent the number of Bundle X, Bundle Y, and Bundle Z ordered, respectively.
3x+y+2z=26(Floral arrangements)2x+3y+z=23(LED uplights)x+2y+3z=23(Table runners)\begin{aligned} 3x + y + 2z &= 26 \quad \text{(Floral arrangements)} \\ 2x + 3y + z &= 23 \quad \text{(LED uplights)} \\ x + 2y + 3z &= 23 \quad \text{(Table runners)} \end{aligned}
Translate the word problem into a standard system of 3 linear equations in 3 variables.
2
Sum the three equations to find a direct linear combination for (x+y+z)(x + y + z).
(3x+2x+x)+(y+3y+2y)+(2z+z+3z)=26+23+23(3x + 2x + x) + (y + 3y + 2y) + (2z + z + 3z) = 26 + 23 + 23
6x+6y+6z=726x + 6y + 6z = 72
Notice that the sum of coefficients for each variable across all three equations is identical (3+2+1=63 + 2 + 1 = 6).
3
Factor out 6 and solve for the total number of bundles (x+y+z)(x + y + z).
6(x+y+z)=72    x+y+z=126(x + y + z) = 72 \implies x + y + z = 12
Dividing both sides by 6 directly gives the required total quantity without needing to solve for individual variables xx, yy, and zz separately.

Anahtar Kavram

Linear combinations in systems of equations
Tahmini Süre:2m 0s
Soru 199Soru

If 9x+19x=2169^{x+1} - 9^x = 216, what is the value of 4x4^x?

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Cevap: 8

Cevap

The value of 4x4^x is 8.
Factoring 9x9^x from 9x+19x9^{x+1} - 9^x gives 9x(91)=2169^x(9 - 1) = 216, or 89x=2168 \cdot 9^x = 216. Dividing by 8 yields 9x=279^x = 27. Expressing both sides with base 3 gives 32x=333^{2x} = 3^3, so 2x=32x = 3 and x=32x = \frac{3}{2}. Substituting this value into 4x4^x results in 43/2=(4)3=23=84^{3/2} = (\sqrt{4})^3 = 2^3 = 8.

Adım Adım Çözüm

1
Factor out common exponent terms from the left side of the equation
9x(911)=216    89x=2169^x(9^1 - 1) = 216 \implies 8 \cdot 9^x = 216
Using the exponent property am+n=amana^{m+n} = a^m \cdot a^n, rewrite 9x+19^{x+1} as 9x919^x \cdot 9^1 to factor out 9x9^x.
2
Isolate the exponential term
9x=279^x = 27
Dividing both sides of 89x=2168 \cdot 9^x = 216 by 8 yields 2727.
3
Convert both sides to a common prime base of 3
(32)x=33    32x=33    2x=3    x=32(3^2)^x = 3^3 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}
Since bases are equal, exponents must be equal.
4
Evaluate the target expression 4x4^x
43/2=(41/2)3=23=84^{3/2} = (4^{1/2})^3 = 2^3 = 8
Substitute x=32x = \frac{3}{2} into 4x4^x and apply the rule am/n=(an)ma^{m/n} = (\sqrt[n]{a})^m.

Anahtar Kavram

Solving exponential equations by factoring and equating powers with a common base.
Soru 200Soru
If xx is a real number that satisfies the absolute value equation 2x7=3x11|2x - 7| = 3x - 11 what is the value of x23xx^2 - 3x?
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Cevap: 44

Cevap

The value of x23xx^2 - 3x is 44.
Solving the absolute value equation 2x7=3x11|2x - 7| = 3x - 11 yields two potential roots: x=4x = 4 and x=3.6x = 3.6. Checking both in the original equation shows that x=3.6x = 3.6 makes the right-hand side negative (0.2-0.2), making it an extraneous solution. The only valid solution is x=4x = 4. Substituting x=4x = 4 into x23xx^2 - 3x yields 423(4)=1612=44^2 - 3(4) = 16 - 12 = 4.

Adım Adım Çözüm

1
Set up equations based on the definition of absolute value
Case 1: 2x7=3x112x - 7 = 3x - 11; Case 2: 2x7=(3x11)2x - 7 = -(3x - 11)
An absolute value equation A=B|A| = B splits into A=BA = B or A=BA = -B, with the requirement that B0B \ge 0.
2
Solve Case 1: 2x7=3x112x - 7 = 3x - 11
3x2x=117    x=43x - 2x = 11 - 7 \implies x = 4
Isolate the variable xx by algebraic rearrangement.
3
Solve Case 2: 2x7=3x+112x - 7 = -3x + 11
5x=18    x=185=3.65x = 18 \implies x = \frac{18}{5} = 3.6
Isolate xx for the negative case.
4
Check both potential solutions in the original equation 2x7=3x11|2x - 7| = 3x - 11
For x=4x = 4: 2(4)7=1=1|2(4) - 7| = |1| = 1 and 3(4)11=13(4) - 11 = 1 (Valid).
For x=3.6x = 3.6: 2(3.6)7=0.2=0.2|2(3.6) - 7| = |0.2| = 0.2 but 3(3.6)11=0.23(3.6) - 11 = -0.2 (Extraneous).
The right side 3x113x - 11 must be non-negative. Since 3(3.6)11=0.2<03(3.6) - 11 = -0.2 < 0, x=3.6x = 3.6 is extraneous.
5
Evaluate the target expression x23xx^2 - 3x using the valid root x=4x = 4
423(4)=1612=44^2 - 3(4) = 16 - 12 = 4
Substitute the single valid root into the requested expression.

Anahtar Kavram

Absolute Value Linear Equations and Extraneous Solution Checking
ÖncekiSayfa 10 / 11Sonraki
Algebra and Functions Alıştırma Soruları — GMAT — Sayfa 10 | Examkin