Tüm alıştırma soruları

Soru 2081Soru →

Let $x$ and $y$ be non-zero integers such that $x^5 y^2 < 0$ and $x + y$ is an odd integer. Which of the following statements MUST be true? Select all that apply.

Geçerli olan tümünü seçin

x

is a negative integer

xy

is an even integer

x^2 + y^2

is an odd integer

(-x)^y

is guaranteed to be a positive integer

If

x < y

, then

-x < -y

Cevabı ve açıklamayı göster

Cevap: $x$ is a negative integer; $xy$ is an even integer; $x^2 + y^2$ is an odd integer

Cevap

The statements '

x

is a negative integer', '

xy

is an even integer', and '

x^2 + y^2

is an odd integer' MUST be true.

The statement specifying that '

x

is a negative integer' is true because

y^2

is strictly positive for any non-zero integer

y

, forcing

x^5 < 0

and thus

x < 0

. The statement '

xy

is an even integer' is true because

x + y

being odd requires one variable to be even and the other to be odd, making their product even. The statement '

x^2 + y^2

is an odd integer' is true because the square of an even number is even and the square of an odd number is odd, and their sum is always odd.

Adım Adım Çözüm

1

Determine the sign of

x

from the given inequality

x^5 y^2 < 0

.

x

must be negative (

x < 0

).

Since

y

is a non-zero integer,

y^2 > 0

. Dividing the inequality by

y^2

gives

x^5 < 0

, which means

x

must be negative.

2

Analyze the parity (even/odd nature) of

x

and

y

using

x + y

is odd.

One of

x, y

is even and the other is odd.

The sum of two integers is odd if and only if one integer is even and the other is odd.

3

Evaluate the statement '

xy

is an even integer'.

The product

xy

is always even.

The product of an even integer and an odd integer is always even.

4

Evaluate the statement '

x^2 + y^2

is an odd integer'.

The sum of squares

x^2 + y^2

is always odd.

Squaring an even integer yields an even integer, and squaring an odd integer yields an odd integer. Adding an even number and an odd number yields an odd number.

Anahtar Kavram

Even-Odd Parity Rules and Exponent Sign Properties

Soru 2082Soru →

An agricultural facility uses three conveyor belts— $X$ , $Y$ , and $Z$ —to transfer harvested grain into a storage silo. Working together at their respective constant rates, Belt $X$ and Belt $Y$ can fill the empty silo in $6\text{ hours}$ . Working together at their respective constant rates, Belt $Y$ and Belt $Z$ can fill the empty silo in $8\text{ hours}$ . If Belt $X$ operates at twice the rate of Belt $Z$ , how many hours would it take Belt $Y$ working alone at its constant rate to fill the empty silo?

10\text{ hours}

12\text{ hours}

16\text{ hours}

18\text{ hours}

24\text{ hours}

Cevabı ve açıklamayı göster

Cevap: $12\text{ hours}$

Cevap

12\text{ hours}

The correct answer is

12\text{ hours}

. Subtracting the combined rate equation for Belts

Y

and

Z

(

r_Y + r_Z = 1/8

) from the equation for Belts

X

and

Y

(

r_X + r_Y = 1/6

) yields

r_X - r_Z = 1/24

. Since Belt

X

works at twice the rate of Belt

Z

(

r_X = 2r_Z

), substituting gives

r_Z = 1/24

. Substituting

r_Z

back into

r_Y + r_Z = 1/8

gives

r_Y = 1/8 - 1/24 = 1/12

. Therefore, Belt

Y

operating alone takes

12\text{ hours}

to fill the silo.

Adım Adım Çözüm

1

Define the work rates of each conveyor belt.

Let

r_X

,

r_Y

, and

r_Z

be the fraction of the silo filled per hour by Belts

X

,

Y

, and

Z

, respectively.

Establishing rates per unit of time allows linear combination of work performed.

2

Set up equations based on the given combined times and rate relationships.

r_X + r_Y = \frac{1}{6}

,

r_Y + r_Z = \frac{1}{8}

, and

r_X = 2r_Z

.

Combined rates equal the reciprocal of the total time required for combined work.

3

Subtract the second equation from the first to isolate

r_X - r_Z

.

(r_X + r_Y) - (r_Y + r_Z) = \frac{1}{6} - \frac{1}{8} \implies r_X - r_Z = \frac{4 - 3}{24} = \frac{1}{24}

.

Eliminating

r_Y

gives a direct linear relationship between

r_X

and

r_Z

.

4

Substitute

r_X = 2r_Z

into

r_X - r_Z = \frac{1}{24}

to solve for

r_Z

.

2r_Z - r_Z = \frac{1}{24} \implies r_Z = \frac{1}{24}

.

Determines the individual rate of Belt

Z

.

5

Substitute

r_Z = \frac{1}{24}

back into the equation

r_Y + r_Z = \frac{1}{8}

to find

r_Y

.

r_Y = \frac{1}{8} - \frac{1}{24} = \frac{3}{24} - \frac{1}{24} = \frac{2}{24} = \frac{1}{12}

.

Finds the individual rate of Belt

Y

.

6

Calculate the time required for Belt

Y

alone.

\text{Time} = \frac{1}{r_Y} = 12\text{ hours}.

The time to complete one full job is the reciprocal of the individual rate.

Anahtar Kavram

Work Rates and Combined Rate Equations

Soru 2083Soru →

A quadratic function is defined by $f(x) = ax^2 + bx + c$ , where $a$ , $b$ , and $c$ are real constants with $a > 0$ . If the vertex of the parabola $y = f(x)$ lies in the third quadrant of the $xy$ -plane, which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

b > 0

The equation

f(x) = 0

has two distinct real solutions.

c

must be negative.

The sum of the two solutions of

f(x) = 0

is positive.

The discriminant

b^2 - 4ac

is less than zero.

Cevabı ve açıklamayı göster

Cevap: $b > 0$ ; The equation $f(x) = 0$ has two distinct real solutions.

Cevap

The statements '

b > 0

' and 'The equation

f(x) = 0

has two distinct real solutions' must be true.

The vertex of the parabola

y = ax^2 + bx + c

is

(h, k) = \left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)

. Because the vertex is in the third quadrant,

h < 0

and

k < 0

. With

a > 0

, the inequality

-\frac{b}{2a} < 0

directly requires

b > 0

. Furthermore, because

a > 0

(parabola opens upward) and the minimum value

k < 0

lies below the x-axis, the graph must cross the x-axis at two distinct points, establishing that

f(x) = 0

has two distinct real solutions.

Adım Adım Çözüm

1

Analyze the coordinates of the vertex

(h, k)

relative to the third quadrant.

The third quadrant requires

h < 0

and

k < 0

.

Points in Quadrant III have negative x-coordinates and negative y-coordinates.

2

Evaluate the sign of

b

using the x-coordinate formula

h = -\frac{b}{2a}

.

b > 0

.

Since

h < 0

and

a > 0

,

-\frac{b}{2a} < 0 \implies \frac{b}{2a} > 0 \implies b > 0

.

3

Determine the number of real solutions using the vertex y-coordinate

k

and orientation

a > 0

.

The discriminant

b^2 - 4ac > 0

, giving two distinct real solutions.

For an upward-opening parabola (

a > 0

), having a vertex below the x-axis (

k < 0

) guarantees the curve crosses the x-axis twice.

4

Test counterexamples for

c

and evaluate the root sum

-\frac{b}{a}

.

c

is not strictly constrained in sign, and the root sum is negative.

The function

f(x) = (x+2)^2 - 1 = x^2 + 4x + 3

has vertex

(-2, -1) \in \text{QIII}

with

c = 3 > 0

. Also, since

a > 0

and

b > 0

, root sum

-\frac{b}{a} < 0

.

Anahtar Kavram

Parabola Vertex and Quadratic Discriminant Analysis

Soru 2084Soru →

For all real numbers

x

and

y

such that

x \neq y

,

x \neq -y

, and

x^2 + y^2 \neq 0

, which of the following expressions is equivalent to

\frac{x^4 - y^4}{x^3 - x^2 y + x y^2 - y^3} - \frac{2 x y}{x + y}

\frac{x^2 + y^2}{x + y}

\frac{(x - y)^2}{x + y}

\frac{x + y - 2xy}{1 + x + y}

\frac{x^2 - y^2}{x + y}

\frac{x + y}{x - y}

Cevabı ve açıklamayı göster

Cevap: $\frac{x^2 + y^2}{x + y}$

Cevap

\frac{x^2 + y^2}{x + y}

Factoring

x^4 - y^4

as

(x - y)(x + y)(x^2 + y^2)

and

x^3 - x^2 y + x y^2 - y^3

as

(x - y)(x^2 + y^2)

simplifies the first term to

x + y

. Combining

x + y

with

-\frac{2xy}{x+y}

over the common denominator

(x + y)

gives

\frac{(x+y)^2 - 2xy}{x+y} = \frac{x^2 + y^2}{x+y}

.

Adım Adım Çözüm

1

Factor the numerator and denominator of the first rational expression

Numerator:

x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) = (x - y)(x + y)(x^2 + y^2)

. Denominator:

x^3 - x^2 y + x y^2 - y^3 = x^2(x - y) + y^2(x - y) = (x - y)(x^2 + y^2)

.

Factoring allows for cancellation of common factors in rational expressions.

2

Simplify the first rational expression by canceling common terms

\frac{(x - y)(x + y)(x^2 + y^2)}{(x - y)(x^2 + y^2)} = x + y

.

Since

x \neq y

and

x^2 + y^2 \neq 0

, the common terms

(x - y)

and

(x^2 + y^2)

are non-zero and can be divided out.

3

Subtract the second expression using a common denominator

(x + y) - \frac{2xy}{x + y} = \frac{(x + y)^2}{x + y} - \frac{2xy}{x + y} = \frac{(x + y)^2 - 2xy}{x + y}

.

Combining terms under the common denominator

(x + y)

enables algebraic reduction.

4

Expand the squared binomial in the numerator and simplify like terms

\frac{x^2 + 2xy + y^2 - 2xy}{x + y} = \frac{x^2 + y^2}{x + y}.

Expanding

(x + y)^2 = x^2 + 2xy + y^2

allows the

+2xy

and

-2xy

terms to cancel out.

Anahtar Kavram

Simplifying rational expressions using polynomial factoring (difference of squares and grouping) and common denominators.

Tahmini Süre:2m 0s

Soru 2085Soru →

A sequence of 40 numbers $a_1, a_2, \dots, a_{40}$ is defined by the formula $a_k = 15 + \frac{3k}{8}$ for each integer $k$ from 1 to 40. Each term $a_k$ is rounded to the nearest integer to form a new sequence $b_1, b_2, \dots, b_{40}$ . (Note: numbers ending in $.5$ are rounded up to the next integer.) What is the value of $\sum_{k=1}^{40} b_k - \sum_{k=1}^{40} a_k$ ?

2.5

0.0

5.0

-2.5

1.0

Cevabı ve açıklamayı göster

Cevap: $2.5$

Cevap

The sum of the rounded sequence exceeds the sum of the exact sequence by

2.5

.

Evaluating

a_k = 15 + \frac{3k}{8}

for

k = 1, 2, \dots, 8

yields fractional parts of

0.375, 0.75, 0.125, 0.5, 0.875, 0.25, 0.625,

and

0.0

. The individual rounding errors

(b_k - a_k)

for these terms are

-0.375, +0.25, -0.125, +0.5, +0.125, -0.25, +0.375,

and

0.0

. Summing these errors yields

+0.5

per 8-term period. For 40 terms (5 full periods), the total error sum is

5 \times 0.5 = 2.5

.

Adım Adım Çözüm

1

Analyze the fractional part of

a_k = 15 + \frac{3k}{8}

over one complete 8-term period.

The fractional parts for

k = 1, 2, \dots, 8

are

0.375, 0.75, 0.125, 0.5, 0.875, 0.25, 0.625, 0.0

respectively.

Since

\frac{3k}{8}

repeats its fractional pattern modulo 8, examining one period reveals the periodic rounding behavior.

2

Calculate the rounding error

(b_k - a_k)

for each term in the 8-term cycle.

The differences

(b_k - a_k)

for

k=1

to

8

are:

-0.375, +0.25, -0.125, +0.5, +0.125, -0.25, +0.375, 0.0

.

Rounding to the nearest integer shifts each number by its distance to that integer. Half-integers like

0.5

round up, yielding a

+0.5

difference.

3

Sum the rounding errors over one 8-term cycle.

(-0.375 + 0.375) + (0.25 - 0.25) + (-0.125 + 0.125) + 0.5 + 0.0 = +0.5

.

Symmetric fractional pairs cancel out, leaving only the

+0.5

error from the half-integer term.

4

Multiply the single-cycle error sum by the number of full cycles in 40 terms.

Since

40 = 5 \times 8

, total difference

= 5 \times 0.5 = 2.5

.

The 40 terms consist of 5 identical 8-term periodic cycles.

Anahtar Kavram

Periodic error analysis in sequence rounding and summation

Soru 2086Soru →

A civil engineering laboratory prepares a composite soil mixture using three material grades: Grade $X$ , Grade $Y$ , and Grade $Z$ . Initially, the ratio of Grade $X$ to Grade $Y$ by weight is $1 : 2$ , and the ratio of Grade $Y$ to Grade $Z$ by weight is $3 : 4$ . A technician adds $30\text{ kg}$ of Grade $X$ to the batch, causing Grade $X$ to constitute exactly $\frac{1}{4}$ of the total weight of the new mixture. Which of the following statements about the batch must be true? Select all such statements.

Geçerli olan tümünü seçin

The initial total weight of the mixture before Grade

X

was added was

306\text{ kg}

.

The weight of Grade

Z

in the mixture is

144\text{ kg}

.

In the final mixture, the ratio of Grade

Y

to Grade

Z

by weight is

3 : 4

.

The initial weight of Grade

X

in the mixture was

72\text{ kg}

.

Grade

Y

accounts for

35\%

of the total weight of the final mixture.

Cevabı ve açıklamayı göster

Cevap: The initial total weight of the mixture before Grade $X$ was added was $306\text{ kg}$ .; The weight of Grade $Z$ in the mixture is $144\text{ kg}$ .; In the final mixture, the ratio of Grade $Y$ to Grade $Z$ by weight is $3 : 4$ .

Cevap

The correct statements are those indicating that the initial total weight of the mixture was 306 kg, the weight of Grade Z in the mixture is 144 kg, and the ratio of Grade Y to Grade Z in the final mixture is 3 : 4.

The unified ratio of X : Y : Z is 3 : 6 : 8. Setting up the fraction of Grade X after adding 30 kg gives (3k + 30) / (17k + 30) = 1/4, which yields k = 18. This gives an initial total weight of 17 * 18 = 306 kg, a Grade Z weight of 8 * 18 = 144 kg, and an unchanged Grade Y to Grade Z ratio of 108 : 144 = 3 : 4. Therefore, the statements asserting the initial total weight as 306 kg, the Grade Z weight as 144 kg, and the final Y to Z ratio as 3 : 4 are all correct.

Adım Adım Çözüm

1

Unify the two given ratio relationships into a single 3-part ratio.

Grade X : Grade Y = 1 : 2 = 3 : 6, and Grade Y : Grade Z = 3 : 4 = 6 : 8. Unified ratio Grade X : Grade Y : Grade Z = 3 : 6 : 8.

Matching the term for Grade Y across both ratios enables setting up unified algebraic variable parts.

2

Define initial quantities using a common multiplier k.

Initial weight of X = 3k, Y = 6k, Z = 8k. Initial total weight = 3k + 6k + 8k = 17k.

Expressing quantities in terms of k allows setting up an equation after adding material.

3

Formulate and solve the equation based on the addition of Grade X.

(3k + 30) / (17k + 30) = 1/4 => 4(3k + 30) = 17k + 30 => 12k + 120 = 17k + 30 => 5k = 90 => k = 18.

Setting the new weight of X over the new total weight equal to 1/4 determines the exact value of k.

4

Evaluate the individual component weights and statements.

Initial total weight = 17(18) = 306 kg. Initial X = 3(18) = 54 kg. Weight of Y = 6(18) = 108 kg. Weight of Z = 8(18) = 144 kg. Final total weight = 306 + 30 = 336 kg. Final Y : Z ratio = 108 : 144 = 3 : 4.

Determining all numerical values allows verifying which statements are true.

Anahtar Kavram

Combining three-variable ratio streams into a unified ratio and solving linear rate/proportion equations upon single-component addition.

Soru 2087Soru →

A sequence of 24 positive numbers $a_1, a_2, \dots, a_{24}$ is defined by $a_n = \frac{120}{n(n+1)}$ for each integer $n$ from $1$ to $24$ . Let $T = \sum_{n=1}^{24} a_n$ be the exact sum of all 24 terms. Which of the following statements must be true? Select all that apply.

Geçerli olan tümünü seçin

The value of

T

rounded to the nearest integer is

115

.

The sum of the first 4 terms,

\sum_{n=1}^{4} a_n

, represents more than

80\%

of the total sum

T

.

The sum of the first 2 terms,

a_1 + a_2

, is equal to

100

.

The percent error in estimating the total sum

T

by using an upper bound of

120

is exactly

4.0\%

.

For the sequence term formula

a_n = f(n)

, the value of

f(1 + 2)

is equal to

f(1) + f(2)

.

Cevabı ve açıklamayı göster

Cevap: The value of $T$ rounded to the nearest integer is $115$ .; The sum of the first 4 terms, $\sum_{n=1}^{4} a_n$ , represents more than $80\%$ of the total sum $T$ .

Cevap

The statements confirming that

T

rounded to the nearest integer is

115

, and that the sum of the first 4 terms represents more than

80\%

of

T

, are both correct.

The exact sum of the telescoping sequence simplifies to

T = 120 \left(1 - \frac{1}{25}\right) = 115.2

. Rounding

115.2

to the nearest integer gives

115

. Furthermore, the partial sum of the first four terms is

120 \left(1 - \frac{1}{5}\right) = 96

, which accounts for

\frac{96}{115.2} = \frac{5}{6} \approx 83.33\%

of the total sum, exceeding

80\%

.

Adım Adım Çözüm

1

Decompose the sequence formula using partial fractions.

a_n = \frac{120}{n(n+1)} = 120 \left( \frac{1}{n} - \frac{1}{n+1} \right)

.

Rewriting the terms as partial fractions converts the sum into a telescoping series.

2

Calculate the exact total sum

T

.

T = 120 \left[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \dots + \left(\frac{1}{24} - \frac{1}{25}\right) \right] = 120 \left(1 - \frac{1}{25}\right) = 120 \times 0.96 = 115.2

.

All intermediate terms cancel out, leaving only the first and last components.

3

Evaluate the statement regarding rounding

T

to the nearest integer.

115.2

rounded to the nearest integer is

115

, making the rounding statement true.

Since the decimal part

.2

is less than

.5

, the number rounds down to

115

.

4

Evaluate the sum of the first 4 terms and compare its percentage to

T

.

\sum_{n=1}^{4} a_n = 120 \left(1 - \frac{1}{5}\right) = 96

. The percentage is

\frac{96}{115.2} = \frac{5}{6} \approx 83.33\% > 80\%

.

Comparing

83.33\%

to

80\%

confirms that the partial sum statement is true.

Anahtar Kavram

Telescoping Series Summation and Percent Estimation

Soru 2088Soru →

At a manufacturing facility, the ratio of the daily output of Machine $X$ to Machine $Y$ is $4 : 5$ , and the ratio of the daily output of Machine $Y$ to Machine $Z$ is $3 : 2$ . If the daily output of Machine $X$ is decreased by $10\%$ and the daily output of Machine $Z$ is increased by $25\%$ , what is the new ratio of the daily output of Machine $X$ to the daily output of Machine $Z$ ?

6 : 5

27 : 25

108 : 125

9 : 10

48 : 125

Cevabı ve açıklamayı göster

Cevap: $108 : 125$

Cevap

The new ratio of the daily output of Machine

X

to Machine

Z

is

108 : 125

.

The unified ratio of the outputs of the three machines is

X : Y : Z = 12 : 15 : 10

. Decreasing Machine

X

's output by

10\%

changes its relative units to

12 \times 0.90 = 10.8

. Increasing Machine

Z

's output by

25\%

changes its relative units to

10 \times 1.25 = 12.5

. Comparing the updated values yields

10.8 : 12.5

, which simplifies to

108 : 125

.

Adım Adım Çözüm

1

Unify the two separate ratios into a single three-part ratio

X : Y : Z

.

Since

X : Y = 4 : 5

and

Y : Z = 3 : 2

, multiply

X : Y

by

3

to get

12 : 15

and multiply

Y : Z

by

5

to get

15 : 10

. Thus,

X : Y : Z = 12 : 15 : 10

.

Machine

Y

is the common element linking both ratios, so its ratio component must be equalized.

2

Calculate the updated values for

X

and

Z

after applying their respective percentage changes.

Machine

X_{new} = 12 \times (1 - 0.10) = 12 \times 0.90 = 10.8

. Machine

Z_{new} = 10 \times (1 + 0.25) = 10 \times 1.25 = 12.5

.

Applying a

10\%

decrease scales a quantity by

0.90

, while a

25\%

increase scales it by

1.25

.

3

Form the ratio

X_{new} : Z_{new}

and convert to lowest integer terms.

10.8 : 12.5 = \frac{10.8}{12.5} = \frac{108}{125}

, which gives the ratio

108 : 125

.

Multiplying both terms by

10

eliminates decimals, yielding the coprime integer ratio

108 : 125

.

Anahtar Kavram

Three-Part Ratio Unification and Relative Percentage Modification

Tahmini Süre:1m 30s

Soru 2089Soru →

A research laboratory operates two types of automated centrifuges, Model X and Model Y. A single operating cycle of Model X processes $40$ biological samples and consumes $10$ kilowatt-hours (kWh) of electricity. A single operating cycle of Model Y processes $25$ biological samples and consumes $15$ kWh of electricity. On a given day, the laboratory processed a total of $775$ biological samples and consumed $325$ kWh of electricity using only these two models. What is the total number of operating cycles completed by Model X and Model Y combined on that day?

10

15

25

30

35

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The total number of combined operating cycles completed by Model X and Model Y is 25.

The correct answer is 25. Setting up the linear system

40x + 25y = 775

and

10x + 15y = 325

and solving via elimination gives

x = 10

cycles for Model X and

y = 15

cycles for Model Y. Adding these together gives

10 + 15 = 25

total cycles.

Adım Adım Çözüm

1

Define variables and set up the system of linear equations based on total samples and total electricity consumption.

Let

x

be the number of cycles for Model X and

y

be the number of cycles for Model Y.
System of equations:
1)

40x + 25y = 775

(Sample constraint)
2)

10x + 15y = 325

(Energy constraint)

Modeling the word problem as two linear equations in two variables allows for algebraic elimination.

2

Simplify both equations by dividing by their greatest common factors.

Divide Equation 1 by 5:

8x + 5y = 155

Divide Equation 2 by 5:

2x + 3y = 65

Simplifying coefficients reduces computational complexity and minimizes arithmetic errors.

3

Eliminate variable

x

by multiplying the simplified second equation by 4 and subtracting the simplified first equation.

4 \times (2x + 3y = 65) \Rightarrow 8x + 12y = 260

Subtract

(8x + 5y = 155)

:

(8x - 8x) + (12y - 5y) = 260 - 155

7y = 105 \Rightarrow y = 15

Equalizing the coefficients of

x

enables solving directly for

y

.

4

Substitute

y = 15

back into

2x + 3y = 65

to find

x

.

2x + 3(15) = 65 \Rightarrow 2x + 45 = 65 \Rightarrow 2x = 20 \Rightarrow x = 10

Evaluating

x

completes the solution for individual cycle counts.

5

Calculate the requested combined total

x + y

.

x + y = 10 + 15 = 25

The question specifically requests the sum of operating cycles of both models combined.

Anahtar Kavram

Systems of Linear Equations in Two Variables

Tahmini Süre:1m 45s

Soru 2090Soru →

For all non-zero real numbers $x$ and $y$ such that $|x| \neq |y|$ and $x^2 + 2xy + 2y^2 \neq 0$ , consider the algebraic expression:

Q(x, y) = \frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} + \frac{2x^3 y - 2xy^3}{x^2 - y^2}

Which of the following expressions are equivalent to $Q(x, y)$ ? Indicate all such expressions.

Geçerli olan tümünü seçin

(x - y)^2 + 2xy + y^2

\frac{x^3 + 2xy^2}{x}

\frac{(x^2 + 2y^2)^2 - 4x^2 y^2}{x^2 - 2xy + 2y^2}

\frac{x^4 - 4y^4}{x^2 + 2y^2}

\frac{x^4 + 2x^2 y^2 + 2y^4}{x^2 + y^2}

Cevabı ve açıklamayı göster

Cevap: $(x - y)^2 + 2xy + y^2$ ; \frac{x^3 + 2xy^2}{x}

Cevap

The expressions equivalent to

Q(x, y)

are

(x - y)^2 + 2xy + y^2

and

\frac{x^3 + 2xy^2}{x}

.

First, simplify

Q(x, y)

by factoring each term. Using Sophie Germain's identity on the numerator of the first term gives

x^4 + 4y^4 = (x^2 + 2y^2)^2 - (2xy)^2 = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)

. Dividing this by

(x^2 + 2xy + 2y^2)

results in

x^2 - 2xy + 2y^2

. For the second term, factoring out

2xy

gives

\frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy

. Adding the two simplified terms yields

Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2

.

Evaluating the options for equivalence:
- The expression

(x - y)^2 + 2xy + y^2

expands to

x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2

.
- The expression

\frac{x^3 + 2xy^2}{x}

factors as

\frac{x(x^2 + 2y^2)}{x} = x^2 + 2y^2

.
Therefore, both of these expressions are equivalent to

Q(x, y)

.

Adım Adım Çözüm

1

Simplify the first algebraic fraction using Sophie Germain's identity.

\frac{x^4 + 4y^4}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2y^2)^2 - (2xy)^2}{x^2 + 2xy + 2y^2} = \frac{(x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)}{x^2 + 2xy + 2y^2} = x^2 - 2xy + 2y^2

Completing the square on

x^4 + 4y^4

allows it to be factored into the product of two quadratic expressions.

2

Simplify the second algebraic fraction by factoring out common factors.

\frac{2x^3 y - 2xy^3}{x^2 - y^2} = \frac{2xy(x^2 - y^2)}{x^2 - y^2} = 2xy

The term

(x^2 - y^2)

cancels out since

x \neq \pm y

.

3

Combine the simplified terms to find the closed-form expression for

Q(x, y)

.

Q(x, y) = (x^2 - 2xy + 2y^2) + 2xy = x^2 + 2y^2

The

-2xy

and

+2xy

terms sum to zero.

4

Test each option for equivalence to

x^2 + 2y^2

.

The option

(x - y)^2 + 2xy + y^2

expands to

x^2 - 2xy + y^2 + 2xy + y^2 = x^2 + 2y^2

. The option

\frac{x^3 + 2xy^2}{x}

simplifies directly to

x^2 + 2y^2

.

Both expressions reduce identically to

x^2 + 2y^2

for all non-zero

x

and

y

.

Anahtar Kavram

Simplifying complex algebraic expressions using polynomial factoring identities (Sophie Germain identity and difference of squares) and algebraic reduction.

Tahmini Süre:2m 0s

Soru 2091Soru →

A commercial facility prepares a fruit blend by mixing fruit concentrate with water in a ratio of $3:7$ by volume. After $15\text{ liters}$ of water evaporate from the mixture during processing, the ratio of fruit concentrate to water in the remaining mixture becomes $1:2$ . What was the total volume, in liters, of the original mixture before evaporation?

Cevabı ve açıklamayı göster

Cevap: 150

Cevap

The total volume of the original mixture before evaporation was 150 liters.

Represent the initial concentrate volume as

3x

liters and the initial water volume as

7x

liters, making the initial total volume

10x

liters. Evaporating

15

liters of water leaves

7x - 15

liters of water while the concentrate remains

3x

liters. Setting the ratio

\frac{3x}{7x - 15}

equal to

\frac{1}{2}

yields

6x = 7x - 15

, so

x = 15

. Substituting

x = 15

into the total volume expression

10x

gives

10(15) = 150

liters.

Adım Adım Çözüm

1

Define initial component volumes using ratio multiplier x

Concentrate volume =

3x

, Water volume =

7x

, Total volume =

10x

The given initial ratio of concentrate to water is

3:7

.

2

Formulate equation based on water evaporation and the new ratio

\frac{3x}{7x - 15} = \frac{1}{2}

Evaporation reduces only the water volume by 15 liters, establishing a new ratio of

1:2

.

3

Solve the algebraic equation for x

6x = 7x - 15 \implies x = 15

Cross-multiplication simplifies the proportional relationship into a linear equation.

4

Calculate original total volume

10 \times 15 = 150

liters

The original total volume is represented by

10x

.

Anahtar Kavram

Solving component adjustment problems using ratio multipliers

Soru 2092Soru →

A sequence of numbers $a_1, a_2, a_3, \dots$ is defined by $a_1 = 5$ and $a_{n+1} = a_n^2 - 2$ for all positive integers $n \ge 1$ . Which of the following is the value of the sum $\sum_{n=1}^{4} \frac{1}{a_n}$ , rounded to the nearest hundredth?

0.20

0.24

0.25

0.32

0.43

Cevabı ve açıklamayı göster

Cevap: 0.25

Cevap

0.25

Evaluating the recurrence relation yields

a_1 = 5

,

a_2 = 23

,

a_3 = 527

, and

a_4 = 277,727

. Summing their reciprocals produces

\frac{1}{5} + \frac{1}{23} + \frac{1}{527} + \frac{1}{277,727} \approx 0.20 + 0.043478 + 0.001898 + 0.000004 = 0.24538

. Rounding

0.24538

to the nearest hundredth yields 0.25.

Adım Adım Çözüm

1

Calculate the first four terms of the defined sequence using the recurrence relation

a_{n+1} = a_n^2 - 2

.

a_1 = 5

,

a_2 = 5^2 - 2 = 23

,

a_3 = 23^2 - 2 = 527

, and

a_4 = 527^2 - 2 = 277,727

.

The recurrence rule determines each subsequent term from the preceding term.

2

Compute the sum of reciprocals

\sum_{n=1}^{4} \frac{1}{a_n} = \frac{1}{5} + \frac{1}{23} + \frac{1}{527} + \frac{1}{277,727}

.

\frac{1}{5} = 0.2

,

\frac{1}{23} \approx 0.043478

,

\frac{1}{527} \approx 0.001898

, and

\frac{1}{277,727} \approx 0.0000036

. Sum

\approx 0.24538

.

Converting each fraction term to decimal form allows straightforward addition.

3

Round the calculated sum

0.24538

to the nearest hundredth.

Since the thousandths digit is

5

,

0.24538

rounds up to

0.25

.

Standard rounding rules dictate rounding up when the digit to the right of the target decimal place is

5

or greater.

Anahtar Kavram

Defined sequence terms evaluation, reciprocal sum estimation, and decimal rounding

Tahmini Süre:1m 30s

Soru 2093Soru →

A community theater sold a total of $250$ tickets for a performance, consisting of adult tickets for $\$15$ each and student tickets for $\$10$ each. The total revenue collected from ticket sales was $\$3,100$ . Let $a$ represent the number of adult tickets sold and $s$ represent the number of student tickets sold. Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

The number of student tickets sold was

130

.

The number of adult tickets sold was

130

.

The total revenue generated from adult ticket sales was

\$1,800

.

The ratio of adult tickets sold to student tickets sold was

13

to

12

.

More adult tickets were sold than student tickets.

Cevabı ve açıklamayı göster

Cevap: The number of student tickets sold was $130$ .; The total revenue generated from adult ticket sales was $\$1,800$ .

Cevap

The correct statements are that the number of student tickets sold was 130, and the total revenue generated from adult ticket sales was $1,800.

The system of equations

a + s = 250

and

15a + 10s = 3100

uniquely yields

a = 120

adult tickets and

s = 130

student tickets. Therefore, the statement that 130 student tickets were sold is true. Furthermore, multiplying 120 adult tickets by their price of

15 per ticket gives a total adult ticket revenue of

1,800, making that statement true as well.

Adım Adım Çözüm

1

Set up a system of two linear equations in terms of

a

and

s

.

Equation 1 (total tickets):

a + s = 250

; Equation 2 (total revenue):

15a + 10s = 3100

.

The sum of the ticket quantities equals total tickets, and the sum of monetary contributions equals total revenue.

2

Express

s

in terms of

a

using Equation 1 and substitute into Equation 2.

s = 250 - a \implies 15a + 10(250 - a) = 3100 \implies 15a + 2500 - 10a = 3100

.

Using substitution eliminates variable

s

to solve for

a

.

3

Solve for

a

and then find

s

.

5a = 600 \implies a = 120

;

s = 250 - 120 = 130

.

Dividing

600

by

5

gives

a = 120

, and subtracting

120

from

250

gives

s = 130

.

4

Evaluate the statements against the values

a = 120

and

s = 130

.

s = 130

is true. Revenue from adult tickets is

120 \times 15 = \$1,800

, which is also true.

Comparing calculated values directly confirms which options state accurate quantitative properties.

Anahtar Kavram

Solving 2x2 Systems of Linear Equations using Substitution or Elimination in Word Problems

Soru 2094Soru →

Consider the system of linear equations in two variables

x

and

y

shown below:

\begin{aligned} 3x + 2y &= 16 \\ x - 4y &= -4 \end{aligned}

If

(x, y)

is the unique solution to the system, what is the value of

\frac{y}{x}

?

\frac{1}{2}

2

4

\frac{2}{5}

6

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{2}$

Cevap

The value of

\frac{y}{x}

is

\frac{1}{2}

.

Solving the system by substitution gives

x = 4

and

y = 2

. Dividing

y

by

x

gives

\frac{2}{4} = \frac{1}{2}

, which is the target ratio.

Adım Adım Çözüm

1

Isolate

x

in the second equation.

x = 4y - 4

Expressing

x

in terms of

y

allows for substitution into the first equation.

2

Substitute

x = 4y - 4

into the first equation.

3(4y - 4) + 2y = 16 \implies 12y - 12 + 2y = 16 \implies 14y = 28 \implies y = 2

Solving the single-variable equation determines the value of

y

.

3

Substitute

y = 2

back into the isolated expression for

x

.

x = 4(2) - 4 = 4

Determines the corresponding value of

x

.

4

Compute the required ratio

\frac{y}{x}

.

\frac{y}{x} = \frac{2}{4} = \frac{1}{2}

Evaluates the final expression requested by the prompt.

Anahtar Kavram

Solving systems of linear equations using substitution or elimination to evaluate a combined expression.

Soru 2095Soru →

A chemical processing plant uses two storage tanks, Tank $X$ and Tank $Y$ . Initially, the ratio of the volume of liquid in Tank $X$ to Tank $Y$ is $3 : 7$ . After $16\text{ liters}$ of liquid are transferred from Tank $Y$ to Tank $X$ , and an additional $8\text{ liters}$ of liquid are added to Tank $X$ from an external supply, the ratio of the volume of liquid in Tank $X$ to Tank $Y$ becomes $6 : 5$ . What was the original volume, in liters, of liquid in Tank $Y$ ?

24

49

56

80

96

Cevabı ve açıklamayı göster

Cevap: 56

Cevap

56 liters

By setting the initial volumes of Tank X and Tank Y to 3k and 7k respectively, the modified volumes after transfer and addition are (3k + 24) and (7k - 16). Setting their ratio equal to 6/5 yields 5(3k + 24) = 6(7k - 16), which simplifies to 27k = 216, or k = 8. Multiplying 8 by 7 gives the original volume of Tank Y as 56 liters.

Adım Adım Çözüm

1

Define initial quantities using a common ratio multiplier

k

.

Let the initial volume of Tank

X

be

3k

liters and the initial volume of Tank

Y

be

7k

liters.

The given initial ratio of

X : Y

is

3 : 7

.

2

Express the new volumes after the liquid transfers.

Tank

X

volume becomes

3k + 16 + 8 = 3k + 24

liters. Tank

Y

volume becomes

7k - 16

liters.

Transferring

16\text{ liters}

from Tank

Y

to Tank

X

decreases Tank

Y

by

16

and increases Tank

X

by

16

. Adding an extra

8\text{ liters}

to Tank

X

brings its total increase to

24\text{ liters}

.

3

Set up the proportion with the new ratio and solve for

k

.

\begin{aligned} \frac{3k + 24}{7k - 16} &= \frac{6}{5} \\[6pt] 5(3k + 24) &= 6(7k - 16) \\[6pt] 15k + 120 &= 42k - 96 \\[6pt] 216 &= 27k \\[6pt] k &= 8 \end{aligned}

The new ratio of Tank

X

to Tank

Y

is given as

6 : 5

.

4

Calculate the original volume of Tank

Y

.

Original volume of Tank

Y = 7k = 7 \times 8 = 56\text{ liters}

.

Tank

Y

initially contained

7k

liters.

Anahtar Kavram

Algebraic setup of part-to-part ratios undergoing quantitative adjustments

Tahmini Süre:1m 30s

Soru 2096Soru →

A sequence of numbers $a_1, a_2, a_3, \dots$ is defined by $a_1 = 50$ and $a_{n+1} = \frac{1}{3} a_n + 15$ for all positive integers $n \ge 1$ . Each term $a_n$ is rounded to the nearest integer to produce a secondary sequence $b_n$ . What is the value of $\sum_{n=1}^{4} b_n$ ?

129

130

131

132

133

Cevabı ve açıklamayı göster

Cevap: 132

Cevap

132

Evaluating each term yields

a_1 = 50

,

a_2 = 31.6667...

,

a_3 = 25.5556...

, and

a_4 = 23.5185...

. Rounding each term individually to the nearest integer gives

b_1 = 50

,

b_2 = 32

,

b_3 = 26

, and

b_4 = 24

. Their sum is

50 + 32 + 26 + 24 = 132

.

Adım Adım Çözüm

1

Calculate the exact value of the first term

a_1

and its rounded value

b_1

.

a_1 = 50

, which is an integer. Thus,

b_1 = 50

.

Given initial value.

2

Calculate the exact value of

a_2

using the recursive formula

a_2 = \frac{1}{3}a_1 + 15

and round to nearest integer.

a_2 = \frac{1}{3}(50) + 15 = \frac{50}{3} + 15 = 31.6667...

. Rounding to the nearest integer gives

b_2 = 32

.

Apply sequence relation and rounding rule.

3

Calculate the exact value of

a_3 = \frac{1}{3}a_2 + 15

and round to nearest integer.

a_3 = \frac{1}{3}\left(\frac{95}{3}\right) + 15 = \frac{95}{9} + 15 = \frac{230}{9} = 25.5556...

. Rounding to the nearest integer gives

b_3 = 26

.

Apply sequence relation and rounding rule.

4

Calculate the exact value of

a_4 = \frac{1}{3}a_3 + 15

and round to nearest integer.

a_4 = \frac{1}{3}\left(\frac{230}{9}\right) + 15 = \frac{230}{27} + 15 = \frac{635}{27} = 23.5185...

. Rounding to the nearest integer gives

b_4 = 24

.

Apply sequence relation and rounding rule.

5

Sum the four rounded terms

b_1 + b_2 + b_3 + b_4

.

\sum_{n=1}^{4} b_n = 50 + 32 + 26 + 24 = 132

.

Add the individual rounded terms.

Anahtar Kavram

Recursive sequence evaluation and term-by-term rounding vs total sum rounding.

Soru 2097Soru →

A commercial print shop operates two types of high-speed printers, Printer M and Printer N. Operating simultaneously for $5$ hours, $2$ units of Printer M and $3$ units of Printer N print a total of $5,500$ pages. Operating simultaneously for $4$ hours, $5$ units of Printer M and $2$ units of Printer N print a total of $6,600$ pages. What is the hourly page output of a single Printer M?

Cevabı ve açıklamayı göster

Cevap: 250

Cevap

The hourly page output of a single Printer M is 250 pages per hour.

Dividing each total page output by the corresponding number of hours produces the simplified linear system:

2m + 3n = 1100

and

5m + 2n = 1650

, where

m

and

n

are the hourly rates of Printer M and Printer N. Multiplying the first equation by

2

gives

4m + 6n = 2200

, and multiplying the second equation by

3

gives

15m + 6n = 4950

. Subtracting the two equations eliminates

n

, giving

11m = 2750

, which simplifies to

m = 250

pages per hour.

Adım Adım Çözüm

1

Define variables and write initial algebraic equations based on time and rate.

Let

m

represent the hourly page output of Printer M and

n

represent the hourly page output of Printer N. The total outputs give

5(2m + 3n) = 5500

and

4(5m + 2n) = 6600

.

Total page output equals total operating time multiplied by the combined hourly output rate.

2

Simplify the system by dividing each equation by its respective number of hours.

First equation:

2m + 3n = 1100

. Second equation:

5m + 2n = 1650

.

Simplifying yields a standard system of linear equations representing the combined hourly rate.

3

Eliminate variable

n

to solve for

m

.

Multiply

2m + 3n = 1100

by

2

to get

4m + 6n = 2200

. Multiply

5m + 2n = 1650

by

3

to get

15m + 6n = 4950

. Subtracting the first modified equation from the second yields

11m = 2750

, so

m = 250

.

Equating the coefficients of

n

allows direct elimination of

n

when subtracting the equations.

Anahtar Kavram

Solving a 2x2 system of linear equations using the method of elimination

Tahmini Süre:1m 30s

Soru 2098Soru →

For all real numbers

x \neq -5

, the algebraic expression

\frac{x^3 + 125}{x^2 - 5x + 25} - \frac{x^2 - 25}{x + 5}

simplifies to a single constant value. What is the value of this constant?

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

The simplified value of the expression for all valid real numbers x is 10.

Factoring the numerators reveals that the first term reduces to

x + 5

and the second term reduces to

x - 5

. Subtracting

(x - 5)

from

(x + 5)

yields

(x + 5) - (x - 5) = 10

, which is constant for all valid values of

x

.

Adım Adım Çözüm

1

Factor the numerator of the first rational term using the sum of cubes identity.

x^3 + 125 = (x + 5)(x^2 - 5x + 25)

The sum of cubes formula

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

applies with

a = x

and

b = 5

.

2

Cancel the non-zero factor

(x^2 - 5x + 25)

from the first fraction.

\frac{(x + 5)(x^2 - 5x + 25)}{x^2 - 5x + 25} = x + 5

The quadratic factor

x^2 - 5x + 25

has a negative discriminant (

(-5)^2 - 4(1)(25) = -75 < 0

), so it is never zero for any real number

x

.

3

Factor the numerator of the second rational term using the difference of squares identity.

x^2 - 25 = (x - 5)(x + 5)

The difference of squares formula

a^2 - b^2 = (a - b)(a + b)

applies with

a = x

and

b = 5

.

4

Cancel the common factor

(x + 5)

from the second fraction.

\frac{(x - 5)(x + 5)}{x + 5} = x - 5

Given

x \neq -5

, the factor

x + 5

is non-zero and can be canceled.

5

Subtract the two simplified terms.

(x + 5) - (x - 5) = x + 5 - x + 5 = 10

Distribute the negative sign to both terms in

(x - 5)

and combine like terms.

Anahtar Kavram

Factoring sum of cubes and difference of squares to simplify rational expressions

Soru 2099Soru →

A courier travels along a straight route from Office P to Office Q at a constant speed $r_1$ , and immediately returns along the exact same route from Office Q to Office P at a constant speed $r_2$ . The ratio of the outbound speed to the return speed is $r_1 : r_2 = 3 : 2$ . Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

The time spent on the return trip is

50\%

greater than the time spent on the outbound trip.

The ratio of the time spent on the outbound trip to the time spent on the return trip is

3 : 2

.

The average speed for the entire round trip is equal to

80\%

of the outbound speed.

The average speed for the entire round trip is equal to the arithmetic average of the outbound speed and the return speed.

The outbound trip accounts for

60\%

of the total travel time for the round trip.

Cevabı ve açıklamayı göster

Cevap: The time spent on the return trip is $50\%$ greater than the time spent on the outbound trip.; The average speed for the entire round trip is equal to $80\%$ of the outbound speed.

Cevap

The statement that the return trip time is 50 percent greater than the outbound trip time, and the statement that the average speed for the entire round trip is equal to 80 percent of the outbound speed.

The return trip time is

50\%

greater than the outbound trip time because speed and time are inversely proportional over equal distances (

t_1 : t_2 = 2 : 3

). Additionally, the overall average speed is

\frac{12k}{5} = 2.4k

, which is precisely

80\%

of the outbound speed (

3k

).

Adım Adım Çözüm

1

Express travel times in terms of distance

D

and rate multiplier

k

.

Let outbound speed

r_1 = 3k

and return speed

r_2 = 2k

. Outbound time

t_1 = \frac{D}{3k}

and return time

t_2 = \frac{D}{2k}

. The ratio of times

t_1 : t_2 = \frac{D/3k}{D/2k} = 2 : 3

.

Time equals distance divided by rate, establishing an inverse relationship between speed and time for constant distance.

2

Evaluate the relative difference between return time and outbound time.

Return time

t_2 = 1.5 t_1

, which means

t_2

is

50\%

greater than

t_1

.

A factor of

1.5

represents a

50\%

increase over the base value

t_1

.

3

Calculate the average speed

R_{\text{avg}}

for the round trip.

Total distance is

2D

, and total time is

t_1 + t_2 = \frac{D}{3k} + \frac{D}{2k} = \frac{5D}{6k}

. Thus,

R_{\text{avg}} = \frac{2D}{\frac{5D}{6k}} = \frac{12k}{5} = 2.4k

.

Average speed is defined as total distance divided by total elapsed time.

4

Compare

R_{\text{avg}}

to the outbound speed

r_1

and the arithmetic mean of rates.

\frac{R_{\text{avg}}}{r_1} = \frac{2.4k}{3k} = 0.80 = 80\%

. The arithmetic mean is

\frac{3k + 2k}{2} = 2.5k \neq 2.4k

. Outbound time fraction is

\frac{t_1}{t_1 + t_2} = \frac{2}{5} = 40\%

.

Unweighted arithmetic averaging fails for rates over equal distances because more time is spent traveling at the lower speed.

Anahtar Kavram

Inverse proportion between speed and time, harmonic mean for round-trip average speed, and part-to-whole time ratios.

Tahmini Süre:2m 0s

Soru 2100Soru →

Working alone at their respective constant rates, Machine $A$ can complete a production order in $6\text{ hours}$ , Machine $B$ in $8\text{ hours}$ , and Machine $C$ in $12\text{ hours}$ . All three machines start working together on the order at 9:00 AM. At 10:00 AM, Machine $A$ breaks down and stops working, while Machines $B$ and $C$ continue working together at their constant rates until the order is completed. At what time will the production order be completed?

11:40 AM

12:00 PM

1:00 PM

1:48 PM

2:12 PM

Cevabı ve açıklamayı göster

Cevap: 1:00 PM

Cevap

The production order will be completed at 1:00 PM.

The option specifying 1:00 PM is correct because during the first hour (9:00 AM to 10:00 AM), all three machines work together and complete

\frac{1}{6} + \frac{1}{8} + \frac{1}{12} = \frac{3}{8}

of the job. This leaves

\frac{5}{8}

of the job remaining at 10:00 AM. Machines B and C work together at a rate of

\frac{1}{8} + \frac{1}{12} = \frac{5}{24}

of the job per hour. Dividing the remaining work (

\frac{5}{8}

) by this combined rate (

\frac{5}{24}

) gives exactly 3 hours. Adding 3 hours to 10:00 AM yields 1:00 PM.

Adım Adım Çözüm

1

Calculate individual work rates for each machine.

Machine A rate =

\frac{1}{6}

order/hr, Machine B rate =

\frac{1}{8}

order/hr, Machine C rate =

\frac{1}{12}

order/hr.

Work rate is defined as job completed per unit of time.

2

Calculate the work completed by all three machines from 9:00 AM to 10:00 AM (1 hour).

Combined rate =

\frac{1}{6} + \frac{1}{8} + \frac{1}{12} = \frac{4 + 3 + 2}{24} = \frac{9}{24} = \frac{3}{8}

of the order.

All three machines work together for exactly 1 hour.

3

Determine the remaining fraction of the order after 10:00 AM.

Remaining work =

1 - \frac{3}{8} = \frac{5}{8}

of the order.

Subtract completed work from the total work (1 whole order).

4

Calculate the combined rate of Machines B and C.

Combined rate of B and C =

\frac{1}{8} + \frac{1}{12} = \frac{3 + 2}{24} = \frac{5}{24}

order/hr.

Machine A stops, leaving only B and C working.

5

Calculate the additional time needed to complete the remaining work.

Time =

\frac{5/8}{5/24} = \frac{5}{8} \times \frac{24}{5} = 3\text{ hours}

.

Divide remaining work by the combined rate of the remaining active machines.

6

Add the additional time to the breakdown time (10:00 AM).

Completion time = 10:00 AM + 3 hours = 1:00 PM.

The remaining work begins at 10:00 AM when Machine A breaks down.

Anahtar Kavram

Combined Work Rates and Staggered Work Times

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