Algebra Alıştırma Soruları — GRE General Test — Sayfa 12

Soru 221Soru →

A parabola defined by the quadratic function $f(x) = ax^2 + bx + c$ , where $a, b,$ and $c$ are real constants with $a > 0$ , has its vertex at a minimum value of $-16$ . The distance between the two $x$ -intercepts of the parabola is $8$ . If $f(1) = -7$ and the $x$ -coordinate of the vertex is positive, what is the value of $f(-2)$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The value of

f(-2)

is 20.

By converting the parabola into vertex form

f(x) = a(x - h)^2 - 16

, the

x

-intercepts are found at

h \pm \frac{4}{\sqrt{a}}

. Equating their difference to

8

yields

a = 1

. Substituting

f(1) = -7

gives

(1 - h)^2 = 9

, which yields

h = 4

under the condition

h > 0

. Evaluating

f(-2) = (-2 - 4)^2 - 16

produces

20

.

Adım Adım Çözüm

1

Express the quadratic function in vertex form using the minimum value

f(x) = a(x - h)^2 - 16

, where

(h, -16)

is the vertex.

Since

a > 0

, the parabola opens upwards and its minimum value occurs at the vertex

y

-coordinate,

k = -16

.

2

Determine the leading coefficient

a

using the distance between

x

-intercepts

a = 1

Setting

f(x) = 0

yields

a(x - h)^2 - 16 = 0 \implies (x - h)^2 = \frac{16}{a} \implies x = h \pm \frac{4}{\sqrt{a}}

. The distance between roots is

\frac{8}{\sqrt{a}} = 8

, which gives

\sqrt{a} = 1 \implies a = 1

.

3

Determine the vertex

x

-coordinate

h

using the point

f(1) = -7

h = 4

Substituting

a = 1

and

x = 1

into the vertex form gives

(1 - h)^2 - 16 = -7 \implies (1 - h)^2 = 9

. Taking square roots gives

1 - h = 3 \implies h = -2

or

1 - h = -3 \implies h = 4

. Since

h > 0

, we select

h = 4

.

4

Evaluate

f(-2)

using the fully specified function

f(-2) = 20

With

f(x) = (x - 4)^2 - 16

, substituting

x = -2

yields

f(-2) = (-2 - 4)^2 - 16 = (-6)^2 - 16 = 36 - 16 = 20

.

Anahtar Kavram

Quadratic Vertex Form, Root Separation, and Evaluation

Soru 222Soru →

For all real numbers $x$ and $y$ , the custom binary operation $\odot$ is defined by $x \odot y = x + y - xy$ . Which of the following statements must be true for all real numbers $a$ , $b$ , and $c$ ? Select all such statements.

Geçerli olan tümünü seçin

a \odot b = b \odot a

a \odot 1 = 1

a \odot 0 = 0

(a \odot b) \odot c = a \odot (b \odot c)

a \odot a = a^2

Cevabı ve açıklamayı göster

Cevap: $a \odot b = b \odot a$ ; $a \odot 1 = 1$ ; $(a \odot b) \odot c = a \odot (b \odot c)$

Cevap

The statements

a \odot b = b \odot a

,

a \odot 1 = 1

, and

(a \odot b) \odot c = a \odot (b \odot c)

must be true for all real numbers

a

,

b

, and

c

.

The custom operation is commutative (

a \odot b = b \odot a

), maps any real number operated with

1

to

1

(

a \odot 1 = 1

), and is associative (

(a \odot b) \odot c = a \odot (b \odot c)

). These three properties hold universally for all real numbers.

Adım Adım Çözüm

1

Evaluate

a \odot b

and

b \odot a

to test commutativity.

a \odot b = a + b - ab

and

b \odot a = b + a - ba

. Since addition and multiplication of real numbers are commutative,

a + b - ab = b + a - ba

.

Verify if order of operands alters the result.

2

Evaluate

a \odot 1

using the operation definition.

a \odot 1 = a + 1 - a(1) = a + 1 - a = 1

.

Test the behavior of operating with the constant

1

.

3

Evaluate

a \odot 0

to verify the zero property statement.

a \odot 0 = a + 0 - a(0) = a

. This equals

a

, not

0

for general values of

a

.

Check if operating with

0

results in

0

.

4

Evaluate both sides of

(a \odot b) \odot c = a \odot (b \odot c)

to test associativity.

Left side:

(a \odot b) \odot c = (a + b - ab) \odot c = (a + b - ab) + c - (a + b - ab)c = a + b + c - ab - ac - bc + abc

. Right side:

a \odot (b \odot c) = a \odot (b + c - bc) = a + (b + c - bc) - a(b + c - bc) = a + b + c - bc - ab - ac + abc

. Both sides are identical.

Verify whether grouping alters the result.

5

Evaluate

a \odot a

.

a \odot a = a + a - a(a) = 2a - a^2

. This is not equal to

a^2

except when

2a - a^2 = a^2 \implies 2a^2 - 2a = 0 \implies a=0

or

a=1

.

Determine if the self-operation yields

a^2

for all real numbers.

Anahtar Kavram

Evaluating algebraic properties (commutativity, identity, associativity) of custom binary operations.

Soru 223Soru →

A rectangular plot of land has a perimeter of 56 meters and an area of 180 square meters. A border of uniform width $x$ meters is constructed around the inside edge of the plot, reducing the remaining inner area to 84 square meters. What is the value of $x$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The border width x is 2 meters.

Solving the system

L + W = 28

and

LW = 180

gives outer dimensions of 18 and 10 meters. The inner area equation

(18 - 2x)(10 - 2x) = 84

simplifies to

x^2 - 14x + 24 = 0

, which factors into

(x - 2)(x - 12) = 0

. Since

2x

must be less than the total width of 10 meters (

x < 5

), the extraneous root

x = 12

is discarded, leaving

x = 2

.

Adım Adım Çözüm

1

Determine the outer dimensions of the plot.

Length = 18 meters, Width = 10 meters

The semi-perimeter is L + W = 56 / 2 = 28, and the area is L * W = 180. Forming the quadratic equation t^2 - 28t + 180 = 0 yields roots t = 18 and t = 10.

2

Write the quadratic expression for the inner area in terms of x.

(18 - 2x)(10 - 2x) = 84

A uniform border of width x along all four sides reduces both the length and width by 2x.

3

Expand and simplify the quadratic equation into standard form.

x^2 - 14x + 24 = 0

Expanding yields 180 - 56x + 4x^2 = 84. Subtracting 84 gives 4x^2 - 56x + 96 = 0, which divides by 4 to give x^2 - 14x + 24 = 0.

4

Factor the quadratic equation to solve for possible values of x.

x = 2 or x = 12

Factoring x^2 - 14x + 24 = 0 gives (x - 2)(x - 12) = 0.

5

Eliminate the extraneous root based on physical geometry constraints.

x = 2

The total width is 10 meters, so 2x must be less than 10 (x < 5). Thus, x = 12 is physically impossible and discarded.

Anahtar Kavram

Quadratic Equations, Factoring, and Extraneous Solutions

Soru 224Soru →

In the $xy$ -plane, line $L$ passes through the points $(t, t+2)$ and $(3t-1, 2t+7)$ , where $t$ is a constant. Line $M$ is perpendicular to line $L$ and passes through the point $(4, -1)$ . If the $y$ -intercept of line $M$ is $11$ , what is the value of $t$ ?

-16

-\frac{14}{5}

\frac{8}{5}

-\frac{2}{7}

14

Cevabı ve açıklamayı göster

Cevap: $-16$

Cevap

-16

The slope of line M is computed from its

y

-intercept

(0, 11)

and the point

(4, -1)

as

\frac{11 - (-1)}{0 - 4} = -3

. Because line L is perpendicular to line M, the slope of line L is the negative reciprocal of

-3

, which is

\frac{1}{3}

. Calculating the slope of line L using points

(t, t+2)

and

(3t-1, 2t+7)

gives

\frac{(2t+7)-(t+2)}{(3t-1)-t} = \frac{t+5}{2t-1}

. Setting

\frac{t+5}{2t-1} = \frac{1}{3}

leads to

3t + 15 = 2t - 1

, which simplifies to

t = -16

.

Adım Adım Çözüm

1

Determine the slope of line M using its given points.

Line M passes through

(4, -1)

and its

y

-intercept

(0, 11)

. The slope

m_M = \frac{11 - (-1)}{0 - 4} = \frac{12}{-4} = -3

.

The slope of a line through

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Determine the required slope of line L.

Since line L is perpendicular to line M, its slope

m_L = -\frac{1}{m_M} = -\frac{1}{-3} = \frac{1}{3}

.

Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.

3

Express the slope of line L in terms of

t

and solve for

t

.

Using points

(t, t+2)

and

(3t-1, 2t+7)

,

m_L = \frac{(2t+7) - (t+2)}{(3t-1) - t} = \frac{t+5}{2t-1}

. Setting

\frac{t+5}{2t-1} = \frac{1}{3}

yields

3(t+5) = 1(2t-1) \implies 3t + 15 = 2t - 1 \implies t = -16

.

Equating the algebraic slope expression to the numerical slope allows solving for the unknown parameter

t

.

Anahtar Kavram

Perpendicular Slopes and Coordinate Line Equations

Tahmini Süre:2m 0s

Soru 225Soru →

Which of the following values of $x$ satisfy the inequality $|6 - 3x| > 9$ ? Select all that apply.

Geçerli olan tümünü seçin

-4

-2

1

3

6

Cevabı ve açıklamayı göster

Cevap: $-4$ ; $-2$ ; $6$

Cevap

The values

-4

,

-2

, and

6

satisfy the inequality

|6 - 3x| > 9

.

The absolute value inequality

|6 - 3x| > 9

is equivalent to

6 - 3x > 9

or

6 - 3x < -9

. Solving these yields

x < -1

or

x > 5

. Among the given choices,

-4

,

-2

, and

6

fall within these solution ranges.

Adım Adım Çözüm

1

Set up the compound linear inequalities from the absolute value inequality

|6 - 3x| > 9

.

6 - 3x > 9

or

6 - 3x < -9

An absolute value inequality of the form

|u| > c

(where

c > 0

) splits into

u > c

or

u < -c

.

2

Solve the first inequality

6 - 3x > 9

.

-3x > 3 \implies x < -1

Subtracting

6

gives

-3x > 3

. Dividing both sides by

-3

reverses the inequality sign to yield

x < -1

.

3

Solve the second inequality

6 - 3x < -9

.

-3x < -15 \implies x > 5

Subtracting

6

gives

-3x < -15

. Dividing both sides by

-3

reverses the inequality sign to yield

x > 5

.

4

Combine the solution sets and evaluate the given options.

The solution set consists of all real numbers where

x < -1

or

x > 5

. Therefore,

-4

,

-2

, and

6

are valid solutions.

Values

-4

and

-2

are strictly less than

-1

, while

6

is strictly greater than

5

.

Anahtar Kavram

Solving absolute value inequalities and reversing inequality signs when multiplying or dividing by negative numbers.

Soru 226Soru →

A logistics company uses two delivery vehicles, Vehicle P and Vehicle Q, to transport cargo between two warehouses that are $240$ miles apart. Vehicle P travels at a constant average speed of $r$ miles per hour ( $r > 0$ ). Vehicle Q travels at a constant average speed that is $20$ miles per hour faster than that of Vehicle P. In addition to driving time, Vehicle P requires a flat setup time of $1$ hour before departure, while Vehicle Q requires a flat setup time of $2$ hours before departure.

Let $T_P(r)$ and $T_Q(r)$ represent the total elapsed time in hours (including setup time) required for Vehicle P and Vehicle Q to complete the trip, respectively.

Which of the following statements are true? Select all such statements.

Geçerli olan tümünü seçin

The total elapsed time for Vehicle P, in hours, as a function of its speed

r

, is given by

T_P(r) = \frac{240 + r}{r}

.

Vehicle P and Vehicle Q take the exact same total elapsed time to complete the trip when

r = 60

miles per hour.

For all positive speeds

r > 0

, Vehicle Q always requires strictly less total elapsed time than Vehicle P to complete the trip.

If

r = 30

miles per hour, the travel time (excluding setup time) for Vehicle Q is half the travel time (excluding setup time) for Vehicle P.

Doubling the speed of Vehicle P from

r = 40

miles per hour to

r = 80

miles per hour reduces its total elapsed time

T_P(r)

by

50\%

.

Cevabı ve açıklamayı göster

Cevap: The total elapsed time for Vehicle P, in hours, as a function of its speed $r$ , is given by $T_P(r) = \frac{240 + r}{r}$ .; Vehicle P and Vehicle Q take the exact same total elapsed time to complete the trip when $r = 60$ miles per hour.

Cevap

The correct statements are the algebraic expression for Vehicle P's total time as a function of speed and the equal total elapsed time condition at a speed of 60 miles per hour.

The expression for Vehicle P's total time accurately combines

1

hour of setup time with

\frac{240}{r}

driving hours to get

\frac{240+r}{r}

. Setting

T_P(r) = T_Q(r)

yields the quadratic equation

r^2 + 20r - 4800 = 0

, which correctly solves to

r = 60

mph, at which point both vehicles require exactly

5

hours total.

Adım Adım Çözüm

1

Model total elapsed time functions

T_P(r)

and

T_Q(r)

using setup time plus travel time.

T_P(r) = 1 + \frac{240}{r} = \frac{240 + r}{r}

and

T_Q(r) = 2 + \frac{240}{r + 20}

.

Total elapsed time is the sum of fixed pre-departure setup overhead and variable driving time.

2

Determine the speed

r

where total elapsed times are equal by setting

T_P(r) = T_Q(r)

.

1 + \frac{240}{r} = 2 + \frac{240}{r + 20} \implies \frac{240}{r} - \frac{240}{r + 20} = 1 \implies 240 \cdot 20 = r(r + 20) \implies r^2 + 20r - 4800 = 0

. Factoring gives

(r - 60)(r + 80) = 0

, so

r = 60

mph.

Solving the rational equation identifies the exact breakeven speed where higher travel efficiency balances extra setup time.

3

Evaluate the remaining candidate assertions against the derived model.

For

r > 60

,

T_Q(r) > T_P(r)

, invalidating the claim that Vehicle Q is always faster. At

r = 40

,

T_P(40) = 7

hours and

T_P(80) = 4

hours, giving a decrease of

\frac{3}{7} \approx 42.86\%

, invalidating the

50\%

reduction claim.

Fixed setup costs distort simple constant-proportion and percentage changes in total time.

Anahtar Kavram

Algebraic Modeling of Combined Time with Fixed Overhead and Variable Rates

Soru 227Soru →

If $x$ is an integer that satisfies both $|2x - 1| \le 9$ and $\frac{x + 2}{-3} > -2$ , how many possible values of $x$ exist?

1

4

7

8

9

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

There are 8 possible integer values for x.

Solving

|2x - 1| \le 9

yields

-4 \le x \le 5

. Solving

\frac{x + 2}{-3} > -2

requires flipping the inequality sign when multiplying by

-3

, which gives

x < 4

. Combining both inequalities yields

-4 \le x < 4

. The integers satisfying this compound inequality are

-4, -3, -2, -1, 0, 1, 2, 3

, making a total of 8 possible integer values.

Adım Adım Çözüm

1

Solve the absolute value inequality

|2x - 1| \le 9

-9 \le 2x - 1 \le 9 \implies -8 \le 2x \le 10 \implies -4 \le x \le 5

An absolute value inequality of the form

|u| \le k

expands to

-k \le u \le k

.

2

Solve the linear inequality

\frac{x + 2}{-3} > -2

x + 2 < 6 \implies x < 4

Multiplying both sides of an inequality by a negative quantity (

-3

) requires reversing the inequality sign from

>

to

<

.

3

Find the intersection of the two solution sets

-4 \le x < 4

The integer

x

must satisfy both

-4 \le x \le 5

and

x < 4

simultaneously.

4

Count the integer values satisfying

-4 \le x < 4

The integers are

-4, -3, -2, -1, 0, 1, 2, 3

, which totals 8 values.

Counting all integers from

-4

up to (but not including)

4

gives 8 valid integers.

Anahtar Kavram

Solving systems of linear inequalities involving absolute values and negative multipliers

Tahmini Süre:1m 30s

Soru 228Soru →

A water purification facility uses a primary filtration system and a secondary filtration system to process untreated water. The primary system operates at a constant rate that is $25\%$ faster than the secondary system. Working together at their normal constant rates, both systems can process a full reservoir of $36,000$ gallons in $8$ hours.

On a day when the primary system operates at only $80\%$ of its normal rate due to maintenance while the secondary system operates at its normal rate, both systems work together for $6$ hours. At that point, the primary system is shut down completely. How many additional hours will it take the secondary system, working alone at its normal rate, to process the remainder of the reservoir?

4.5

hours

5.0

hours

6.0

hours

7.2

hours

8.0

hours

Cevabı ve açıklamayı göster

Cevap: $6.0$ hours

Cevap

6.0

hours

The correct answer of

6.0

hours is derived by establishing that the combined normal processing rate is

4,500

gal/hr. Since the primary system is

25\%

faster than the secondary system (

P = 1.25S

), the secondary rate is

2,000

gal/hr and the primary rate is

2,500

gal/hr. At

80\%

efficiency, the primary system operates at

2,000

gal/hr, making the joint rate

4,000

gal/hr. In

6

hours,

24,000

gallons are processed, leaving

12,000

gallons. The secondary system working alone at

2,000

gal/hr processes the remaining volume in exactly

6.0

hours.

Adım Adım Çözüm

1

Determine the combined normal operating rate and set up individual rates.

Combined rate =

4,500

gal/hr; Secondary rate =

2,000

gal/hr; Primary rate =

2,500

gal/hr.

The combined rate is

\frac{36,000\text{ gallons}}{8\text{ hours}} = 4,500\text{ gal/hr}

. Let

S

be the secondary rate. The primary rate is

1.25S

. Thus,

S + 1.25S = 2.25S = 4,500

, yielding

S = 2,000\text{ gal/hr}

and

P = 2,500\text{ gal/hr}

.

2

Calculate the reduced primary rate and the total water processed in the first 6 hours.

24,000

gallons processed in the first 6 hours.

During maintenance, the primary system operates at

80\%

of

2,500\text{ gal/hr}

, which is

0.80 \times 2,500 = 2,000\text{ gal/hr}

. The combined rate during this period is

2,000 + 2,000 = 4,000\text{ gal/hr}

. Over 6 hours, the volume processed is

4,000 \times 6 = 24,000\text{ gallons}

.

3

Find the remaining volume of water to process.

12,000

gallons remaining.

Subtract the volume processed from the total reservoir capacity:

36,000 - 24,000 = 12,000\text{ gallons}

.

4

Calculate the time required for the secondary system alone to process the remainder.

6.0

hours.

Divide the remaining volume by the normal secondary system rate

S = 2,000\text{ gal/hr}

:

\frac{12,000\text{ gallons}}{2,000\text{ gal/hr}} = 6.0\text{ hours}

.

Anahtar Kavram

Linear rate modeling and percentage rate adjustment in combined work problems

Tahmini Süre:2m 30s

Soru 229Soru →

Consider the quadratic equation $x^2 - 2(k + 1)x + (k^2 + 4k + 1) = 0$ , where $k$ is a real constant. Let $\alpha$ and $\beta$ be the real roots of this equation. If $\alpha^2 + \beta^2 = 18$ , what is the sum of all possible values of $k$ ?

-2\sqrt{2}

0

2\sqrt{2}

2 - 2\sqrt{3}

-4\sqrt{2}

Cevabı ve açıklamayı göster

Cevap: $-2\sqrt{2}$

Cevap

The sum of all possible real values of

k

is

-2\sqrt{2}

.

Using Vieta's formulas,

\alpha + \beta = 2(k + 1)

and

\alpha\beta = k^2 + 4k + 1

. Expanding

\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

gives

2k^2 + 2

. Setting

2k^2 + 2 = 18

yields

k^2 = 8

, so

k = \pm 2\sqrt{2}

. Furthermore, for the roots

\alpha

and

\beta

to be real numbers, the discriminant

\Delta = [-2(k+1)]^2 - 4(k^2+4k+1) = -8k

must be non-negative, requiring

k \le 0

. Only

k = -2\sqrt{2}

satisfies this requirement. Hence, the sum of all valid values of

k

is

-2\sqrt{2}

.

Adım Adım Çözüm

1

Apply Vieta's formulas to express the sum and product of roots in terms of

k

.

\alpha + \beta = 2(k + 1)

and

\alpha\beta = k^2 + 4k + 1

.

For any quadratic equation

x^2 - bx + c = 0

, the sum of roots is

b

and the product of roots is

c

.

2

Express

\alpha^2 + \beta^2

in terms of

k

using the algebraic identity

(\alpha + \beta)^2 - 2\alpha\beta

.

\alpha^2 + \beta^2 = [2(k + 1)]^2 - 2(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 2k^2 - 8k - 2 = 2k^2 + 2

.

Expanding and simplifying allows us to relate the given sum of squares to the parameter

k

.

3

Set the simplified expression equal to

18

and solve for

k

.

2k^2 + 2 = 18 \implies 2k^2 = 16 \implies k^2 = 8 \implies k = \pm 2\sqrt{2}

.

This yields two candidate real values for the parameter

k

.

4

Analyze the discriminant

\Delta

to enforce the condition that roots

\alpha

and

\beta

are real numbers.

\Delta = [-2(k + 1)]^2 - 4(1)(k^2 + 4k + 1) = 4(k^2 + 2k + 1) - 4(k^2 + 4k + 1) = -8k

. For real roots,

\Delta \ge 0 \implies -8k \ge 0 \implies k \le 0

.

A quadratic equation has real roots if and only if its discriminant is non-negative.

5

Filter candidate values of

k

against the condition

k \le 0

.

k = -2\sqrt{2}

satisfies

k \le 0

(yielding

\Delta = 16\sqrt{2} > 0

), while

k = 2\sqrt{2}

fails

k \le 0

(yielding

\Delta = -16\sqrt{2} < 0

). Thus,

k = -2\sqrt{2}

is the unique valid value.

Values of

k

that make the discriminant negative produce non-real complex roots and must be discarded as extraneous.

Anahtar Kavram

Quadratic Vieta's formulas combined with discriminant non-negativity constraint for real roots

Tahmini Süre:2m 30s

Soru 230Soru →

A theater sold a total of $500$ tickets for an evening performance, consisting of VIP tickets priced at $\$80$ each and General Admission tickets priced at $\$50$ each. On the day of the show, a promotional discount of $20\%$ was applied to all General Admission tickets, while VIP ticket prices remained unchanged. If the total revenue collected from ticket sales was $\$31,000$ , which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

More than 70% of the total revenue was generated from VIP ticket sales.

The number of VIP tickets sold was 50 greater than the number of General Admission tickets sold.

The ratio of VIP tickets sold to General Admission tickets sold was 11 to 7.

Applying the 20% discount to General Admission tickets reduced the total potential revenue by 20%.

The number of General Admission tickets sold was 250.

Cevabı ve açıklamayı göster

Cevap: More than 70% of the total revenue was generated from VIP ticket sales.; The number of VIP tickets sold was 50 greater than the number of General Admission tickets sold.

Cevap

The statements confirming that more than 70% of total revenue came from VIP ticket sales and that 50 more VIP tickets were sold than General Admission tickets are correct.

The system of equations

V + G = 500

and

80V + 40G = 31,000

yields

V = 275

VIP tickets and

G = 225

General Admission tickets. The VIP revenue is

22,000, which is approximately 70.97% of the total

31,000 revenue (greater than 70%). Additionally, the difference

275 - 225 = 50

confirms that 50 more VIP tickets were sold than General Admission tickets.

Adım Adım Çözüm

1

Define variables and determine the discounted price of General Admission tickets.

Let

V

be the number of VIP tickets and

G

be the number of General Admission tickets. The discounted price for General Admission tickets is

\$50 \times (1 - 0.20) = \$40

.

Establishing correct variable representations and effective unit prices is required to build the revenue model.

2

Set up and solve the system of linear equations.

V + G = 500

and

80V + 40G = 31,000

. Substituting

G = 500 - V

gives

80V + 40(500 - V) = 31,000 \implies 40V = 11,000 \implies V = 275

. Thus,

G = 225

.

Solving the linear system determines the exact quantity of each ticket type sold.

3

Calculate revenue shares and verify each statement.

VIP revenue =

275 \times \$80 = \$22,000

. Revenue share of VIP =

\frac{22,000}{31,000} \approx 70.97\% > 70\%

. Ticket difference =

275 - 225 = 50

. Ratio

V:G = 275:225 = 11:9

.

Evaluating each calculated metric against the given statements determines which statements must be true.

Anahtar Kavram

Linear word problems involving systems of equations and percentage modifications.

Tahmini Süre:2m 0s

Soru 231Soru →

A manufacturing plant operates two automated production lines, Assembly Line X and Assembly Line Y. Assembly Line Y operates at a standard rate that is $25\%$ faster than the standard rate of Assembly Line X. During a specific shift, Assembly Line X operated for $4$ hours at its standard rate, after which its processing rate decreased by $20\%$ for an additional $2$ hours due to maintenance. Assembly Line Y began operating $1$ hour after Line X started; it operated at its standard rate for $4$ hours, and then operated for another $2$ hours at $80\%$ of its standard rate. If the two lines produced a combined total of $2,772$ units during this shift, what was the standard operating rate of Assembly Line X, in units per hour?

Cevabı ve açıklamayı göster

Cevap: 220

Cevap

The standard operating rate of Assembly Line X is 220 units per hour.

Assembly Line X produces

4r

units in its first 4 hours and

1.6r

units in its final 2 hours, totaling

5.6r

units. Assembly Line Y produces

4 \times 1.25r = 5r

units in its first 4 hours and

2 \times (0.80 \times 1.25r) = 2r

units in its final 2 hours, totaling

7r

units. The sum of their outputs is

5.6r + 7r = 12.6r = 2,772

. Dividing

2,772

by

12.6

gives

r = 220

.

Adım Adım Çözüm

1

Set up rate expressions for both assembly lines using a single variable

Standard rate of Line X =

r

; Standard rate of Line Y =

1.25r

Line Y is 25% faster than Line X, so its rate is

r + 0.25r = 1.25r

.

2

Calculate the total work done by Assembly Line X

Line X output =

4(r) + 2(0.80r) = 5.6r

units

Line X worked 4 hours at 100% rate and 2 hours at 80% rate.

3

Calculate the total work done by Assembly Line Y

Line Y output =

4(1.25r) + 2(0.80 \times 1.25r) = 5r + 2r = 7r

units

Line Y worked 4 hours at full rate

1.25r

and 2 hours at 80% of

1.25r

, which equals rate

r

.

4

Equate combined production to 2,772 units and solve for

r

5.6r + 7r = 12.6r = 2,772 \implies r = 220

Dividing the total combined output by the total rate multiplier 12.6 yields the baseline standard rate.

Anahtar Kavram

Algebraic Modeling of Staggered Work and Variable Production Rates

Tahmini Süre:2m 30s

Soru 232Soru →

For all positive real numbers $x$ and $y$ , the custom operation $\star$ is defined by $x \star y = \frac{x^2 + y^2}{xy}$ . The function $g$ is defined by $g(t) = 4(t \star 3) - 7$ . If $g(k) = 3$ , where $k > 3$ , what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of

k

is 6.

Solving

g(k) = 3

yields

4(k \star 3) - 7 = 3

, so

k \star 3 = \frac{5}{2}

. Replacing

k \star 3

with

\frac{k^2+9}{3k}

results in the quadratic equation

2k^2 - 15k + 18 = 0

. The roots are

1.5

and

6

. Because

k

must be strictly greater than

3

, the only valid solution is

6

.

Adım Adım Çözüm

1

Isolate the custom operation term in the function equation

k \star 3 = \frac{5}{2}

Setting

g(k) = 4(k \star 3) - 7 = 3

allows us to find the required value of

k \star 3

by adding 7 to both sides and dividing by 4.

2

Express the custom operation algebraically using its definition

\frac{k^2 + 9}{3k} = \frac{5}{2}

By definition,

x \star y = \frac{x^2 + y^2}{xy}

, so

k \star 3 = \frac{k^2 + 3^2}{3(k)} = \frac{k^2 + 9}{3k}

.

3

Form and solve the quadratic equation

k = 1.5 \text{ or } k = 6

Cross-multiplying gives

2(k^2 + 9) = 5(3k)

, which simplifies to

2k^2 - 15k + 18 = 0

. Factoring as

(2k - 3)(k - 6) = 0

yields two roots:

k = 1.5

and

k = 6

.

4

Apply the domain constraint

k = 6

The problem specifies that

k > 3

. Since

1.5 \le 3

,

k = 1.5

is rejected, leaving

k = 6

as the unique correct answer.

Anahtar Kavram

Custom Binary Symbol Operations and Composite Functions

Soru 233Soru →

An agricultural facility uses two automated irrigation systems, System P and System Q, to water a field. System P pumps water at a constant rate of $120$ gallons per hour after requiring an initial setup overhead of $30$ minutes ( $0.5$ hours). System Q pumps water at a constant rate of $180$ gallons per hour after requiring an initial setup overhead of $45$ minutes ( $0.75$ hours). Both systems begin their setup process at the exact same time and operate continuously until a combined total of $1,605$ gallons of water has been pumped. Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

The total elapsed time from when setup began until the target volume was reached is

6

hours.

System Q pumped

285

more gallons of water than System P.

System P was actively pumping water for exactly

5

hours and

15

minutes.

System Q accounted for more than

60\%

of the total water pumped.

The ratio of the volume of water pumped by System P to that pumped by System Q is

44:63

.

Cevabı ve açıklamayı göster

Cevap: The total elapsed time from when setup began until the target volume was reached is $6$ hours.; System Q pumped $285$ more gallons of water than System P.; The ratio of the volume of water pumped by System P to that pumped by System Q is $44:63$ .

Cevap

The correct statements are: the total elapsed time from when setup began is 6 hours, System Q pumped 285 more gallons than System P, and the ratio of water pumped by System P to System Q is 44:63.

Solving the combined linear rate model yields a total elapsed time of 6 hours. With this duration, System P operates for 5.5 hours producing 660 gallons, and System Q operates for 5.25 hours producing 945 gallons. This confirms that the total time is 6 hours, System Q produces 285 more gallons than System P (945 - 660 = 285), and the ratio of System P's volume to System Q's volume is 660:945, which simplifies to 44:63.

Adım Adım Çözüm

1

Define the variable for total time and establish time expressions for active pumping.

Let

t

be the total elapsed time in hours since both systems began setup (

t \geq 0.75

). System P actively pumps for

(t - 0.5)

hours, and System Q actively pumps for

(t - 0.75)

hours.

Setup overhead delays the start of pumping, so active pumping duration equals total elapsed time minus setup time.

2

Set up and solve the linear rate equation for combined total volume.

120(t - 0.5) + 180(t - 0.75) = 1605 \implies 120t - 60 + 180t - 135 = 1605 \implies 300t - 195 = 1605 \implies 300t = 1800 \implies t = 6

hours.

The sum of the volumes produced by both systems must equal the target total of 1,605 gallons.

3

Calculate individual pumping times and volumes produced by each system.

System P: active time

= 5.5

hours, volume

= 120 \times 5.5 = 660

gallons. System Q: active time

= 5.25

hours, volume

= 180 \times 5.25 = 945

gallons.

Individual volumes are required to evaluate statements regarding volume differences, percentages, and ratios.

4

Evaluate each given statement against the calculated values.

Elapsed time is

6

hours (True). Volume difference is

945 - 660 = 285

gallons (True). Active pumping time for P is

5.5

hours, not

5.25

hours (False). Percentage for Q is

945 / 1605 \approx 58.88\% < 60\%

(False). Ratio P to Q is

660 : 945 = 44 : 63

(True).

Determines which of the statements must be selected.

Anahtar Kavram

Linear Modeling with Combined Work Rates and Staggered Start Times

Soru 234Soru →

In the $xy$ -plane, line $L_1$ is given by the equation $ax + 3y = 12$ , where $a$ is a constant. Line $L_2$ is perpendicular to line $L_1$ and passes through the point $(4, -1)$ . If the $x$ -intercept of line $L_2$ is double its $y$ -intercept, and both intercepts of line $L_2$ are non-zero, what is the value of $a$ ?

-6

-\frac{3}{2}

\frac{3}{2}

3

6

Cevabı ve açıklamayı göster

Cevap: $-6$

Cevap

-6

Rearranging

L_1: ax + 3y = 12

into slope-intercept form gives

y = -\frac{a}{3}x + 4

, yielding a slope of

m_1 = -\frac{a}{3}

. The perpendicular line

L_2

has a slope of

m_2 = \frac{3}{a}

. Using the point

(4, -1)

, the equation for

L_2

is

y + 1 = \frac{3}{a}(x - 4)

, which simplifies to

y = \frac{3}{a}x - \frac{12+a}{a}

. Setting

x = 0

gives the

y

-intercept

y_0 = -\frac{12+a}{a}

, and setting

y = 0

gives the

x

-intercept

x_0 = \frac{12+a}{3}

. Equating

x_0 = 2 y_0

yields

\frac{12+a}{3} = -2\frac{12+a}{a}

. Canceling the non-zero term

(12+a)

gives

\frac{1}{3} = -\frac{2}{a}

, which leads to

a = -6

.

Adım Adım Çözüm

1

Determine the slope of line

L_1

and line

L_2

Slope of

L_1

is

m_1 = -\frac{a}{3}

, so the slope of perpendicular line

L_2

is

m_2 = \frac{3}{a}

.

Perpendicular lines have slopes that are negative reciprocals of each other.

2

Write the equation of line

L_2

using point-slope form

y - (-1) = \frac{3}{a}(x - 4) \implies y = \frac{3}{a}x - \frac{12}{a} - 1 = \frac{3}{a}x - \left(\frac{12 + a}{a}\right)

.

Line

L_2

passes through

(4, -1)

with slope

m_2 = \frac{3}{a}

.

3

Find the

x

-intercept and

y

-intercept of line

L_2

The

y

-intercept is

y_0 = -\frac{12 + a}{a}

. Setting

y = 0

gives

0 = \frac{3}{a}x - \frac{12 + a}{a} \implies x_0 = \frac{12 + a}{3}

.

Intercepts occur where

x = 0

(for

y

-intercept) and

y = 0

(for

x

-intercept).

4

Set up and solve the given relation

x_0 = 2 y_0

\frac{12 + a}{3} = 2 \left( -\frac{12 + a}{a} \right)

. Since the intercepts are non-zero,

12 + a \neq 0

. Dividing both sides by

(12 + a)

gives

\frac{1}{3} = -\frac{2}{a} \implies a = -6

.

The problem states the

x

-intercept is double the

y

-intercept.

Anahtar Kavram

Perpendicular Slopes and Line Intercepts

Tahmini Süre:2m 0s

Soru 235Soru →

Let $f(x) = x^2 + bx + c$ , where $b$ and $c$ are integers. The equation $f(x) = 0$ has two distinct real roots, $\alpha$ and $\beta$ . If $\alpha + \beta = \alpha\beta$ and $c > 0$ , which of the following statements must be true? Indicate all such statements.

Geçerli olan tümünü seçin

The constant

c

is strictly greater than 4.

Both roots

\alpha

and

\beta

are positive.

The vertex of the parabola

y = f(x)

lies in Quadrant IV of the xy-plane.

The coefficient

b

is positive.

The difference between the two roots,

|\alpha - \beta|

, can equal 2.

Cevabı ve açıklamayı göster

Cevap: The constant $c$ is strictly greater than 4.; Both roots $\alpha$ and $\beta$ are positive.; The vertex of the parabola $y = f(x)$ lies in Quadrant IV of the xy-plane.

Cevap

The statements asserting that the constant

c

is strictly greater than 4, both roots are positive, and the vertex lies in Quadrant IV must all be true.

By Vieta's formulas,

\alpha + \beta = -b

and

\alpha\beta = c

. Equating sum and product gives

b = -c

. The discriminant

\Delta = c^2 - 4c > 0

requires

c > 4

because

c > 0

. Since sum and product of the roots equal

c > 0

, both roots are positive. The vertex coordinates

(c/2, -c(c-4)/4)

have a positive x-value and a negative y-value, placing the vertex in Quadrant IV.

Adım Adım Çözüm

1

Apply Vieta's formulas and establish the relationship between coefficients

\alpha + \beta = -b

and

\alpha\beta = c

. Setting them equal gives

-b = c \implies b = -c

.

The stem specifies that the sum of the roots equals their product.

2

Analyze the discriminant for distinct real roots

\Delta = b^2 - 4c = (-c)^2 - 4c = c(c - 4) > 0

. Since

c > 0

,

c - 4 > 0 \implies c > 4

.

Two distinct real roots require a strictly positive discriminant.

3

Determine the signs of the roots

\alpha + \beta = c > 0

and

\alpha\beta = c > 0

, implying

\alpha > 0

and

\beta > 0

.

If the sum and product of two real numbers are both positive, both numbers must be positive.

4

Find the location of the parabola's vertex

x_v = \frac{c}{2} > 0

and

y_v = -\frac{c(c-4)}{4} < 0

, placing the vertex in Quadrant IV.

A point with a positive x-coordinate and negative y-coordinate resides in the fourth quadrant.

5

Evaluate the remaining options regarding

b

and

|\alpha - \beta|

b = -c < -4

(negative), and

|\alpha - \beta| = 2 \implies c = 2 + 2\sqrt{2}

, which is not an integer.

These evaluations disprove the statements that

b

is positive and that

|\alpha - \beta|

can equal 2.

Anahtar Kavram

Quadratic Root Properties and Vieta's Formulas

Soru 236Soru →

A laboratory processes two types of chemical samples, Type A and Type B. Processing each Type A sample requires $40$ minutes and costs $\$30$ , while processing each Type B sample requires $30$ minutes and costs $\$50$ . On a given day, the laboratory spent a total of $20$ hours processing these two types of samples at a total cost of $\$1,450$ . How many more Type B samples were processed than Type A samples?

5

10

15

20

35

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

5 more Type B samples were processed than Type A samples.

The correct answer is 5. By defining

x

as the number of Type A samples and

y

as the number of Type B samples, we construct the time equation

40x + 30y = 1,200

and cost equation

30x + 50y = 1,450

. Solving this system yields

x = 15

and

y = 20

. The difference

y - x = 20 - 15 = 5

.

Adım Adım Çözüm

1

Define variables and convert units to maintain consistency.

Let

x

be the number of Type A samples and

y

be the number of Type B samples. Total time available is

20 \text{ hours} \times 60 \text{ minutes/hour} = 1,200 \text{ minutes}

.

Time specifications for individual samples are given in minutes, so total time must also be in minutes.

2

Set up a system of two linear equations representing total time and total cost.

Time equation:

40x + 30y = 1,200 \implies 4x + 3y = 120

. Cost equation:

30x + 50y = 1,450 \implies 3x + 5y = 145

.

The total processing time is the sum of time spent on each sample type, and total cost is the sum of costs for each sample type.

3

Solve the system of equations using elimination.

Multiply the time equation by 3:

12x + 9y = 360

. Multiply the cost equation by 4:

12x + 20y = 580

. Subtracting the first from the second gives

11y = 220 \implies y = 20

. Substitute

y = 20

into

4x + 3(20) = 120 \implies 4x = 60 \implies x = 15

.

Eliminating

x

allows direct calculation of

y

, which then yields

x

.

4

Calculate the difference requested by the problem.

Difference =

y - x = 20 - 15 = 5

.

The question specifically asks for how many more Type B samples were processed than Type A samples.

Anahtar Kavram

Formulating and solving systems of linear equations from real-world rate and budget constraints.

Tahmini Süre:1m 45s

Soru 237Soru →

For all real numbers $x$ , the function $g$ is defined by $g(x) = cx + 5$ , where $c$ is a constant. The custom operation $\diamond$ is defined for all real numbers $a$ and $b$ by $a \diamond b = g(a+b) - g(a-b)$ . If $3 \diamond 1 = 12$ , what is the value of $g(4)$ ?

Cevabı ve açıklamayı göster

Cevap: 29

Cevap

29

Applying the function definition

g(x) = cx + 5

to the custom operation yields

a \diamond b = [c(a+b)+5] - [c(a-b)+5] = 2bc

. Substituting

a=3

and

b=1

into

3 \diamond 1 = 12

gives

2(1)c = 12

, so

c = 6

. Consequently,

g(x) = 6x + 5

, and evaluating at

x = 4

yields

g(4) = 6(4) + 5 = 29

.

Adım Adım Çözüm

1

Substitute

(a+b)

and

(a-b)

into the function definition

g(x) = cx + 5

to simplify

a \diamond b

.

a \diamond b = [c(a+b) + 5] - [c(a-b) + 5] = 2bc

.

Applying the definition of the custom binary operation in terms of function

g

eliminates the constant term

5

.

2

Use the given equality

3 \diamond 1 = 12

to determine the constant

c

.

2(1)c = 12 \implies 2c = 12 \implies c = 6

.

Plugging

a=3

and

b=1

into

2bc = 12

yields an equation in terms of

c

.

3

Evaluate

g(4)

using

c = 6

.

g(4) = 6(4) + 5 = 29

.

Substituting

x = 4

into

g(x) = 6x + 5

calculates the required numerical value.

Anahtar Kavram

Functions and Custom Symbol Operations

Tahmini Süre:1m 30s

Soru 238Soru →

In the $xy$ -plane, line $m$ has an $x$ -intercept of $-6$ and a $y$ -intercept of $3$ . Line $k$ is perpendicular to line $m$ and passes through the point $(4, 9)$ . What is the $x$ -intercept of line $k$ ?

Cevabı ve açıklamayı göster

Cevap: 8.5

Cevap

8.5

First, determine the slope of line

m

using the given intercepts

(-6,0)

and

(0,3)

, yielding

\frac{3-0}{0-(-6)} = \frac{1}{2}

. Since line

k

is perpendicular to line

m

, its slope is the negative reciprocal of

\frac{1}{2}

, which is

-2

. Using the point-slope equation with point

(4,9)

, the equation of line

k

is

y - 9 = -2(x - 4)

, simplifying to

y = -2x + 17

. Finding the

x

-intercept by setting

y = 0

yields

0 = -2x + 17

, giving

x = 8.5

.

Adım Adım Çözüm

1

Calculate the slope of line

m

Slope of line

m

is

\frac{1}{2}

Line

m

passes through the points

(-6,0)

and

(0,3)

.

2

Determine the slope of perpendicular line

k

Slope of line

k

is

-2

Perpendicular lines have negative reciprocal slopes.

3

Derive the equation of line

k

y = -2x + 17

Use point-slope form with given point

(4,9)

and slope

-2

.

4

Solve for the

x

-intercept of line

k

x = 8.5

Set

y = 0

in the linear equation

y = -2x + 17

.

Anahtar Kavram

Perpendicular lines, slope calculation from intercepts, and line equations

Soru 239Soru →

A commercial bakery uses two automated ovens, Oven X and Oven Y, to bake identical orders of bread. Working alone at its constant rate, Oven X bakes a full order of bread in $8$ hours, while Oven Y, working alone at its constant rate, bakes the same order in $12$ hours. Both ovens begin baking a full order together at 8:00 a.m. At 10:00 a.m., Oven X shuts down due to a maintenance alert, and Oven Y continues working alone at its constant rate until the order is completed. How many total hours, from 8:00 a.m. until completion, does it take to finish the order?

Cevabı ve açıklamayı göster

Cevap: 9

Cevap

The total time required from 8:00 a.m. to complete the order is

9

hours.

Working together for

2

hours at a combined rate of

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

per hour completes

\frac{5}{12}

of the order. The remaining

\frac{7}{12}

of the order takes Oven Y

7

hours to complete at its rate of

\frac{1}{12}

per hour. Adding the initial

2

hours gives a total time of

9

hours.

Adım Adım Çözüm

1

Determine the individual hourly work rates.

Oven X completes

\frac{1}{8}

of the job per hour, and Oven Y completes

\frac{1}{12}

of the job per hour.

Work rate is the reciprocal of the total time required to complete one full job.

2

Calculate the fraction of the job completed in the first

2

hours.

Combined rate is

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

job per hour. In

2

hours, they complete

2 \times \frac{5}{24} = \frac{5}{12}

of the job.

Both ovens work simultaneously for

2

hours before Oven X stops.

3

Determine the remaining fraction of the job.

1 - \frac{5}{12} = \frac{7}{12}

of the job remains.

The full order represents

1

whole unit of work.

4

Find the additional time required for Oven Y to finish the remaining job alone.

\text{Time} = \frac{7/12}{1/12} = 7

hours.

Time equals remaining work divided by Oven Y's individual work rate.

5

Calculate the total time elapsed from start to completion.

2 + 7 = 9

hours.

The total time includes the

2

hours of combined work plus the

7

hours Oven Y worked alone.

Anahtar Kavram

Combined Work Rates and Modeling Staggered Work

Soru 240Soru →

A technology company manages data storage across three servers: Server X, Server Y, and Server Z. Initially, the amount of data stored on Server Y is $25\%$ greater than the amount stored on Server X, and Server Z stores $40$ gigabytes less data than Server Y. During a system reorganization, the data on Server X increases by $20\%$ , the data on Server Y decreases by $20\%$ , and the data on Server Z increases by $60$ gigabytes. If the total amount of data stored across all three servers after the reorganization is $10\%$ greater than the total initial amount of data, what was the initial amount of data, in gigabytes, stored on Server X?

120

150

160

192

200

Cevabı ve açıklamayı göster

Cevap: 160

Cevap

160 gigabytes

Let

x

be the initial storage on Server X in gigabytes. Then Server Y initially stores

1.25x

gigabytes and Server Z stores

1.25x - 40

gigabytes. The initial total storage is

T_1 = x + 1.25x + (1.25x - 40) = 3.5x - 40

. After reorganization, Server X stores

1.20x

, Server Y stores

0.80(1.25x) = 1.00x

, and Server Z stores

(1.25x - 40) + 60 = 1.25x + 20

. The new total storage is

T_2 = 1.20x + 1.00x + 1.25x + 20 = 3.45x + 20

. Since

T_2

is

10\%

greater than

T_1

, we have

3.45x + 20 = 1.10(3.5x - 40)

. Expanding the right side gives

3.45x + 20 = 3.85x - 44

. Rearranging terms yields

0.40x = 64

, which simplifies to

x = 160

gigabytes.

Adım Adım Çözüm

1

Define variables for the initial storage on each server in terms of the initial storage on Server X,

x

.

Server X =

x

, Server Y =

1.25x

, Server Z =

1.25x - 40

.

Server Y is 25% greater than Server X (

1 + 0.25 = 1.25

), and Server Z is 40 gigabytes less than Server Y.

2

Sum the initial storage values to find the total initial storage

T_1

.

T_1 = x + 1.25x + (1.25x - 40) = 3.5x - 40

.

Combining like terms gives the overall starting storage equation.

3

Express the storage on each server after the reorganization in terms of

x

.

Server X' =

1.20x

, Server Y' =

0.80(1.25x) = 1.00x

, Server Z' =

(1.25x - 40) + 60 = 1.25x + 20

.

Server X increases by 20%, Server Y decreases by 20% of its initial value, and Server Z gains 60 gigabytes.

4

Sum the updated storage values to find the new total storage

T_2

.

T_2 = 1.20x + 1.00x + (1.25x + 20) = 3.45x + 20

.

Adding the three updated amounts gives the new total expression.

5

Set up and solve the equation representing the

10\%

overall increase (

T_2 = 1.10 T_1

).

3.45x + 20 = 1.10(3.5x - 40) \implies 3.45x + 20 = 3.85x - 44 \implies 64 = 0.40x \implies x = 160

.

Solving the linear algebraic model yields the value for initial storage on Server X.

Anahtar Kavram

Linear algebraic modeling of dynamic multi-variable systems involving percentage changes.

Tahmini Süre:2m 30s

Algebra

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