Arithmetic Alıştırma Soruları — GRE General Test — Sayfa 15

Soru 281Soru →

If $m$ and $n$ are negative integers such that $m < n$ , which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

2^m < 2^n

\left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n

m^2 < n^2

\sqrt{m^2} = m

2^{m+n} = 2^m + 2^n

Cevabı ve açıklamayı göster

Cevap: $2^m < 2^n$ ; \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n

Cevap

The statements

2^m < 2^n

and \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n must be true.

For the statement involving base

2

, since

2 > 1

, the exponential function is strictly increasing, so

m < n

guarantees

2^m < 2^n

. For the statement involving base

\frac{1}{2}

, since

0 < \frac{1}{2} < 1

, the function is strictly decreasing, meaning a smaller input

m

produces a larger output, so \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n.

Adım Adım Çözüm

1

Analyze the expression

2^m < 2^n

for base greater than 1

Since

b = 2 > 1

, raising

2

to a larger exponent yields a larger value. Because

m < n

,

2^m < 2^n

is always true.

Exponential functions with a base

b > 1

are strictly increasing.

2

Analyze the expression \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n for fractional base between 0 and 1

Since

b = \frac{1}{2}

is between

0

and

1

, raising

\frac{1}{2}

to a smaller exponent yields a larger value. Because

m < n

, \left(\frac{1}{2}\right)^m > \left(\frac{1}{2}\right)^n is always true.

Exponential functions with a base

0 < b < 1

are strictly decreasing.

3

Evaluate the remaining algebraic statements using counterexamples

For

m = -3

and

n = -2

:

m^2 = 9 > 4 = n^2

, so

m^2 < n^2

is false.

\sqrt{m^2} = \sqrt{9} = 3 \neq -3

, so

\sqrt{m^2} = m

is false.

2^{-5} = \frac{1}{32} \neq \frac{1}{8} + \frac{1}{4} = \frac{3}{8}

, so

2^{m+n} = 2^m + 2^n

is false.

A single counterexample disproves that a statement MUST be true.

Anahtar Kavram

Monotonicity of exponential functions and properties of square roots of negative bases

Soru 282Soru →

Working alone at its constant rate, Printer $X$ can complete a printing job in $4\text{ hours}$ . Working alone at its constant rate, Printer $Y$ can complete the same printing job in $6\text{ hours}$ . Printer $X$ begins working on the job alone and works for $1\text{ hour}$ . At that point, Printer $Y$ joins Printer $X$ , and both printers work together at their respective constant rates until the job is completed. What is the total time, in hours, required to complete the entire job from start to finish?

1.8\text{ hours}

2.4\text{ hours}

2.8\text{ hours}

3.2\text{ hours}

3.5\text{ hours}

Cevabı ve açıklamayı göster

Cevap: $2.8\text{ hours}$

Cevap

2.8\text{ hours}

The correct answer is

2.8\text{ hours}

. Printer

X

works alone for

1\text{ hour}

at a rate of

\frac{1}{4}

job per hour, completing

\frac{1}{4}

of the total job. This leaves

\frac{3}{4}

of the job unfinished. When Printer

Y

joins, their combined rate is

\frac{1}{4} + \frac{1}{6} = \frac{5}{12}

job per hour. Dividing the remaining

\frac{3}{4}

of the job by

\frac{5}{12}

yields

\frac{3}{4} \times \frac{12}{5} = \frac{9}{5} = 1.8\text{ hours}

for the joint work phase. Adding the initial

1\text{ hour}

of solo work yields a total of

1 + 1.8 = 2.8\text{ hours}

.

Adım Adım Çözüm

1

Calculate individual work rates and the portion of the job completed in the first hour

Printer

X

's rate is

\frac{1}{4}

job/hour and Printer

Y

's rate is

\frac{1}{6}

job/hour. In the first hour, Printer

X

completes

1 \times \frac{1}{4} = \frac{1}{4}

of the job.

Printer

X

works alone for the first hour before Printer

Y

joins.

2

Determine the remaining fraction of the job

Remaining job =

1 - \frac{1}{4} = \frac{3}{4}

.

The entire job is represented by

1

, so subtracting the completed portion yields the remaining portion.

3

Calculate the combined rate of both printers working together

Combined rate =

\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}

job/hour.

Rates add when workers or machines work simultaneously.

4

Calculate the time required for both printers to finish the remaining job and find total time

Time together =

\frac{3/4}{5/12} = \frac{3}{4} \times \frac{12}{5} = \frac{9}{5} = 1.8\text{ hours}

. Total time =

1 + 1.8 = 2.8\text{ hours}

.

Time equals remaining work divided by combined rate, plus the

1\text{ hour}

already elapsed.

Anahtar Kavram

Combined Work Rates and Multi-Stage Work Problems

Alternatif Yöntem

Convert the job into arbitrary work units. Let the job equal

12\text{ units}

(the LCM of

4

and

6

). Printer

X

produces

12 / 4 = 3\text{ units/hour}

and Printer

Y

produces

12 / 6 = 2\text{ units/hour}

. In the first hour, Printer

X

produces

3\text{ units}

, leaving

12 - 3 = 9\text{ units}

. Working together, their combined rate is

3 + 2 = 5\text{ units/hour}

. The remaining

9\text{ units}

take

9 / 5 = 1.8\text{ hours}

. Total time is

1 + 1.8 = 2.8\text{ hours}

.

Tahmini Süre:1m 30s

Soru 283Soru →

If $x$ is a real number such that $\frac{4^{x+2} - 4^x}{15} = 2^{x+3}$ , what is the value of $x$ ?

1

2

3

5

6

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

Factoring out

4^x

from the numerator gives

4^x(4^2 - 1) = 15 \cdot 4^x

. Dividing by 15 simplifies the left-hand side to

4^x

. Expressing

4^x

as

2^{2x}

allows setting

2^{2x} = 2^{x+3}

. Equating the exponents

2x = x + 3

yields

x = 3

.

Adım Adım Çözüm

1

Factor the numerator of the left-hand side

4^{x+2} - 4^x = 4^x(4^2 - 1) = 4^x(16 - 1) = 15 \cdot 4^x

Factoring out the common exponential term

4^x

simplifies the subtraction.

2

Simplify the fraction on the left-hand side

\frac{15 \cdot 4^x}{15} = 4^x

The factor of 15 in the numerator cancels with 15 in the denominator.

3

Rewrite

4^x

in terms of base 2

4^x = (2^2)^x = 2^{2x}

Both sides must have a common base to equate their exponents.

4

Set the exponential expressions equal and solve for

x

2^{2x} = 2^{x+3} \implies 2x = x + 3 \implies x = 3

Since the bases are equal and positive (base 2), their exponents must be equal.

Anahtar Kavram

Factoring common exponential terms and equating exponents with identical bases

Tahmini Süre:1m 30s

Soru 284Soru →

A logistics facility utilizes two automated sorting lines, Line A and Line B, to process incoming shipments. Line A processes shipments at a constant rate that is 20 percent greater than the constant rate of Line B. When both lines operate simultaneously, they process a combined total of 3,300 shipments in 3 hours. Working alone at its constant rate, how many hours will it take Line B to process 2,250 shipments?

Cevabı ve açıklamayı göster

Cevap: 4.5

Cevap

4.5 hours

Let

r

represent the rate of Line B in shipments per hour. Because Line A processes 20% faster than Line B, Line A's rate is

1.20r

. Together, their combined processing rate is

r + 1.20r = 2.20r

shipments per hour. Operating for 3 hours, the total shipments processed is

3 \times 2.20r = 6.60r = 3,300

. Solving for

r

gives

r = 500

shipments per hour. To process 2,250 shipments alone, Line B requires

\frac{2,250}{500} = 4.5

hours.

Adım Adım Çözüm

1

Define the relationship between the individual processing rates.

If Line B processes at rate

r

shipments per hour, Line A processes at

1.20r

shipments per hour.

Line A's rate is 20 percent greater than Line B's rate.

2

Find the combined processing rate.

Combined rate =

r + 1.20r = 2.20r

shipments per hour.

When working together, individual rates add up.

3

Determine Line B's rate (

r

).

3 \times 2.20r = 3,300 \implies 6.60r = 3,300 \implies r = 500

shipments per hour.

Total work equals combined rate multiplied by time.

4

Calculate the time required for Line B to complete 2,250 shipments.

\text{Time} = \frac{2,250}{500} = 4.5

hours.

Time taken equals total work divided by the individual rate.

Anahtar Kavram

Combined Rates and Ratio Relationships

Soru 285Soru →

If $n$ is a real number such that $\frac{27^n + 27^n + 27^n}{3^{n+2}} = 243$ , what is the value of $n$ ?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

3

Rewriting

27^n + 27^n + 27^n

as

3 \cdot (3^3)^n = 3^{3n+1}

allows the left-hand side to simplify to

\frac{3^{3n+1}}{3^{n+2}} = 3^{2n-1}

. Equating this to

243 = 3^5

gives

2n - 1 = 5

, which solves to

n = 3

.

Adım Adım Çözüm

1

Express repeated addition in the numerator as multiplication.

27^n + 27^n + 27^n = 3 \cdot 27^n

Adding three identical quantities is equivalent to multiplying one quantity by 3.

2

Convert base 27 to base 3 and apply exponent multiplication.

3 \cdot (3^3)^n = 3^1 \cdot 3^{3n} = 3^{3n+1}

Since

27 = 3^3

, using the power rule

(a^b)^c = a^{bc}

and product rule

a^b \cdot a^c = a^{b+c}

converts the numerator to a single power of 3.

3

Simplify the fraction using the quotient rule of exponents.

\frac{3^{3n+1}}{3^{n+2}} = 3^{(3n+1) - (n+2)} = 3^{2n-1}

Dividing exponential terms with the same base requires subtracting the exponent in the denominator from the exponent in the numerator.

4

Rewrite 243 with base 3 and equate exponents across the equal sign.

3^{2n-1} = 3^5 \implies 2n - 1 = 5

Since

243 = 3^5

, two exponential expressions with the same base are equal if and only if their exponents are equal.

5

Solve the linear equation for

n

.

2n = 6 \implies n = 3

Adding 1 to both sides yields

2n = 6

, and dividing by 2 yields

n = 3

.

Anahtar Kavram

Combining repeated addition of exponential terms and converting expressions to a common base using exponent rules.

Soru 286Soru →

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 287Soru →

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 288Soru →

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 289Soru →

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 290Soru →

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 291Soru →

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 292Soru →

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 293Soru →

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 294Soru →

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 295Soru →

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 296Soru →

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 297Soru →

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

Soru 298Soru →

A liquid chemical solution is composed of Chemical A, Chemical B, and Water in the volume ratio $2 : 3 : 5$ , respectively. If $40\text{ liters}$ of Chemical A and $20\text{ liters}$ of Water are added to the mixture, the ratio of Chemical A to Water becomes $1 : 2$ . What was the total volume, in liters, of the original solution?

120

200

300

420

600

Cevabı ve açıklamayı göster

Cevap: $600$

Cevap

600\text{ liters}

Let the original volumes of Chemical A, Chemical B, and Water be

2x

,

3x

, and

5x

liters, respectively. The total volume of the original solution is

2x + 3x + 5x = 10x

liters. After adding

40\text{ liters}

of Chemical A and

20\text{ liters}

of Water, the ratio of Chemical A to Water is given as

1 : 2

. Setting up the proportion

\frac{2x + 40}{5x + 20} = \frac{1}{2}

and cross-multiplying yields

2(2x + 40) = 5x + 20

, which simplifies to

4x + 80 = 5x + 20

. Solving for

x

gives

x = 60

. Substituting

x = 60

into the total volume formula gives

10(60) = 600\text{ liters}

. Thus,

600

is the correct answer.

Adım Adım Çözüm

1

Represent the initial volumes of each component in terms of a common multiplier

x

.

Chemical A

= 2x

, Chemical B

= 3x

, Water

= 5x

. Total volume

= 2x + 3x + 5x = 10x

.

Ratios

2 : 3 : 5

mean the actual quantities are proportional to these ratio parts.

2

Set up an equation based on the new volumes of Chemical A and Water.

\frac{2x + 40}{5x + 20} = \frac{1}{2}

Adding

40\text{ L}

of Chemical A and

20\text{ L}

of Water changes their ratio to

1 : 2

.

3

Cross-multiply and solve for

x

.

2(2x + 40) = 1(5x + 20) \implies 4x + 80 = 5x + 20 \implies x = 60.

Solving the linear equation yields the common ratio unit value.

4

Calculate the total original volume using

x = 60

.

\text{Total Volume} = 10x = 10(60) = 600\text{ liters}.

The total solution consists of

10

parts in total.

Anahtar Kavram

Three-part ratio algebra and proportion adjustments

Tahmini Süre:1m 30s

Soru 299Soru →

An event production company initially allocates its total equipment budget among Stage Lights, Ambient LED Strips, and Spotlight Towers in the ratio $4 : 5 : 3$ , respectively. Due to venue price adjustments, the budget allocated to Stage Lights is increased by $25\%$ , while the budget allocated to Spotlight Towers is reduced by $20\%$ . If the total overall budget remains unchanged, what is the new ratio of the budget allocated to Stage Lights to Ambient LED Strips to Spotlight Towers?

25 : 23 : 12

25 : 25 : 12

25 : 17 : 12

35 : 22 : 3

5 : 4 : 3

Cevabı ve açıklamayı göster

Cevap: $25 : 23 : 12$

Cevap

The new ratio of the budget allocated to Stage Lights to Ambient LED Strips to Spotlight Towers is

25 : 23 : 12

.

By setting the initial allocations to

4x

,

5x

, and

3x

, the total budget is

12x

. The new allocation for Stage Lights becomes

4x \times 1.25 = 5x

, and for Spotlight Towers it becomes

3x \times 0.80 = 2.4x

. To keep the total budget at

12x

, the Ambient LED Strips allocation must be

12x - 5x - 2.4x = 4.6x

. The ratio

5 : 4.6 : 2.4

scales up to

50 : 46 : 24

, which simplifies to

25 : 23 : 12

.

Adım Adım Çözüm

1

Represent the initial component allocations and the total budget in terms of a common variable.

Let the initial allocations be Stage Lights

= 4x

, Ambient LED Strips

= 5x

, and Spotlight Towers

= 3x

. Total budget

= 4x + 5x + 3x = 12x

.

Establishing algebraic quantities allows precise tracking of individual changes relative to the constant total.

2

Calculate the updated allocations for Stage Lights and Spotlight Towers after their respective percentage changes.

New Stage Lights

= 4x \times (1 + 0.25) = 5x

.
New Spotlight Towers

= 3x \times (1 - 0.20) = 2.4x

.

Applying the specified percentage shifts directly to their corresponding original component values.

3

Determine the new allocation for Ambient LED Strips using the fixed total budget condition.

New Ambient LED Strips

= 12x - (5x + 2.4x) = 12x - 7.4x = 4.6x

.

Because the total overall budget remains unchanged at

12x

, the sum of all three new allocations must equal

12x

.

4

Write the new ratio and simplify to integer terms.

Ratio

= 5x : 4.6x : 2.4x = 5 : 4.6 : 2.4 = 50 : 46 : 24 = 25 : 23 : 12

.

Multiplying by

10

clears decimals, and dividing by

2

reduces the ratio to its simplest integer form.

Anahtar Kavram

Multi-part ratio adjustment with percentage changes under a constant total constraint

Tahmini Süre:1m 30s

Soru 300Soru →

A coffee roaster blends Arabica, Robusta, and Liberica beans in the ratio $4 : 3 : 2$ by weight to prepare a master batch. To adjust the flavor profile, the roaster adds $15\text{ kg}$ of Robusta beans and $15\text{ kg}$ of Liberica beans to the batch without changing the amount of Arabica beans. In the modified batch, Arabica beans constitute exactly $\frac{1}{3}$ of the total weight. Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

The initial weight of Arabica beans in the master batch was

40\text{ kg}

.

In the modified batch, the ratio of Robusta beans to Liberica beans is

9 : 7

.

In the initial master batch, Robusta beans constituted

\frac{1}{3}

of the total weight.

The total weight of the batch increased by

25\%

after the additions.

In the modified batch, Liberica beans account for

35\%

of the total weight.

Cevabı ve açıklamayı göster

Cevap: The initial weight of Arabica beans in the master batch was $40\text{ kg}$ .; In the modified batch, the ratio of Robusta beans to Liberica beans is $9 : 7$ .; In the initial master batch, Robusta beans constituted $\frac{1}{3}$ of the total weight.

Cevap

The correct statements are those asserting that the initial weight of Arabica beans was 40 kg, that the modified ratio of Robusta to Liberica beans is 9 : 7, and that Robusta beans constituted 1/3 of the total weight in the initial batch.

Solving the proportion equation reveals that the scaling factor for the initial ratio is 10, giving an initial batch weight of 90 kg with 40 kg Arabica, 30 kg Robusta, and 20 kg Liberica. Consequently, Arabica originally weighed 40 kg, Robusta originally made up 30/90 = 1/3 of the batch, and the new Robusta-to-Liberica ratio is 45 : 35 = 9 : 7. All three of these statements are mathematically true.

Adım Adım Çözüm

1

Represent initial component weights and total weight using a multiplier variable.

For ratio

4 : 3 : 2

, Arabica =

4x

, Robusta =

3x

, Liberica =

2x

, and initial total weight

W = 9x

.

Ratios define proportional components in terms of a common scalar variable.

2

Set up the equation for the modified batch based on the new total and Arabica ratio.

Added weight =

15 + 15 = 30\text{ kg}

. New total weight =

9x + 30

. Arabica weight remains

4x

. Since Arabica is

\frac{1}{3}

of the new total:

\frac{4x}{9x + 30} = \frac{1}{3} \implies 12x = 9x + 30 \implies 3x = 30 \implies x = 10

.

Equating the unchanged part to the given fraction of the new total allows solving for the scale factor.

3

Calculate all initial and modified quantities.

Initial total

W = 90\text{ kg}

. Initial Arabica =

40\text{ kg}

, Robusta =

30\text{ kg}

, Liberica =

20\text{ kg}

. Modified total =

120\text{ kg}

. Modified Robusta =

45\text{ kg}

, Modified Liberica =

35\text{ kg}

.

Determining exact values enables verification of each statement.

4

Evaluate each given statement.

Initial Arabica =

40\text{ kg}

(True). Modified Robusta : Liberica =

45 : 35 = 9 : 7

(True). Initial Robusta fraction =

\frac{30}{90} = \frac{1}{3}

(True). Percent weight increase =

\frac{30}{90} = 33.33\%

(False). Modified Liberica percentage =

\frac{35}{120} = 29.17\%

(False).

Direct comparison of calculated quantities with statement claims identifies all correct choices.

Anahtar Kavram

Ratio scale factors and part-to-whole proportions under quantity modifications

Tahmini Süre:2m 0s

Arithmetic

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