Arithmetic Practice Questions — GRE General Test — Page 14

If $x$ is a positive real number such that $x^{\frac{3}{4}} = 27$ , what is the value of $x^{-\frac{1}{2}}$ ?

\frac{1}{9}

9

-\frac{1}{9}

\frac{1}{3}

\frac{1}{27}

Show answer & explanation

Answer: $\frac{1}{9}$

Answer

\frac{1}{9}

Raising both sides of

x^{\frac{3}{4}} = 27

to the power of

\frac{4}{3}

gives

x = (3^3)^{\frac{4}{3}} = 3^4 = 81

. Substituting

x = 81

into

x^{-\frac{1}{2}}

gives

81^{-\frac{1}{2}} = \frac{1}{\sqrt{81}} = \frac{1}{9}

.

Step-by-Step Solution

1

Solve for

x

in the equation

x^{\frac{3}{4}} = 27

.

x = 27^{\frac{4}{3}} = (3^3)^{\frac{4}{3}} = 3^4 = 81

Raise both sides to the power of

\frac{4}{3}

to isolate

x

.

2

Evaluate

x^{-\frac{1}{2}}

for

x = 81

.

81^{-\frac{1}{2}} = \frac{1}{81^{\frac{1}{2}}} = \frac{1}{\sqrt{81}} = \frac{1}{9}

Apply the negative exponent rule

a^{-n} = \frac{1}{a^n}

and the fractional exponent rule

a^{\frac{1}{2}} = \sqrt{a}

.

Key Concept

Fractional and Negative Exponents

Estimated Time:1m 30s

Question 262Question →

A chemical processing tank receives two liquid solutions, Solution $X$ and Solution $Y$ , from separate inlet pipes.

- Solution $X$ contains chemical $A$ and water in a volume ratio of $3:2$ and enters the tank at a constant rate of $150$ liters per hour.
- Solution $Y$ contains chemical $A$ and water in a volume ratio of $1:4$ and enters the tank at a constant rate of $250$ liters per hour.

Both inlet pipes run simultaneously into an initially empty tank for $4$ hours. After $4$ hours, the inlet pipes are shut off. To adjust the mixture, pure chemical $A$ is added to the tank at a constant rate of $50$ liters per hour, while water is continuously drained from the tank at a constant rate of $30$ liters per hour.

How many hours must this adjustment process run until the volume of chemical $A$ in the tank is equal to the volume of water in the tank?

Show answer & explanation

Answer: 6

Answer

6

To find the time

t

when the volumes of chemical A and water in the tank are equal, first determine the initial quantities contributed by both solutions during the 4-hour filling period. Solution X provides

150 \times 4 = 600

liters total, containing

\frac{3}{5} \times 600 = 360

liters of chemical A and

\frac{2}{5} \times 600 = 240

liters of water. Solution Y provides

250 \times 4 = 1000

liters total, containing

\frac{1}{5} \times 1000 = 200

liters of chemical A and

\frac{4}{5} \times 1000 = 800

liters of water. Adding these amounts yields

360 + 200 = 560

liters of chemical A and

240 + 800 = 1040

liters of water. In the adjustment phase of

t

hours, chemical A increases at

50

L/hr to

560 + 50t

, while water decreases at

30

L/hr to

1040 - 30t

. Equating the two expressions gives

560 + 50t = 1040 - 30t

, which simplifies to

80t = 480

, resulting in

t = 6

hours.

Step-by-Step Solution

1

Determine the volumes of chemical A and water supplied by Solution X during the first 4 hours.

Solution X delivers 600 liters in total, consisting of 360 liters of chemical A and 240 liters of water.

Solution X flows at 150 L/hr for 4 hours (150 * 4 = 600 L) with a 3:2 chemical A to water ratio, meaning chemical A represents 3/5 of the total volume and water represents 2/5.

2

Determine the volumes of chemical A and water supplied by Solution Y during the first 4 hours.

Solution Y delivers 1000 liters in total, consisting of 200 liters of chemical A and 800 liters of water.

Solution Y flows at 250 L/hr for 4 hours (250 * 4 = 1000 L) with a 1:4 chemical A to water ratio, meaning chemical A represents 1/5 of the total volume and water represents 4/5.

3

Calculate the total initial quantities of chemical A and water present in the tank prior to the adjustment phase.

Total chemical A = 560 liters; Total water = 1040 liters.

Sum the quantities from both solutions: Chemical A = 360 + 200 = 560 L; Water = 240 + 800 = 1040 L.

4

Formulate linear expressions representing the total volume of chemical A and water after t hours of adjustment.

Chemical A volume = 560 + 50t; Water volume = 1040 - 30t.

Pure chemical A is added at 50 L/hr, increasing its total volume, while water is drained at 30 L/hr, reducing its total volume.

5

Set the two component volume expressions equal to each other and solve for t.

t = 6 hours.

Solving 560 + 50t = 1040 - 30t leads to 80t = 480, which yields t = 6.

Key Concept

Multi-stream mixture rate integration and ratio equality modeling

Question 263Question →

A positive integer $N$ has exactly $12$ positive divisors. If the greatest common divisor of $N$ and $35$ is $7$ , and $N$ is a multiple of $6$ , which of the following values could be equal to $N$ ? Indicate all such values.

Select all that apply

84

126

210

294

441

Show answer & explanation

Answer: $84$ ; $126$ ; $294$

Answer

The positive integer

N

could be equal to

84

,

126

, or

294

.

The integer

N

must contain the prime factors

2

,

3

, and

7

, but not

5

. For

N

to have exactly

12

divisors, its prime factorization exponent set

\{a,b,c\}

must satisfy

(a+1)(b+1)(c+1) = 12

, which restricts the exponents to a permutation of

\{1, 1, 2\}

. Evaluating the three permutations yields

84

,

126

, and

294

.

Step-by-Step Solution

1

Analyze the conditions given for

N

.

Since

N

is a multiple of

6

,

2 \mid N

and

3 \mid N

. Since

\gcd(N, 35) = 7

,

7 \mid N

and

5 \nmid N

. Thus,

N

must have prime factors

2, 3, 7

and no prime factor of

5

.

Establishing the prime factors of

N

based on divisibility and GCD conditions.

2

Determine the prime factorization form and divisor count.

Let

N = 2^a \cdot 3^b \cdot 7^c

, where

a \ge 1, b \ge 1, c \ge 1

. The number of divisors is given by

(a+1)(b+1)(c+1) = 12

.

The total number of positive divisors of a prime-factored integer

\prod p_i^{e_i}

is

\prod (e_i+1)

.

3

Find all valid exponent combinations

(a, b, c)

.

The factors of

12

into three integers each

\ge 2

are

2 \times 2 \times 3

. Therefore, the set of exponents

\{a, b, c\}

must be a permutation of

\{1, 1, 2\}

.

Each exponent increment

(e+1)

must be at least

2

since every prime

2, 3, 7

is present.

4

Calculate the possible numerical values of

N

.

Case 1:

2^2 \cdot 3^1 \cdot 7^1 = 84

.
Case 2:

2^1 \cdot 3^2 \cdot 7^1 = 126

.
Case 3:

2^1 \cdot 3^1 \cdot 7^2 = 294

.

Evaluating all three possible permutations of exponents for prime bases

2, 3,

and

7

.

Key Concept

Prime Factorization, Divisor Count Formula, and GCD Constraints

Question 264Question →

If $x$ and $y$ are real numbers such that $x < 0 < y$ and $x^2 > y^2$ , which of the following statements must be true? Select all such statements.

Select all that apply

x^3 < y^3

\sqrt{x^2} + x = 0

x + y < 0

\sqrt{x^2 + y^2} = |x| + y

(-x)^2 = -x^2

Show answer & explanation

Answer: $x^3 < y^3$ ; $\sqrt{x^2} + x = 0$ ; $x + y < 0$

Answer

The correct statements are

x^3 < y^3

,

\sqrt{x^2} + x = 0

, and

x + y < 0

.

The statement

x^3 < y^3

is correct because cubing a negative number keeps it negative while cubing a positive number keeps it positive. The statement

\sqrt{x^2} + x = 0

is correct because

\sqrt{x^2} = |x| = -x

for negative numbers. The statement

x + y < 0

is correct because

x^2 > y^2

implies

|x| > y

, meaning the negative component

x

has a larger absolute magnitude than the positive component

y

.

Step-by-Step Solution

1

Analyze the signs of odd powers for

x < 0 < y

x^3 < 0

and

y^3 > 0

, which guarantees

x^3 < y^3

.

Odd powers preserve the original sign of the base.

2

Apply the definition of principal square roots to negative values

\sqrt{x^2} = |x| = -x

, so

\sqrt{x^2} + x = -x + x = 0

.

The square root symbol

\sqrt{}

denotes the principal (non-negative) root.

3

Compare absolute values using

x^2 > y^2

|x| > y \implies -x > y \implies x + y < 0

.

Since

x < 0

, its magnitude

|x|

is

-x

, which dominates the positive value

y

.

4

Evaluate the false options against exponent and radical rules

\sqrt{x^2 + y^2} \neq |x| + y

due to non-distributivity of roots, and

(-x)^2 = x^2 \neq -x^2

.

Radicals do not distribute over sums, and even powers eliminate negative signs.

Key Concept

Properties of real exponents, radical expressions, and absolute values for negative bases

Estimated Time:1m 30s

Question 265Question →

A municipal authority allocates water from a central reservoir to three sectors: Agriculture, Industry, and Residential. Initially, the ratio of the volume of water allocated to Agriculture to that of Industry is $5 : 3$ , and the ratio of the volume allocated to Industry to that of Residential is $4 : 5$ . During a drought, the total supply is reallocated such that Agriculture's allocation is decreased by $20\%$ , Industry's allocation is decreased by $10\%$ , and Residential's allocation is increased by $12\%$ . After these adjustments, what is the ratio of Agriculture's new allocation to Residential's new allocation?

20 : 21

5 : 7

4 : 3

21 : 20

15 : 14

Show answer & explanation

Answer: $20 : 21$

Answer

The ratio of Agriculture's new allocation to Residential's new allocation is

20 : 21

.

To find the new ratio, first unify the given ratios Agriculture : Industry (

5 : 3

) and Industry : Residential (

4 : 5

) by finding a common multiplier for Industry (12). This yields an overall initial ratio of

20 : 12 : 15

. Applying a

20\%

decrease to Agriculture gives

20 \times 0.80 = 16

, and applying a

12\%

increase to Residential gives

15 \times 1.12 = 16.8

. The ratio of the new allocations is

16 : 16.8

, which simplifies to

160 : 168 = 20 : 21

.

Step-by-Step Solution

1

Express the three initial allocations in a unified three-part ratio.

Agriculture : Industry =

5 : 3 = 20 : 12

, Industry : Residential =

4 : 5 = 12 : 15

. Therefore, Agriculture : Industry : Residential =

20 : 12 : 15

.

Industry is the common element linking both ratios, so its ratio component must be equalized (LCM of 3 and 4 is 12).

2

Assign algebraic representations to the initial allocations based on the unified ratio.

Let Agriculture's initial allocation be

20x

, Industry's initial allocation be

12x

, and Residential's initial allocation be

15x

.

This allows for exact percentage calculations on consistent base values.

3

Calculate the updated allocations after applying the specified percentage adjustments.

Agriculture's new allocation =

20x \times (1 - 0.20) = 16x

. Residential's new allocation =

15x \times (1 + 0.12) = 16.8x

.

A

20\%

decrease reduces a quantity to

80\%

of its original value, and a

12\%

increase expands it to

112\%

of its original value.

4

Compute and simplify the ratio of Agriculture's new allocation to Residential's new allocation.

\frac{\text{Agriculture}_{\text{new}}}{\text{Residential}_{\text{new}}} = \frac{16x}{16.8x} = \frac{160}{168} = \frac{20}{21}.

Multiplying both terms by 10 eliminates decimals, and dividing both by their greatest common divisor (8) yields the simplified integer ratio

20 : 21

.

Key Concept

Combining relative ratios through a common quantity and calculating proportional adjustments.

Alternative Method

Instead of setting a variable

x

, assume a concrete initial volume for Industry equal to

12

units. Consequently, Agriculture is

20

units and Residential is

15

units. Agriculture's new volume is

20 - 4 = 16

units, and Residential's new volume is

15 + 1.8 = 16.8

units. The ratio

\frac{16}{16.8} = \frac{20}{21}

is obtained immediately.

Estimated Time:2m 0s

Question 266Question →

If $k$ and $m$ are integers such that $(-k)^5 m < 0$ and $k - m$ is an odd integer, which of the following expressions must be a positive even integer?

k^2 m^2

(-k)^2 m

(k - m)^2

-k m

k^3 + m^3

Show answer & explanation

Answer: $k^2 m^2$

Answer

The expression

k^2 m^2

must be a positive even integer.

Simplifying

(-k)^5 m < 0

gives

-k^5 m < 0 \implies k^5 m > 0

. This confirms

k \neq 0

and

m \neq 0

, so both

k^2

and

m^2

are positive integers, making

k^2 m^2 > 0

. Additionally,

k - m

being odd requires one variable to be even and the other odd. Squaring an even integer yields an even integer, and multiplying by any integer keeps it even. Hence, the expression

k^2 m^2

is guaranteed to be a positive even integer.

Step-by-Step Solution

1

Analyze the sign condition

(-k)^5 m < 0

.

Since

(-k)^5 = -k^5

, the inequality becomes

-k^5 m < 0

, which means

k^5 m > 0

. This implies that neither

k

nor

m

is zero, and both

k

and

m

have the same sign (either both positive or both negative).

Raising a negative quantity to an odd power retains the negative sign.

2

Analyze the parity condition

k - m

is odd.

The difference between two integers is odd if and only if one integer is even and the other is odd.

Even minus odd (or odd minus even) produces an odd result.

3

Evaluate the sign and parity of

k^2 m^2

.

Since

k \neq 0

and

m \neq 0

,

k^2 > 0

and

m^2 > 0

, so

k^2 m^2 > 0

(positive). Since one of

k

or

m

is even, its square is also even, so the product

k^2 m^2

must be even. Thus,

k^2 m^2

is guaranteed to be a positive even integer.

The product of non-zero squares is positive, and any integer multiple of an even number is even.

Key Concept

Parity rules under subtraction/multiplication and sign rules under odd powers.

Question 267Question →

If $x = 4$ , what is the value of $\sqrt{9^x + 9^x + 9^x + 9^x}$ ?

81

162

324

648

1,296

Show answer & explanation

Answer: 162

Answer

162

Combining the four identical terms under the square root gives

\sqrt{4 \cdot 9^4}

. Splitting the root using product rules yields

\sqrt{4} \cdot \sqrt{9^4} = 2 \cdot 9^2 = 2 \cdot 81 = 162

.

Step-by-Step Solution

1

Substitute

x = 4

into the given radical expression.

The expression becomes

\sqrt{9^4 + 9^4 + 9^4 + 9^4}

.

Direct substitution of the given variable value.

2

Combine the four identical terms under the radical.

9^4 + 9^4 + 9^4 + 9^4 = 4 \cdot 9^4

.

Repeated addition of

4

identical terms is equivalent to multiplication by

4

.

3

Apply the product rule for square roots,

\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}

.

\sqrt{4 \cdot 9^4} = \sqrt{4} \cdot \sqrt{9^4} = 2 \cdot 9^2.

Both

4

and

9^4

are perfect squares.

4

Evaluate the numerical value.

2 \cdot 81 = 162.

Simplifying the arithmetic expression.

Key Concept

Combining like terms under radical sign and applying product rules of square roots

Estimated Time:1m 15s

Question 268Question →

A sequence of 50 numerical measurements $x_1, x_2, \dots, x_{50}$ is collected. Each measurement $x_i$ is rounded to the nearest tenth to produce a rounded value $r_i$ . The sum of the 50 rounded values, $\sum_{i=1}^{50} r_i$ , is equal to $250.0$ . If $S$ represents the true sum of the unrounded measurements $\sum_{i=1}^{50} x_i$ , what is the maximum possible percent error of the rounded sum relative to the true sum $S$ , rounded to the nearest hundredth of a percent?

0.99\%

1.00\%

1.01\%

2.00\%

2.04\%

Show answer & explanation

Answer: $1.01\%$

Answer

The maximum possible percent error of the rounded sum relative to the true sum is

1.01\%

.

When rounding numbers to the nearest tenth, the maximum error for each number is

0.05

. For 50 numbers, the maximum possible error in the sum is

50 \times 0.05 = 2.5

. The true sum

S

therefore lies in the range

[247.5, 252.5]

. To maximize the percent error relative to

S

, defined as

\frac{|250.0 - S|}{S} \times 100\%

, we use the maximum numerator

2.5

and the smallest possible denominator

S = 247.5

. This yields

\frac{2.5}{247.5} \times 100\% \approx 1.01\%

.

Step-by-Step Solution

1

Determine the maximum rounding error for a single term

Maximum error per measurement is

|x_i - r_i| \le 0.05

When rounding to the nearest tenth, any value within

0.05

of the rounded value rounds to that tenth.

2

Calculate the maximum cumulative error for the sequence sum

Maximum total error

= 50 \times 0.05 = 2.5

The maximum difference between the true sum

S

and the rounded sum

250.0

occurs when all individual rounding errors accumulate in the same direction.

3

Find the range of possible true sum values

S

247.5 \le S \le 252.5

Subtracting and adding the maximum error of

2.5

from the rounded sum

250.0

establishes the bounds for

S

.

4

Set up and maximize the percent error expression

Max percent error occurs at minimum

S = 247.5

, giving

\frac{2.5}{247.5} \times 100\% \approx 1.0101\%

Percent error relative to

S

is given by

\frac{|250.0 - S|}{S} \times 100\%

. To maximize this ratio, we divide the maximum numerator

2.5

by the smallest positive denominator

S = 247.5

.

Key Concept

Error propagation in sequence sums and optimizing percent error base values

Question 269Question →

Let $a$ , $b$ , and $c$ be non-zero integers such that $\frac{a}{b} < 0$ , $a^3 b c > 0$ , and $a + b$ is an odd integer. Which of the following statements must be true? Select all such statements.

Select all that apply

c < 0

a b

is an even integer

a c > 0

a^2 + c

is an even integer

b + c < 0

Show answer & explanation

Answer: $c < 0$ ; $a b$ is an even integer

Answer

The statements '

c < 0

' and '

a b

is an even integer' must be true.

The statement '

c < 0

' must be true because

\frac{a}{b} < 0

forces

a b < 0

, and since

a^3 b c = a^2 (a b) c

with

a^2 > 0

,

c

must be negative to yield a positive product. The statement '

a b

is an even integer' must be true because an odd sum

a + b

requires one variable to be even and the other to be odd, making their product even.

Step-by-Step Solution

1

Determine the sign relationship between

a

and

b

.

a

and

b

have opposite signs, so

a b < 0

.

The quotient

\frac{a}{b} < 0

implies the numerator and denominator have different signs.

2

Determine the sign of

c

using

a^3 b c > 0

.

c < 0

.

Rewrite

a^3 b c

as

a^2 \cdot (a b) \cdot c

. Since

a \neq 0

,

a^2 > 0

. Since

a b < 0

, the product

a^2 (a b) < 0

. For the entire product

a^2 (a b) c

to be positive,

c

must be negative.

3

Analyze the parity of

a

and

b

from

a + b

being odd.

One of

a

or

b

is even and the other is odd, so

a b

must be even.

An odd sum of two integers requires one even and one odd addend. The product of an even integer and any integer is always even.

4

Test the remaining options for counterexamples.

The statements '

a c > 0

', '

a^2 + c

is an even integer', and '

b + c < 0

' can be false under valid assignments.

For example, if

a = 2

,

b = -1

, and

c = -3

, then

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

,

a^3 b c = 8(-1)(-3) = 24 > 0

, and

a + b = 1

(odd). Here,

a c = -6 < 0

,

a^2 + c = 4 - 3 = 1

(odd), and

b + c = -4 < 0

, but setting

b = 5, a = -2, c = -1

gives

b + c = 4 > 0

.

Key Concept

Deducing sign and parity properties of integers

Estimated Time:1m 30s

Question 270Question →

On the real number line, the distance between a real number $k$ and $-3$ is strictly less than $7$ , and the distance between $k$ and $5$ is at least $4$ . Which of the following inequalities represents the complete set of all possible values of $k$ ?

-10 < k \le 1

-10 \le k \le 1

-10 < k < 1

-4 < k \le 1

1 \le k < 4

Show answer & explanation

Answer: $-10 < k \le 1$

Answer

-10 < k \le 1

The correct inequality

-10 < k \le 1

properly combines the strict bound from the distance to

-3

(which gives

-10 < k < 4

) with the non-strict bound from the distance to

5

(which gives

k \le 1

or

k \ge 9

). Taking their intersection gives

-10 < k \le 1

.

Step-by-Step Solution

1

Express the first condition using absolute value notation and solve for

k

.

|k - (-3)| < 7 \implies |k + 3| < 7 \implies -7 < k + 3 < 7 \implies -10 < k < 4

.

Distance on a number line between

x

and

y

is given by

|x - y|

.

2

Express the second condition using absolute value notation and solve for

k

.

|k - 5| \ge 4 \implies k - 5 \le -4

or

k - 5 \ge 4 \implies k \le 1

or

k \ge 9

.

The phrase 'at least 4' means greater than or equal to 4.

3

Find the overlap (intersection) of the two solution sets.

(-10 < k < 4) \cap (k \le 1 \text{ or } k \ge 9) = -10 < k \le 1

.

Since

k

must be less than

4

, the region

k \ge 9

contains no valid solutions, leaving only

-10 < k \le 1

.

Key Concept

Absolute value as distance on the number line and solving compound absolute value inequalities.

Estimated Time:1m 30s

Question 271Question →

A biomanufacturing facility operates two harvesting lines, Line A and Line B, to collect a refined protein compound suspended in a liquid growth medium into a single storage tank.

- Line A processes liquid at a constant rate of $120\text{ liters per hour}$ , and its output contains protein and medium in a volume ratio of $1 : 4$ .
- Line B processes liquid at a constant rate of $180\text{ liters per hour}$ , and its output contains protein and medium in a volume ratio of $1 : 9$ .

Both lines run simultaneously for exactly $5\text{ hours}$ into the empty storage tank. Which of the following statements about the resulting liquid mixture in the storage tank must be true? Indicate all such statements.

Select all that apply

The total volume of protein collected in the storage tank is

210\text{ liters}

.

Protein accounts for exactly

14\%

of the total liquid volume in the storage tank.

The ratio of total protein to total medium in the storage tank is

7 : 43

.

The total volume of protein collected in the storage tank is

250\text{ liters}

.

Protein accounts for exactly

15\%

of the total liquid volume in the storage tank.

Show answer & explanation

Answer: The total volume of protein collected in the storage tank is $210\text{ liters}$ .; Protein accounts for exactly $14\%$ of the total liquid volume in the storage tank.; The ratio of total protein to total medium in the storage tank is $7 : 43$ .

Answer

The correct statements are those asserting that the total volume of protein collected is 210 liters, that protein accounts for exactly 14% of the total liquid volume, and that the ratio of total protein to total medium is 7 to 43.

The total protein collected is

210\text{ liters}

(

120\text{ L}

from Line A and

90\text{ L}

from Line B). Dividing this by the overall volume of

1500\text{ liters}

gives

14\%

. Subtracting protein from total volume gives

1290\text{ liters}

of medium, yielding a protein-to-medium ratio of

210 : 1290 = 7 : 43

. Therefore, the three statements asserting

210\text{ liters}

of protein, a

14\%

concentration, and a

7:43

ratio are all correct.

Step-by-Step Solution

1

Calculate the total liquid volume produced by each line in 5 hours.

Line A volume =

120\text{ L/hr} \times 5\text{ hr} = 600\text{ liters}

. Line B volume =

180\text{ L/hr} \times 5\text{ hr} = 900\text{ liters}

. Total combined volume =

600 + 900 = 1500\text{ liters}

.

Total volume per line is the product of its constant flow rate and duration.

2

Convert the part-to-part ratios to part-to-whole fractions to find the protein volume from each line.

Line A ratio

1:4

means protein is

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

of the volume. Protein A =

\frac{1}{5} \times 600 = 120\text{ L}

. Line B ratio

1:9

means protein is

\frac{1}{1+9} = \frac{1}{10}

of the volume. Protein B =

\frac{1}{10} \times 900 = 90\text{ L}

. Total protein =

120 + 90 = 210\text{ liters}

.

Ratios of

a:b

correspond to a component fraction of

\frac{a}{a+b}

of the total mixture.

3

Determine the percentage concentration of protein in the final mixture.

\text{Percentage} = \frac{210\text{ L}}{1500\text{ L}} \times 100\% = 14\%

.

The overall concentration is total protein volume divided by overall mixture volume.

4

Determine the simplified ratio of total protein to total medium.

Total medium volume =

1500 - 210 = 1290\text{ L}

. Ratio of protein to medium =

210 : 1290 = 7 : 43

.

Dividing both parts of

210 : 1290

by their greatest common divisor (

30

) yields

7 : 43

.

Key Concept

Combining rates and converting part-to-part ratios to part-to-whole fractions

Question 272Question →

Let $m$ and $n$ be integers such that $m < 0$ , $n > 0$ , $(-1)^m = -1$ , and $m + n$ is an even integer. Which of the following expressions must be a positive even integer?

m \cdot n

(-m)^n

n - m

(m + n)^2

m^2 - n

Show answer & explanation

Answer: $n - m$

Answer

The expression

n - m

must be a positive even integer.

The expression

n - m

subtracts a negative odd integer from a positive odd integer, which equals adding two positive odd integers. The sum of two positive odd integers is always a positive even integer.

Step-by-Step Solution

1

Determine the parity and sign of

m

.

Since

m < 0

and

(-1)^m = -1

,

m

must be a negative odd integer.

An odd exponent on

-1

yields

-1

.

2

Determine the parity and sign of

n

.

Since

n > 0

and

m + n

is even,

n

must be a positive odd integer.

The sum of two integers is even if and only if both integers have the same parity. Since

m

is odd,

n

must also be odd.

3

Evaluate the sign and parity of

n - m

.

n - m = n + (-m)

. Since

n \ge 1

and

-m \ge 1

,

n - m \ge 2

(strictly positive). Also, odd minus odd is always even.

Combining the sign rules (

n > 0

and

-m > 0

) with the even-odd subtraction rule confirms

n - m

is always a positive even integer.

Key Concept

Even-Odd Properties and Sign Rules

Question 273Question →

Three water pumps, $P$ , $Q$ , and $R$ , operate at constant individual rates. The ratio of the rate of pump $P$ to the rate of pump $Q$ is $2 : 3$ . When all three pumps operate simultaneously, their combined rate is $3$ times the rate of pump $P$ alone. If pump $R$ working alone can drain a full reservoir in $24$ hours, how many hours would pump $Q$ working alone take to drain the same full reservoir?

8

hours

12

hours

16

hours

32

hours

56

hours

Show answer & explanation

Answer: $8$ hours

Answer

8 hours

The correct answer is 8 hours. By setting the rate of pump Q as

\frac{3}{2}

times the rate of pump P, the combined rate of all three pumps is

r_P + \frac{3}{2} r_P + r_R = \frac{5}{2} r_P + r_R

. Setting this equal to

3 r_P

shows that pump R's rate is

\frac{1}{2} r_P

. Since pump R takes 24 hours (

r_R = \frac{1}{24}

), pump P's rate is

\frac{1}{12}

(taking 12 hours), and pump Q's rate is

\frac{3}{2} \times \frac{1}{12} = \frac{1}{8}

(taking 8 hours).

Step-by-Step Solution

1

Express the rates of pumps P and Q in terms of a common variable.

Let

r_P

,

r_Q

, and

r_R

be the rates of pumps

P

,

Q

, and

R

in reservoirs per hour. Given

r_P : r_Q = 2 : 3

, we have

r_Q = \frac{3}{2} r_P

.

Relating pump rates using the given ratio simplifies the system of equations to one variable.

2

Set up the combined rate equation and solve for

r_R

in terms of

r_P

.

r_P + r_Q + r_R = 3 r_P \implies r_P + \frac{3}{2} r_P + r_R = 3 r_P \implies \frac{5}{2} r_P + r_R = 3 r_P \implies r_R = \frac{1}{2} r_P

.

The total rate is the sum of individual rates, allowing us to express pump R's rate in terms of pump P's rate.

3

Calculate

r_P

and

r_Q

using the given rate for pump R.

Since pump

R

takes

24

hours alone,

r_R = \frac{1}{24}

. Therefore,

\frac{1}{2} r_P = \frac{1}{24} \implies r_P = \frac{1}{12}

. Then

r_Q = \frac{3}{2} \times \frac{1}{12} = \frac{1}{8}

.

Knowing pump R's explicit numerical rate allows finding the numerical rates for pumps P and Q.

4

Determine the time required for pump Q alone to drain the reservoir.

\text{Time for } Q = \frac{1}{r_Q} = \frac{1}{1/8} = 8 \text{ hours}.

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

Key Concept

Combined Work Rates and Ratio Relationships

Estimated Time:2m 0s

Question 274Question →

A data processing center uses two server clusters, Cluster $X$ and Cluster $Y$ , operating at constant individual processing rates. The ratio of the rate of Cluster $X$ to the rate of Cluster $Y$ is $3 : 5$ . Cluster $X$ alone can process a standard dataset of size $D$ gigabytes in 20 hours.

If Cluster $X$ and Cluster $Y$ work together for 4 hours at their initial rates, and then Cluster $X$ 's rate is increased by $33\frac{1}{3}\%$ while Cluster $Y$ 's rate is increased by $20\%$ , how many additional hours will it take for the two clusters working together at their new rates to complete the remaining portion of dataset $D$ ?

Show answer & explanation

Answer: 2.8

Answer

It will take 2.8 additional hours for the two clusters working together at their new rates to complete the remaining portion of dataset

D

.

Representing Cluster

X

's rate as

3k

and Cluster

Y

's rate as

5k

establishes the dataset size

D = 20 \times 3k = 60k

. During the first 4 hours, both clusters process

4 \times (3k + 5k) = 32k

GB, leaving

28k

GB remaining. After rate increases, Cluster

X

's rate becomes

4k

and Cluster

Y

's rate becomes

6k

, resulting in a new combined rate of

10k

. Dividing the remaining

28k

GB by

10k

GB/hr yields

2.8

hours.

Step-by-Step Solution

1

Define variables for the initial rates and dataset size based on the given ratio.

Let the processing rate of Cluster

X

be

r_X = 3k

GB/hr and Cluster

Y

be

r_Y = 5k

GB/hr for some constant

k > 0

. Since Cluster

X

alone completes dataset

D

in 20 hours,

D = 3k \times 20 = 60k

GB.

Relating the ratio of individual rates to total work defines all quantities in terms of a single parameter

k

.

2

Calculate the amount of work finished during the initial joint operation.

The combined initial rate is

r_X + r_Y = 3k + 5k = 8k

GB/hr. Working together for 4 hours completes

8k \times 4 = 32k

GB.

Working simultaneously means their processing rates add together.

3

Find the remaining work and the updated processing rates after adjustments.

Remaining dataset volume =

60k - 32k = 28k

GB. Cluster

X

's new rate =

3k \times \left(1 + \frac{1}{3}\right) = 4k

GB/hr. Cluster

Y

's new rate =

5k \times 1.20 = 6k

GB/hr. New combined rate =

4k + 6k = 10k

GB/hr.

Modifying the individual rates changes the combined throughput for the remaining task.

4

Compute the additional time required to process the remaining dataset.

Additional time =

\frac{28k \text{ GB}}{10k \text{ GB/hr}} = 2.8

hours.

Dividing the remaining work volume by the new combined rate gives the exact required time.

Key Concept

Combined work rates, ratio proportionality, and percentage rate adjustments.

Question 275Question →

If $a$ and $b$ are real numbers such that $|a - 3| = 5$ and $|2b + 1| = 9$ , what is the minimum possible value of $|a - b|$ ?

1

2

3

4

6

Show answer & explanation

Answer: 3

Answer

The minimum possible value of

|a - b|

is

3

.

Solving

|a - 3| = 5

gives two possible values for

a

:

a = 8

and

a = -2

. Solving

|2b + 1| = 9

gives two possible values for

b

:

b = 4

and

b = -5

. Evaluating the distance

|a - b|

for all four pairs

(a,b)

gives

|8 - 4| = 4

,

|8 - (-5)| = 13

,

|-2 - 4| = 6

, and

|-2 - (-5)| = 3

. The minimum possible value is

3

.

Step-by-Step Solution

1

Solve the absolute value equation

|a - 3| = 5

for all possible values of

a

.

a - 3 = 5 \implies a = 8

or

a - 3 = -5 \implies a = -2

. Thus,

a \in \{-2, 8\}

.

An absolute value equation

|x| = k

splits into two linear equations:

x = k

and

x = -k

.

2

Solve the absolute value equation

|2b + 1| = 9

for all possible values of

b

.

2b + 1 = 9 \implies 2b = 8 \implies b = 4

or

2b + 1 = -9 \implies 2b = -10 \implies b = -5

. Thus,

b \in \{-5, 4\}

.

An absolute value equation

|2b + 1| = 9

has two cases:

2b + 1 = 9

and

2b + 1 = -9

.

3

Calculate

|a - b|

for all four possible pairs of

(a, b)

.

For

(8, 4): |8 - 4| = 4

.
For

(8, -5): |8 - (-5)| = 13

.
For

(-2, 4): |-2 - 4| = 6

.
For

(-2, -5): |-2 - (-5)| = 3

.

To find the minimum possible value of

|a - b|

, every valid combination of

a

and

b

must be tested.

4

Identify the minimum value among the calculated absolute differences.

The minimum calculated value is

3

.

Comparing

4, 13, 6,

and

3

yields

3

as the smallest value.

Key Concept

Solving absolute value equations and finding distances between points on the real number line

Estimated Time:1m 30s

Question 276Question →

If $x$ and $y$ are real numbers such that $x < -1$ and $0 < y < 1$ , which of the following statements must be true? Select all such statements.

Select all that apply

x^2 > y^2

\sqrt{x^2 y^4} = -x y^2

x^3 > y^3

\sqrt{x^2} + y = x + y

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

Show answer & explanation

Answer: $x^2 > y^2$ ; $\sqrt{x^2 y^4} = -x y^2$

Answer

The correct statements are the inequality asserting that the square of the first variable is greater than the square of the second, and the identity simplifying the square root of the product of the powers to negative the product of the first variable and the square of the second.

The statement comparing squared values is correct because any real number less than -1 has an absolute value greater than 1, so its square is strictly greater than 1, while any positive number less than 1 has a square strictly less than 1. The radical identity statement is correct because taking the square root of

x^2

yields

|x|

, which evaluates to

-x

when

x

is negative.

Step-by-Step Solution

1

Analyze the given bounds for both variables.

For the first variable,

x < -1

, which implies

|x| > 1

,

x

is negative,

x^2 > 1

, and

x^3 < -1

. For the second variable,

0 < y < 1

, which implies

y > 0

,

y^2 < 1

, and

y^3 > 0

.

Establishing explicit bounds on magnitudes and signs is necessary to evaluate powers and absolute value roots.

2

Evaluate the inequality comparing the squared terms.

Since

x^2 > 1

and

y^2 < 1

, it follows directly that

x^2 > y^2

.

Transitive comparison across the threshold value of 1 proves the inequality holds.

3

Simplify the radical expression

\sqrt{x^2 y^4}

.

\sqrt{x^2 y^4} = \sqrt{x^2} \cdot \sqrt{y^4} = |x| \cdot y^2

. Since

x < 0

,

|x| = -x

, so the expression simplifies to

-x y^2

.

The principal square root of

x^2

must equal the absolute value

|x|

, which requires a sign flip when

x

is negative.

4

Verify remaining candidate expressions for potential fallacies.

Odd powers preserve negative signs so

x^3 < y^3

;

\sqrt{x^2}

equals

-x \neq x

; and expanding

(x+y)^2

produces a nonzero cross-term

2xy

.

Eliminating false options confirms that only two statements are universally true.

Key Concept

Principal square roots and even/odd power behaviors under negative variable bounds

Question 277Question →

Three consecutive integers $a$ , $b$ , and $c$ satisfy $a < b < c$ . If $a + b + c = -9$ and $a \cdot b \cdot c < 0$ , what is the value of $(-1)^a + (-1)^b + (-1)^c$ ?

Show answer & explanation

Answer: 1

Answer

The value of the expression is 1.

The sum of three consecutive integers

a + b + c = 3b = -9

determines

b = -3

, making

a = -4

and

c = -2

. The product

(-4)(-3)(-2) = -24

is negative, confirming the given condition. Applying exponent sign rules, raising

-1

to an even integer power yields

1

, while raising

-1

to an odd integer power yields

-1

. Thus,

(-1)^{-4} = 1

,

(-1)^{-3} = -1

, and

(-1)^{-2} = 1

. Summing these three terms gives

1 + (-1) + 1 = 1

.

Step-by-Step Solution

1

Find the values of integers

a

,

b

, and

c

.

a = -4

,

b = -3

,

c = -2

Three consecutive integers centered at

b

sum to

3b = -9

, so

b = -3

.

2

Check the sign condition of the product

a \cdot b \cdot c

.

(-4)(-3)(-2) = -24 < 0

The product of three negative numbers is negative.

3

Evaluate

(-1)^n

for each integer power.

(-1)^{-4} = 1

,

(-1)^{-3} = -1

,

(-1)^{-2} = 1

Negative one raised to an even integer power is 1; raised to an odd integer power is -1.

4

Sum the three evaluated terms.

1 + (-1) + 1 = 1

Addition of the resulting values.

Key Concept

Even-odd exponent rules for negative bases and sign rules for product of signed integers.

Question 278Question →

A specialty paint manufacturing facility creates a custom dye by mixing three liquid concentrates: Red, Yellow, and Blue. In the formulation, the ratio of the volume of Red concentrate to Yellow concentrate is $2 : 3$ , and the ratio of the volume of Yellow concentrate to Blue concentrate is $4 : 5$ . If a single storage vat contains $140\text{ liters}$ of this fully mixed custom dye, how many liters of Yellow concentrate are in the vat?

32\text{ liters}

48\text{ liters}

60\text{ liters}

70\text{ liters}

84\text{ liters}

Show answer & explanation

Answer: $48\text{ liters}$

Answer

48\text{ liters}

of Yellow concentrate

To find the volume of Yellow concentrate, first combine the given ratios (Red : Yellow = 2 : 3 and Yellow : Blue = 4 : 5) into a unified ratio by expressing Yellow with a common term. Multiplying the first ratio by 4 gives Red : Yellow = 8 : 12, and multiplying the second ratio by 3 gives Yellow : Blue = 12 : 15. The combined ratio Red : Yellow : Blue is 8 : 12 : 15, yielding a total of 8 + 12 + 15 = 35 parts. Yellow accounts for 12 out of 35 parts. Multiplying 12/35 by the total volume of 140 liters yields (12/35) * 140 = 48 liters.

Step-by-Step Solution

1

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

Red : Yellow =

2 : 3 = 8 : 12

and Yellow : Blue =

4 : 5 = 12 : 15

. Thus, Red : Yellow : Blue =

8 : 12 : 15

.

Yellow is the common component in both ratios. Finding a common multiple for Yellow's ratio term (LCM of 3 and 4 is 12) allows us to express all three quantities in a unified ratio scale.

2

Calculate the total number of ratio parts and determine Yellow's fraction of the total mixture.

Total parts =

8 + 12 + 15 = 35

. Yellow's fraction of the total volume =

\frac{12}{35}

.

To convert a part-to-part ratio into a part-to-whole ratio, divide the target component's ratio parts by the sum of all ratio parts.

3

Multiply Yellow's fraction by the total volume of the mixture.

Yellow volume =

\frac{12}{35} \times 140 = 12 \times 4 = 48\text{ liters}

.

Multiplying the part-to-whole fraction by the total volume yields the exact volume of Yellow concentrate present.

Key Concept

Combining pairwise ratios into a unified three-part ratio to determine part-to-whole proportions.

Estimated Time:1m 30s

Question 279Question →

Let $r$ and $s$ be integers such that $(-1)^{r+s} = -1$ and $r^2 s + r$ is an odd integer. Which of the following statements must be true? Select all such statements.

Select all that apply

r - s

is an odd integer.

r + 2s

is an even integer.

r^2 + s^2

is an odd integer.

s(r + 1)

is an odd integer.

r^2 s + s

is an even integer.

Show answer & explanation

Answer: $r - s$ is an odd integer.; $r^2 + s^2$ is an odd integer.; $r^2 s + s$ is an even integer.

Answer

The statements that must be true are '

r - s

is an odd integer', '

r^2 + s^2

is an odd integer', and '

r^2 s + s

is an even integer'.

From

(-1)^{r+s} = -1

, the sum

r+s

must be odd, meaning

r

and

s

have opposite parity. Factoring

r^2 s + r

yields

r(rs + 1) = \text{odd}

, which requires both

r

and

rs + 1

to be odd. Hence,

r

is odd, which forces

s

to be even. Testing the options shows that subtracting an even number from an odd number gives an odd number, adding the squares of an odd and an even number gives an odd number, and multiplying any integer by the even number

s

gives an even number.

Step-by-Step Solution

1

Determine the parity of

r + s

from

(-1)^{r+s} = -1

r + s

is an odd integer

For

(-1)^k = -1

, the exponent

k

must be an odd integer. Therefore,

r + s

is odd, which implies that one variable is even and the other is odd.

2

Analyze the given expression

r^2 s + r

r

is odd and

s

is even

Factor

r^2 s + r

as

r(rs + 1)

. For the product of two integers to be odd, both factors must be odd. Thus,

r

must be odd. Since

r + s

is odd and

r

is odd,

s

must be even. (Verification: if

s

is even and

r

is odd,

rs + 1

is even + 1 = odd, so

r(rs + 1)

is odd

\times

odd = odd).

3

Evaluate each given statement using

r = \text{odd}

and

s = \text{even}

Statements '

r - s

is an odd integer', '

r^2 + s^2

is an odd integer', and '

r^2 s + s

is an even integer' are true.

1)

\text{odd} - \text{even} = \text{odd}

(True).
2)

\text{odd} + 2(\text{even}) = \text{odd} + \text{even} = \text{odd}

(False for even).
3)

(\text{odd})^2 + (\text{even})^2 = \text{odd} + \text{even} = \text{odd}

(True).
4)

(\text{even})(\text{odd} + 1) = \text{even} \times \text{even} = \text{even}

(False for odd).
5)

r^2 s + s = s(r^2 + 1) = \text{even} \times \text{even} = \text{even}

(True).

Key Concept

Parity rules for integer addition, multiplication, and exponents

Question 280Question →

A commercial bakery prepares a specialty grain blend using oats, wheat, and rye. Initially, the ratio of oats to wheat to rye by weight in the blend is $4 : 3 : 2$ . After $15\text{ kg}$ of oats and $15\text{ kg}$ of rye are added to the mixture while the amount of wheat remains unchanged, the ratio of oats to rye in the new blend becomes $3 : 2$ . Which of the following statements about the final grain blend must be true? Select all such statements.

Select all that apply

The weight of wheat in the final blend is

22.5\text{ kg}

.

The total weight of the final grain blend is

97.5\text{ kg}

.

The ratio of wheat to rye in the final grain blend is

3 : 4

.

Oats account for exactly

50\%

of the total weight of the final grain blend.

The total weight of the blend increased by

50\%

from its initial weight.

Show answer & explanation

Answer: The weight of wheat in the final blend is $22.5\text{ kg}$ .; The total weight of the final grain blend is $97.5\text{ kg}$ .; The ratio of wheat to rye in the final grain blend is $3 : 4$ .

Answer

The weight of wheat in the final blend is

22.5\text{ kg}

, the total weight of the final grain blend is

97.5\text{ kg}

, and the ratio of wheat to rye in the final grain blend is

3 : 4

.

Solving the ratio proportion equation yields a multiplier constant of

x = 7.5

. Substituting

x = 7.5

yields final weights of

45\text{ kg}

for oats,

22.5\text{ kg}

for wheat, and

30\text{ kg}

for rye. The weight of wheat is indeed

22.5\text{ kg}

, the total final weight is

45 + 22.5 + 30 = 97.5\text{ kg}

, and the ratio of wheat to rye is

22.5 : 30 = 3 : 4

. Thus, these three statements are correct.

Step-by-Step Solution

1

Define variables using the initial ratio.

Let the initial weight of oats be

4x

, wheat be

3x

, and rye be

2x

.

Expressing quantities in terms of a common ratio multiplier

x

ensures proportional relationships are preserved.

2

Set up an equation using the updated quantities and given new ratio.

\frac{4x + 15}{2x + 15} = \frac{3}{2}

15 kg was added to both oats and rye, establishing a new ratio of 3 to 2 between oats and rye.

3

Solve for the multiplier

x

.

2(4x + 15) = 3(2x + 15) \implies 8x + 30 = 6x + 45 \implies 2x = 15 \implies x = 7.5

.

Cross-multiplying yields the unique value for the ratio constant

x

.

4

Calculate the final weight of each component and the total weight.

Initial oats =

30\text{ kg}

, final oats =

45\text{ kg}

. Wheat (unchanged) =

22.5\text{ kg}

. Initial rye =

15\text{ kg}

, final rye =

30\text{ kg}

. Final total weight =

45 + 22.5 + 30 = 97.5\text{ kg}

.

Evaluating each component confirms all specific properties of the final mixture.

5

Verify each offered statement against calculated values.

Wheat weight is

22.5\text{ kg}

(True). Total weight is

97.5\text{ kg}

(True). Ratio of Wheat to Rye is

22.5 : 30 = 3 : 4

(True). Oats percentage is

\frac{45}{97.5} \approx 46.15\%

(False). Weight percent increase is

\frac{30}{67.5} \approx 44.44\%

(False).

Comparing calculated facts directly determines which statements must be true.

Key Concept

Ratio adjustment and algebraic formulation of multi-part mixture problems

Arithmetic

Answer

Step-by-Step Solution

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

Alternative Method

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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