Arithmetic Practice Questions — GRE General Test — Page 12

Question 221Question →

Three water pumps, $P_1$ , $P_2$ , and $P_3$ , operate at constant individual rates to fill a large reservoir. The ratio of the pumping rate of $P_1$ to that of $P_2$ is $3 : 4$ , and the ratio of the pumping rate of $P_2$ to that of $P_3$ is $3 : 5$ .

At 8:00 AM, all three pumps begin filling an empty reservoir together. At 10:00 AM, pump $P_1$ shuts down, while $P_2$ and $P_3$ continue operating at their original rates. At 11:00 AM, the operating rate of $P_2$ is decreased by $25\%$ , and the operating rate of $P_3$ is increased by $25\%$ . The two remaining pumps continue at these adjusted rates until the reservoir is completely full at 1:00 PM.

If pump $P_3$ were to fill the empty reservoir working alone at its original constant rate, how many hours would it take?

Show answer & explanation

Answer: 9.1

Answer

It would take pump

P_3

exactly

9.1

hours (or

9.1

when entered numerically) to fill the empty reservoir alone at its original constant rate.

By unifying the given ratios

r_1 : r_2 = 3 : 4

and

r_2 : r_3 = 3 : 5

, we obtain the relative rates

r_1 = 9k

,

r_2 = 12k

, and

r_3 = 20k

. Summing the work done across the three intervals (8–10 AM at rate

41k

, 10–11 AM at rate

32k

, and 11 AM–1 PM at rate

34k

) yields a total capacity of

182k

. Dividing total work

182k

by

P_3

's original rate of

20k

gives

9.1

hours.

Step-by-Step Solution

1

Unify the individual pumping rate ratios into a single ratio

r_1 : r_2 : r_3

.

r_1 = 9k

,

r_2 = 12k

,

r_3 = 20k

Aligning the ratio of

P_1:P_2 = 3:4 = 9:12

and

P_2:P_3 = 3:5 = 12:20

establishes a common scale factor

k

.

2

Calculate the work completed in each of the three time intervals.

W_1 = 82k

,

W_2 = 32k

,

W_3 = 68k

Multiply the duration of each interval by the sum of active rates during that period.

3

Sum the total work completed to get total capacity and divide by

P_3

's original rate.

Total capacity

= 182k

; Time

= \frac{182k}{20k} = 9.1

hours

The total work equals the sum of work done across all three phases, and time is work divided by rate.

Key Concept

Multi-stage work rates, compound ratio unification, and percentage rate adjustments

Estimated Time:3m 0s

Question 222Question →

Let $a$ and $b$ be non-zero real numbers such that $|a + b| = 3|a - b|$ . What is the value of $\left|\frac{a^2 + b^2}{ab}\right|$ ?

\frac{3}{2}

2

\frac{5}{2}

3

\frac{10}{3}

Show answer & explanation

Answer: $\frac{5}{2}$

Answer

The value of

\left|\frac{a^2 + b^2}{ab}\right|

is

\frac{5}{2}

.

Squaring both sides of the given equation

|a + b| = 3|a - b|

eliminates the absolute values to give

(a + b)^2 = 9(a - b)^2

. Expanding and simplifying yields

2a^2 - 5ab + 2b^2 = 0

, which factors as

(2a - b)(a - 2b) = 0

. Thus,

b = 2a

or

a = 2b

. In either case, substituting into

\left|\frac{a^2 + b^2}{ab}\right|

reduces the expression to

\frac{5}{2}

.

Step-by-Step Solution

1

Square both sides of the absolute value equality

(a + b)^2 = 9(a - b)^2

Since both sides of

|a + b| = 3|a - b|

are non-negative real numbers, squaring preserves equality and eliminates the absolute value signs.

2

Expand both algebraic expressions and collect like terms

a^2 + 2ab + b^2 = 9a^2 - 18ab + 9b^2 \implies 8a^2 - 20ab + 8b^2 = 0 \implies 2a^2 - 5ab + 2b^2 = 0

Expanding the binomial squares allows combining like terms to solve for the relationship between

a

and

b

.

3

Factor the quadratic expression in terms of

a

and

b

(2a - b)(a - 2b) = 0 \implies b = 2a \text{ or } a = 2b

Factoring determines the exact proportional relationship between

a

and

b

.

4

Substitute the relation into the target expression

\left|\frac{a^2 + (2a)^2}{a(2a)}\right| = \left|\frac{5a^2}{2a^2}\right| = \frac{5}{2}

Substituting

b = 2a

allows

a^2

to cancel completely, leaving a constant numerical value.

Key Concept

Properties of Real Numbers and Absolute Value Equations

Estimated Time:2m 0s

Question 223Question →

A high-precision optical sensor measures a time interval as $T = \frac{0.00064 \times (2.5 \times 10^8)}{1.6 \times 10^{-1}}$ nanoseconds. Which of the following values are equivalent to $T$ ? Select all such values.

Select all that apply

1.0 \times 10^6

100,000

\frac{4.0 \times 10^7}{40}

(2.5 \times 10^3) \times (4.0 \times 10^2)

1.0 \times 10^{-6}

Show answer & explanation

Answer: $1.0 \times 10^6$ ; $\frac{4.0 \times 10^7}{40}$ ; $(2.5 \times 10^3) \times (4.0 \times 10^2)$

Answer

The values equivalent to

T

are

1.0 \times 10^6

,

\frac{4.0 \times 10^7}{40}

, and

(2.5 \times 10^3) \times (4.0 \times 10^2)

.

Evaluating

T

gives

\frac{(6.4 \times 10^{-4}) \times (2.5 \times 10^8)}{1.6 \times 10^{-1}} = \frac{1.6 \times 10^5}{1.6 \times 10^{-1}} = 1.0 \times 10^6

. The expression representing

1.0 \times 10^6

directly matches

T

. The quotient

\frac{4.0 \times 10^7}{40}

simplifies to

\frac{4.0 \times 10^7}{4.0 \times 10^1} = 1.0 \times 10^6

, which matches

T

. The product

(2.5 \times 10^3) \times (4.0 \times 10^2)

simplifies to

10.0 \times 10^5 = 1.0 \times 10^6

, which also matches

T

.

Step-by-Step Solution

1

Convert the decimal

0.00064

into scientific notation

0.00064 = 6.4 \times 10^{-4}

Expressing all terms in scientific notation simplifies exponent operations.

2

Simplify the numerator of the expression for

T

(6.4 \times 10^{-4}) \times (2.5 \times 10^8) = (6.4 \times 2.5) \times 10^{-4 + 8} = 16.0 \times 10^4 = 1.6 \times 10^5

Multiply coefficients directly and add exponents for product of powers with equal base.

3

Divide the numerator by the denominator

1.6 \times 10^{-1}

T = \frac{1.6 \times 10^5}{1.6 \times 10^{-1}} = \left(\frac{1.6}{1.6}\right) \times 10^{5 - (-1)} = 1.0 \times 10^6 = 1,000,000

Subtract the denominator exponent from the numerator exponent:

5 - (-1) = 6

.

4

Evaluate the choices to check equivalence to

1.0 \times 10^6

1.0 \times 10^6

,

\frac{4.0 \times 10^7}{40} = 1.0 \times 10^6

, and

(2.5 \times 10^3) \times (4.0 \times 10^2) = 10.0 \times 10^5 = 1.0 \times 10^6

are all equal to

T

.

Matching each simplified expression to the calculated value of

T

determines the correct options.

Key Concept

Operations with Decimals and Exponents in Scientific Notation

Question 224Question →

Two positive integers $x$ and $y$ have a greatest common divisor (GCD) of $12$ and a least common multiple (LCM) of $360$ . Which of the following could be the value of $x$ ? Indicate all such values.

Select all that apply

24

48

60

90

120

Show answer & explanation

Answer: $24$ ; $60$ ; $120$

Answer

The possible values for

x

are

24

,

60

, and

120

.

Any valid value of

x

must be a multiple of

\gcd(x, y) = 12

and a divisor of

\text{lcm}(x, y) = 360

. Expressing

12

as

2^2 \cdot 3^1

and

360

as

2^3 \cdot 3^2 \cdot 5^1

,

x

must be of the form

2^a \cdot 3^b \cdot 5^c

with

2 \leq a \leq 3

,

1 \leq b \leq 2

, and

0 \leq c \leq 1

. The values

24

,

60

, and

120

meet all exponent constraints.

Step-by-Step Solution

1

Find the prime factorizations of the given GCD and LCM.

\gcd(x, y) = 12 = 2^2 \cdot 3^1 \cdot 5^0

and

\text{lcm}(x, y) = 360 = 2^3 \cdot 3^2 \cdot 5^1

.

Prime factorization allows analysis of exponent constraints for each prime factor.

2

Determine the constraints on any valid integer

x

.

Any valid value of

x

must be a multiple of

12

and a factor of

360

. Specifically,

x = 2^a \cdot 3^b \cdot 5^c

where

2 \leq a \leq 3

,

1 \leq b \leq 2

, and

0 \leq c \leq 1

.

The GCD defines the minimum exponent for each prime factor, and the LCM defines the maximum exponent.

3

Test each provided option against the prime exponent bounds.

24 = 2^3 \cdot 3^1

satisfies the bounds (

a=3, b=1, c=0

).

48 = 2^4 \cdot 3^1

violates

a \leq 3

.

60 = 2^2 \cdot 3^1 \cdot 5^1

satisfies the bounds (

a=2, b=1, c=1

).

90 = 2^1 \cdot 3^2 \cdot 5^1

violates

a \geq 2

.

120 = 2^3 \cdot 3^1 \cdot 5^1

satisfies the bounds (

a=3, b=1, c=1

).

Options satisfying all exponent inequalities are valid possible values of

x

.

Key Concept

Prime exponent properties of GCD and LCM

Question 225Question →

Let $p$ , $q$ , and $r$ be integers such that $p < 0 < q < r$ . If $p$ is an odd integer, $q$ is an even integer, and $r$ is an odd integer, which of the following expressions must be negative? Select all that apply.

Select all that apply

p^q(q - r)

p(p - q)^r

(p - r)(-p)^r

p^r(r - p)^q

(-p)^q(q - p)

Show answer & explanation

Answer: $p^q(q - r)$ ; $(p - r)(-p)^r$ ; $p^r(r - p)^q$

Answer

The expressions that must be negative are

p^q(q - r)

,

(p - r)(-p)^r

, and

p^r(r - p)^q

.

Expressions

p^q(q - r)

,

(p - r)(-p)^r

, and

p^r(r - p)^q

evaluate to the product of a positive factor and a negative factor in every case, making their values strictly negative.

Step-by-Step Solution

1

Analyze the given signs and parities of variables

p < 0

(negative, odd),

q > 0

(positive, even),

r > 0

(positive, odd), with

q < r

.

Establishing the domain and sign/parity properties of each variable is essential before evaluating exponential and subtractive terms.

2

Evaluate the sign of

p^q(q - r)

p^q > 0

because an even exponent yields a positive result for non-zero bases. Since

q < r

,

(q - r) < 0

. Thus,

\text{positive} \times \text{negative} = \text{negative}

.

Demonstrates that this expression is guaranteed to be negative.

3

Evaluate the sign of

p(p - q)^r

p < 0

.

p - q < 0

, and raising a negative number to an odd exponent

r

gives a negative result. Thus,

p(p - q)^r = \text{negative} \times \text{negative} = \text{positive}

.

Shows that this expression is positive, so it cannot be negative.

4

Evaluate the sign of

(p - r)(-p)^r

p - r < 0

because subtracting a positive number from a negative number is negative.

-p > 0

, so

(-p)^r > 0

. Thus,

\text{negative} \times \text{positive} = \text{negative}

.

Demonstrates that this expression is guaranteed to be negative.

5

Evaluate the sign of

p^r(r - p)^q

p^r < 0

because a negative number raised to an odd exponent is negative.

r - p > 0

, so

(r - p)^q > 0

. Thus,

\text{negative} \times \text{positive} = \text{negative}

.

Demonstrates that this expression is guaranteed to be negative.

6

Evaluate the sign of

(-p)^q(q - p)

-p > 0 \implies (-p)^q > 0

.

q - p > 0

. Thus,

\text{positive} \times \text{positive} = \text{positive}

.

Shows that this expression is always positive.

Key Concept

Sign rules for bases raised to even vs. odd powers, and order of operations with signed quantities.

Question 226Question →

In Year 1, a non-profit organization received all of its funding from two sources: private donations and government grants. Private donations accounted for $60\%$ of the total funding, and government grants accounted for the remaining $40\%$ . In Year 2, private donations increased by $25\%$ compared to Year 1, while government grants decreased by $15\%$ compared to Year 1. Which of the following statements must be true? Select all such statements.

Select all that apply

The total funding received by the organization in Year 2 was

9\%

greater than the total funding received in Year 1.

In Year 2, private donations accounted for more than

68\%

of the total funding received in Year 2.

In Year 2, government grants accounted for

34\%

of the total funding received in Year 2.

In Year 2, private donations accounted for

75\%

of the total funding received in Year 2.

The ratio of private donations to government grants in Year 2 was

25\%

greater than the ratio of private donations to government grants in Year 1.

Show answer & explanation

Answer: The total funding received by the organization in Year 2 was $9\%$ greater than the total funding received in Year 1.; In Year 2, private donations accounted for more than $68\%$ of the total funding received in Year 2.

Answer

The correct statements are that the total funding in Year 2 was

9\%

greater than in Year 1, and that private donations accounted for more than

68\%

of the total funding in Year 2.

Total funding in Year 2 is

0.60(1.25) + 0.40(0.85) = 0.75 + 0.34 = 1.09

times the Year 1 total, which represents a

9\%

increase. Furthermore, private donations in Year 2 represent

\frac{0.75}{1.09} \approx 68.81\%

of the Year 2 total funding, which is greater than

68\%

.

Step-by-Step Solution

1

Define variables for Year 1 funding components relative to total Year 1 funding (

T_1

).

Private donations in Year 1:

P_1 = 0.60 T_1

. Government grants in Year 1:

G_1 = 0.40 T_1

.

Establishing Year 1 amounts in terms of total funding

T_1

provides the baseline for percentage changes.

2

Calculate Year 2 funding amounts after applying the respective percent changes.

P_2 = 1.25 \times 0.60 T_1 = 0.75 T_1

.

G_2 = 0.85 \times 0.40 T_1 = 0.34 T_1

.

Increasing private donations by

25\%

multiplies

P_1

by

1.25

; decreasing government grants by

15\%

multiplies

G_1

by

0.85

.

3

Compute total Year 2 funding (

T_2

) and compare it with

T_1

.

T_2 = P_2 + G_2 = 0.75 T_1 + 0.34 T_1 = 1.09 T_1

. Net change is a

9\%

increase.

Adding the components yields total Year 2 funding, confirming that

T_2

is

1.09

times

T_1

.

4

Calculate component shares as percentages of total Year 2 funding (

T_2

).

Share of private donations in Year 2 =

\frac{0.75 T_1}{1.09 T_1} \approx 68.81\%

. Share of government grants in Year 2 =

\frac{0.34 T_1}{1.09 T_1} \approx 31.19\%

.

Percentage share of a total requires dividing each component by the new total funding (

1.09 T_1

), not the original baseline (

T_1

).

Key Concept

Weighted percentage change and percentage base shift

Question 227Question →

At a pharmaceutical manufacturing facility, a liquid solution undergoes a three-stage purification process to remove a specific chemical compound. Stage I removes $20\%$ of the compound present at the start of Stage I. Stage II removes $25\%$ of the compound present at the start of Stage II. Stage III removes $30\%$ of the compound present at the start of Stage III. If $16.8\text{ grams}$ of the compound remain in the solution after Stage III, how many grams of the compound were present in the solution immediately before Stage I?

28.0\text{ grams}

33.6\text{ grams}

40.0\text{ grams}

56.0\text{ grams}

67.2\text{ grams}

Show answer & explanation

Answer: $40.0\text{ grams}$

Answer

40.0 grams

To find the initial quantity, we determine the compound fraction remaining after each stage. Stage I leaves

80\%

, Stage II leaves

75\%

, and Stage III leaves

70\%

. The overall fraction remaining after all three stages is

0.80 \times 0.75 \times 0.70 = 0.42

. Given that

0.42

of the original amount equals

16.8\text{ grams}

, dividing

16.8

by

0.42

gives the initial mass of

40.0\text{ grams}

.

Step-by-Step Solution

1

Determine the remaining fraction of the compound after each individual stage.

Stage I leaves

1 - 0.20 = 0.80

of its input; Stage II leaves

1 - 0.25 = 0.75

of its input; Stage III leaves

1 - 0.30 = 0.70

of its input.

Percentage decreases must be represented as multiplicative factors of the base amount entering each stage.

2

Calculate the combined multi-stage remaining multiplier.

Combined multiplier

= 0.80 \times 0.75 \times 0.70 = 0.42

(or

42\%

).

Successive percentage reductions compound multiplicatively, not additively.

3

Set up an equation relating the initial mass

X

to the final remaining mass.

0.42 X = 16.8\text{ grams}

.

The final mass is equal to the initial mass multiplied by the overall remaining decimal fraction.

4

Solve for the initial mass

X

.

X = \frac{16.8}{0.42} = 40.0\text{ grams}

.

Dividing the final amount by

0.42

yields the exact starting quantity.

Key Concept

Successive Percent Reductions and Base Shifts

Alternative Method

Work backwards from the final amount stage by stage: Before Stage III, mass

= 16.8 \div 0.70 = 24.0\text{ g}

. Before Stage II, mass

= 24.0 \div 0.75 = 32.0\text{ g}

. Before Stage I, mass

= 32.0 \div 0.80 = 40.0\text{ g}

.

Estimated Time:2m 0s

Question 228Question →

If $A = 0.00048 \times 10^{-3}$ and $B = 1.2 \times 10^{-5}$ , what is the value of $\frac{A + B}{4 \times 10^{-8}}$ expressed in scientific notation?

3.12 \times 10^2

1.5 \times 10^3

4.2 \times 10^2

3.12 \times 10^{-4}

3.12 \times 10^1

Show answer & explanation

Answer: $3.12 \times 10^2$

Answer

3.12 \times 10^2

Converting

A = 0.00048 \times 10^{-3}

to powers of 10 gives

4.8 \times 10^{-7}

, which equals

0.048 \times 10^{-5}

. Adding

B = 1.2 \times 10^{-5}

yields

(0.048 + 1.2) \times 10^{-5} = 1.248 \times 10^{-5}

. Dividing by

4 \times 10^{-8}

gives

\frac{1.248}{4} \times 10^{-5 - (-8)} = 0.312 \times 10^3 = 3.12 \times 10^2

.

Step-by-Step Solution

1

Express

A

in standard scientific notation and then match its exponent to

B

's exponent.

A = 0.00048 \times 10^{-3} = 4.8 \times 10^{-7} = 0.048 \times 10^{-5}

To perform addition between numbers in scientific notation, their exponents must be equal.

2

Add

A

and

B

.

A + B = 0.048 \times 10^{-5} + 1.2 \times 10^{-5} = 1.248 \times 10^{-5}

Combine the coefficients once the powers of 10 match.

3

Divide

A + B

by

4 \times 10^{-8}

and express the final result in scientific notation.

\frac{1.248 \times 10^{-5}}{4 \times 10^{-8}} = \left(\frac{1.248}{4}\right) \times 10^{-5 - (-8)} = 0.312 \times 10^3 = 3.12 \times 10^2

Divide the coefficients and subtract the exponent in the denominator from the exponent in the numerator.

Key Concept

Decimals, Place Value, and Operations in Scientific Notation

Question 229Question →

In 2024, a municipal water treatment plant processed water using two filtration systems: System A and System B. System A processed $60\%$ of the plant's total water volume, and System B processed the remaining $40\%$ . In 2025, the volume of water processed by System A increased by $20\%$ , while the volume of water processed by System B decreased by $15\%$ . What was the net percentage change in the total volume of water processed by the plant from 2024 to 2025?

An increase of

6\%

An increase of

5\%

A decrease of

1\%

An increase of

2.5\%

An increase of

3.5\%

Show answer & explanation

Answer: An increase of $6\%$

Answer

An increase of

6\%

To find the net percentage change, calculate the weighted contribution of each component change relative to the initial total. Assuming a total volume of

100

units in 2024, System A processed

60

units and System B processed

40

units. In 2025, System A processed

60 \times 1.20 = 72

units, and System B processed

40 \times 0.85 = 34

units. The new total volume is

72 + 34 = 106

units. The net change from

100

to

106

represents an increase of

6\%

.

Step-by-Step Solution

1

Assume a convenient baseline total volume for 2024.

Let the total volume in 2024 be

100

units. System A processes

60

units and System B processes

40

units.

Choosing

100

simplifies percentage calculations.

2

Calculate the volume processed by each system in 2025.

System A:

60 \times (1 + 0.20) = 72

units.
System B:

40 \times (1 - 0.15) = 34

units.

Apply the respective percentage increase and decrease to each system's 2024 volume.

3

Calculate the total volume processed in 2025 and the net percentage change.

Total 2025 volume =

72 + 34 = 106

units.
Net change =

106 - 100 = +6

units, which is an increase of

\frac{6}{100} \times 100\% = 6\%

.

Compare the new total volume to the original baseline total volume.

Key Concept

Weighted Percent Change

Estimated Time:1m 30s

Question 230Question →

A renewable energy facility stores electricity in three separate battery modules: Module $X$ , Module $Y$ , and Module $Z$ . The ratio of the initial energy stored in Module $X$ to Module $Y$ is $2 : 3$ , and the ratio of the initial energy stored in Module $Y$ to Module $Z$ is $5 : 8$ . Each module discharges its stored energy at a constant individual rate. Operating alone, Module $Y$ can completely discharge its initial stored energy in $5$ hours. Module $X$ discharges energy at a rate $33\frac{1}{3}\%$ greater than Module $Y$ , and Module $Z$ discharges energy at a rate $25\%$ less than the combined discharge rate of Modules $X$ and $Y$ . If all three modules begin discharging simultaneously, how many hours will it take to completely discharge the total initial energy stored across all three modules?

3.5 hours

4 hours

5.25 hours

6 hours

7.5 hours

Show answer & explanation

Answer: 4 hours

Answer

4 hours

To find the total time required, first unify the two given ratios. Since

X : Y = 2 : 3

and

Y : Z = 5 : 8

, express both ratios with a common value for

Y

(

15

). This yields

X : Y : Z = 10 : 15 : 24

, so total energy is

10k + 15k + 24k = 49k

. Next, determine rates in terms of

k

: Module

Y

discharges

15k

in

5

hours, so its rate is

3k

units/hr. Module

X

discharges at a rate

33\frac{1}{3}\%

greater than Module

Y

, giving

3k \times \frac{4}{3} = 4k

units/hr. The combined rate of

X

and

Y

is

4k + 3k = 7k

units/hr. Module

Z

discharges at

25\%

less than this combined rate, giving

7k \times 0.75 = 5.25k

units/hr. The simultaneous discharge rate of all three modules is

4k + 3k + 5.25k = 12.25k

units/hr. Dividing total energy

49k

by

12.25k

gives exactly 4 hours.

Step-by-Step Solution

1

Determine the unified initial energy ratio among all three modules.

Energy ratio Module

X

: Module

Y

: Module

Z

=

10 : 15 : 24

, giving a total initial energy of

49k

units.

Module

X : Y = 2 : 3 = 10 : 15

and Module

Y : Z = 5 : 8 = 15 : 24

. Combining these gives

X : Y : Z = 10 : 15 : 24

for a common constant

k

.

2

Calculate the individual discharge rate of Module Y.

Discharge rate of Module

Y

=

3k

energy units per hour.

Module

Y

has

15k

energy units and discharges completely in

5

hours, so

\text{Rate}_Y = \frac{15k}{5} = 3k

.

3

Determine the discharge rates of Module X and Module Z.

Discharge rate of Module

X

=

4k

units per hour; discharge rate of Module

Z

=

5.25k

units per hour.

Module

X

rate is

33\frac{1}{3}\%

greater than Module

Y

:

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

. Combined rate of

X

and

Y

is

4k + 3k = 7k

. Module

Z

rate is

25\%

less than this combined rate:

\text{Rate}_Z = 7k \times 0.75 = 5.25k

.

4

Calculate total combined discharge rate and total time required.

Combined rate =

12.25k

units per hour; Total time =

4

hours.

Total rate

= 4k + 3k + 5.25k = 12.25k

units per hour. Total time

= \frac{\text{Total Energy}}{\text{Total Rate}} = \frac{49k}{12.25k} = 4

hours.

Key Concept

Combining multi-part ratios into a unified scale and calculating combined work rates with percentage adjustments

Question 231Question →

The greatest common divisor of a positive integer $n$ and $120$ is $24$ , and their least common multiple is $2520$ . How many positive divisors does $n$ have?

6

9

12

24

48

Show answer & explanation

Answer: 24

Answer

24

First, write the prime factorizations of

120

, the

\text{gcd}

, and the

\text{lcm}

:
-

120 = 2^3 \cdot 3^1 \cdot 5^1

-

\text{gcd}(n, 120) = 24 = 2^3 \cdot 3^1

-

\text{lcm}(n, 120) = 2520 = 2^3 \cdot 3^2 \cdot 5^1 \cdot 7^1

For any prime

p

,

\min(v_p(n), v_p(120)) = v_p(\text{gcd})

and

\max(v_p(n), v_p(120)) = v_p(\text{lcm})

:
- For

p=2

:

\min(v_2(n), 3) = 3

and

\max(v_2(n), 3) = 3 \implies v_2(n) = 3

.
- For

p=3

:

\min(v_3(n), 1) = 1

and

\max(v_3(n), 1) = 2 \implies v_3(n) = 2

.
- For

p=5

:

\min(v_5(n), 1) = 0 \implies v_5(n) = 0

.
- For

p=7

:

\max(v_7(n), 0) = 1 \implies v_7(n) = 1

.

Thus,

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

. The number of positive divisors of

n

is

(3+1)(2+1)(1+1) = 4 \cdot 3 \cdot 2 = 24

. Therefore, the option specifying 24 is correct.

Step-by-Step Solution

1

Find the prime factorizations of

120

,

\text{gcd}(n, 120)

, and

\text{lcm}(n, 120)

.

120 = 2^3 \cdot 3^1 \cdot 5^1

,

\text{gcd} = 24 = 2^3 \cdot 3^1

, and

\text{lcm} = 2520 = 2^3 \cdot 3^2 \cdot 5^1 \cdot 7^1

.

Expressing given quantities in terms of prime factors allows exact determination of the exponents in

n

.

2

Determine the prime factorization of

n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d

.

a = 3

,

b = 2

,

c = 0

, and

d = 1

, giving

n = 2^3 \cdot 3^2 \cdot 7^1 = 504

.

Since

\min(a, 3) = 3

and

\max(a, 3) = 3

,

a=3

. Since

\min(b, 1) = 1

and

\max(b, 1) = 2

,

b=2

. Since

\min(c, 1) = 0

,

c=0

. Since

\max(d, 0) = 1

,

d=1

.

3

Calculate the number of positive divisors of

n

.

(3 + 1)(2 + 1)(1 + 1) = 4 \cdot 3 \cdot 2 = 24

.

The total number of positive divisors for an integer

p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}

is

(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)

.

Key Concept

Prime Factorization, Greatest Common Divisor (GCD), Least Common Multiple (LCM), and Divisor Counting

Estimated Time:1m 30s

Question 232Question →

A logistics company operates a fleet of $14$ vans consisting solely of Model $X$ and Model $Y$ vans.

- The ratio of the cargo volume capacity of one Model $X$ van to one Model $Y$ van is $3 : 5$ .
- The ratio of the energy consumption rate (in kWh per mile) of one Model $X$ van to one Model $Y$ van is $4 : 5$ .
- When fully loaded, a Model $X$ van carries $30\text{ m}^3$ of cargo and a Model $Y$ van carries $50\text{ m}^3$ of cargo.
- The entire fleet of $14$ vans, when fully loaded, transports a total of $620\text{ m}^3$ of cargo in one trip.
- A Model $X$ van consumes $2\text{ kWh}$ of energy per mile.

Which of the following statements must be true? Select all such statements.

Select all that apply

There are exactly

4

Model

X

vans in the fleet.

Model

X

vans account for

\frac{3}{8}

of the total cargo capacity transported by the fleet.

The total energy consumed by the entire fleet per mile is

33\text{ kWh}

.

The average energy consumption rate per van in the fleet is

2.25\text{ kWh}

per mile.

If the cargo capacity of each Model

Y

van is increased by

20\%

, the total cargo capacity of the entire fleet would increase by

20\%

.

Show answer & explanation

Answer: There are exactly $4$ Model $X$ vans in the fleet.; The total energy consumed by the entire fleet per mile is $33\text{ kWh}$ .

Answer

The correct statements are that there are exactly 4 Model X vans in the fleet and that the total energy consumed by the entire fleet per mile is 33 kWh.

The statement specifying that there are exactly 4 Model X vans is correct because solving the system

n_X + n_Y = 14

and

30 n_X + 50 n_Y = 620

gives

n_X = 4

. The statement asserting that total fleet energy consumption per mile is 33 kWh is also correct because

4(2\text{ kWh}) + 10(2.5\text{ kWh}) = 33\text{ kWh}

.

Step-by-Step Solution

1

Determine the number of Model X and Model Y vans using fleet totals.

Let

n_X

be the number of Model X vans and

n_Y

be the number of Model Y vans. Given

n_X + n_Y = 14

, substitute

n_Y = 14 - n_X

into the total cargo equation:

30 n_X + 50(14 - n_X) = 620 \implies 700 - 20 n_X = 620 \implies 20 n_X = 80 \implies n_X = 4

. Thus,

n_Y = 10

.

This establishes the exact composition of the fleet.

2

Evaluate the fraction of cargo carried by Model X vans.

Total cargo carried by Model X vans =

4 \times 30 = 120\text{ m}^3

. Fleet total cargo =

620\text{ m}^3

. Fraction =

\frac{120}{620} = \frac{6}{31}

.

Shows that the fleet cargo share depends on the van counts, not just the single-van ratio of 3 to 5.

3

Calculate individual consumption rates and total fleet consumption per mile.

The consumption rate ratio of Model X to Model Y is

4 : 5

. Given Model X rate is

2\text{ kWh/mile}

, Model Y rate is

2 \times \frac{5}{4} = 2.5\text{ kWh/mile}

. Total fleet consumption per mile =

4(2) + 10(2.5) = 8 + 25 = 33\text{ kWh}

.

Applies unit rates to the known number of each van model to find total rate.

4

Calculate the true weighted average consumption rate per van.

Average consumption rate per van =

\frac{\text{Total energy per mile}}{\text{Total vans}} = \frac{33\text{ kWh}}{14\text{ vans}} \approx 2.36\text{ kWh/mile}

.

Demonstrates why taking an unweighted mean

(2.0 + 2.5)/2 = 2.25

is incorrect when group sizes are unequal.

5

Analyze the impact of a 20% increase in Model Y capacity.

Initial Model Y total capacity =

10 \times 50 = 500\text{ m}^3

. A

20\%

increase adds

0.20 \times 500 = 100\text{ m}^3

. Percent increase in total fleet capacity =

\frac{100}{620} \approx 16.13\%

.

A percentage change in a part does not equal the percentage change in the whole.

Key Concept

Weighted averages and component vs. whole relationships in multi-part ratios and rates.

Question 233Question →

A sequence $a_1, a_2, a_3, \dots$ is defined by $a_1 = 3$ and $a_{n+1} = \sqrt{a_n + 12}$ for all integers $n \geq 1$ . If $S_{100} = \sum_{k=1}^{100} a_k$ , what is the value of $S_{100}$ rounded to the nearest integer?

396

398

399

400

401

Show answer & explanation

Answer: 399

Answer

399

The sequence terms approach the fixed limit

L = 4

defined by

L = \sqrt{L + 12}

. Each term

a_k

is strictly less than 4, with difference

\epsilon_k = 4 - a_k

. Evaluating the first few differences yields

\epsilon_1 = 1

,

\epsilon_2 \approx 0.127

,

\epsilon_3 \approx 0.016

,

\epsilon_4 \approx 0.002

, and subsequent terms are negligible. The total sum

S_{100} = 100 \times 4 - \sum_{k=1}^{100} \epsilon_k \approx 400 - 1.145 = 398.855

. Rounding

398.855

to the nearest integer gives 399.

Step-by-Step Solution

1

Determine the limit/fixed point of the recursive sequence.

Solving

L = \sqrt{L + 12}

yields

L^2 - L - 12 = 0 \implies (L - 4)(L + 3) = 0

. Since terms are positive,

L = 4

.

Understanding the upper limit

L = 4

helps set up a upper bound for all terms in the sequence.

2

Calculate the first few terms and their deficits

\epsilon_k = 4 - a_k

from the limit 4.

\epsilon_1 = 4 - 3 = 1

;

\epsilon_2 = 4 - \sqrt{15} \approx 0.12702

;

\epsilon_3 = 4 - \sqrt{15.87298} \approx 0.01591

;

\epsilon_4 \approx 0.00199

;

\epsilon_5 \approx 0.00025

.

Because convergence is extremely fast (each error term shrinks by a factor greater than 8), only the first few deficit terms contribute meaningfully to the sum.

3

Sum the total deficit

\sum_{k=1}^{100} \epsilon_k

.

Total deficit

\approx 1 + 0.12702 + 0.01591 + 0.00199 + 0.00025 + \dots \approx 1.145

.

Adding all error terms gives the precise deviation of

S_{100}

from

100 \times 4 = 400

.

4

Compute

S_{100}

and round to the nearest integer.

S_{100} = 400 - 1.145 = 398.855

. Rounded to the nearest integer,

398.855

rounds to

399

.

Since

398.855 \geq 398.5

, rounding to the nearest integer yields 399.

Key Concept

Fixed-point sequence convergence, summation bounding, and rounding

Estimated Time:2m 30s

Question 234Question →

A manufacturing facility produces precision metal components. Each component has a mass of $8.4 \times 10^{-4}$ kilograms. If a shipment container holds a batch of these components with a total mass of $1.05$ kilograms, how many components are in the container?

Show answer & explanation

Answer: 1250

Answer

The container holds 1,250 components.

To find the number of components, divide the total batch mass by the mass per component:

\frac{1.05 \text{ kg}}{8.4 \times 10^{-4} \text{ kg}}

. Expressing

\frac{1}{10^{-4}}

as

10^4

transforms the expression into

\frac{1.05}{8.4} \times 10^4

. Dividing

1.05

by

8.4

yields

0.125

. Finally,

0.125 \times 10^4 = 1250

.

Step-by-Step Solution

1

Set up the ratio of total mass to single component mass

Number of components =

\frac{1.05}{8.4 \times 10^{-4}}

Dividing total mass by individual component mass gives the total count.

2

Apply exponent rules to move the power of 10 to the numerator

\frac{1.05}{8.4} \times 10^4

Since

\frac{1}{10^{-4}} = 10^4

, shifting the negative exponent to the numerator changes its sign.

3

Divide the decimal coefficients

\frac{1.05}{8.4} = 0.125

Simplifying

\frac{105}{840}

reduces to

\frac{1}{8} = 0.125

.

4

Evaluate the product with the place value shift

0.125 \times 10,000 = 1250

Multiplying by

10^4

shifts the decimal point 4 places to the right.

Key Concept

Division with scientific notation and decimal place value adjustment

Estimated Time:1m 30s

Question 235Question →

A commercial property was purchased at an initial value of $V$ dollars. In its first year, the value of the property increased by $25\%$ . In its second year, the property's value decreased by $20\%$ relative to its value at the end of the first year. In its third year, the value of the property increased by $15\%$ relative to its value at the end of the second year. If the value of the property at the end of the third year was $\$276,000$ , what was the initial purchase value $V$ , in dollars, of the property?

Show answer & explanation

Answer: 240000

Answer

240000

Each percent change applies to the value at the end of the previous period. A

25\%

increase scales the value by

1.25

, a

20\%

decrease scales it by

0.80

, and a

15\%

increase scales it by

1.15

. The net multiplier is

1.25 \times 0.80 \times 1.15 = 1.15

. Solving

1.15V = 276,000

gives

V = 240,000

.

Step-by-Step Solution

1

Convert percent changes to growth and decay multiplier factors

Year 1 factor = 1.25, Year 2 factor = 0.80, Year 3 factor = 1.15

Percent changes are compounded by multiplying successive growth/decay factors.

2

Multiply the factors sequentially to express the final value in terms of initial value V

Final Value = V × 1.25 × 0.80 × 1.15 = 1.15V

Combining successive multipliers simplifies the relationship between original and final amounts.

3

Set the algebraic expression equal to $276,000 and solve for V

V = 276,000 / 1.15 = 240,000

Dividing the final value by the overall multiplier determines the original base value.

Key Concept

Successive Percent Changes and Base Shift

Estimated Time:1m 30s

Question 236Question →

A chemical synthesis plant uses two reaction vessels, Vessel 1 and Vessel 2, to produce a liquid solution composed entirely of Compound X, Compound Y, and water.

- Vessel 1 produces the solution at a constant rate of $60$ liters per hour, with Compound X, Compound Y, and water in the volume ratio $2 : 1 : 3$ , respectively.
- Vessel 2 produces the solution at a constant rate of $90$ liters per hour, with Compound X, Compound Y, and water in the volume ratio $1 : 3 : 2$ , respectively.

If both vessels operate simultaneously to fill an initially empty storage tank, which of the following statements must be true regarding the mixture in the storage tank after operating for any duration $H$ hours ( $H > 0$ )? Select all such statements.

Select all that apply

Compound Y accounts for exactly

\frac{11}{30}

of the total volume of solution in the storage tank.

The ratio of the volume of Compound X to the volume of water in the storage tank is

7 : 12

.

The volume of Compound X in the storage tank is greater than the volume of Compound Y in the storage tank.

Water constitutes strictly more than

40\%

of the total volume of solution in the storage tank.

Compound X represents exactly

\frac{7}{18}

of the total volume of solution in the storage tank.

Show answer & explanation

Answer: Compound Y accounts for exactly $\frac{11}{30}$ of the total volume of solution in the storage tank.; The ratio of the volume of Compound X to the volume of water in the storage tank is $7 : 12$ .

Answer

The statements confirming that Compound Y accounts for exactly 11/30 of the total solution volume and that the ratio of Compound X to water is 7:12 are both correct.

The total volume rate of solution produced by both vessels combined is 150 liters per hour. Within this total, Compound X is produced at 35 liters per hour, Compound Y at 55 liters per hour, and water at 60 liters per hour. Consequently, Compound Y accounts for 55 / 150 = 11/30 of the total volume, and the ratio of Compound X to water is 35 : 60 = 7 : 12. Both of these statements must be true.

Step-by-Step Solution

1

Calculate the hourly output rates for each component from Vessel 1.

Total rate = 60 L/hr. Ratio X:Y:Water = 2:1:3 (sum of parts = 6). Compound X = 60 * (2/6) = 20 L/hr; Compound Y = 60 * (1/6) = 10 L/hr; Water = 60 * (3/6) = 30 L/hr.

Decompose the total production rate into component rates using part-to-whole fractions.

2

Calculate the hourly output rates for each component from Vessel 2.

Total rate = 90 L/hr. Ratio X:Y:Water = 1:3:2 (sum of parts = 6). Compound X = 90 * (1/6) = 15 L/hr; Compound Y = 90 * (3/6) = 45 L/hr; Water = 90 * (2/6) = 30 L/hr.

Decompose the second vessel's production rate into component rates.

3

Combine the component rates from both vessels.

Total solution rate = 60 + 90 = 150 L/hr. Compound X = 20 + 15 = 35 L/hr; Compound Y = 10 + 45 = 55 L/hr; Water = 30 + 30 = 60 L/hr.

Find the overall composition delivered to the storage tank per hour.

4

Evaluate each candidate statement against the combined component rates.

Fraction of Y = 55 / 150 = 11/30 (True). Ratio X to Water = 35 : 60 = 7 : 12 (True). X (35 L/hr) > Y (55 L/hr) (False). Water percentage = 60 / 150 = 40% (Not strictly greater than 40%, so False). Fraction of X = 35 / 150 = 7/30 != 7/18 (False).

Determine which statements are mathematically true for any operating duration H > 0.

Key Concept

Weighted rate aggregation and multi-component ratio analysis

Question 237Question →

If $x$ is a real number such that $0 < x < 1$ , which of the following expressions is equivalent to $\sqrt{\frac{x^{-2} - 1}{x^{-2} + 2 + x^2}}$ ?

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

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

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

\frac{\sqrt{1-x^2}}{(1+x)^2}

\frac{1-x}{1+x}

Show answer & explanation

Answer: $\frac{\sqrt{1-x^2}}{1+x^2}$

Answer

The expression

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

is equivalent to the given radical expression.

Expressing

x^{-2}

as

\frac{1}{x^2}

allows the numerator to be rewritten as

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

and the denominator as

\frac{(1+x^2)^2}{x^2}

. Dividing these fractions cancels out

x^2

, leaving

\frac{1-x^2}{(1+x^2)^2}

under the radical. Taking the square root of the numerator and denominator separately gives

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

.

Step-by-Step Solution

1

Rewrite negative exponents as fractions in both the numerator and denominator.

Numerator:

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

. Denominator:

x^{-2} + 2 + x^2 = \frac{1}{x^2} + 2 + x^2 = \frac{1 + 2x^2 + x^4}{x^2}

.

Converting negative exponents into positive fractional exponents allows common denominators to be established.

2

Factor the perfect square trinomial in the denominator.

The numerator of the denominator expression is

1 + 2x^2 + x^4 = (1 + x^2)^2

. Thus, the entire denominator is

\frac{(1+x^2)^2}{x^2}

.

Recognizing

1 + 2x^2 + x^4

as

(1+x^2)^2

simplifies taking the square root.

3

Simplify the quotient inside the radical.

\frac{\frac{1-x^2}{x^2}}{\frac{(1+x^2)^2}{x^2}} = \frac{1-x^2}{(1+x^2)^2}.

Canceling the common factor of

x^2

in the denominators reduces the nested fraction.

4

Apply the square root rule

\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

.

\sqrt{\frac{1-x^2}{(1+x^2)^2}} = \frac{\sqrt{1-x^2}}{\sqrt{(1+x^2)^2}} = \frac{\sqrt{1-x^2}}{1+x^2}.

Since

1+x^2 > 0

for all real

x

,

\sqrt{(1+x^2)^2} = 1+x^2

.

Key Concept

Simplifying radical expressions containing negative powers and algebraic fractions

Estimated Time:2m 0s

Question 238Question →

Three decorative light signals flash at regular intervals of $15$ seconds, $20$ seconds, and $36$ seconds, respectively. If all three signals flash simultaneously at 12:00 PM, how many times will all three signals flash simultaneously between 12:01 PM and 1:00 PM, inclusive?

Show answer & explanation

Answer: 20

Answer

20

The three signals flash together every 180 seconds (3 minutes), which is the least common multiple of 15, 20, and 36. Within the 60-minute interval from 12:01 PM to 1:00 PM inclusive, simultaneous flashes occur at minute 3, 6, 9, ..., up to minute 60, resulting in exactly 20 simultaneous flashes.

Step-by-Step Solution

1

Find the prime factorization of each interval in seconds

15 = 3 \cdot 5

,

20 = 2^2 \cdot 5

,

36 = 2^2 \cdot 3^2

Prime factorization allows systematic calculation of the least common multiple.

2

Calculate the LCM of 15, 20, and 36

\text{LCM} = 2^2 \cdot 3^2 \cdot 5 = 180

seconds

The LCM gives the minimum period of time after which all three signals flash together.

3

Convert the period into minutes and find the frequency in 60 minutes

180\text{ seconds} = 3\text{ minutes}

;

60 \div 3 = 20

flashes

The timeframe between 12:01 PM and 1:00 PM inclusive contains 60 minutes, yielding 20 multiples of 3 minutes.

Key Concept

Least Common Multiple (LCM) for periodic events

Question 239Question →

Let $x$ and $y$ be integers such that $x < 0 < y$ . If $x$ and $y$ satisfy all of the following conditions:

1. $(-1)^{x y + x} = -1$
2. $(-1)^{x^2 y + y} = 1$
3. $y^2 - x^2 = 19$

What is the value of $x + y$ ?

Show answer & explanation

Answer: 1

Answer

The value of

x + y

is 1.

Condition 1 dictates that

(-1)^{x y + x} = -1

, meaning

x y + x = x(y + 1)

is odd. For the product

x(y+1)

to be odd, both

x

and

y+1

must be odd, which means

x

is odd and

y

is even. Condition 3 factors as

(y - x)(y + x) = 19

. Given

x < 0 < y

, we know

y - x > y + x

. Since 19 is prime, its unique positive factor pair requires

y - x = 19

and

y + x = 1

. Solving this system yields

y = 10

and

x = -9

, which satisfies all sign and parity constraints. Thus,

x + y = -9 + 10 = 1

.

Step-by-Step Solution

1

Analyze parity requirements from Condition 1

x

is odd and

y

is even

Since

(-1)^{x(y+1)} = -1

, the exponent

x(y+1)

must be odd, requiring both

x

and

y+1

to be odd.

2

Check consistency with Condition 2

Condition 2 is satisfied

x^2 y + y = y(x^2 + 1)

is always even when

y

is even, making

(-1)^{x^2 y + y} = 1

true.

3

Factor difference of squares and set up system using sign rules

y - x = 19

and

y + x = 1

Since 19 is prime and

x < 0 < y

,

y - x > y + x > 0

, forcing the factor pair to be 19 and 1.

4

Solve for

x

and

y

and sum them

x = -9

,

y = 10

, giving

x + y = 1

Adding the system yields

2y = 20 \implies y = 10

, and substituting into

y + x = 1

yields

x = -9

.

Key Concept

Even-Odd Exponent Rules and Sign Properties of Integers

Question 240Question →

If

x

is a real number such that

\sqrt{\frac{25^{x+1} - 25^x}{5^{x+2} + 5^{x+1}}} = 250

what is the value of

x

?

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

Factoring out common powers in the numerator and denominator yields

25^x(25-1) = 24 \cdot 5^{2x}

and

5^{x+1}(5+1) = 6 \cdot 5^{x+1}

. Simplifying their ratio inside the square root gives

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{x-1}

. Taking the square root gives

2 \cdot 5^{(x-1)/2}

. Setting this equal to 250 yields

5^{(x-1)/2} = 125 = 5^3

. Equating the exponents gives

(x-1)/2 = 3

, which solves to

x = 7

.

Step-by-Step Solution

1

Factor the numerator and express terms with a common base of 5

25^{x+1} - 25^x = 25^x(25 - 1) = 24 \cdot (5^2)^x = 24 \cdot 5^{2x}

Factoring out

25^x

simplifies the difference into a single term with base 5.

2

Factor the denominator using base 5

5^{x+2} + 5^{x+1} = 5^{x+1}(5 + 1) = 6 \cdot 5^{x+1}

Factoring out the common power

5^{x+1}

simplifies the sum into a single term.

3

Simplify the fraction inside the square root

\frac{24 \cdot 5^{2x}}{6 \cdot 5^{x+1}} = 4 \cdot 5^{2x - (x+1)} = 4 \cdot 5^{x-1}

Dividing coefficients (24/6 = 4) and applying exponent rules for division (

a^m / a^n = a^{m-n}

).

4

Take the square root of the simplified expression

\sqrt{4 \cdot 5^{x-1}} = \sqrt{4} \cdot \sqrt{5^{x-1}} = 2 \cdot 5^{\frac{x-1}{2}}

Using radical rules

\sqrt{ab} = \sqrt{a}\sqrt{b}

and

\sqrt{a^k} = a^{k/2}

.

5

Set the simplified radical expression equal to 250 and solve for x

2 \cdot 5^{\frac{x-1}{2}} = 250 \implies 5^{\frac{x-1}{2}} = 125 \implies 5^{\frac{x-1}{2}} = 5^3 \implies \frac{x-1}{2} = 3 \implies x = 7

Dividing both sides by 2 gives

5^{(x-1)/2} = 125 = 5^3

. Equating exponents yields

(x-1)/2 = 3

, so

x = 7

.

Key Concept

Exponent rules, base conversion, factoring exponential terms, and radical simplification

Estimated Time:2m 0s

Arithmetic

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Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution