Arithmetic Practice Questions — GRE General Test — Page 6

Question 101Question →

An asset management firm divides its total portfolio into three distinct funds: Fund X, Fund Y, and Fund Z. Initially, $\frac{1}{3}$ of the total portfolio value is in Fund X, and $\frac{2}{5}$ is in Fund Y, with the remaining fraction in Fund Z. Over the course of one year, the value of Fund X increases by $\frac{1}{5}$ , the value of Fund Y decreases by $\frac{1}{4}$ , and the value of Fund Z increases by $\frac{1}{2}$ . At the end of the year, what fraction of the portfolio's total value is contained in Fund Z?

\frac{4}{11}

\frac{2}{5}

\frac{4}{7}

\frac{4}{15}

\frac{15}{16}

Show answer & explanation

Answer: $\frac{4}{11}$

Answer

The fraction of the portfolio's total final value contained in Fund Z is

\frac{4}{11}

.

Fund Z grows to

\frac{2}{5}

of the original portfolio value, while the overall portfolio value increases to

\frac{11}{10}

of the original total. Comparing Fund Z's final value to the new total yields

\frac{2/5}{11/10} = \frac{4}{11}

.

Step-by-Step Solution

1

Calculate the initial fraction of the total portfolio allocated to Fund Z.

The initial allocation for Fund Z is

1 - \left(\frac{1}{3} + \frac{2}{5}\right) = 1 - \frac{11}{15} = \frac{4}{15}

.

The sum of all initial fund allocations must equal 1.

2

Determine the final value of each fund relative to the initial total portfolio value

V

.

Fund X:

\frac{1}{3}V \times \left(1 + \frac{1}{5}\right) = \frac{2}{5}V

. Fund Y:

\frac{2}{5}V \times \left(1 - \frac{1}{4}\right) = \frac{3}{10}V

. Fund Z:

\frac{4}{15}V \times \left(1 + \frac{1}{2}\right) = \frac{2}{5}V

.

Multiply each initial fund share by its respective growth or reduction factor.

3

Calculate the new total portfolio value relative to the initial value

V

.

Total final value =

\frac{2}{5}V + \frac{3}{10}V + \frac{2}{5}V = \frac{4}{10}V + \frac{3}{10}V + \frac{4}{10}V = \frac{11}{10}V

.

Sum the final values of all three individual funds.

4

Compute the final fraction of the portfolio contained in Fund Z.

Fraction =

\frac{\frac{2}{5}V}{\frac{11}{10}V} = \frac{2}{5} \times \frac{10}{11} = \frac{4}{11}

.

Divide the final value of Fund Z by the new total portfolio value.

Key Concept

Rational number operations and shifting base values in multi-step fraction problems

Estimated Time:2m 0s

Question 102Question →

If $k$ is a positive integer such that $k$ is divisible by $15$ and $k^2$ is divisible by $360$ , what is the minimum possible number of positive divisors of $k$ ?

6

8

12

18

24

Show answer & explanation

Answer: 12

Answer

The minimum possible number of positive divisors of

k

is 12.

The minimum possible number of positive divisors occurs when

k

has the smallest possible prime exponents satisfying all divisibility conditions. Since

15 = 3 \times 5

divides

k

,

k

must have prime factors

3^1

and

5^1

. For

k^2

to be divisible by

360 = 2^3 \times 3^2 \times 5^1

,

k^2

must have at least

2^3

, which requires

k

to have at least

2^2 = 4

. Thus, the minimal prime factorization of

k

is

k = 2^2 \times 3^1 \times 5^1 = 60

. Using the divisor count formula

(e_1 + 1)(e_2 + 1)(e_3 + 1)

, we get

(2+1)(1+1)(1+1) = 12

positive divisors.

Step-by-Step Solution

1

Find the prime factorizations of 15 and 360.

15 = 3^1 \times 5^1

and

360 = 2^3 \times 3^2 \times 5^1

.

Expressing numbers in terms of prime factors reveals the required prime exponents for divisibility.

2

Determine the minimum exponents of the prime factors required for

k

.

Since

15

divides

k

,

k

must contain at least

3^1

and

5^1

. For

k^2

to be divisible by

360 = 2^3 \times 3^2 \times 5^1

,

k^2

must contain at least

2^3

, which means

k

must contain at least

2^2

. The required factors

3^2

and

5^1

in

k^2

are automatically provided since

(3^1)^2 = 3^2

and

(5^1)^2 = 5^2

. Thus, the minimal

k

is

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

.

Taking the minimum required exponent for each prime factor minimizes the total number of divisors of

k

.

3

Calculate the number of positive divisors of the minimal

k

.

The number of divisors for

k = 2^2 \times 3^1 \times 5^1

is

(2 + 1)(1 + 1)(1 + 1) = 3 \times 2 \times 2 = 12

.

The divisor count formula adds 1 to each prime exponent and multiplies the results.

Key Concept

Properties of Integer Divisibility and Divisor Counting

Estimated Time:1m 30s

Question 103Question →

What is the value of $\frac{4.8 \times 10^{-3}}{1.2 \times 10^{-7}}$ expressed in scientific notation?

4.0 \times 10^4

4.0 \times 10^{-10}

4.0 \times 10^{-4}

3.6 \times 10^4

4.0 \times 10^3

Show answer & explanation

Answer: $4.0 \times 10^4$

Answer

4.0 \times 10^4

Dividing

4.8 \times 10^{-3}

by

1.2 \times 10^{-7}

requires dividing the coefficient

4.8

by

1.2

to obtain

4.0

, and subtracting the exponent of the denominator from that of the numerator:

-3 - (-7) = 4

. This yields

4.0 \times 10^4

.

Step-by-Step Solution

1

Divide the numerical coefficients.

4.8 \div 1.2 = 4.0

When dividing numbers in scientific notation, divide the coefficients independently from the powers of ten.

2

Apply exponent rules to divide the powers of ten.

10^{-3} \div 10^{-7} = 10^{-3 - (-7)} = 10^4

Subtract the exponent of the denominator from the exponent of the numerator.

3

Combine the coefficient and power of ten into scientific notation.

4.0 \times 10^4

The coefficient

4.0

is between

1

and

10

, so the expression is in proper standard scientific notation form.

Key Concept

Division of numbers in scientific notation

Question 104Question →

The price of an item was increased by $25\%$ and then the new price was decreased by $20\%$ . Which of the following statements must be true regarding the final price of the item relative to its initial price? Select all such statements.

Select all that apply

The final price of the item is equal to its initial price.

The overall net percent change in the price of the item is

0\%

.

The ratio of the final price to the initial price is

1

to

1

.

The final price of the item is

5\%

greater than its initial price.

The final price of the item is

5\%

less than its initial price.

Show answer & explanation

Answer: The final price of the item is equal to its initial price.; The overall net percent change in the price of the item is $0\%$ .; The ratio of the final price to the initial price is $1$ to $1$ .

Answer

The statements confirming that the final price equals the initial price, that the overall net percent change is 0%, and that the ratio of the final price to initial price is 1 to 1 are all correct.

Let

P

be the original price. After a

25\%

increase, the price becomes

1.25P

. Decreasing this intermediate price by

20\%

yields

1.25P \times 0.80 = 1.00P

. Because the final price is identical to the initial price, the final price is equal to the initial price, the net percentage change is

0\%

, and the ratio of final to initial price is

1:1

.

Step-by-Step Solution

1

Represent the initial price with a variable

Let the initial price of the item be

P

.

Using a variable makes tracking successive percent changes straightforward.

2

Apply the

25\%

increase

Increased price

= P \times (1 + 0.25) = 1.25P

.

A

25\%

increase corresponds to multiplying the base by

1.25

.

3

Apply the

20\%

decrease to the new base

Final price

= 1.25P \times (1 - 0.20) = 1.25P \times 0.80 = 1.00P

.

The

20\%

decrease applies to the updated base value of

1.25P

.

4

Evaluate each statement based on the final price

1.00P

The final price equals the initial price (

P

), the net percent change is

0\%

, and the ratio of final to initial price is

1:1

.

All three true statements directly reflect that the final price is equal to the initial price.

Key Concept

Successive Percentage Changes and Base Shift

Estimated Time:45s

Question 105Question →

If $x$ is an odd integer and $y$ is an even integer, which of the following expressions must result in an even integer? Select all that apply.

Select all that apply

xy + x

xy + y

(x + 1)(y + 1)

x^2 + y

x(y + 3)

Show answer & explanation

Answer: $xy + y$ ; $(x + 1)(y + 1)$

Answer

The expressions that must be even are

xy + y

and

(x + 1)(y + 1)

.

The expression

xy + y

is guaranteed to be even because

xy

is even (product of odd and even) and adding another even integer

y

produces an even sum. The expression

(x + 1)(y + 1)

is also guaranteed to be even because adding 1 to an odd integer

x

creates an even integer

x + 1

, and multiplying an even number by any integer always yields an even product.

Step-by-Step Solution

1

Analyze fundamental parity rules for addition and multiplication of integers.

Recall that

\text{odd} \times \text{even} = \text{even}

,

\text{odd} \times \text{odd} = \text{odd}

,

\text{even} + \text{even} = \text{even}

, and

\text{odd} + \text{even} = \text{odd}

.

Parity rules govern the even/odd behavior of combined expressions.

2

Evaluate the expression

xy + y

.

Since

x

is odd and

y

is even,

xy

is even. Then

\text{even} + y\,(\text{even}) = \text{even}

.

Adding two even terms always results in an even number.

3

Evaluate the expression

(x + 1)(y + 1)

.

Since

x

is odd,

x + 1

is even. The product of an even integer and any integer (

y + 1

) is always even.

An even factor guarantees an even product.

Key Concept

Even and Odd Parity Rules under Arithmetic Operations

Question 106Question →

A laboratory container holds a solution composed solely of alcohol, acid, and water. Initially, alcohol accounts for $\frac{3}{8}$ of the solution's total volume, and acid accounts for $\frac{1}{4}$ of the total volume. In a two-step process, a chemist first removes $\frac{1}{3}$ of the alcohol present and $\frac{1}{2}$ of the acid present, with no water removed. Next, the chemist adds pure water until water accounts for $\frac{3}{5}$ of the solution's new total volume. What fraction of the final solution's total volume is alcohol?

\frac{4}{15}

\frac{1}{4}

\frac{4}{9}

\frac{2}{5}

\frac{2}{3}

Show answer & explanation

Answer: $\frac{4}{15}$

Answer

The fraction of the final solution's total volume that is alcohol is

\frac{4}{15}

.

The correct fraction

\frac{4}{15}

is derived by calculating the remaining alcohol volume as

\frac{1}{4}

of the initial total volume

V

, and determining the final total volume

V_{\text{final}} = \frac{15}{16}V

using the constant non-water volume of

\frac{3}{8}V

. Dividing

\frac{1}{4}V

by

\frac{15}{16}V

yields

\frac{4}{15}

.

Step-by-Step Solution

1

Determine initial volume fractions for all three components.

Let

V

be the initial total volume. Alcohol is

\frac{3}{8}V

, acid is

\frac{1}{4}V = \frac{2}{8}V

, and water is

1 - (\frac{3}{8} + \frac{2}{8}) = \frac{3}{8}V

.

Establishing the initial baseline volumes allows accurate tracking through the multi-step changes.

2

Calculate component volumes after the removal step.

Remaining alcohol =

\frac{3}{8}V \times (1 - \frac{1}{3}) = \frac{1}{4}V

. Remaining acid =

\frac{1}{4}V \times (1 - \frac{1}{2}) = \frac{1}{8}V

. Water remains

\frac{3}{8}V

.

Removing specified fractions of individual components changes their absolute volumes.

3

Calculate non-water volume and final total volume after water is added.

Total non-water volume =

\frac{1}{4}V + \frac{1}{8}V = \frac{3}{8}V

. Since water becomes

\frac{3}{5}

of the final volume

V_{\text{final}}

, non-water is

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

of

V_{\text{final}}

. Thus,

\frac{2}{5} V_{\text{final}} = \frac{3}{8}V \implies V_{\text{final}} = \frac{5}{2} \times \frac{3}{8}V = \frac{15}{16}V

.

Adding only water keeps the non-water volume constant, providing a fixed reference point to find the new total volume.

4

Compute the final fraction of alcohol in the solution.

\text{Alcohol Fraction} = \frac{\text{Alcohol Volume}}{V_{\text{final}}} = \frac{\frac{1}{4}V}{\frac{15}{16}V} = \frac{1}{4} \times \frac{16}{15} = \frac{4}{15}.

The required quantity is the part-to-whole ratio of remaining alcohol to the final total volume.

Key Concept

Multi-step fraction operations involving component-wise removals and fixed non-water volumes in liquid mixtures.

Estimated Time:2m 30s

Question 107Question →

What is the largest two-digit positive integer $n$ such that when $n$ is divided by $4$ , the remainder is $3$ , and when $n$ is divided by $5$ , the remainder is $2$ ?

Show answer & explanation

Answer: 87

Answer

The largest two-digit positive integer satisfying both remainder conditions is 87.

Any integer satisfying both remainder requirements must be of the form

n = 20m + 7

for an integer

m

. Testing values for

m

shows that

m = 4

yields

n = 87

, which is the largest two-digit integer fitting this rule.

Step-by-Step Solution

1

Set up modular arithmetic equations for the given remainder conditions.

n \equiv 3 \pmod 4

and

n \equiv 2 \pmod 5

Translate the verbal description of remainders into mathematical congruence relations.

2

Combine the congruence relations to find the general form of

n

.

n = 20m + 7

for non-negative integers

m

The least common multiple of

4

and

5

is

20

, meaning solutions repeat every

20

units.

3

Find the maximum integer

m

that produces a two-digit integer.

For

m = 4

,

n = 87

. For

m = 5

,

n = 107

.

Two-digit integers are strictly less than

100

.

Key Concept

Simultaneous Remainders and Divisibility Cycles

Estimated Time:1m 30s

Question 108Question →

If $n$ is a negative odd integer, which of the following expressions MUST be a positive even integer?

n^2 + 1

-n^2 + 1

(n + 1)^2

2n - 1

n^3 + 1

Show answer & explanation

Answer: $n^2 + 1$

Answer

The expression

n^2 + 1

must be a positive even integer.

Squaring any negative odd integer

n

produces a positive odd integer (

n^2 \ge 1

). Adding

1

to a positive odd integer always yields a positive even integer (

n^2 + 1 \ge 2

).

Step-by-Step Solution

1

Analyze the parity and sign of

n^2

Since

n

is negative and odd, any even power of

n

(such as

n^2

) produces a positive odd integer.

Negative times negative yields positive, and odd times odd yields odd.

2

Analyze the effect of adding

1

to

n^2

n^2 + 1

is the sum of a positive odd integer and

1

, which results in a positive even integer.

Adding

1

to any odd integer results in an even integer, and adding

1

to a positive integer maintains positivity.

Key Concept

Parity and sign rules for exponents and addition

Question 109Question →

A water storage tank is initially filled to $\frac{4}{5}$ of its total capacity with water. First, $\frac{1}{4}$ of the water in the tank is drained for irrigation. Next, an amount of water equal to $\frac{1}{3}$ of the remaining water in the tank is added back. Finally, $\frac{3}{8}$ of the water currently in the tank is removed for domestic use. Which of the following statements about the volume of water in the tank must be true? Select all such statements.

Select all that apply

After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations.

The final volume of water in the tank is equal to

\frac{5}{8}

of the initial volume of water.

The total volume of water removed from the tank during the first and third operations combined is equal to

\frac{1}{2}

of the total capacity of the tank.

The volume of water removed in the first step is equal to

\frac{1}{4}

of the total capacity of the tank.

The final volume of water in the tank is equal to

\frac{2}{5}

of the total capacity of the tank.

Show answer & explanation

Answer: After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations.; The final volume of water in the tank is equal to $\frac{5}{8}$ of the initial volume of water.; The total volume of water removed from the tank during the first and third operations combined is equal to $\frac{1}{2}$ of the total capacity of the tank.

Answer

The correct statements are: (1) After the second operation, the volume of water in the tank is equal to the initial volume of water before any operations; (2) The final volume of water in the tank is equal to 5/8 of the initial volume of water; and (3) The total volume of water removed from the tank during the first and third operations combined is equal to 1/2 of the total capacity of the tank.

The statement regarding the volume after the second operation is correct because adding 1/3 of the remaining 3/5 capacity adds 1/5 capacity, returning the total volume to 4/5 capacity. The statement comparing the final volume to the initial volume is correct because (1/2) divided by (4/5) equals 5/8. The statement regarding total water removed in the first and third operations is correct because 1/5 capacity plus 3/10 capacity equals 1/2 capacity.

Step-by-Step Solution

1

Define total tank capacity as

T

and find initial water volume.

Initial volume

V_0 = \frac{4}{5}T

.

The problem states the tank is initially

\frac{4}{5}

full.

2

Calculate water remaining and removed after Operation 1.

Removed =

\frac{1}{4} \times \frac{4}{5}T = \frac{1}{5}T

. Remaining

V_1 = \frac{4}{5}T - \frac{1}{5}T = \frac{3}{5}T

.

Draining

\frac{1}{4}

of existing water leaves

\frac{3}{4}

of the existing water.

3

Calculate water remaining after Operation 2.

Added =

\frac{1}{3} \times \frac{3}{5}T = \frac{1}{5}T

. New volume

V_2 = \frac{3}{5}T + \frac{1}{5}T = \frac{4}{5}T

.

Adding

\frac{1}{3}

of the remaining volume increases it by a factor of

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

.

4

Calculate water remaining and removed after Operation 3.

Removed =

\frac{3}{8} \times \frac{4}{5}T = \frac{3}{10}T

. Final volume

V_3 = \frac{4}{5}T \times \left(1 - \frac{3}{8}\right) = \frac{4}{5}T \times \frac{5}{8} = \frac{1}{2}T

.

Removing

\frac{3}{8}

of current water leaves

\frac{5}{8}

of that volume.

5

Evaluate each given statement against calculated values.

V_2 = V_0 = \frac{4}{5}T

(True).

\frac{V_3}{V_0} = \frac{\frac{1}{2}T}{\frac{4}{5}T} = \frac{5}{8}

(True). Total removed =

\frac{1}{5}T + \frac{3}{10}T = \frac{1}{2}T

(True).

Direct comparison with calculated step outcomes confirms these three statements are valid.

Key Concept

Sequential Fraction Multiplication and Part-to-Whole Relationships

Estimated Time:1m 45s

Question 110Question →

An investment fund had an initial value of $\$4{}000$ . During the first year, the value of the fund increased by $15\%$ . During the second year, the new value of the fund increased by $10\%$ . What was the total value of the fund at the end of the second year?

\$5{}060

\$5{}000

\$4{}600

\$4{}840

\$5{}290

Show answer & explanation

Answer: $\$5{}060$

Answer

The total value of the fund at the end of the second year was

\$5{}060

.

To find the final value after successive percent increases, compute the growth sequentially. After Year 1, the fund grows to

\$4{}000 \times 1.15 = \$4{}600

. In Year 2, the

10\%

increase applies to the new amount of

\$4{}600

, resulting in

\$4{}600 \times 1.10 = \$5{}060

. Thus,

\$5{}060

is the correct final value.

Step-by-Step Solution

1

Calculate the fund's value at the end of the first year.

\$4{}000 \times (1 + 0.15) = \$4{}000 \times 1.15 = \$4{}600

A

15\%

increase on the initial

\$4{}000

base increases the value to

115\%

of its original amount.

2

Calculate the fund's value at the end of the second year using the updated base value.

\$4{}600 \times (1 + 0.10) = \$4{}600 \times 1.10 = \$5{}060

The second year's

10\%

increase applies to the new base value of

\$4{}600

, not the initial

\$4{}000

.

Key Concept

Successive Percent Changes and Base Shift

Question 111Question →

In a manufacturing facility, three automated machines—Machine A, Machine B, and Machine C—produce components at constant individual rates. Machine A working alone completes $\frac{3}{10}$ of a standard daily order in $3$ hours. Machine B working alone completes $\frac{2}{5}$ of the order in $6$ hours. Machine C working alone completes $\frac{1}{4}$ of the order in $5$ hours.

Initially, all three machines work together for $3$ hours. At the end of $3$ hours, Machine A malfunctions and stops, while Machines B and C continue working together without interruption until the daily order is finished. How many total hours does it take from the start to complete the entire daily order?

Show answer & explanation

Answer: 6

Answer

The total time required from the start to complete the entire daily order is 6 hours.

Each machine's rate per hour is found by dividing the given fraction by hours: Machine A is

\frac{1}{10}

, B is

\frac{1}{15}

, and C is

\frac{1}{20}

. Summing these gives an initial combined rate of

\frac{13}{60}

per hour. In the first 3 hours, the three machines complete

3 \times \frac{13}{60} = \frac{13}{20}

of the order, leaving

1 - \frac{13}{20} = \frac{7}{20}

of the order remaining. When Machine A stops, the remaining combined rate of B and C is

\frac{1}{15} + \frac{1}{20} = \frac{7}{60}

per hour. Dividing the remaining

\frac{7}{20}

by

\frac{7}{60}

yields 3 additional hours. Adding the initial 3 hours gives a total duration of 6 hours.

Step-by-Step Solution

1

Determine individual hourly work rates for each machine.

Machine A rate =

\frac{1}{10}

order/hr, Machine B rate =

\frac{1}{15}

order/hr, Machine C rate =

\frac{1}{20}

order/hr.

Divide the fraction of the job completed by the duration in hours to find the unit rate for each machine.

2

Calculate the combined rate of all three machines and the fraction completed in the first 3 hours.

Combined rate =

\frac{13}{60}

order/hr; Work completed in 3 hours =

\frac{13}{20}

, leaving

\frac{7}{20}

of the order unfinished.

Sum the three rates using a common denominator of 60, then multiply by 3 hours to find the total work done during the first stage.

3

Calculate the combined rate of Machines B and C, and find the additional time required to complete the remaining fraction.

Combined rate of B and C =

\frac{7}{60}

order/hr; Additional time needed = 3 hours.

Divide the remaining fraction

\frac{7}{20}

by the combined rate

\frac{7}{60}

to determine the remaining hours needed.

4

Add the initial duration and additional duration to obtain the total time.

Total time =

3 + 3 = 6

hours.

The question asks for the total elapsed time from the start of the job.

Key Concept

Adding rational fractions with different denominators to solve multi-stage work and rate problems.

Question 112Question →

A technology company allocates its quarterly research budget among three key projects: Project Alpha, Project Beta, and Project Gamma. Initially, Project Alpha receives $\frac{2}{5}$ of the total budget. Of the remaining budget, Project Beta is allocated $\frac{4}{7}$ , and Project Gamma receives the rest. During a mid-quarter review, $\frac{1}{6}$ of Project Alpha's allocated budget and $\frac{1}{4}$ of Project Beta's allocated budget are transferred to Project Gamma. After these transfers, Project Gamma's final budget is what fraction of the total initial quarterly research budget?

\frac{5}{21}

\frac{13}{50}

\frac{43}{105}

\frac{83}{165}

\frac{16}{105}

Show answer & explanation

Answer: $\frac{43}{105}$

Answer

\frac{43}{105}

Project Alpha receives

\frac{2}{5}

of the total budget, leaving

\frac{3}{5}

. Project Beta gets

\frac{4}{7}

of

\frac{3}{5}

, which is

\frac{12}{35}

. Project Gamma's initial share is

\frac{3}{5} - \frac{12}{35} = \frac{9}{35}

. Transferring

\frac{1}{6}

of Alpha's share contributes

\frac{1}{15}

, and transferring

\frac{1}{4}

of Beta's share contributes

\frac{3}{35}

. Adding these to Gamma's initial share yields

\frac{9}{35} + \frac{3}{35} + \frac{1}{15} = \frac{12}{35} + \frac{1}{15} = \frac{36 + 7}{105} = \frac{43}{105}

.

Step-by-Step Solution

1

Determine Project Alpha's initial allocation and the remaining budget fraction.

Project Alpha gets

\frac{2}{5}

of the total budget

B

. The remaining fraction is

1 - \frac{2}{5} = \frac{3}{5} B

.

Subtractions from the whole yield the remaining unallocated portion.

2

Calculate the initial allocations for Project Beta and Project Gamma.

Project Beta receives

\frac{4}{7} \times \frac{3}{5} B = \frac{12}{35} B

. Project Gamma receives the rest of the remaining budget:

\frac{3}{5} B - \frac{12}{35} B = \frac{21}{35} B - \frac{12}{35} B = \frac{9}{35} B

.

Project Beta's share is a fraction of the remaining budget, not the total budget.

3

Calculate the transferred amounts from Project Alpha and Project Beta to Project Gamma.

Transfer from Alpha =

\frac{1}{6} \times \frac{2}{5} B = \frac{1}{15} B

. Transfer from Beta =

\frac{1}{4} \times \frac{12}{35} B = \frac{3}{35} B

.

Transfers are fractional parts of each project's individual initial allocation.

4

Sum Project Gamma's initial share and the two transferred amounts.

Gamma's final fraction =

\frac{9}{35} + \frac{1}{15} + \frac{3}{35} = \frac{12}{35} + \frac{1}{15} = \frac{36}{105} + \frac{7}{105} = \frac{43}{105}

.

Finding the least common denominator (

105

) allows exact addition of the rational numbers.

Key Concept

Multi-step sequential operations with fractions and rational numbers

Question 113Question →

A technology company's annual revenue increased by $40\%$ from 2021 to 2022, and then decreased by $15\%$ from 2022 to 2023. If the company's annual revenue in 2023 was $\$476,000$ , what was its annual revenue in 2021?

\$336,000

\$380,800

\$400,000

\$404,600

\$560,000

Show answer & explanation

Answer: $\$400,000$

Answer

\$400,000

The correct answer is

\$400,000

. Expressing the multi-step percentage change relative to the initial 2021 revenue

R

gives an intermediate 2022 value of

1.40R

and a final 2023 value of

1.40R \times 0.85 = 1.19R

. Solving

1.19R = 476,000

yields

R = 400,000

.

Step-by-Step Solution

1

Define the unknown variable and express successive percentage changes algebraically.

Let

R

be the 2021 annual revenue. The 2022 revenue is

R \times (1 + 0.40) = 1.40R

. The 2023 revenue is

1.40R \times (1 - 0.15) = 1.40R \times 0.85

.

Sequential percentage adjustments apply to the newly resulting base values rather than the initial value.

2

Calculate the net multiplier for the two-year change.

1.40 \times 0.85 = 1.19

. Thus, the 2023 revenue equals

1.19R

.

Multiplying the growth factor (

1.40

) by the decay factor (

0.85

) gives the overall percentage relationship between 2021 and 2023.

3

Solve for the initial revenue

R

using the given 2023 revenue.

1.19R = 476,000 \implies R = \frac{476,000}{1.19} = 400,000

.

Dividing the final amount by the combined multiplier determines the original starting revenue.

Key Concept

Successive Percent Change and Base Shift

Question 114Question →

Let $n$ be a positive integer that is a factor of $180$ . If $n$ is divisible by $6$ but is not divisible by $4$ , what is the maximum possible number of positive divisors of $n$ ?

6

8

12

18

24

Show answer & explanation

Answer: 12

Answer

12

The prime factorization of

180

is

2^2 \times 3^2 \times 5^1

. Any factor

n

of

180

is of the form

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

. For

n

to be divisible by

6

, we must have

a \ge 1

and

b \ge 1

. For

n

not to be divisible by

4

, we must have

a < 2

, which forces

a = 1

. To maximize the total number of divisors

(a+1)(b+1)(c+1)

, we select the largest possible values for

b

and

c

, which are

b = 2

and

c = 1

. This yields

n = 2^1 \times 3^2 \times 5^1 = 90

, giving

(1+1)(2+1)(1+1) = 12

positive divisors.

Step-by-Step Solution

1

Find the prime factorization of 180.

180 = 2^2 \times 3^2 \times 5^1

Any positive factor

n

of

180

must have the form

n = 2^a \times 3^b \times 5^c

, where

0 \le a \le 2

,

0 \le b \le 2

, and

0 \le c \le 1

.

2

Apply the divisibility conditions to determine the possible values of the exponent

a

.

a = 1

Since

n

is divisible by

6 = 2 \times 3

,

a \ge 1

and

b \ge 1

. Since

n

is not divisible by

4 = 2^2

,

a < 2

. Thus,

a

must equal

1

.

3

Maximize the divisor count formula

(a+1)(b+1)(c+1)

using the available ranges for

b

and

c

.

b = 2

and

c = 1

To maximize the number of positive divisors, choose the maximum allowable values for

b

(

b=2

) and

c

(

c=1

).

4

Calculate the maximum number of positive divisors.

(1+1)(2+1)(1+1) = 2 \times 3 \times 2 = 12

For

n = 2^1 \times 3^2 \times 5^1 = 90

, the total number of positive divisors is

12

.

Key Concept

Divisor count formula and prime factor constraints

Estimated Time:1m 30s

Question 115Question →

A charity foundation allocates its annual grant budget among three sectors: Healthcare, Education, and Environmental Conservation. First, $\frac{2}{5}$ of the total annual budget is allocated to Healthcare. Next, $\frac{5}{9}$ of the remaining budget is allocated to Education. The final remainder of $\$120,000$ is allocated to Environmental Conservation. What was the foundation's total annual grant budget?

\$32,000

\$240,000

\$360,000

\$450,000

\$2,700,000

Show answer & explanation

Answer: $\$450,000$

Answer

\$450,000

The option specifying

\$450,000

is correct because allocating

\frac{2}{5}

leaves

\frac{3}{5}

of the budget. Taking

\frac{5}{9}

of

\frac{3}{5}

gives

\frac{1}{3}

of the total. The combined allocations leave

1 - (\frac{2}{5} + \frac{1}{3}) = \frac{4}{15}

of the total budget, which equals

\$120,000

. Dividing

\$120,000

by

\frac{4}{15}

yields

\$450,000

.

Step-by-Step Solution

1

Determine the fraction of the total budget remaining after the Healthcare allocation.

Since Healthcare receives

\frac{2}{5}

of the total budget, the remaining fraction is

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

.

Sequential fraction operations require updating the remaining whole after each step.

2

Calculate the fraction of the total budget allocated to Education.

Education receives

\frac{5}{9}

of the remaining

\frac{3}{5}

, which is

\frac{5}{9} \times \frac{3}{5} = \frac{15}{45} = \frac{1}{3}

of the total budget.

To find a fraction of a remaining fraction, multiply the two fractions together.

3

Find the final remaining fraction allocated to Environmental Conservation.

Subtract both allocations from the whole:

1 - \left(\frac{2}{5} + \frac{1}{3}\right) = 1 - \frac{11}{15} = \frac{4}{15}

.

The final dollar amount corresponds to the fraction of the total budget left over.

4

Solve for the total annual grant budget.

\frac{4}{15} \times \text{Total} = \$120,000 \implies \text{Total} = \$120,000 \times \frac{15}{4} = \$450,000

.

Dividing the part by its corresponding fractional proportion yields the total whole.

Key Concept

Sequential Fraction Operations and Finding the Whole from a Fractional Part

Question 116Question →

Let $m$ and $n$ be positive integers such that $72^2 \cdot m = n^3$ . Which of the following statements MUST be true? Select all that apply.

Select all that apply

m

is divisible by

9

n

is divisible by

36

m

must be a perfect square

n

is divisible by

54

The product

m \cdot n

is divisible by

108

Show answer & explanation

Answer: $m$ is divisible by $9$ ; $n$ is divisible by $36$ ; The product $m \cdot n$ is divisible by $108$

Answer

The statements that

m

is divisible by

9

,

n

is divisible by

36

, and the product

m \cdot n

is divisible by

108

MUST be true.

Analyzing prime factorizations shows that

n^3 = 2^6 \cdot 3^4 \cdot m

. For the right-hand side to be a perfect cube, the power of

3

must be elevated to at least

6

, requiring

m

to contain

3^2 = 9

as a factor. Consequently,

n^3

contains at least

2^6 \cdot 3^6 = 36^3

, forcing

n

to be a multiple of

36

. Finally, multiplying

m = 9k^3

by

n = 36k

yields

324k^4

, which is a multiple of

108

for all integer values of

k

.

Step-by-Step Solution

1

Express

72^2

in terms of its prime factorization.

72 = 2^3 \cdot 3^2 \implies 72^2 = (2^3 \cdot 3^2)^2 = 2^6 \cdot 3^4

Prime factorization allows us to analyze the exponents required for

n^3

to be a perfect cube.

2

Determine the minimum prime factor requirements for

m

and

n

.

2^6 \cdot 3^4 \cdot m = n^3 \implies

exponent of

2

in

m

must be

\ge 0

(a multiple of 3), exponent of

3

in

m

must be

\ge 2

(since

4 + 2 = 6

is a multiple of 3).

Every prime factor in a perfect cube must have an exponent divisible by 3.

3

Establish general algebraic expressions for

m

and

n

.

m = 9k^3

and

n = 36k

for any positive integer

k

.

This captures all possible integer solutions for

m

and

n

.

4

Test each statement using the general forms

m = 9k^3

and

n = 36k

.

m

is divisible by

9

(

9k^3

is a multiple of

9

);

n

is divisible by

36

(

36k

is a multiple of

36

);

m \cdot n = 324k^4 = 108(3k^4)

, which is divisible by

108

. Counterexamples show

m

need not be a square (

m=72

) and

n

need not be divisible by

54

(

n=36

).

Verifies which properties hold universally versus which can fail.

Key Concept

Prime Factor Exponents in Perfect Powers and Divisibility Rules

Question 117Question →

An employee's hourly wage increased from $\$20.00$ to $\$25.00$ . What was the percent increase in the employee's hourly wage?

Show answer & explanation

Answer: 25

Answer

The percent increase in the employee's hourly wage is

25\%

.

The percent increase is calculated using the formula

\text{Percent Increase} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%

. Substituting the given values gives

\frac{25 - 20}{20} \times 100\% = \frac{5}{20} \times 100\% = 25\%

.

Step-by-Step Solution

1

Calculate the absolute increase in hourly wage

Increase =

\$25.00 - \$20.00 = \$5.00

Percent change is calculated relative to the absolute change from the initial value.

2

Calculate the percent change using the original wage as the base

\text{Percent Increase} = \frac{\$5.00}{\$20.00} \times 100\% = 25\%

The formula for percent change is

\frac{\text{Amount of Change}}{\text{Original Amount}} \times 100\%

.

Key Concept

Percent Change Formula

Question 118Question →

If $m$ is an even integer and $n$ is an odd integer, what is the value of $(-1)^m + (-1)^n$ ?

Show answer & explanation

Answer: 0

Answer

0

Since

m

is an even integer,

(-1)^m = 1

. Since

n

is an odd integer,

(-1)^n = -1

. Adding

1

and

-1

yields

0

.

Step-by-Step Solution

1

Evaluate

(-1)^m

for an even integer

m

(-1)^m = 1

Raising

-1

to an even integer power always results in

1

.

2

Evaluate

(-1)^n

for an odd integer

n

(-1)^n = -1

Raising

-1

to an odd integer power always results in

-1

.

3

Sum the two evaluated expressions

1 + (-1) = 0

Combining

1

and

-1

equals

0

.

Key Concept

Exponent sign rules for even and odd powers

Question 119Question →

On the real number line, the point $P$ corresponds to the real number $x$ . If $|x - 3| = 7$ , which of the following is a possible value of $x + 5$ ?

-9

-4

1

9

12

Show answer & explanation

Answer: $1$

Answer

The correct answer is

1

.

Solving

|x - 3| = 7

produces two possible values for

x

:

x = 10

and

x = -4

. Substituting

x = -4

into

x + 5

yields

-4 + 5 = 1

, which is listed among the choices.

Step-by-Step Solution

1

Set up the two linear equations defined by the absolute value equation

|x - 3| = 7

.

The two cases are

x - 3 = 7

and

x - 3 = -7

.

By definition of absolute value,

|k| = 7

means

k = 7

or

k = -7

.

2

Solve each case for

x

.

From

x - 3 = 7

,

x = 10

. From

x - 3 = -7

,

x = -4

.

Adding

3

to both sides of each equation isolates

x

.

3

Evaluate

x + 5

for both solutions of

x

.

If

x = 10

,

x + 5 = 15

. If

x = -4

,

x + 5 = 1

.

Substitute each possible value of

x

into the expression

x + 5

.

Key Concept

Absolute Value as Distance on the Number Line

Estimated Time:45s

Question 120Question →

In a community library, $\frac{4}{15}$ of the total book collection consists of fiction books, $\frac{2}{5}$ consists of non-fiction books, and the remaining 240 books are children's books. If $\frac{3}{8}$ of the fiction books are hardcovers and the rest are softcovers, how many softcover fiction books are in the library?

Show answer & explanation

Answer: 120

Answer

The total number of softcover fiction books in the library is 120.

To find the number of softcover fiction books, first calculate the total number of books in the library. The combined fraction of fiction and non-fiction books is

\frac{4}{15} + \frac{2}{5} = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}

. Therefore, children's books represent the remaining

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

of the collection. Since

\frac{1}{3}

of the total equals 240 books, the total collection is

240 \times 3 = 720

books. Next, compute the total number of fiction books:

\frac{4}{15} \times 720 = 192

. Finally, since

\frac{3}{8}

of fiction books are hardcovers, the remaining fraction of softcovers is

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

. Multiplying

\frac{5}{8} \times 192

gives 120 softcover fiction books.

Step-by-Step Solution

1

Find the combined fraction of fiction and non-fiction books, then determine the fraction of children's books.

Combined fraction =

\frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}

. Children's books fraction =

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

.

The sum of all non-overlapping fractions comprising the entire collection must equal 1.

2

Calculate the total number of books in the library.

Total collection =

240 \div \frac{1}{3} = 720

books.

The 240 children's books represent exactly

\frac{1}{3}

of the whole collection.

3

Determine the total count of fiction books.

Fiction books =

\frac{4}{15} \times 720 = 192

books.

Fiction books make up

\frac{4}{15}

of the total 720 books.

4

Calculate the number of softcover fiction books.

Softcover fiction books =

\left(1 - \frac{3}{8}\right) \times 192 = \frac{5}{8} \times 192 = 120

books.

If

\frac{3}{8}

of the fiction subset are hardcovers, the remaining

\frac{5}{8}

of that subset are softcovers.

Key Concept

Solving multi-step rational number word problems by combining fractions and determining fractional parts of a subset.

Estimated Time:1m 30s

Arithmetic

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