Arithmetic Practice Questions — GRE General Test — Page 9

Question 161Question →

If $p$ , $p + 2$ , and $p + 4$ are all prime numbers, what is the value of $p^2 + 5$ ?

6

9

14

19

30

Show answer & explanation

Answer: 14

Answer

The value of

p^2 + 5

is 14.

For any three consecutive odd integers

p

,

p + 2

, and

p + 4

, exactly one of them must be a multiple of 3. Because all three expressions represent prime numbers, the term divisible by 3 must be equal to 3 (since 3 is the only prime divisible by 3). Setting

p = 3

gives

p + 2 = 5

and

p + 4 = 7

, both of which are prime. Any choice of

p > 3

forces either

p + 2

or

p + 4

to be a multiple of 3 greater than 3, making it composite. Thus

p = 3

is uniquely determined, and evaluating

p^2 + 5

gives

3^2 + 5 = 14

.

Step-by-Step Solution

1

Analyze the possible remainders when the prime

p

is divided by 3.

Any positive integer

p

can be written in one of three forms:

3k

,

3k + 1

, or

3k + 2

for some integer

k

.

Dividing any integer by 3 leaves a remainder of 0, 1, or 2.

2

Test each remainder case for the expressions

p

,

p + 2

, and

p + 4

.

If

p = 3k + 1

, then

p + 2 = 3k + 3 = 3(k + 1)

, which is divisible by 3. If

p = 3k + 2

, then

p + 4 = 3k + 6 = 3(k + 2)

, which is divisible by 3.

Among any three consecutive odd numbers of the form

p, p+2, p+4

, exactly one of them must be a multiple of 3.

3

Deduce the unique value of

p

.

The only way all three numbers

p

,

p + 2

, and

p + 4

can be prime is if the one divisible by 3 is equal to 3 itself, which forces

p = 3

.

The only prime number divisible by 3 is 3 itself; any larger multiple of 3 is composite.

4

Evaluate the target expression

p^2 + 5

using

p = 3

.

3^2 + 5 = 9 + 5 = 14

.

Substitute the uniquely determined value

p = 3

into the given algebraic expression.

Key Concept

Divisibility properties of consecutive odd integers and prime number definitions

Estimated Time:1m 30s

Question 162Question →

On the real number line, $p$ and $q$ are real numbers such that $|3 - 2p| \le 5$ and $|q + 4| = 2$ . What is the maximum possible value of $|p^2 - q|$ ?

22

18

14

10

7

Show answer & explanation

Answer: 22

Answer

22

Solving

|3 - 2p| \le 5

gives

-5 \le 3 - 2p \le 5 \implies -1 \le p \le 4

, so

p^2

can range from

0

up to

16

. Solving

|q + 4| = 2

yields

q = -2

or

q = -6

. To maximize

|p^2 - q|

, we combine the maximum possible value of

p^2

(

16

) with

q = -6

, obtaining

|16 - (-6)| = 22

.

Step-by-Step Solution

1

Solve the absolute value inequality

|3 - 2p| \le 5

for

p

.

-1 \le p \le 4

-5 \le 3 - 2p \le 5 \implies -8 \le -2p \le 2

. Dividing by

-2

and reversing the inequality direction gives

-1 \le p \le 4

.

2

Determine the range of possible values for

p^2

.

0 \le p^2 \le 16

Since

p

spans from

-1

to

4

(which includes

0

), the minimum square is

0^2 = 0

and the maximum square is

4^2 = 16

.

3

Solve the absolute value equation

|q + 4| = 2

for

q

.

q = -2

or

q = -6

q + 4 = 2 \implies q = -2

, and

q + 4 = -2 \implies q = -6

.

4

Find the combination of

p^2

and

q

that maximizes

|p^2 - q|

.

Maximum value is 22

Pairing

p^2 = 16

with

q = -6

yields

|16 - (-6)| = 22

, which is greater than

|16 - (-2)| = 18

.

Key Concept

Absolute value inequalities, real number bounds, and distance optimization.

Question 163Question →

A chemical laboratory starts with a liquid solution having a total volume of $3.2 \times 10^{-4}$ cubic meters. The solution is partitioned equally into $400$ identical micro-vials. Subsequently, heat treatment causes the volume of liquid in each micro-vial to evaporate, reducing its volume by $75\%$ . What is the final volume of liquid remaining in a single micro-vial, expressed in scientific notation?

2.0 \times 10^{-7}

cubic meters

8.0 \times 10^{-7}

cubic meters

6.0 \times 10^{-7}

cubic meters

2.0 \times 10^{-5}

cubic meters

2.0 \times 10^{-6}

cubic meters

Show answer & explanation

Answer: $2.0 \times 10^{-7}$ cubic meters

Answer

2.0 \times 10^{-7}

cubic meters

Dividing the initial total volume of

3.2 \times 10^{-4}

cubic meters by

400

(

4 \times 10^2

) gives an initial volume of

8.0 \times 10^{-7}

cubic meters per micro-vial. Reducing this volume by

75\%

means

25\%

of the liquid remains. Multiplying

8.0 \times 10^{-7}

by

0.25

yields

2.0 \times 10^{-7}

cubic meters.

Step-by-Step Solution

1

Express the number of micro-vials in scientific notation.

400 = 4 \times 10^2

Converting into scientific notation simplifies division involving powers of 10.

2

Calculate the initial volume per micro-vial before evaporation.

\frac{3.2 \times 10^{-4}}{4 \times 10^2} = \left(\frac{3.2}{4}\right) \times 10^{-4 - 2} = 0.8 \times 10^{-6} = 8.0 \times 10^{-7}

cubic meters

Equal partitioning requires dividing the total volume by the total number of containers.

3

Determine the remaining fraction of liquid after a

75\%

volume reduction.

Remaining fraction

= 100\% - 75\% = 25\% = 0.25

A reduction by

75\%

leaves

25\%

of the liquid volume in the vial.

4

Multiply the volume per micro-vial by the remaining fraction.

8.0 \times 10^{-7} \times 0.25 = 2.0 \times 10^{-7}

cubic meters

Applying the remaining fraction yields the final volume in standard scientific notation format

a \times 10^n

where

1 \le a < 10

.

Key Concept

Operations with Decimals, Place Value, and Scientific Notation

Estimated Time:2m 0s

Question 164Question →

What is the value of $\frac{2^7 + 2^7}{2^4}$ ?

1

8

16

128

1024

Show answer & explanation

Answer: 16

Answer

16

Combining the numerator yields

2^7 + 2^7 = 2(2^7) = 2^8

. Dividing

2^8

by

2^4

using the quotient rule gives

2^{8-4} = 2^4 = 16

.

Step-by-Step Solution

1

Simplify the numerator by factoring out common terms

2^7 + 2^7 = 2 \times 2^7 = 2^{1+7} = 2^8

Adding two identical exponential terms is equivalent to multiplying one term by 2.

2

Apply the quotient rule of exponents

\frac{2^8}{2^4} = 2^{8-4} = 2^4

When dividing exponential expressions with the same base, subtract the exponent of the denominator from the exponent of the numerator.

3

Evaluate the power

2^4 = 16

Compute the numerical value of 2 raised to the 4th power.

Key Concept

Combining like exponential terms and applying the quotient rule of exponents.

Question 165Question →

A municipal library system allocated a total annual acquisition budget of $B$ dollars between fiction and non-fiction titles. In the first half of the year, $60\%$ of the total budget was spent, with $70\%$ of that spent amount allocated to fiction books and the remainder to non-fiction books. In the second half of the year, the remaining budget was fully spent such that equal dollar amounts were allocated to fiction and non-fiction books. What percentage of the total annual budget $B$ was spent on fiction books?

42\%

50\%

62\%

72\%

82\%

Show answer & explanation

Answer: $62\%$

Answer

The total percentage of budget

B

spent on fiction books is

62\%

.

To find the total percentage of budget

B

spent on fiction books, we analyze the two periods separately relative to the total budget

B

. In the first half,

60\%

of

B

is spent, and

70\%

of this spent amount goes to fiction, giving

0.70 \times 0.60 B = 0.42 B

(

42\%

of

B

). In the second half, the remaining

40\%

of

B

is spent equally between fiction and non-fiction, so fiction receives

50\%

of

40\%

, which is

0.50 \times 0.40 B = 0.20 B

(

20\%

of

B

). Adding both periods yields

42\% + 20\% = 62\%

of the total annual budget

B

.

Step-by-Step Solution

1

Calculate the percentage of total budget

B

spent on fiction in the first half of the year.

0.70 \times 0.60 B = 0.42 B

, which is

42\%

of

B

.

The library spent

60\%

of budget

B

, and

70\%

of that spent portion went to fiction books.

2

Determine the remaining budget percentage for the second half of the year.

100\% - 60\% = 40\%

of

B

.

The remaining budget is the total budget minus the portion spent during the first half.

3

Calculate the percentage of total budget

B

spent on fiction in the second half of the year.

0.50 \times 0.40 B = 0.20 B

, which is

20\%

of

B

.

The remaining

40\%

budget was split equally between fiction and non-fiction.

4

Sum the fiction expenditures from both halves of the year.

42\% + 20\% = 62\%

of

B

.

Adding the contributions from both periods gives the total fraction of budget

B

allocated to fiction books.

Key Concept

Multi-stage percentage change and part-of-a-part base shift calculations.

Estimated Time:1m 30s

Question 166Question →

If $n$ is a positive integer that is divisible by $6$ but not divisible by $4$ , and $n^2$ has exactly $15$ positive divisors, what is the value of $n$ ?

6

12

18

24

36

Show answer & explanation

Answer: $18$

Answer

The value of

n

is

18

.

Because

n

is divisible by

6

, it must have prime factors

2

and

3

. Since

n

is not divisible by

4

, the exponent of

2

in

n

is exactly

1

, meaning

n^2

has

2^2

as a factor. The divisor count formula for

n^2

requires

(2+1)(2b+1) = 15

, which gives

2b+1 = 5

, so

b=2

. Thus,

n = 2^1 \cdot 3^2 = 18

.

Step-by-Step Solution

1

Analyze the prime factorization structure of

n

based on given divisibility conditions.

Since

n

is divisible by

6 = 2 \cdot 3

, its prime factorization must contain at least one factor of

2

and one factor of

3

. Since

n

is not divisible by

4 = 2^2

, the exponent of

2

in the prime factorization of

n

must be exactly

1

. Thus,

n = 2^1 \cdot 3^b \cdot k

, where

k

contains no factors of

2

or

3

.

Establishing the exponent of

2

narrows down the search space for prime factor exponents.

2

Express

n^2

in terms of its prime factors and write the formula for its number of positive divisors.

n^2 = 2^2 \cdot 3^{2b} \cdot k^2

. The number of positive divisors of

n^2

is given by

d(n^2) = (2 + 1)(2b + 1) \cdot d(k^2) = 3(2b + 1) \cdot d(k^2) = 15

.

The total number of positive divisors of a number

p_1^{a_1} p_2^{a_2} \dots

is

(a_1 + 1)(a_2 + 1) \dots

.

3

Solve for

b

and determine if

k

has any additional prime factors.

Dividing

15

by

3

gives

(2b + 1) \cdot d(k^2) = 5

. Since

5

is prime, we must have

d(k^2) = 1

(meaning

k = 1

) and

2b + 1 = 5

, which yields

2b = 4 \implies b = 2

.

Determining the exponent of

3

fixes the exact prime factorization of

n

.

4

Calculate

n

.

n = 2^1 \cdot 3^2 = 18

.

Multiplying the prime factors together yields the target integer.

Key Concept

Prime Factorization and Divisor Count Formula

Estimated Time:1m 30s

Question 167Question →

At a research institute, an annual equipment budget is distributed among three departments. First, $\frac{2}{5}$ of the total budget is allocated to the Biology department. Next, $\frac{3}{8}$ of the remaining budget is allocated to the Chemistry department. The Physics department receives the remaining balance of $\$45,000$ . What was the institute's total annual equipment budget?

$100,000

$112,500

$120,000

$180,000

$200,000

Show answer & explanation

Answer: $120,000

Answer

$120,000

Subtracting the Biology department's share of

\frac{2}{5}

leaves

\frac{3}{5}

of the total budget. The Chemistry department receives

\frac{3}{8}

of this remainder, leaving

\frac{5}{8}

of the remainder for the Physics department. Multiplying these fractions yields

\frac{5}{8} \times \frac{3}{5} = \frac{3}{8}

of the total budget for Physics. Equating

\frac{3}{8}

of the total budget to

\$45,000

gives a total budget of

\$120,000

.

Step-by-Step Solution

1

Determine the fraction of the budget remaining after the Biology allocation.

The remaining fraction is

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

of the total budget.

The Biology department receives

\frac{2}{5}

of the overall budget.

2

Calculate the fraction of the total budget allocated to Physics.

The Physics department receives

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

of the remaining budget, which equals

\frac{5}{8} \times \frac{3}{5} = \frac{3}{8}

of the total budget.

The Chemistry department receives

\frac{3}{8}

of the remaining budget, leaving

\frac{5}{8}

of that remainder for Physics.

3

Solve for the total budget

B

.

\frac{3}{8}B = 45,000 \implies B = 45,000 \times \frac{8}{3} = 120,000.

The Physics allocation of

\$45,000

represents

\frac{3}{8}

of the total budget

B

.

Key Concept

Sequential Fraction of Remainder Calculations

Estimated Time:1m 30s

Question 168Question →

A commercial trucking fleet monitored its vehicle fuel efficiency (in miles per gallon) and average fuel price (in dollars per gallon) over a two-year period. In Year 1, vehicle fuel efficiency increased by $25\%$ , while the price of fuel increased by $15\%$ . In Year 2, vehicle fuel efficiency increased by an additional $20\%$ relative to Year 1, while the price of fuel decreased by $10\%$ relative to Year 1. Assuming the total annual miles driven per vehicle remained constant throughout the two-year period, by what percent did the total annual fuel cost per vehicle change from the beginning of Year 1 to the end of Year 2?

31\%

decrease

69\%

decrease

40\%

decrease

28\%

decrease

45\%

increase

Show answer & explanation

Answer: $31\%$ decrease

Answer

A

31\%

decrease

The total fuel cost is given by

\text{Cost} = \text{Distance} \times \frac{\text{Price}}{\text{Efficiency}}

. After Year 1, efficiency becomes

1.25 E_0

and price becomes

1.15 P_0

. After Year 2, efficiency increases to

1.25 E_0 \times 1.20 = 1.50 E_0

, while price changes to

1.15 P_0 \times 0.90 = 1.035 P_0

. The new cost ratio is

\frac{1.035}{1.50} = 0.69

, meaning the new cost is

69\%

of the original cost. This represents a net decrease of

100\% - 69\% = 31\%

.

Step-by-Step Solution

1

Formulate the fuel cost relationship in terms of distance, efficiency, and price.

Initial cost

C_0 = D \times \frac{P_0}{E_0}

, where

D

is distance driven,

P_0

is initial price per gallon, and

E_0

is initial fuel efficiency in miles per gallon.

Total fuel used is

\frac{D}{E_0}

, so multiplying by price per gallon gives total cost.

2

Calculate the updated fuel efficiency

E_2

and price

P_2

at the end of Year 2 using successive percentage changes.

E_2 = E_0 \times 1.25 \times 1.20 = 1.50 E_0

and

P_2 = P_0 \times 1.15 \times 0.90 = 1.035 P_0

.

Compounding changes sequentially: Year 1 efficiency factor is

1.25

, Year 2 factor is

1.20

; Year 1 price factor is

1.15

, Year 2 factor is

0.90

.

3

Compute the final fuel cost

C_2

as a fraction of the initial cost

C_0

.

C_2 = D \times \frac{1.035 P_0}{1.50 E_0} = \frac{1.035}{1.50} \times C_0 = 0.69 C_0

.

Substituting the expressions for

P_2

and

E_2

into the fuel cost formula.

4

Determine the net percentage change from

C_0

to

C_2

.

\text{Percent Change} = \frac{0.69 C_0 - C_0}{C_0} \times 100\% = -31\%

, representing a

31\%$ decrease.

Subtracting

1.00

from

0.69

yields

-0.31

, which corresponds to a

31\%

reduction.

Key Concept

Successive Percentage Changes and Rate Relationships

Question 169Question →

A positive integer $n$ has a prime factorization of the form $2^x \times 3^y \times 7^z$ , where $x$ , $y$ , and $z$ are positive integers. The greatest common divisor of $n$ and $420$ is $84$ , and the least common multiple of $n$ and $420$ is $8,820$ . What is the total number of positive integer factors of $n$ ?

8

12

18

27

36

Show answer & explanation

Answer: $27$

Answer

The total number of positive integer factors of

n

is

27

.

By prime factorizing

420

,

84

, and

8,820

, we find that

420 = 2^2 \times 3^1 \times 5^1 \times 7^1

,

\gcd(n, 420) = 2^2 \times 3^1 \times 7^1

, and

\text{lcm}(n, 420) = 2^2 \times 3^2 \times 5^1 \times 7^2

. Comparing the minimum and maximum powers for each prime factor shows that

n = 2^2 \times 3^2 \times 7^2

. Applying the divisor counting formula gives

(2+1)(2+1)(2+1) = 27

.

Step-by-Step Solution

1

Express all given values in their prime factorized forms.

420 = 2^2 \times 3^1 \times 5^1 \times 7^1

,

\gcd(n, 420) = 84 = 2^2 \times 3^1 \times 7^1

, and

\text{lcm}(n, 420) = 8,820 = 2^2 \times 3^2 \times 5^1 \times 7^2

.

Finding the prime factorizations allows comparison of prime exponent bounds for GCD and LCM.

2

Determine the exponents

x

,

y

, and

z

for

n = 2^x \times 3^y \times 7^z

.

For prime

2

:

\max(x, 2) = 2

and

\min(x, 2) = 2 \implies x = 2

.
For prime

3

:

\max(y, 1) = 2

and

\min(y, 1) = 1 \implies y = 2

.
For prime

7

:

\max(z, 1) = 2

and

\min(z, 1) = 1 \implies z = 2

.

GCD takes the minimum exponent of each prime factor, while LCM takes the maximum exponent.

3

Calculate the total number of positive factors of

n = 2^2 \times 3^2 \times 7^2

.

Total factors

= (x + 1)(y + 1)(z + 1) = (2 + 1)(2 + 1)(2 + 1) = 3 \times 3 \times 3 = 27

.

The number of positive divisors of

p_1^{a} p_2^{b} \dots p_k^{k}

is

(a+1)(b+1)\dots(k+1)

.

Key Concept

Relationship between prime factorization, greatest common divisor (GCD), least common multiple (LCM), and the number of positive divisors.

Question 170Question →

Consider the expression $N = \frac{0.00048 \times (2.5 \times 10^7)}{1.2 \times 10^{-2}}$ . Which of the following expressions are equal to $N$ ? Select all such values.

Select all that apply

1.0 \times 10^6

(4.0 \times 10^4) \times 25

1.0 \times 10^2

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

0.00012 \times 10^9

Show answer & explanation

Answer: $1.0 \times 10^6$ ; $(4.0 \times 10^4) \times 25$ ; $\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}$

Answer

The expressions equal to

N

are

1.0 \times 10^6

,

(4.0 \times 10^4) \times 25

, and

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

.

Evaluating

N

simplifies to

1.0 \times 10^6

. The expression stating

1.0 \times 10^6

is an exact scientific notation match. The expression

(4.0 \times 10^4) \times 25

computes to

40,000 \times 25 = 1,000,000

. The fraction

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

evaluates to

1.0 \times 10^{-3 - (-9)} = 1.0 \times 10^6

.

Step-by-Step Solution

1

Express all numbers in the stem's numerator in standard scientific notation

0.00048 = 4.8 \times 10^{-4}

and the numerator becomes

(4.8 \times 10^{-4}) \times (2.5 \times 10^7)

Standardizing numbers into scientific notation simplifies operations involving powers of 10

2

Multiply the coefficients and combine the powers of 10 in the numerator

(4.8 \times 2.5) \times 10^{-4 + 7} = 12 \times 10^3 = 1.2 \times 10^4

Coefficients are multiplied directly while exponents are added according to index laws

3

Divide the simplified numerator by the denominator

\frac{1.2 \times 10^4}{1.2 \times 10^{-2}} = \left(\frac{1.2}{1.2}\right) \times 10^{4 - (-2)} = 1.0 \times 10^6 = 1,000,000

Subtracting a negative exponent in the denominator is equivalent to adding its positive value

4

Evaluate each given option to determine equivalence to

1.0 \times 10^6

The expressions representing

1.0 \times 10^6

are

1.0 \times 10^6

,

(4.0 \times 10^4) \times 25

, and

\frac{5.0 \times 10^{-3}}{5.0 \times 10^{-9}}

Direct comparison verifies which options match the calculated value of

N

Key Concept

Operations with Decimals and Scientific Notation

Question 171Question →

If $x$ is a real number such that $x^4 = 16$ , which of the following values could be equal to $x^3$ ? Select all that apply.

Select all that apply

-8

-4

4

8

16

Show answer & explanation

Answer: $-8$ ; $8$

Answer

The values

-8

and

8

are the possible values of

x^3

.

Solving

x^4 = 16

gives two real solutions,

x = 2

and

x = -2

, because raising any real number to an even power yields a non-negative result. Cubing each solution gives

2^3 = 8

and

(-2)^3 = -8

. Thus, both

-8

and

8

are correct values for

x^3

.

Step-by-Step Solution

1

Find all real solutions for

x

in the equation

x^4 = 16

.

x = 2

or

x = -2

.

Taking the fourth root of both sides gives

|x| = \sqrt[4]{16} = 2

, so

x

can be positive or negative.

2

Calculate

x^3

for the positive root

x = 2

.

2^3 = 8

.

Cubing a positive number yields a positive result.

3

Calculate

x^3

for the negative root

x = -2

.

(-2)^3 = -8

.

Cubing a negative number yields a negative result because an odd exponent preserves the sign.

Key Concept

Even and odd power rules for positive and negative real bases

Question 172Question →

If $N = 2^4 \times 3^3 \times 5^2$ , how many positive integer factors of $N$ are divisible by $6$ but not divisible by $12$ ?

Show answer & explanation

Answer: 9

Answer

The number of positive integer factors of

N

divisible by 6 but not by 12 is 9.

Any factor of

N = 2^4 \times 3^3 \times 5^2

is of the form

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

. For the factor to be divisible by

6

, we must have

a \ge 1

and

b \ge 1

. For it to not be divisible by

12 = 2^2 \times 3^1

, we must have

a < 2

. Therefore,

a

must equal

1

. The possible choices for

a

are

1

value (

1

), for

b

are

3

values (

1, 2, 3

), and for

c

are

3

values (

0, 1, 2

). Multiplying these options gives

1 \times 3 \times 3 = 9

.

Step-by-Step Solution

1

Express the prime factorization structure of a factor of

N

.

Any factor of

N

takes the form

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

with bounds

0 \le a \le 4

,

0 \le b \le 3

, and

0 \le c \le 2

.

Divisors of a number are formed by taking prime factors with exponents between zero and their maximum powers in the original number.

2

Determine exponent constraints for divisibility by 6 and non-divisibility by 12.

Divisibility by

6

requires

a \ge 1

and

b \ge 1

. Non-divisibility by

12

requires

a < 2

. Thus,

a = 1

exactly.

A factor must contain at least one factor of 2 and one factor of 3 to be a multiple of 6, but containing two or more factors of 2 makes it a multiple of 12.

3

Count combinations of choices for the exponents.

Exponent

a

has

1

option (

a = 1

),

b

has

3

options (

b \in \{1, 2, 3\}

), and

c

has

3

options (

c \in \{0, 1, 2\}

). Total

= 1 \times 3 \times 3 = 9

.

Applying the fundamental counting principle by multiplying the number of choices for independent prime factor exponents.

Key Concept

Counting Divisors with Prime Factorization and Divisibility Constraints

Question 173Question →

At the beginning of Year 1, an investor divided a sum of money between two portfolio accounts, Account X and Account Y, such that the initial balance of Account X was $25\%$ greater than the initial balance of Account Y.

Over a two-year period:
- Account X earned compound interest at a constant annual rate of $20\%$ per year.
- Account Y decreased in value by $10\%$ during Year 1, and then increased in value by $r\%$ during Year 2.

If the total combined value of both accounts at the end of Year 2 was $32\%$ greater than the total combined initial balance at the beginning of Year 1, what is the value of $r$ ?

22

25

27

30

36

Show answer & explanation

Answer: $30$

Answer

The value of

r

is

30

.

The correct value is

30

. Account X starts at

125\%

of Account Y's initial value (

125

vs

100

, totaling

225

). Compounding Account X at

20\%

per year for 2 years yields

125 \times 1.20^2 = 180

. Account Y drops

10\%

to

90

in Year 1. For the overall total to reach

225 \times 1.32 = 297

, Account Y must reach

297 - 180 = 117

at the end of Year 2. The percentage increase from

90

to

117

is

\frac{117 - 90}{90} \times 100\% = \frac{27}{90} \times 100\% = 30\%

.

Step-by-Step Solution

1

Define initial account balances using a suitable variable base

Let initial balance of Account Y be

Y_0 = 100

units. Initial balance of Account X is

X_0 = 1.25 \times 100 = 125

units. Combined initial balance is

T_0 = 125 + 100 = 225

units.

Establishing a standard numerical base simplifies tracking multi-account percentage changes.

2

Calculate the value of Account X at the end of Year 2

X_2 = 125 \times (1 + 0.20)^2 = 125 \times 1.44 = 180

units.

Account X compounds annually at

20\%

for 2 full years.

3

Express the value of Account Y at the end of Year 2 in terms of

r

After Year 1:

Y_1 = 100 \times (1 - 0.10) = 90

units. After Year 2:

Y_2 = 90 \times \left(1 + \frac{r}{100}\right)

units.

Account Y first loses

10\%

of its initial value, creating a new base of

90

units for the Year 2 growth rate.

4

Determine the required total combined final value

T_2 = 225 \times (1 + 0.32) = 225 \times 1.32 = 297

units.

The overall portfolio value grew by

32\%

relative to the initial combined balance of

225

units.

5

Set up and solve the combined equation for

r

180 + 90 \left(1 + \frac{r}{100}\right) = 297 \implies 90 \left(1 + \frac{r}{100}\right) = 117 \implies 1 + \frac{r}{100} = 1.30 \implies r = 30

.

Equating the sum of individual final account balances to the target portfolio balance yields the exact value of

r

.

Key Concept

Successive percent changes on shifting bases combined with compound interest calculations across multiple accounts.

Question 174Question →

An electronics retailer evaluated its annual revenue over a four-year period. In Year 2, the revenue increased by $25\%$ compared to Year 1. In Year 3, the revenue decreased by $20\%$ compared to Year 2. In Year 4, the revenue increased by $15\%$ compared to Year 3.

Which of the following statements regarding the retailer's revenue must be true? Select all such statements.

Select all that apply

The revenue in Year 3 is equal to the revenue in Year 1.

The overall percentage change in revenue from Year 1 to Year 4 is a

15\%

increase.

The revenue in Year 2 is

25\%

greater than the revenue in Year 3.

The overall percentage change in revenue from Year 1 to Year 3 is a

5\%

increase.

The percent increase from Year 3 to Year 4 is greater than the percent decrease from Year 2 to Year 3.

Show answer & explanation

Answer: The revenue in Year 3 is equal to the revenue in Year 1.; The overall percentage change in revenue from Year 1 to Year 4 is a $15\%$ increase.; The revenue in Year 2 is $25\%$ greater than the revenue in Year 3.

Answer

The statements asserting that Year 3 revenue equals Year 1 revenue, that the overall revenue change from Year 1 to Year 4 is a 15% increase, and that Year 2 revenue is 25% greater than Year 3 revenue are all true.

Let the revenue in Year 1 be

R

. Revenue in Year 2 is

1.25R

. Revenue in Year 3 is

1.25R \times 0.80 = 1.00R

, which equals the revenue in Year 1. Revenue in Year 4 is

1.00R \times 1.15 = 1.15R

, representing a net

15\%

increase from Year 1. Furthermore, Year 2 revenue (

1.25R

) compared to Year 3 revenue (

1.00R

) is calculated as

\frac{1.25R - 1.00R}{1.00R} = 0.25

, or a

25\%

increase. Therefore, the first three statements are mathematically true.

Step-by-Step Solution

1

Define variables and calculate Year 2 revenue relative to Year 1

Let Year 1 revenue be

R

. Year 2 revenue =

R \times (1 + 0.25) = 1.25R

.

A 25% increase corresponds to a multiplier of 1.25.

2

Calculate Year 3 revenue relative to Year 2 and Year 1

Year 3 revenue =

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

.

A 20% decrease applies to the new base value of

1.25R

.

3

Calculate Year 4 revenue relative to Year 3 and Year 1

Year 4 revenue =

1.00R \times (1 + 0.15) = 1.15R

.

A 15% increase applies to Year 3 revenue, which equals

R

.

4

Evaluate each statement against the calculated values

Year 3 revenue (

1.00R

) equals Year 1 revenue (

R

). Year 4 revenue (

1.15R

) is 15% greater than Year 1 revenue (

R

). Year 2 revenue (

1.25R

) is 25% greater than Year 3 revenue (

1.00R

).

Determines which statements hold true mathematically.

Key Concept

Successive Percent Changes and Base Value Shifting

Question 175Question →

Let $r$ and $s$ be rational numbers such that $0 < r < \frac{1}{2}$ and $1 < s < 2$ . Which of the following inequalities MUST be true? Select all that apply.

Select all that apply

r \cdot s < 1

\frac{s}{r} > 2

s - r > 1

\frac{r}{s} > \frac{1}{4}

r + s > \frac{3}{2}

Show answer & explanation

Answer: $r \cdot s < 1$ ; \frac{s}{r} > 2

Answer

The correct inequalities that must be true are

r \cdot s < 1

and

\frac{s}{r} > 2

.

The inequality stating that the product of

r

and

s

is less than

1

is guaranteed because multiplying the maximum possible bounds (

\frac{1}{2}

and

2

) gives

1

. The inequality stating that the ratio of

s

to

r

is greater than

2

is guaranteed because the numerator is strictly greater than

1

and the denominator is strictly less than

\frac{1}{2}

.

Step-by-Step Solution

1

Analyze

r \cdot s < 1

Since

r < \frac{1}{2}

and

s < 2

, and both are positive rational numbers, multiplying their upper limits gives

r \cdot s < \left(\frac{1}{2}\right)(2) = 1

. This statement MUST be true.

Properties of inequality multiplication for positive rational numbers.

2

Analyze

\frac{s}{r} > 2

Since

s > 1

and

r < \frac{1}{2}

, taking the ratio gives

\frac{s}{r} > \frac{1}{1/2} = 2

. This statement MUST be true.

Dividing a larger positive number by a fraction less than

1/2

amplifies the quotient.

3

Test counterexamples for remaining options

For

s - r > 1

, pick

s = 1.2

and

r = 0.4 \Rightarrow s - r = 0.8 \le 1

(False). For

\frac{r}{s} > \frac{1}{4}

, pick

r = 0.1

and

s = 1.6 \Rightarrow \frac{r}{s} = \frac{1}{16} < \frac{1}{4}

(False). For

r + s > \frac{3}{2}

, pick

r = 0.1

and

s = 1.1 \Rightarrow r + s = 1.2 < 1.5

(False).

A single valid counterexample disproves a 'must be true' statement.

Key Concept

Properties of Rational Numbers and Inequalities

Question 176Question →

Let $n$ be an integer such that $-15 \le n \le 15$ . How many integer values of $n$ satisfy both of the following conditions?

1. $(-1)^{n^2 + n + 1} < 0$
2. $(n - 3)(n + 4)^2 < 0$

Show answer & explanation

Answer: 17

Answer

17

Condition 1 is satisfied by every integer because

n^2 + n = n(n+1)

is always even (as the product of two consecutive integers), which makes

n^2 + n + 1

always odd and

(-1)^{\text{odd}} = -1 < 0

. Condition 2 requires

(n - 3)(n + 4)^2 < 0

. Since

(n+4)^2 > 0

for all

n \neq -4

, this reduces to

n < 3

while excluding

n = -4

. Within the range

-15 \le n \le 15

, there are 18 integers strictly less than 3, and removing

n = -4

leaves 17 valid values.

Step-by-Step Solution

1

Determine the parity of the exponent

n^2 + n + 1

n^2 + n + 1

is always odd for any integer

n

, making

(-1)^{n^2 + n + 1} = -1 < 0

unconditionally true.

The product of consecutive integers

n(n+1)

is always even, so adding 1 results in an odd number.

2

Solve the sign inequality

(n - 3)(n + 4)^2 < 0

n < 3

with

n \neq -4

.

A squared expression is strictly positive except when its base is zero. At

n = -4

, the product becomes 0, violating the strict inequality.

3

Count the integer solutions in the range

-15 \le n \le 15

17 integers satisfy both conditions.

There are 18 integers less than 3 in the interval

[-15, 15]

, and excluding

n = -4

gives

18 - 1 = 17

.

Key Concept

Parity of consecutive integer products and sign rules for squared terms in inequalities

Estimated Time:2m 0s

Question 177Question →

In a given fiscal quarter, a retail company increased the unit price of its primary product by $p\%$ while the total quantity of the product sold decreased by $p\%$ , where $p > 0$ . As a result of these two concurrent changes, the total revenue generated from the product experienced a net decrease of exactly $4\%$ . Which of the following statements must be true? Select all that apply.

Select all that apply

The value of

p

is

20

.

The final revenue generated from the product is equal to

96\%

of the initial revenue.

If the unit price had instead decreased by

p\%

and the quantity sold had increased by

p\%

, total revenue would still have decreased by

4\%

.

The net percentage change in total revenue depends directly on the initial unit price of the product.

The sequential percentage changes of

+p\%

and

-p\%

offset each other, resulting in zero net change in total revenue.

Show answer & explanation

Answer: The value of $p$ is $20$ .; The final revenue generated from the product is equal to $96\%$ of the initial revenue.; If the unit price had instead decreased by $p\%$ and the quantity sold had increased by $p\%$ , total revenue would still have decreased by $4\%$ .

Answer

The correct statements are that the value of p is 20, the final revenue is 96% of the initial revenue, and reversing the direction of the percentage changes would still result in a 4% decrease.

The correct options accurately identify that

p = 20

, that the final revenue is

96\%

of the initial value, and that reversing the roles of increase and decrease produces the exact same net multiplier of

0.96

. Using the difference of squares identity

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

, the combined multiplier for price and quantity shifts is

1 - (p/100)^2 = 0.96

. Solving gives

p^2 = 400

, so

p = 20

. Additionally,

(1 - 0.20)(1 + 0.20) = 0.96

, showing symmetry in net outcome.

Step-by-Step Solution

1

Express the final revenue as a function of initial revenue, price change, and quantity change.

Final Revenue = Initial Revenue

\times (1 + \frac{p}{100}) \times (1 - \frac{p}{100})

Total revenue is Price

\times

Quantity, so independent percentage changes multiply.

2

Apply the difference of squares formula and set equal to the given net revenue decrease.

(1 + \frac{p}{100})(1 - \frac{p}{100}) = 1 - \frac{p^2}{10000} = 1 - 0.04 = 0.96

A

4\%

decrease leaves

96\%

or

0.96

of the original value.

3

Solve the algebraic equation for

p

.

\frac{p^2}{10000} = 0.04 \implies p^2 = 400 \implies p = 20

Isolating

p

demonstrates that

p = 20

.

4

Evaluate alternative scenario with price decrease and quantity increase.

(1 - \frac{20}{100})(1 + \frac{20}{100}) = (0.80)(1.20) = 0.96

Multiplication is commutative, yielding the exact same

4\%

net decrease.

Key Concept

Successive Percent Changes and Base Shift Multipliers

Question 178Question →

A laboratory technician prepares a cleaning solution by mixing a concentrated formula and distilled water in a ratio of $2 : 7$ by volume. If the technician uses $14$ liters of the concentrated formula, how many liters of distilled water are required?

4

35

49

63

98

Show answer & explanation

Answer: $49$

Answer

49

liters of distilled water

The ratio of concentrated formula to distilled water is given as

2 : 7

. Since

14

liters of concentrated formula represents

2

ratio parts (

14 = 2 \times 7

), each ratio part equals

7

liters. Multiplying the

7

ratio parts of distilled water by

7

liters gives

7 \times 7 = 49

liters.

Step-by-Step Solution

1

Set up the proportion relating concentrated formula to distilled water

\frac{\text{Concentrated Formula}}{\text{Distilled Water}} = \frac{2}{7} = \frac{14}{x}

The given ratio is part-to-part (

2

parts formula for every

7

parts water).

2

Solve for the unknown volume

x

using cross-multiplication

2 \cdot x = 7 \cdot 14 \implies 2x = 98

Cross-multiplying equates the products of the diagonal terms in a proportion.

3

Divide by

2

to isolate

x

x = 49

Isolating

x

provides the required amount of distilled water in liters.

Key Concept

Direct Part-to-Part Proportions

Question 179Question →

An investment fund managed two separate accounts, Account X and Account Y, over a two-year period.

- At the start of Year 1, Account X held $\$100,000$ and Account Y held $\$150,000$ .
- During Year 1, Account X increased in value by $20\%$ , while Account Y decreased in value by $x\%$ .
- During Year 2, Account X decreased in value by $x\%$ , while Account Y increased in value by $20\%$ .
- At the end of Year 2, the combined total value of both accounts was $\$243,000$ .

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

Select all that apply

The value of

x

is equal to

19

.

At the end of Year 2, Account X experienced a net decrease of

2.8\%

relative to its initial value.

At the end of Year 1, Account Y had a greater dollar value than Account X.

At the end of Year 2, the overall percentage decrease in the value of Account Y from its initial value was greater than that of Account X.

The overall percent change of the combined portfolio over the two-year period was a decrease of

7.0\%

.

Show answer & explanation

Answer: The value of $x$ is equal to $19$ .; At the end of Year 2, Account X experienced a net decrease of $2.8\%$ relative to its initial value.; At the end of Year 1, Account Y had a greater dollar value than Account X.

Answer

The correct statements are those asserting that

x = 19

, that Account X experienced a net decrease of

2.8\%

relative to its initial value, and that at the end of Year 1 Account Y had a greater dollar value than Account X.

The statement specifying that

x = 19

is correct because solving the combined account equation

\$300,000(1 - x/100) = \$243,000

yields

1 - x/100 = 0.81

, giving

x = 19

. The statement regarding Account X experiencing a net decrease of

2.8\%

is correct because

1.20 \times 0.81 = 0.972

, which corresponds to a

2.8\%

decrease from initial value. The statement asserting that Account Y had a greater dollar value at the end of Year 1 than Account X is correct because Account Y was worth

\$121,500

compared to

\$120,000

for Account X.

Step-by-Step Solution

1

Express the value of each account at the end of Year 1 in terms of

x

.

Account X value =

\$100,000 \times (1 + 0.20) = \$120,000

. Account Y value =

\$150,000 \times (1 - x/100)

.

Calculate the intermediate values after Year 1 percentage changes.

2

Express the value of each account at the end of Year 2 and sum them to form an equation for the combined total.

Account X end of Year 2 =

\$120,000(1 - x/100)

. Account Y end of Year 2 =

\$150,000(1 - x/100)(1 + 0.20) = \$180,000(1 - x/100)

. Combined total =

(\$120,000 + \$180,000)(1 - x/100) = \$300,000(1 - x/100) = \$243,000

.

Use the given final total value of

\$243,000

to solve for

x

.

3

Solve for

x

.

1 - x/100 = 243,000 / 300,000 = 0.81 \implies x/100 = 0.19 \implies x = 19

.

Find the numerical value of

x

.

4

Evaluate each statement against the calculated values.

1.

x = 19

is true.
2. Net change multiplier for Account X =

1.20 \times 0.81 = 0.972

, representing a

2.8\%

decrease (true).
3. End of Year 1 values: Account X =

\$120,000

, Account Y =

\$150,000 \times 0.81 = \$121,500

. Since

\$121,500 > \$120,000

, Account Y was greater (true).
4. Account Y net change multiplier =

0.81 \times 1.20 = 0.972

, identical to Account X, so percentage decreases are equal (false).
5. Overall combined percent change =

(243,000 - 250,000)/250,000 = -7,000/250,000 = -2.8\%

(false).

Determine which statements are mathematically true.

Key Concept

Successive Percentage Changes and Base Shifts in Multi-Asset Portfolios

Question 180Question →

A positive integer $n$ yields a remainder of $7$ when divided by $12$ . What is the remainder when the expression $5n + 3$ is divided by $6$ ?

0

1

2

3

5

Show answer & explanation

Answer: $2$

Answer

The remainder when

5n + 3

is divided by

6

is

2

.

Since

n

leaves a remainder of

7

when divided by

12

, we can write

n = 12k + 7

for some integer

k \ge 0

. Substituting this into

5n + 3

yields

5(12k + 7) + 3 = 60k + 38

. Factoring out

6

gives

6(10k + 6) + 2

. Because

6(10k + 6)

is divisible by

6

, the remainder of the expression when divided by

6

is

2

.

Step-by-Step Solution

1

Express

n

in terms of the division algorithm for divisor

12

.

n = 12k + 7

for some non-negative integer

k

An integer that leaves a remainder of

7

when divided by

12

can be represented as

12k + 7

.

2

Substitute the expression for

n

into

5n + 3

and simplify.

5(12k + 7) + 3 = 60k + 35 + 3 = 60k + 38

Algebraic expansion allows us to analyze the entire expression modulo

6

.

3

Determine the remainder of

60k + 38

when divided by

6

.

60k + 38 = 6(10k + 6) + 2

Since

60k + 36

is an exact multiple of

6

, the leftover term

2

is the remainder.

Key Concept

Properties of Integers and Remainder Arithmetic

Estimated Time:1m 30s

Arithmetic

Answer

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Common Mistakes

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