Arithmetic Practice Questions — GRE General Test — Page 13

Question 241Question →

A cyclist completes a journey consisting of three distinct segments: an uphill segment, a flat segment, and a downhill segment. The ratio of the distances of the uphill, flat, and downhill segments is $2 : 3 : 5$ , respectively. The cyclist's average speed on the flat segment is twice her average speed on the uphill segment, and her average speed on the downhill segment is three times her average speed on the uphill segment. If the cyclist's overall average speed for the entire journey is $30$ miles per hour, what is her average speed, in miles per hour, on the flat segment?

Show answer & explanation

Answer: 31

Answer

31

The correct average speed on the flat segment is 31 miles per hour. Setting up segment distances as

2x

,

3x

, and

5x

(total distance

10x

) and segment speeds as

v

,

2v

, and

3v

, the segment times are

t_1 = \frac{2x}{v}

,

t_2 = \frac{3x}{2v}

, and

t_3 = \frac{5x}{3v}

. The total travel time is

T = \frac{31x}{6v}

. Dividing total distance

10x

by total time

T

yields an overall average speed of

\frac{60v}{31} = 30

. Solving for

v

gives

v = 15.5

miles per hour. Thus, the average speed on the flat segment is

2v = 31

miles per hour.

Step-by-Step Solution

1

Define segment distances using ratio multipliers.

Distances are

d_1 = 2x

,

d_2 = 3x

, and

d_3 = 5x

, giving total distance

D = 10x

.

The distances of the three segments are in the ratio

2 : 3 : 5

.

2

Express segment speeds relative to the uphill speed

v

.

Uphill speed is

v

, flat speed is

2v

, and downhill speed is

3v

.

The problem states flat speed is twice uphill speed, and downhill speed is three times uphill speed.

3

Calculate the time spent on each segment.

t_1 = \frac{2x}{v}

,

t_2 = \frac{3x}{2v}

, and

t_3 = \frac{5x}{3v}

.

Time equals distance divided by speed (

t = \frac{d}{v}

).

4

Calculate total travel time by summing individual segment times.

T = \frac{x}{v} \left(2 + \frac{3}{2} + \frac{5}{3}\right) = \frac{31x}{6v}

.

Combining fractions with a common denominator of 6 gives

\frac{12 + 9 + 10}{6} = \frac{31}{6}

.

5

Relate total distance and total time to the overall average speed.

\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{10x}{\frac{31x}{6v}} = \frac{60v}{31} = 30

.

Overall average speed is defined as total distance divided by total time.

6

Solve for

v

and calculate the flat segment speed

2v

.

v = 15.5

mph, so flat segment speed

= 2(15.5) = 31

mph.

Solving

\frac{60v}{31} = 30

yields

v = 15.5

, making

2v = 31

.

Key Concept

Weighted Average Speed and Multi-Segment Distance-Rate-Time Ratios

Question 242Question →

A sequence of positive real numbers $a_1, a_2, a_3, \dots$ is defined by $a_1 = 3$ and $a_{n+1} = a_n + \frac{2}{a_n}$ for all integers $n \ge 1$ . Which of the following statements must be true? Select all that apply.

Select all that apply

The sequence

a_1, a_2, a_3, \dots

is strictly increasing.

The term

a_{50}

is strictly greater than

14

.

When

a_{50}

is rounded to the nearest integer, the result is

15

.

The sequence of consecutive term differences

d_n = a_{n+1} - a_n

is strictly decreasing for

n \ge 1

.

The term

a_{100}

is strictly greater than

25

.

Show answer & explanation

Answer: The sequence $a_1, a_2, a_3, \dots$ is strictly increasing.; The term $a_{50}$ is strictly greater than $14$ .; The sequence of consecutive term differences $d_n = a_{n+1} - a_n$ is strictly decreasing for $n \ge 1$ .

Answer

The correct statements are that the sequence is strictly increasing, the term

a_{50}

is strictly greater than 14, and the sequence of consecutive term differences

d_n = a_{n+1} - a_n

is strictly decreasing.

The statement asserting strict monotonicity is correct because

a_{n+1} - a_n = \frac{2}{a_n} > 0

for positive terms. The statement regarding

a_{50} > 14

is correct because

a_{50}^2 > 3^2 + 4(49) = 205 > 196

. The statement concerning consecutive differences is correct because

\frac{2}{a_n}

strictly decreases as

a_n

increases.

Step-by-Step Solution

1

Analyze monotonicity of the sequence

a_{n+1} - a_n = \frac{2}{a_n} > 0

for all

n \ge 1

Since

a_1 = 3 > 0

, all terms remain positive, making each term strictly larger than the previous.

2

Analyze the sequence of term differences

d_n = \frac{2}{a_n}

decreases as

a_n

increases

Since

a_n

grows strictly monotonically, its reciprocal strictly decreases, so

d_{n+1} < d_n

.

3

Derive a lower bound for

a_{50}

using quadratic expansion

a_{50}^2 > 205 \implies a_{50} > 14

Expanding

a_{k+1}^2 = a_k^2 + 4 + \frac{4}{a_k^2} > a_k^2 + 4

and summing from

k=1

to

49

gives

a_{50}^2 > 3^2 + 4(49) = 205 > 196 = 14^2

.

4

Derive an upper bound for

a_{50}

to test the rounding statement

a_{50} < 14.45

, so rounding to the nearest integer yields 14

Using

\frac{4}{a_k^2} < \frac{4}{4k+5}

, the sum of error terms is bounded above by

\int_0^{49} \frac{4}{4x+5} dx = \ln(40.2) \approx 3.7

. Thus

a_{50}^2 < 208.7 < 14.5^2

, so

a_{50}

rounds to 14, not 15.

5

Estimate

a_{100}

to test the magnitude bound

a_{100} < \sqrt{409.4} \approx 20.23 < 25

Summing the squared recurrence up to 100 terms gives

a_{100}^2 < 405 + \ln(80.2) \approx 409.4

, showing

a_{100}

cannot exceed 25.

Key Concept

Estimation of non-linear recursive sequences using squared bounds and integral comparison.

Estimated Time:3m 0s

Question 243Question →

Two deep-space radio signals have power measurements of $S_1 = 4.5 \times 10^{-7}$ watts and $S_2 = 3.5 \times 10^{-6}$ watts. A combined signal power is defined as $S_3 = S_1 + S_2$ . When $S_3$ is written in scientific notation as $a \times 10^n$ , where $1 \leq a < 10$ and $n$ is an integer, what is the value of $a + n$ ?

-5.00

-2.05

2.00

9.95

32.50

Show answer & explanation

Answer: $-2.05$

Answer

The value of

a + n

is

-2.05

.

To add numbers in scientific notation, first rewrite them with identical powers of 10. Expressing

4.5 \times 10^{-7}

as

0.45 \times 10^{-6}

allows direct addition with

3.5 \times 10^{-6}

, resulting in

3.95 \times 10^{-6}

. Here

a = 3.95

(which satisfies

1 \leq a < 10

) and

n = -6

. Adding

a + n

yields

3.95 + (-6) = -2.05

.

Step-by-Step Solution

1

Express both quantities with a common power of 10 to allow addition.

S_1 = 4.5 \times 10^{-7} = 0.45 \times 10^{-6}

watts.

Before adding numbers in scientific notation, their powers of 10 must match.

2

Add the coefficients while maintaining the common power of 10.

S_3 = (0.45 + 3.5) \times 10^{-6} = 3.95 \times 10^{-6}

watts.

Distributive property allows adding coefficients once exponents match.

3

Verify proper scientific notation form

a \times 10^n

where

1 \leq a < 10

.

a = 3.95

and

n = -6

.

The coefficient

3.95

satisfies

1 \leq 3.95 < 10

, so no further decimal shift is needed.

4

Calculate the requested sum

a + n

.

3.95 + (-6) = -2.05

.

Adding the coefficient

3.95

to the negative exponent

-6

gives

-2.05

.

Key Concept

Scientific Notation Addition and Place Value Alignment

Question 244Question →

If $x$ and $y$ are positive integers such that $3^{x+2} \cdot 4^y - 3^x \cdot 4^{y+1} = 11,520$ , what is the value of $x + y$ ?

5

6

7

8

10

Show answer & explanation

Answer: 6

Answer

6

Factoring

3^x \cdot 4^y

from the expression

3^{x+2} \cdot 4^y - 3^x \cdot 4^{y+1}

yields

3^x \cdot 4^y (3^2 - 4^1) = 5 \cdot 3^x \cdot 4^y

. Setting this equal to 11,520 and dividing by 5 gives

3^x \cdot 4^y = 2,304

. Prime factorization of 2,304 gives

3^2 \cdot 4^4

, so

x = 2

and

y = 4

. The sum

x + y

is equal to 6.

Step-by-Step Solution

1

Factor out the greatest common exponential factor

3^x \cdot 4^y

from the left side of the equation.

3^x \cdot 4^y (3^2 - 4^1) = 11,520

By exponent rules,

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

and

4^{y+1} = 4^y \cdot 4^1

.

2

Evaluate the constant factor inside the parentheses.

3^2 - 4^1 = 9 - 4 = 5

, so

5 \cdot 3^x \cdot 4^y = 11,520

Simplifying numerical exponents.

3

Divide both sides of the equation by 5.

3^x \cdot 4^y = 2,304

Isolating the variable exponential terms.

4

Determine the prime factorization of 2,304 into powers of 3 and 4.

2,304 = 9 \cdot 256 = 3^2 \cdot 4^4

, which implies

x = 2

and

y = 4

Unique factorization for integer bases.

5

Calculate the sum

x + y

.

x + y = 2 + 4 = 6

Answering the explicit prompt.

Key Concept

Factoring Exponential Expressions and Unique Factorization

Question 245Question →

For two positive integers $m$ and $n$ , the greatest common divisor is $\gcd(m, n) = 15$ and the least common multiple is $\text{lcm}(m, n) = 900$ . If $m = 75$ , what is the total number of positive divisors of $n$ ?

18

9

5

12

3

Show answer & explanation

Answer: 18

Answer

18

The correct answer is 18. First, use the relation

m \cdot n = \gcd(m, n) \cdot \text{lcm}(m, n)

to find

n = \frac{15 \cdot 900}{75} = 180

. Next, write 180 in prime factorized form:

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

. The total number of positive divisors is found by adding 1 to each exponent and multiplying the results:

(2 + 1)(2 + 1)(1 + 1) = 3 \cdot 3 \cdot 2 = 18

.

Step-by-Step Solution

1

Calculate the value of

n

using the fundamental product identity for GCD and LCM.

n = \frac{\gcd(m, n) \cdot \text{lcm}(m, n)}{m} = \frac{15 \cdot 900}{75} = 180

.

For any two positive integers, the product of the integers equals the product of their GCD and LCM.

2

Find the prime factorization of

180

.

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

.

Breaking down 180 into prime powers allows determination of the total count of positive divisors.

3

Apply the divisor counting formula by adding 1 to each prime exponent and multiplying.

(2+1)(2+1)(1+1) = 3 \cdot 3 \cdot 2 = 18

.

If an integer has prime factorization

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

, the number of positive divisors is

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

.

Key Concept

GCD and LCM fundamental identity (

a \cdot b = \gcd(a,b) \cdot \text{lcm}(a,b)

) combined with the prime factorization divisor counting formula.

Estimated Time:1m 30s

Question 246Question →

In Year 1, a software company had a total of 400 subscribers divided between a Basic plan and a Premium plan, with $70\%$ of the subscribers enrolled in the Basic plan. In Year 2, the number of Basic plan subscribers decreased by $20\%$ , while the total number of subscribers across both plans increased by $10\%$ . What was the percentage increase in the number of Premium plan subscribers from Year 1 to Year 2?

24\%

30\%

50\%

80\%

180\%

Show answer & explanation

Answer: $80\%$

Answer

The percentage increase in Premium plan subscribers from Year 1 to Year 2 was

80\%

.

In Year 1,

70\%

of the 400 subscribers were on the Basic plan (

0.70 \times 400 = 280

), which leaves 120 subscribers on the Premium plan. In Year 2, the total subscriber count increased by

10\%

to 440 (

400 \times 1.10 = 440

), while Basic subscribers decreased by

20\%

to 224 (

280 \times 0.80 = 224

). Subtracting 224 Basic subscribers from the total 440 gives 216 Premium subscribers in Year 2. The increase in Premium subscribers is

216 - 120 = 96

. To find the percentage increase, divide the increase of 96 by the original Premium subscriber count of 120, yielding

\frac{96}{120} = 0.80

, or

80\%

.

Step-by-Step Solution

1

Calculate the number of Basic and Premium subscribers in Year 1

Basic subscribers =

0.70 \times 400 = 280

; Premium subscribers =

400 - 280 = 120

Determining the starting baseline values for each category is necessary to evaluate subsequent changes.

2

Calculate the total subscribers and Basic subscribers in Year 2

Total subscribers in Year 2 =

400 \times (1 + 0.10) = 440

; Basic subscribers in Year 2 =

280 \times (1 - 0.20) = 224

Apply the given percentage changes to find the updated group totals in Year 2.

3

Find the number of Premium subscribers in Year 2 and the absolute increase

Premium subscribers in Year 2 =

440 - 224 = 216

; Absolute increase =

216 - 120 = 96

Subtracting the Basic subscriber count from the total count yields the Premium subscriber count for Year 2.

4

Compute the percentage increase for Premium subscribers

Percentage increase =

\frac{96}{120} \times 100\% = 80\%

Divide the absolute increase by the initial Year 1 Premium subscriber count (the correct base value) and convert to a percentage.

Key Concept

Successive percentage changes and base shifting in multi-part totals

Estimated Time:1m 30s

Question 247Question →

If $x$ and $y$ are real numbers such that $|x - 3| \le 2$ and $|y + 1| \le 4$ , which of the following could be the value of $|x - y|$ ? Select all such values.

Select all that apply

-2

0

5

10

12

Show answer & explanation

Answer: $0$ ; $5$ ; $10$

Answer

The possible values of

|x - y|

are 0, 5, and 10.

Solving the inequalities yields

1 \le x \le 5

and

-5 \le y \le 3

. The expression

|x - y|

represents the distance between

x

and

y

on the number line. Because the intervals overlap between 1 and 3,

x

and

y

can be equal, making the minimum distance 0. The maximum distance occurs at the extreme points

x = 5

and

y = -5

, giving a distance of

|5 - (-5)| = 10

. Therefore, any value from 0 to 10 inclusive is possible, making 0, 5, and 10 valid values.

Step-by-Step Solution

1

Solve the absolute value inequality for

x

.

-2 \le x - 3 \le 2 \implies 1 \le x \le 5

Unwrapping

|x - 3| \le 2

gives the bounds for

x

on the real number line.

2

Solve the absolute value inequality for

y

.

-4 \le y + 1 \le 4 \implies -5 \le y \le 3

Unwrapping

|y + 1| \le 4

gives the bounds for

y

on the real number line.

3

Determine the minimum and maximum possible values for

|x - y|

.

Minimum value is 0 (since the intervals

[1, 5]

and

[-5, 3]

overlap at

[1, 3]

). Maximum value is

|5 - (-5)| = 10

. Thus,

0 \le |x - y| \le 10

.

The absolute value

|x - y|

represents the distance between

x

and

y

on the number line, which can take any real value from 0 to 10.

4

Evaluate the given choices against the range

[0, 10]

.

The values 0, 5, and 10 fall within

[0, 10]

, whereas

-2

is impossible for absolute values and

12

exceeds the maximum bound.

Determines which specific options are valid outcomes for

|x - y|

.

Key Concept

Absolute value as distance on the real number line and range of differences between bounded real variables

Estimated Time:1m 30s

Question 248Question →

The positive integer $N$ is divisible by $60$ . If the greatest common divisor of $N$ and $500$ is $100$ , which of the following statements MUST be true? Select all such statements.

Select all that apply

N

is a multiple of

300

.

N

is divisible by

125

.

The prime factorization of

N

contains at least two factors of

2

.

N

is divisible by

9

.

The greatest common divisor of

N

and

250

is

50

.

Show answer & explanation

Answer: $N$ is a multiple of $300$ .; The prime factorization of $N$ contains at least two factors of $2$ .; The greatest common divisor of $N$ and $250$ is $50$ .

Answer

The statements asserting that

N

is a multiple of

300

, that the prime factorization of

N

contains at least two factors of

2

, and that the greatest common divisor of

N

and

250

is

50

must all be true.

Analyzing the prime factorizations:

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

and

500 = 2^2 \cdot 5^3

. Since

N

is divisible by

60

, the prime factorization of

N

must have exponents

a \ge 2

for prime

2

,

b \ge 1

for prime

3

, and

c \ge 1

for prime

5

. Furthermore,

\gcd(N, 500) = 100 = 2^2 \cdot 5^2

. The GCD rule requires taking the minimum exponent for each prime factor:

\min(a, 2) = 2 \implies a \ge 2

, and

\min(c, 3) = 2 \implies c = 2

. Therefore,

N = 2^{a} \cdot 3^{b} \cdot 5^{2}

where

a \ge 2

and

b \ge 1

. This means

N

is divisible by

2^2 \cdot 3^1 \cdot 5^2 = 300

, contains at least two prime factors of

2

, and has

\gcd(N, 250) = \gcd(2^a \cdot 3^b \cdot 5^2, 2^1 \cdot 5^3) = 2^1 \cdot 5^2 = 50

.

Step-by-Step Solution

1

Express

60

and

500

in their prime factorizations.

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

and

500 = 2^2 \cdot 5^3

.

Prime factorization allows us to analyze divisibility and GCD conditions in terms of prime exponents.

2

Apply the condition that

N

is divisible by

60

.

The exponent of

2

in

N

is at least

2

, the exponent of

3

is at least

1

, and the exponent of

5

is at least

1

.

For

N

to be divisible by an integer,

N

must contain at least as many of each prime factor as that integer.

3

Apply the GCD condition

\gcd(N, 500) = 100 = 2^2 \cdot 5^2

.

The exponent of

5

in

N

must be exactly

2

.

Since

500

has

5^3

and the GCD has

5^2

, taking the minimum exponent of

5

between

N

and

500

yields

2

. Thus,

N

has

5^2

and not

5^3

or higher.

4

Evaluate each statement against the established exponent bounds for

N

.

N

has prime factor powers

2^{\ge 2} \cdot 3^{\ge 1} \cdot 5^2

. This guarantees

N

is a multiple of

2^2 \cdot 3^1 \cdot 5^2 = 300

, contains at least two factors of

2

, and yields

\gcd(N, 250) = 2^1 \cdot 5^2 = 50

.

Comparing the prime factor requirements of each choice confirms which statements MUST be true.

Key Concept

Prime factorization rules for divisibility and greatest common divisor (GCD)

Question 249Question →

In Year 1, a retail company's total revenue was generated by two divisions: Division X, which accounted for $40\%$ of total revenue, and Division Y, which accounted for the remaining $60\%$ . From Year 1 to Year 2, revenue from Division X increased by $20\%$ , while revenue from Division Y decreased by $10\%$ . From Year 2 to Year 3, revenue from Division X decreased by $10\%$ , while revenue from Division Y increased by $20\%$ . Which of the following statements must be true? Select all such statements.

Select all that apply

Total company revenue in Year 3 was

8\%

greater than total company revenue in Year 1.

From Year 1 to Year 3, revenue from Division X increased by

8\%

.

In Year 2, Division X accounted for more than

50\%

of total company revenue.

Total company revenue in Year 2 was

10\%

greater than total company revenue in Year 1.

From Year 1 to Year 3, revenue from Division Y increased by

8\%

.

Show answer & explanation

Answer: Total company revenue in Year 3 was $8\%$ greater than total company revenue in Year 1.; From Year 1 to Year 3, revenue from Division X increased by $8\%$ .; From Year 1 to Year 3, revenue from Division Y increased by $8\%$ .

Answer

The correct statements are: total company revenue in Year 3 was 8% greater than in Year 1; revenue from Division X increased by 8% from Year 1 to Year 3; and revenue from Division Y increased by 8% from Year 1 to Year 3.

The statements confirming an 8% overall increase for total company revenue, Division X revenue, and Division Y revenue from Year 1 to Year 3 are all correct. A 20% increase followed by a 10% decrease (and vice versa) multiplies the starting amount by

1.20 \times 0.90 = 1.08

, which represents a net 8% increase on each respective division's initial base.

Step-by-Step Solution

1

Assign a variable to initial total revenue and calculate division revenues for Year 1.

Let total revenue in Year 1 be

R

. Revenue from Division X is

0.40 R

and revenue from Division Y is

0.60 R

.

Establishing initial quantities provides the base for computing percentage changes across subsequent years.

2

Calculate division revenues and total revenue for Year 2.

Division X revenue in Year 2 =

0.40 R \times 1.20 = 0.48 R

. Division Y revenue in Year 2 =

0.60 R \times 0.90 = 0.54 R

. Total Year 2 revenue =

0.48 R + 0.54 R = 1.02 R

.

Applying the respective 20% increase and 10% decrease to Year 1 base values yields Year 2 amounts.

3

Calculate division revenues and total revenue for Year 3.

Division X revenue in Year 3 =

0.48 R \times 0.90 = 0.432 R

. Division Y revenue in Year 3 =

0.54 R \times 1.20 = 0.648 R

. Total Year 3 revenue =

0.432 R + 0.648 R = 1.08 R

.

Applying the respective 10% decrease and 20% increase to Year 2 values yields Year 3 amounts.

4

Evaluate the percent changes from Year 1 to Year 3 for each division and the total company.

Division X change:

\frac{0.432 R - 0.40 R}{0.40 R} = 8\%

. Division Y change:

\frac{0.648 R - 0.60 R}{0.60 R} = 8\%

. Total revenue change:

\frac{1.08 R - 1.00 R}{1.00 R} = 8\%

. Also, Division X proportion in Year 2 was

\frac{0.48 R}{1.02 R} \approx 47.06\%

.

Determines which of the given statements hold true based on accurate relative changes.

Key Concept

Weighted and Successive Percent Change

Estimated Time:2m 0s

Question 250Question →

Let $p$ , $q$ , and $r$ be non-zero integers satisfying the following three conditions:

I. $p^3 q r^2 < 0$
II. $(-p)^q < 0$
III. $(-1)^{p + r} = 1$

Which of the following expressions MUST be negative?

q \cdot p^r

p \cdot (-q)^r

(-p)^r \cdot q

(p - q)^r

q \cdot (r - p)^q

Show answer & explanation

Answer: $q \cdot p^r$

Answer

The expression

q \cdot p^r

MUST be negative.

Condition II establishes that

p > 0

and

q

is odd. Condition I establishes that

q < 0

. Since

p > 0

, raising

p

to any integer exponent

r

results in a strictly positive value (

p^r > 0

). Multiplying this positive value by negative

q

guarantees a negative outcome regardless of the value or sign of

r

.

Step-by-Step Solution

1

Analyze Condition II:

(-p)^q < 0

Base

-p < 0 \implies p > 0

, and exponent

q

must be an odd integer.

A negative number raised to an integer power is negative if and only if the exponent is odd.

2

Analyze Condition I:

p^3 q r^2 < 0

q < 0

(negative odd integer).

Since

p > 0

,

p^3 > 0

. Also

r \neq 0 \implies r^2 > 0

. For the overall product

p^3 q r^2

to be negative,

q

must be negative.

3

Analyze Condition III:

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

p + r

is an even integer, so

p

and

r

have the same parity.

(-1)^k = 1

requires

k

to be an even integer.

4

Evaluate the sign of

q \cdot p^r

q \cdot p^r < 0

for all valid values.

Since

p > 0

, any integer power

p^r > 0

. Multiplying positive

p^r

by negative

q

yields a negative result.

Key Concept

Even-odd exponent rules and negative base sign determination

Question 251Question →

On the real number line, the distance between $x$ and $3$ is equal to twice the distance between $x$ and $-9$ . What is the sum of all possible real values of $x$ ?

-26

-21

-14

-5

22

Show answer & explanation

Answer: $-26$

Answer

The sum of all possible real values of

x

is

-26

.

The distance between

x

and

a

on the real number line is expressed as

|x - a|

. Thus, the condition translates to

|x - 3| = 2|x - (-9)|

, which simplifies to

|x - 3| = 2|x + 9|

. Setting up the two algebraic cases gives

x - 3 = 2(x + 9)

, leading to

x = -21

, and

x - 3 = -2(x + 9)

, leading to

x = -5

. Adding these two solutions yields

(-21) + (-5) = -26

.

Step-by-Step Solution

1

Translate the geometric statement on the number line into an absolute value equation.

The distance between

x

and

3

is

|x - 3|

, and the distance between

x

and

-9

is

|x - (-9)| = |x + 9|

. Therefore,

|x - 3| = 2|x + 9|

.

Distance between two points

a

and

b

on the number line is given by

|a - b|

.

2

Solve Case 1 where the expressions inside the absolute values have the same sign.

x - 3 = 2(x + 9) \implies x - 3 = 2x + 18 \implies x = -21

.

Removing absolute values with identical signs gives a linear equation in

x

.

3

Solve Case 2 where the expressions inside the absolute values have opposite signs.

x - 3 = -2(x + 9) \implies x - 3 = -2x - 18 \implies 3x = -15 \implies x = -5

.

Removing absolute values with opposite signs accounts for the second possible geometric location.

4

Sum the valid real solutions.

(-21) + (-5) = -26

.

The question asks for the sum of all possible real values of

x

.

Key Concept

Absolute Value as Distance on the Real Number Line

Question 252Question →

Three positive integers $x$ , $y$ , and $z$ satisfy $\gcd(x, y) = 12$ , $\gcd(y, z) = 18$ , and $\gcd(x, z) = 30$ , where $\gcd(a, b)$ denotes the greatest common divisor of $a$ and $b$ . What is the minimum possible value of the least common multiple $\text{lcm}(x, y, z)$ ?

90

180

360

540

1080

Show answer & explanation

Answer: 180

Answer

180

The correct answer is 180. Expressing the pairwise GCDs in prime factorized form shows that

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

,

\gcd(y, z) = 2^1 \cdot 3^2

, and

\gcd(x, z) = 2^1 \cdot 3^1 \cdot 5^1

. To satisfy these simultaneously while minimizing the overall LCM, the maximum power of 2 across the numbers must be

2^2

, the maximum power of 3 must be

3^2

, and the maximum power of 5 must be

5^1

. Thus,

\text{lcm}(x, y, z) = 2^2 \cdot 3^2 \cdot 5^1 = 180

.

Step-by-Step Solution

1

Express the given GCD values in their prime factorized forms.

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

,

\gcd(y, z) = 18 = 2^1 \cdot 3^2 \cdot 5^0

, and

\gcd(x, z) = 30 = 2^1 \cdot 3^1 \cdot 5^1

.

Analyzing prime factor exponents determines the minimum required power of each prime in

x

,

y

, and

z

.

2

Determine the minimum necessary exponents for prime 2 in

x

,

y

, and

z

.

Since

\gcd(x, y)

has

2^2

, both

x

and

y

must contain at least

2^2

. Thus, the maximum exponent of 2 across the integers is at least 2.

The least common multiple takes the maximum exponent for each prime factor among all three numbers.

3

Determine the minimum necessary exponents for prime 3 in

x

,

y

, and

z

.

Since

\gcd(y, z)

has

3^2

, both

y

and

z

must contain at least

3^2

. Thus, the maximum exponent of 3 across the integers is at least 2.

To satisfy

\gcd(y, z) = 18

, the prime factor 3 must appear with an exponent of at least 2 in both

y

and

z

.

4

Determine the minimum necessary exponents for prime 5 in

x

,

y

, and

z

.

Since

\gcd(x, z)

has

5^1

, both

x

and

z

must contain at least

5^1

. Thus, the maximum exponent of 5 across the integers is at least 1.

To satisfy

\gcd(x, z) = 30

, the prime factor 5 must appear with an exponent of at least 1 in both

x

and

z

.

5

Calculate the minimum least common multiple

\text{lcm}(x, y, z)

.

\text{lcm}(x, y, z) = 2^2 \cdot 3^2 \cdot 5^1 = 4 \cdot 9 \cdot 5 = 180

. (A valid assignment is

x = 60

,

y = 36

,

z = 90

).

Multiplying the minimum maximum prime factor powers yields the smallest possible value for the LCM.

Key Concept

Prime Factor Exponents in Pairwise GCD and LCM Relations

Estimated Time:2m 0s

Question 253Question →

In January, a coffee roasting facility produced two specialty blends: Roast A and Roast B. Roast A sold for $\$10$ per pound and Roast B sold for $\$30$ per pound. A distributor purchased a total of $100$ pounds of coffee consisting of these two blends at an average price of $\$18$ per pound. In February, the price of Roast A increased by $20\%$ , while the price of Roast B decreased by $4\%$ . If the distributor purchased the exact same quantities of Roast A and Roast B in February as in January, by what percent did the total cost of the order increase?

3.85\%

4\%

8\%

10.4\%

16\%

Show answer & explanation

Answer: $4\%$

Answer

The total cost of the order increased by

4\%

.

The choice stating

4\%

is correct because the total cost in January is

\$1,800

(

60

lbs of Roast A at

\$10

and

40

lbs of Roast B at

\$30

). In February, the price of Roast A rises to

\$12

and Roast B drops to

\$28.80

, bringing the total cost to

\$1,872

. The change of

\$72

represents a

4\%

increase relative to the baseline January cost of

\$1,800

.

Step-by-Step Solution

1

Determine the quantities of Roast A and Roast B purchased in January.

Roast A:

60

lbs, Roast B:

40

lbs.

Let

a

be the pounds of Roast A and

b

be the pounds of Roast B. We are given

a + b = 100

and

10a + 30b = 18 \times 100 = 1800

. Substituting

b = 100 - a

gives

10a + 30(100 - a) = 1800 \implies -20a = -1200 \implies a = 60

and

b = 40

.

2

Calculate the total cost of the order in January.

Total January Cost =

\$1,800

.

Total cost =

(60 \text{ lbs} \times \$10/\text{lb}) + (40 \text{ lbs} \times \$30/\text{lb}) = \$600 + \$1,200 = \$1,800

.

3

Calculate the new prices and total cost in February.

Total February Cost =

\$1,872

.

New price of Roast A =

\$10 \times (1 + 0.20) = \$12/\text{lb}

. New price of Roast B =

\$30 \times (1 - 0.04) = \$28.80/\text{lb}

. Total February Cost =

(60 \times \$12) + (40 \times \$28.80) = \$720 + \$1,152 = \$1,872

.

4

Calculate the percentage increase in total cost from January to February.

Percent Increase =

4\%

.

Percent Change =

\frac{\text{February Cost} - \text{January Cost}}{\text{January Cost}} \times 100\% = \frac{\$1,872 - \$1,800}{\$1,800} \times 100\% = \frac{\$72}{\$1,800} \times 100\% = 4\%

.

Key Concept

Weighted Percent Change and Systems of Equations

Estimated Time:1m 30s

Question 254Question →

On a number line, point $A$ is located at $10$ and point $B$ is located at $2$ . If point $P$ , with coordinate $x > 0$ , is three times as far from point $A$ as it is from point $B$ , what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

4

The distance from

P(x)

to

A(10)

is

|x - 10|

and to

B(2)

is

|x - 2|

. Setting

|x - 10| = 3|x - 2|

gives two equations:

x - 10 = 3x - 6

, which yields

x = -2

, and

x - 10 = -3x + 6

, which yields

x = 4

. Since

x > 0

, the correct value is 4.

Step-by-Step Solution

1

Formulate the distance relationship using absolute value notation.

|x - 10| = 3|x - 2|

The distance between two points

u

and

v

on the real number line is given by

|u - v|

.

2

Split the absolute value equation into two linear equations representing possible cases.

x - 10 = 3(x - 2)

or

x - 10 = -3(x - 2)

The equality

|a| = |b|

implies

a = b

or

a = -b

.

3

Solve each case algebraically.

Case 1 gives

x - 10 = 3x - 6 \implies 2x = -4 \implies x = -2

. Case 2 gives

x - 10 = -3x + 6 \implies 4x = 16 \implies x = 4

.

Standard linear equation solving.

4

Apply the given domain condition

x > 0

.

x = 4

The solution

x = -2

is negative and therefore violates the condition

x > 0

.

Key Concept

Distance on a number line represented by absolute value equations

Question 255Question →

A commercial coffee roasting facility operates three roasters: $X$ , $Y$ , and $Z$ .

- Roaster $X$ processes coffee beans at a constant rate of $60\text{ kg/hr}$ , producing a blend with an Arabica-to-Robusta ratio of $3:2$ by weight.
- Roaster $Y$ processes coffee beans at a constant rate of $90\text{ kg/hr}$ , producing a blend with an Arabica-to-Robusta ratio of $2:1$ by weight.
- Roaster $Z$ processes coffee beans at a constant rate of $150\text{ kg/hr}$ , producing a blend with an Arabica-to-Robusta ratio of $1:4$ by weight.

All three roasters operate simultaneously for $4$ hours to complete a production order. Which of the following statements regarding the production output over this $4$ -hour period must be true? Select all such statements.

Select all that apply

The total weight of Arabica beans processed across all three roasters is

504\text{ kg}

.

Arabica beans account for exactly

42\%

of the total weight of coffee beans processed.

The overall ratio of total Arabica weight to total Robusta weight produced is

21:29

.

The combined hourly production rate of Robusta beans across the three roasters is

150\text{ kg/hr}

.

Roaster

Z

produces a greater total weight of Arabica beans than Roaster

X

.

Show answer & explanation

Answer: The total weight of Arabica beans processed across all three roasters is $504\text{ kg}$ .; Arabica beans account for exactly $42\%$ of the total weight of coffee beans processed.; The overall ratio of total Arabica weight to total Robusta weight produced is $21:29$ .

Answer

The statements confirming that total Arabica weight is

504\text{ kg}

, Arabica accounts for

42\%

of total weight, and the overall ratio of Arabica to Robusta is

21:29

are all correct.

The total production from Roaster

X

is

240\text{ kg}

(

144\text{ kg}

Arabica,

96\text{ kg}

Robusta), from Roaster

Y

is

360\text{ kg}

(

240\text{ kg}

Arabica,

120\text{ kg}

Robusta), and from Roaster

Z

is

600\text{ kg}

(

120\text{ kg}

Arabica,

480\text{ kg}

Robusta). The total Arabica weight is

144 + 240 + 120 = 504\text{ kg}

, which constitutes

\frac{504}{1,200} = 42\%

of the

1,200\text{ kg}

total output. The remaining

696\text{ kg}

is Robusta, giving an Arabica to Robusta ratio of

504:696 = 21:29

.

Step-by-Step Solution

1

Calculate total output and Arabica quantity for each roaster over 4 hours

Roaster

X

:

240\text{ kg}

total,

\frac{3}{5} \times 240 = 144\text{ kg}

Arabica. Roaster

Y

:

360\text{ kg}

total,

\frac{2}{3} \times 360 = 240\text{ kg}

Arabica. Roaster

Z

:

600\text{ kg}

total,

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

Arabica.

Converting hourly rates to total batch weights over

4

hours and applying part-to-whole fractions determined by each ratio.

2

Aggregate total Arabica and Robusta weights across all roasters

Total Arabica =

144 + 240 + 120 = 504\text{ kg}

. Total coffee =

240 + 360 + 600 = 1,200\text{ kg}

. Total Robusta =

1,200 - 504 = 696\text{ kg}

.

Adding individual component weights to get total aggregate weights.

3

Evaluate percentage and ratio statements

Arabica percentage =

\frac{504}{1,200} \times 100\% = 42\%

. Arabica to Robusta ratio =

\frac{504}{696} = \frac{21}{29}

.

Verifying overall proportions and simplifying the fraction by dividing by

24

.

Key Concept

Multi-stage weighted component ratios and rate conversion

Question 256Question →

If $a$ and $b$ are nonzero real numbers such that $a^2 b^3 < 0$ and $\sqrt{a^2 b^4} = -a b^2$ , which of the following statements must be true? Select all such statements.

Select all that apply

a + b < 0

\frac{a^3}{b} > 0

\sqrt{a^4 b^2} = a^2 b

(-a)^3 b^2 < 0

\sqrt{(a + b)^2} = a + b

Show answer & explanation

Answer: $a + b < 0$ ; $\frac{a^3}{b} > 0$

Answer

The statements that must be true are the inequality asserting that the sum of the variables is negative (

a + b < 0

) and the inequality asserting that the quotient of the cubed variable and the second variable is positive (

\frac{a^3}{b} > 0

).

Analyzing the given constraints reveals that both variables are negative. From

a^2 b^3 < 0

, since

a^2 > 0

, we must have

b^3 < 0

, so

b < 0

. Next, from

\sqrt{a^2 b^4} = |a| b^2 = -a b^2

, dividing by

b^2 > 0

gives

|a| = -a

, which implies

a < 0

. Thus,

a < 0

and

b < 0

. The statement asserting

a + b < 0

is true because the sum of two negative numbers is negative. The statement asserting

\frac{a^3}{b} > 0

is true because

a^3 < 0

and

b < 0

, and dividing two negative numbers yields a positive quotient.

Step-by-Step Solution

1

Determine the sign of

b

using the given inequality

a^2 b^3 < 0

.

b < 0

Since

a

is a nonzero real number,

a^2 > 0

. For the product

a^2 b^3

to be negative,

b^3

must be negative, which implies

b < 0

.

2

Determine the sign of

a

using the identity

\sqrt{a^2 b^4} = -a b^2

.

a < 0

Simplify the radical:

\sqrt{a^2 b^4} = \sqrt{a^2} \cdot \sqrt{(b^2)^2} = |a| b^2

. Equating this to

-a b^2

gives

|a| b^2 = -a b^2

. Since

b \neq 0

,

b^2 > 0

, so dividing by

b^2

yields

|a| = -a

. For a nonzero real number,

|a| = -a

implies

a < 0

.

3

Evaluate statement

a + b < 0

.

True

The sum of two negative numbers (

a < 0

and

b < 0

) is always negative.

4

Evaluate statement

\frac{a^3}{b} > 0

.

True

Since

a < 0

,

a^3 < 0

. Dividing the negative quantity

a^3

by the negative quantity

b

yields a positive result.

5

Evaluate statement

\sqrt{a^4 b^2} = a^2 b

.

False

\sqrt{a^4 b^2} = \sqrt{a^4}\sqrt{b^2} = a^2 |b|

. Since

b < 0

,

|b| = -b

, so

\sqrt{a^4 b^2} = -a^2 b

.

6

Evaluate statement

(-a)^3 b^2 < 0

.

False

Since

a < 0

,

-a > 0

, making

(-a)^3 > 0

. Since

b \neq 0

,

b^2 > 0

. The product of two positive numbers is positive, so

(-a)^3 b^2 > 0

.

7

Evaluate statement

\sqrt{(a + b)^2} = a + b

.

False

\sqrt{x^2} = |x|

for any real

x

. Since

a + b < 0

,

\sqrt{(a + b)^2} = |a + b| = -(a + b) \neq a + b

.

Key Concept

Properties of even exponents, odd exponents, and principal square roots of negative variable terms.

Question 257Question →

The prime factorizations of two positive integers $A$ and $B$ are given by $A = 2^a \times 3^5 \times 5^b$ and $B = 2^4 \times 3^c \times 7^2$ , where $a$ , $b$ , and $c$ are positive integers. If the greatest common divisor of $A$ and $B$ is $\gcd(A, B) = 2^2 \times 3^3$ , and their least common multiple is \text{lcm} $(A, B) = 2^4 \times 3^5 \times 5^3 \times 7^2$ , what is the value of $a + b + c$ ?

Show answer & explanation

Answer: 8

Answer

The correct answer is 8.

For any two positive integers expressed in prime factorized form, the greatest common divisor contains each prime factor raised to the minimum of its exponents in the two numbers, while the least common multiple contains each prime factor raised to the maximum of its exponents. For prime factor 2, the GCD has exponent 2, so min(a, 4) = 2, giving a = 2. For prime factor 3, the GCD has exponent 3, so min(5, c) = 3, giving c = 3. For prime factor 5, the LCM has exponent 3, so max(b, 0) = 3, giving b = 3. Adding these values together yields a + b + c = 2 + 3 + 3 = 8.

Step-by-Step Solution

1

Analyze the prime factor

2

min(a, 4) = 2, so a = 2

The greatest common divisor takes the minimum exponent for each prime factor shared between A and B.

2

Analyze the prime factor

3

min(5, c) = 3, so c = 3

The exponent of 3 in the GCD is 3, which must equal the smaller of the two exponents 5 and c.

3

Analyze the prime factor

5

max(b, 0) = 3, so b = 3

The least common multiple takes the maximum exponent for each prime factor present in either number.

4

Calculate the sum

a + b + c

2 + 3 + 3 = 8

Add the solved values of the three unknown prime exponents.

Key Concept

Relating prime factor exponents to GCD (minimum exponent) and LCM (maximum exponent)

Question 258Question →

If

x

is a positive integer such that

\frac{4^x + 4^x + 4^x + 4^x}{2^x + 2^x} = 512

what is the value of

x

?

Show answer & explanation

Answer: 8

Answer

The value of

x

is 8.

Combining four terms of

4^x

yields

4 \cdot 4^x = 4^{x+1} = 2^{2x+2}

. Combining two terms of

2^x

yields

2 \cdot 2^x = 2^{x+1}

. Dividing the numerator by the denominator gives

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

. Since

512 = 2^9

, setting

2^{x+1} = 2^9

gives

x + 1 = 9

, which leads directly to

x = 8

.

Step-by-Step Solution

1

Simplify the numerator by combining identical added terms.

The numerator

4^x + 4^x + 4^x + 4^x

equals

4 \cdot 4^x

, which simplifies to

4^{x+1}

.

Adding four identical quantities is equivalent to multiplying that quantity by 4.

2

Simplify the denominator by combining identical added terms.

The denominator

2^x + 2^x

equals

2 \cdot 2^x

, which simplifies to

2^{x+1}

.

Adding two identical quantities is equivalent to multiplying that quantity by 2.

3

Convert the numerator to base 2 and simplify the fraction.

Since

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

, the fraction becomes

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

.

Converting all terms to a common base allows using the exponent quotient rule

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

.

4

Solve for

x

by equating the simplified power to 512.

Setting

2^{x+1} = 512 = 2^9

yields

x + 1 = 9

, so

x = 8

.

When exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Key Concept

Combining repeated addition of exponential terms and converting powers to a common base using exponent laws (

a^m \cdot a^n = a^{m+n}

and

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

).

Question 259Question →

A textile mill uses two weaving looms, Loom $P$ and Loom $Q$ , to produce a fabric blend composed of silk, wool, and cotton.

- Loom $P$ produces silk, wool, and cotton in the ratio $2 : 3 : 5$ by weight, operating at a constant output rate of $120\text{ kg per hour}$ .
- Loom $Q$ produces silk, wool, and cotton in the ratio $1 : 4 : 3$ by weight, operating at a constant output rate of $160\text{ kg per hour}$ .

If both looms operate simultaneously for 5 hours, what is the ratio of the total weight of wool produced to the total weight of cotton produced in the combined output?

29 : 30

7 : 8

32 : 35

29 : 70

30 : 29

Show answer & explanation

Answer: $29 : 30$

Answer

The ratio of the total weight of wool produced to the total weight of cotton produced is

29 : 30

.

To find the overall ratio of wool to cotton, determine the actual mass of each material produced per hour by each machine. Loom

P

outputs

\frac{3}{10} \times 120 = 36\text{ kg}

of wool and

\frac{5}{10} \times 120 = 60\text{ kg}

of cotton per hour. Loom

Q

outputs

\frac{4}{8} \times 160 = 80\text{ kg}

of wool and

\frac{3}{8} \times 160 = 60\text{ kg}

of cotton per hour. Combining both looms yields

116\text{ kg}

of wool and

120\text{ kg}

of cotton per hour. Over 5 hours, the combined output is

580\text{ kg}

of wool and

600\text{ kg}

of cotton. The ratio of total wool to total cotton is

580 : 600

, which simplifies to

29 : 30

.

Step-by-Step Solution

1

Calculate the hourly production rates of wool and cotton for Loom

P

.

Loom

P

produces fabric at

120\text{ kg/hr}

with ratio parts

2 + 3 + 5 = 10

.
- Wool rate from

P = \frac{3}{10} \times 120 = 36\text{ kg/hr}

- Cotton rate from

P = \frac{5}{10} \times 120 = 60\text{ kg/hr}

Converting the ratio into component rates by multiplying each component's fraction by Loom

P

's total hourly output.

2

Calculate the hourly production rates of wool and cotton for Loom

Q

.

Loom

Q

produces fabric at

160\text{ kg/hr}

with ratio parts

1 + 4 + 3 = 8

.
- Wool rate from

Q = \frac{4}{8} \times 160 = 80\text{ kg/hr}

- Cotton rate from

Q = \frac{3}{8} \times 160 = 60\text{ kg/hr}

Converting the ratio into component rates by multiplying each component's fraction by Loom

Q

's total hourly output.

3

Determine total quantities produced over the 5-hour period.

Total wool

= (36 + 80) \times 5 = 116 \times 5 = 580\text{ kg}

.
Total cotton

= (60 + 60) \times 5 = 120 \times 5 = 600\text{ kg}

.

Combining the hourly outputs from both machines and multiplying by the total duration of 5 hours.

4

Compute and simplify the ratio of total wool to total cotton.

\text{Ratio} = \frac{580}{600} = \frac{29}{30}

, which is expressed as

29 : 30$.

Dividing both terms of the ratio by their greatest common divisor, 20.

Key Concept

Combining weighted rates across multiple ratio-based sub-components

Estimated Time:2m 0s

Question 260Question →

Let $n = 2^a \cdot 3^b \cdot 5^c$ be a positive integer, where $a$ , $b$ , and $c$ are non-negative integers. If the greatest common divisor of $n$ and $360$ is $45$ , and the least common multiple of $n$ and $90$ is $450$ , what is the value of $a + b + c$ ?

2

3

4

5

6

Show answer & explanation

Answer: 4

Answer

The value of

a + b + c

is

4

.

Prime factorizing the given values yields

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

,

45 = 2^0 \cdot 3^2 \cdot 5^1

,

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

, and

450 = 2^1 \cdot 3^2 \cdot 5^2

. Because

\gcd(n, 360) = 45

, taking the minimum exponent of 2 implies

\min(a, 3) = 0

, so

a = 0

. Taking the minimum exponent of 3 implies

b \ge 2

, and for 5 implies

c \ge 1

. Next, using

\text{lcm}(n, 90) = 450

, taking the maximum exponent of 3 gives

\max(b, 2) = 2

, which forces

b = 2

. Taking the maximum exponent of 5 gives

\max(c, 1) = 2

, which forces

c = 2

. Therefore,

a + b + c = 0 + 2 + 2 = 4

.

Step-by-Step Solution

1

Express all given integers in their prime factorizations.

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

,

45 = 3^2 \cdot 5^1 = 2^0 \cdot 3^2 \cdot 5^1

,

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

, and

450 = 2^1 \cdot 3^2 \cdot 5^2

.

Prime factorizations allow determination of exponents using exponent rules for GCD and LCM.

2

Apply the GCD condition

\gcd(n, 360) = 45

.

\min(a, 3) = 0 \implies a = 0

,

\min(b, 2) = 2 \implies b \ge 2

, and

\min(c, 1) = 1 \implies c \ge 1

.

The greatest common divisor takes the minimum exponent for each prime factor shared between the numbers.

3

Apply the LCM condition

\text{lcm}(n, 90) = 450

using

a = 0

.

\max(b, 2) = 2 \implies b \le 2

(so

b = 2

), and

\max(c, 1) = 2 \implies c = 2

.

The least common multiple takes the maximum exponent for each prime factor.

4

Calculate the sum

a + b + c

.

a + b + c = 0 + 2 + 2 = 4

.

Adding the individual prime factor exponents yields the requested sum.

Key Concept

Relating prime factor exponents to GCD (minimum powers) and LCM (maximum powers)

Estimated Time:1m 30s

Arithmetic

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept

Common Mistakes

Answer

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Key Concept