Pre-Algebra Practice Questions — ACT — Page 7

Question 121Question →

What is the value of the expression $\frac{3^4 \cdot 3^2}{3^3}$ ?

8

9

243

729

27

Show answer & explanation

Answer: 27

Answer

27

Evaluating the numerator using the product rule of exponents yields

3^6

. Then, applying the quotient rule of exponents to divide

3^6

by

3^3

yields

3^3

. Evaluating

3^3

gives

27

.

Step-by-Step Solution

1

Simplify the numerator using the product rule of exponents:

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

.

3^4 \cdot 3^2 = 3^{4+2} = 3^6

When multiplying exponential terms with the same base, add their exponents.

2

Simplify the fraction using the quotient rule of exponents:

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

.

\frac{3^6}{3^3} = 3^{6-3} = 3^3

When dividing exponential terms with the same base, subtract the exponent in the denominator from the exponent in the numerator.

3

Evaluate the simplified exponential expression.

3^3 = 3 \cdot 3 \cdot 3 = 27

Evaluate the base raised to the power of 3 by multiplying it by itself three times.

Key Concept

Simplifying expressions using the product and quotient rules of exponents

Estimated Time:45s

Question 122Question →

In a local park, $\frac{2}{5}$ of the trees are maple trees and $\frac{1}{3}$ of the trees are oak trees. The remaining trees are pine trees. What fraction of the trees in the park are pine trees?

\frac{3}{8}

\frac{5}{8}

\frac{4}{15}

\frac{11}{15}

\frac{14}{15}

Show answer & explanation

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

Answer

\frac{4}{15}

To find the fraction of pine trees, first calculate the total fraction occupied by maple and oak trees. Finding a common denominator of

15

for

\frac{2}{5}

and

\frac{1}{3}

gives

\frac{6}{15} + \frac{5}{15} = \frac{11}{15}

. Subtracting this combined fraction from

1

(represented as

\frac{15}{15}

) yields the remaining fraction of pine trees, which is

\frac{4}{15}

.

Step-by-Step Solution

1

Find the sum of the fractions representing maple and oak trees.

\frac{2}{5} + \frac{1}{3} = \frac{6}{15} + \frac{5}{15} = \frac{11}{15}

Before finding the remaining fraction, we need to know the combined fraction of the park occupied by maple and oak trees.

2

Subtract the combined fraction of maple and oak trees from the whole.

1 - \frac{11}{15} = \frac{15}{15} - \frac{11}{15} = \frac{4}{15}

Since the remaining trees are pine trees, subtracting the combined fraction of other trees from the total of

1

yields the fraction of pine trees.

Key Concept

To find a remaining fractional part of a whole, first sum the known fractional parts using a common denominator, and then subtract this sum from 1.

Question 123Question →

At the start of the fiscal year, a city's municipal budget is divided among three departments: Education, Healthcare, and Infrastructure. Education receives $\frac{3}{8}$ of the total budget, Healthcare receives $40\%$ of the total budget, and Infrastructure receives the remainder. Midyear, the total municipal budget is increased by $25\%$ . The budget allocated to Education is increased by $20\%$ of its original amount, and the budget allocated to Healthcare is increased by $15\%$ of its original amount. What percent of the new total municipal budget is allocated to Infrastructure?

Show answer & explanation

Answer: 27.2

Answer

27.2

The correct answer is

27.2\%

. By representing the initial total budget as

1

, we find the initial Infrastructure allocation is

1 - (0.375 + 0.40) = 0.225

. After the budget increase of

25\%

, the new total budget is

1.25

. The new Education allocation is

0.375 \times 1.20 = 0.45

, and the new Healthcare allocation is

0.40 \times 1.15 = 0.46

. The remaining amount for Infrastructure is

1.25 - 0.45 - 0.46 = 0.34

. Expressing

0.34

as a percentage of the new total budget

1.25

gives

\frac{0.34}{1.25} \times 100\% = 27.2\%

.

Step-by-Step Solution

1

Convert the initial budget allocations to decimal shares of the original total budget.

Education share is

0.375

, Healthcare share is

0.40

, and Infrastructure share is

0.225

.

Expressing all initial shares as decimals relative to a total budget of

1.0

makes successive calculations straightforward.

2

Calculate the updated budget amounts for the entire city, Education, and Healthcare.

The new total budget is

1.25

, the new Education allocation is

0.375 \times 1.20 = 0.45

, and the new Healthcare allocation is

0.40 \times 1.15 = 0.46

.

Adjust each allocation by its respective percentage change to find its share relative to the initial total.

3

Compute the remaining budget share left for Infrastructure.

New Infrastructure share is

1.25 - (0.45 + 0.46) = 0.34

of the initial budget.

The sum of the three department budgets must equal the new total budget of

1.25

.

4

Find the percentage of the new total budget represented by the new Infrastructure allocation.

\frac{0.34}{1.25} \times 100\% = 27.2\%

Divide the new Infrastructure share by the new total budget to find the new ratio, then convert to a percentage.

Key Concept

Converting between fractions, decimals, and percentages, and tracking changes across multiple base values.

Alternative Method

Instead of using

1.0

as the base, you can assume an initial total budget of

800

dollars (since

800

is a multiple of

8

, which simplifies

\frac{3}{8}

). Initial Education =

300

, Healthcare =

320

, Infrastructure =

180

. The new total budget is

800 \times 1.25 = 1000

. New Education =

300 \times 1.20 = 360

. New Healthcare =

320 \times 1.15 = 368

. New Infrastructure =

1000 - (360 + 368) = 272

. Percentage =

\frac{272}{1000} \times 100\% = 27.2\%

.

Estimated Time:3m 0s

Question 124Question →

A student earned scores of $85$ , $90$ , $88$ , and $92$ on the first four quizzes of the semester. What score must the student earn on the fifth quiz to have an average (arithmetic mean) score of exactly $90$ for all five quizzes?

1

5

89

90

95

Show answer & explanation

Answer: 95

Answer

The student must earn a score of 95 on the fifth quiz.

To find the score needed on the fifth quiz to achieve an average of

90

, first find the total number of points required. An average of

90

on

5

quizzes requires a total sum of

5 \times 90 = 450

points. The sum of the first four quiz scores is

85 + 90 + 88 + 92 = 355

points. Subtracting the current sum from the required total sum (

450 - 355

) yields

95

, which is the score the student must earn on the fifth quiz.

Step-by-Step Solution

1

Calculate the sum of the student's scores on the first four quizzes.

85 + 90 + 88 + 92 = 355

To find the total number of points the student has earned so far.

2

Calculate the total sum of points required to achieve an average of

90

over five quizzes.

5 \times 90 = 450

The arithmetic mean formula is Mean = Sum / Count, which can be rearranged to Sum = Mean * Count.

3

Subtract the sum of the first four quiz scores from the required total sum of five quiz scores.

450 - 355 = 95

The difference represents the score needed on the fifth quiz to achieve the target average.

Key Concept

Finding a missing value given a target arithmetic mean

Question 125Question →

An investment fund allocates its capital into three portfolios: Real Estate, Technology, and Green Energy. Initially, $\frac{1}{4}$ of the total capital is allocated to Real Estate, and $\frac{2}{5}$ of the total capital is allocated to Technology. The remaining capital is allocated to Green Energy. During the first year, the Real Estate portfolio decreases in value by $16\%$ , the Technology portfolio increases in value by $15\%$ , and the Green Energy portfolio increases in value by $p\%$ . If the total value of the entire investment fund at the end of the first year has increased by $9\%$ of its initial value, what is the value of $p$ ?

10.0

10.5

18.9

20.0

31.4

Show answer & explanation

Answer: $20.0$

Answer

The value of

p

is

20.0

.

The correct answer is

20.0

. By assuming a total capital of

100

units, we find that the Real Estate portfolio starts at

25

units, the Technology portfolio at

40

units, and the Green Energy portfolio at

35

units. The change in Real Estate is

-4

units, and the change in Technology is

+6

units. To reach a total fund increase of

9

units, the change in the Green Energy portfolio must satisfy

-4 + 6 + \text{Change in Green Energy} = 9

, which means the Green Energy portfolio must increase by

7

units. Finding what percent

7

is of

35

yields

\frac{7}{35} = 0.20

, or

20\%

.

Step-by-Step Solution

1

Determine the initial allocation of capital for each portfolio by assuming a total initial capital of

100

units.

Real Estate receives

\frac{1}{4} \times 100 = 25

units. Technology receives

\frac{2}{5} \times 100 = 40

units. The remaining capital allocated to Green Energy is

100 - (25 + 40) = 35

units.

This establishes the weighted share of each portfolio relative to the total capital.

2

Calculate the changes in value for the Real Estate and Technology portfolios based on their given percentage changes.

Real Estate change:

-16\% \text{ of } 25 = -0.16 \times 25 = -4

units. Technology change:

+15\% \text{ of } 40 = 0.15 \times 40 = +6

units.

This determines the absolute change in value contributed by these two portfolios.

3

Calculate the total change in value for the entire fund.

Total fund change:

+9\% \text{ of } 100 = +9

units.

The total change in the fund is the sum of the individual changes in the portfolios.

4

Set up an equation representing the sum of the changes in all three portfolios and solve for

p

.

-4 + 6 + \left(\frac{p}{100} \times 35\right) = 9 \implies 2 + 0.35p = 9 \implies 0.35p = 7 \implies p = 20

.

This solves for the unknown percentage rate of increase required for the Green Energy portfolio.

Key Concept

Weighted average of percentage changes using fractional and decimal representations.

Alternative Method

Instead of choosing an arbitrary starting capital of

100

, you can use decimals directly. Let

C

be the total capital. The weighted changes are

0.25C(-0.16) + 0.40C(0.15) + 0.35C\left(\frac{p}{100}\right) = 0.09C

. Dividing both sides by

C

yields

-0.04 + 0.06 + 0.0035p = 0.09

. Simplifying gives

0.02 + 0.0035p = 0.09 \implies 0.0035p = 0.07 \implies p = 20

.

Estimated Time:3m 0s

Question 126Question →

At a marketing firm, the ratio of designers to writers is $3:4$ , and the ratio of writers to managers is $5:2$ . The firm hires $10$ new designers and $4$ new managers, while the number of writers remains the same. If the new ratio of designers to managers is $2:1$ , how many writers are employed at the firm?

8

20

30

40

120

Show answer & explanation

Answer: 40

Answer

There are 40 writers employed at the firm.

To find the number of writers, we combine the ratios of designers to writers (

3:4

) and writers to managers (

5:2

) by finding a common multiple for the number of writers (20). This gives a combined ratio of

15:20:8

for designers, writers, and managers. Representing these quantities as

15x

,

20x

, and

8x

, we can write the equation for the new ratio of designers to managers:

\frac{15x + 10}{8x + 4} = 2

. Solving this equation gives

15x + 10 = 16x + 8

, which simplifies to

x = 2

. The number of writers is

20(2) = 40

.

Step-by-Step Solution

1

Align the two given ratios to find a common ratio.

The ratio of designers to writers is

3:4 = 15:20

. The ratio of writers to managers is

5:2 = 20:8

. The combined ratio of designers to writers to managers is

15:20:8

.

To analyze the changes across the different categories, we need a single ratio that shares the common term (writers).

2

Express the quantities of employees in terms of a variable

x

.

Let the number of designers be

15x

, writers be

20x

, and managers be

8x

.

This allows us to represent the actual number of employees algebraically.

3

Set up an equation based on the new ratio of designers to managers after the new hires.

\frac{15x + 10}{8x + 4} = \frac{2}{1}

The firm hired 10 designers and 4 managers, and the ratio of these new quantities is

2:1

.

4

Solve the equation for

x

.

15x + 10 = 2(8x + 4) \implies 15x + 10 = 16x + 8 \implies x = 2

.

Solving for

x

gives us the multiplier to find the actual employee counts.

5

Calculate the total number of writers.

W = 20(2) = 40

.

The number of writers is represented by

20x

.

Key Concept

Combining multiple ratios and setting up algebraic equations from changing rates/proportions.

Alternative Method

Express the number of designers and managers in terms of the number of writers,

W

. The number of designers is

\frac{3}{4}W

, and the number of managers is

\frac{2}{5}W

. After hiring, the equation is

\frac{\frac{3}{4}W + 10}{\frac{2}{5}W + 4} = 2

. Solving for

W

yields

\frac{3}{4}W + 10 = \frac{4}{5}W + 8 \implies 2 = \frac{1}{20}W \implies W = 40

.

Estimated Time:2m 0s

Question 127Question →

Four distinct quantities are defined by different exponential, radical, or scientific notation expressions. What is the correct order of these quantities from the smallest value to the largest value?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of the expressions from least to greatest is

5^{20}

, followed by

3^{30}

, then

8.0 \times 10^{14}

, and finally

\sqrt{2^{100}}

.

Simplifying

\sqrt{2^{100}}

yields

2^{50}

. Expressing

5^{20}

as

25^{10}

and

3^{30}

as

27^{10}

shows that

5^{20} < 3^{30}

. By estimating upper and lower bounds relative to powers of

10

, we find that

3^{30} < 30^{10} \approx 5.9 \times 10^{14}

, which is less than

8.0 \times 10^{14}

. Meanwhile,

2^{50} = 1024^5 > (10^3)^5 = 10^{15}

, which is greater than

8.0 \times 10^{14}

. This establishes the sequence

5^{20} < 3^{30} < 8.0 \times 10^{14} < \sqrt{2^{100}}

.

Step-by-Step Solution

1

Simplify the radical expression

\sqrt{2^{100}}

using fractional exponent rules.

\sqrt{2^{100}} = (2^{100})^{1/2} = 2^{50}

Converting the square root to a power of

1/2

simplifies the expression into a single base and exponent.

2

Compare the exponential expressions

5^{20}

and

3^{30}

by expressing them with a common power of

10

.

5^{20} = (5^2)^{10} = 25^{10}

and

3^{30} = (3^3)^{10} = 27^{10}

. Since

25 < 27

, then

25^{10} < 27^{10}

, so

5^{20} < 3^{30}

.

Rewriting the powers to share an exponent of

10

allows direct comparison of the base values.

3

Establish an upper limit for

3^{30}

to compare it with the scientific notation quantity

8.0 \times 10^{14}

.

3^{30} = 27^{10} < 30^{10} = 3^{10} \times 10^{10}

. Since

3^5 = 243

, we have

3^{10} = 243^2 = 59,049

. Therefore,

3^{10} \times 10^{10} = 5.9049 \times 10^{14}

. Since

5.9049 \times 10^{14} < 8.0 \times 10^{14}

, we have

3^{30} < 8.0 \times 10^{14}

.

Using a slightly larger round number base of

30

allows an upper bound calculation that demonstrates

3^{30}

is strictly less than

8.0 \times 10^{14}

.

4

Establish a lower limit for the simplified radical expression

2^{50}

to compare it with

8.0 \times 10^{14}

.

2^{50} = (2^{10})^5 = 1024^5

. Since

1024 > 1000 = 10^3

, then

1024^5 > (10^3)^5 = 10^{15}

. Because

10^{15} = 10 \times 10^{14} > 8.0 \times 10^{14}

, it follows that

2^{50} > 8.0 \times 10^{14}

.

Using the approximation

2^{10} \approx 10^3

shows that

2^{50}

is greater than

10^{15}

, which exceeds the scientific notation value.

5

Combine the individual inequalities to construct the complete chain of comparisons.

5^{20} < 3^{30} < 8.0 \times 10^{14} < \sqrt{2^{100}}

Linking the inequalities using transitive properties determines the correct least-to-greatest order.

Key Concept

Comparing complex numerical expressions containing powers, radicals, and scientific notation by finding common exponents and using base-10 estimation.

Alternative Method

Alternatively, convert all expressions into scientific notation approximations. Note that

5^{20} = 25^{10} = (2.5 \times 10)^{10} = 2.5^{10} \times 10^{10}

. Since

2.5^{10} \approx 9536

, we get

5^{20} \approx 9.5 \times 10^{13}

. Applying a similar approximation to

3^{30} = 27^{10} = 2.7^{10} \times 10^{10} \approx 2.06 \times 10^{14}

. Since

2^{50} = 1.024^5 \times 10^{15} \approx 1.13 \times 10^{15}

, we can compare the coefficients and exponents:

9.5 \times 10^{13} < 2.06 \times 10^{14} < 8.0 \times 10^{14} < 1.13 \times 10^{15}

.

Estimated Time:3m 0s

Question 128Question →

A school district requires a chaperone-to-student ratio of $3:5$ for all field trips. If there are exactly $15$ chaperones registered for an upcoming field trip, what is the maximum number of students who can attend?

9

17

24

25

40

Show answer & explanation

Answer: $25$

Answer

The maximum number of students who can attend is

25

.

The correct answer is

25

. To maintain the required ratio of

3

chaperones to

5

students, we set up the proportion

\frac{3}{5} = \frac{15}{x}

, where

x

is the number of students. Cross-multiplying gives

3x = 75

, and dividing by

3

gives

x = 25

.

Step-by-Step Solution

1

Set up the proportion using the given chaperone-to-student ratio.

\frac{\text{Chaperones}}{\text{Students}} = \frac{3}{5} = \frac{15}{x}

, where

x

represents the maximum number of students.

Establishing a direct proportion allows us to find the unknown quantity while maintaining the required ratio.

2

Solve for the unknown variable

x

by cross-multiplying.

3 \times x = 5 \times 15 \implies 3x = 75

Cross-multiplication is a standard algebraic method to solve proportions.

3

Divide both sides of the equation by

3

to isolate

x

.

x = 25

This yields the final value of students who can attend.

Key Concept

Solving direct proportions from part-to-part ratios

Estimated Time:45s

Question 129Question →

A metal alloy is composed of copper, zinc, and nickel. In the original alloy, the ratio of the weight of copper to zinc is $3:4$ , and the ratio of the weight of zinc to nickel is $4:5$ . A metallurgist melts this $240$ -gram alloy and adds a mixture of pure copper and pure nickel. The weight of the copper added is twice the weight of the nickel added. In the resulting alloy, the weight of copper is equal to the weight of nickel. What is the ratio of the weight of zinc to the total weight of the new alloy?

\frac{4}{15}

\frac{1}{3}

\frac{2}{9}

\frac{2}{7}

\frac{4}{33}

Show answer & explanation

Answer: $\frac{2}{9}$

Answer

The correct ratio of the weight of zinc to the total weight of the new alloy is

\frac{2}{9}

.

The correct answer is

\frac{2}{9}

. To find this, we first establish the joint ratio of the metals as Copper : Zinc : Nickel =

3 : 4 : 5

. For a

240

-gram alloy, this gives

60

g of copper,

80

g of zinc, and

100

g of nickel. Adding

x

grams of nickel and

2x

grams of copper yields a new copper weight of

60 + 2x

and a new nickel weight of

100 + x

. Since these weights are equal in the new alloy,

60 + 2x = 100 + x \implies x = 40

. The new total weight is

240 + 3(40) = 360

grams. The weight of zinc remains

80

grams, so the ratio of zinc to the total weight is

80/360 = \frac{2}{9}

.

Step-by-Step Solution

1

Express the ratio of the three metals in the original alloy.

Copper : Zinc : Nickel =

3 : 4 : 5

Since Copper : Zinc =

3:4

and Zinc : Nickel =

4:5

, the zinc terms are already aligned, allowing us to combine them into a single three-part ratio.

2

Calculate the initial weights of copper, zinc, and nickel in the

240

-gram alloy.

Copper =

60

grams, Zinc =

80

grams, Nickel =

100

grams

The sum of the ratio parts is

3 + 4 + 5 = 12

. Each part represents

240 \text{ grams} / 12 = 20

grams. Multiplying each part by 20 gives the initial weights.

3

Define variables for the added metals and set up the equation for the new weights.

New Copper =

60 + 2x

, New Nickel =

100 + x

Let

x

represent the weight of nickel added in grams. Since the weight of the copper added is twice that of the nickel added,

2x

grams of copper are added.

4

Solve for the value of

x

using the condition that the final weights of copper and nickel are equal.

x = 40

Setting the expressions equal gives

60 + 2x = 100 + x

. Subtracting

x

and

60

from both sides yields

x = 40

grams.

5

Determine the new total weight of the alloy and calculate the final ratio of zinc to the total weight.

New Total Weight =

360

grams, Final Ratio =

\frac{2}{9}

The new total weight is the original weight plus the added metals:

240 + 3(40) = 360

grams. Since no zinc was added, the weight of zinc remains

80

grams. The ratio is

80 / 360 = \frac{2}{9}

.

Key Concept

Combining ratios and setting up algebraic equations to model changing quantities in mixtures.

Question 130Question →

If $a = 2 \times 10^3$ and $b = 3 \times 10^{-4}$ , what is the value of the expression $\sqrt{\frac{a^3 \cdot b^2}{1.8}}$ ?

Show answer & explanation

Answer: 20

Answer

20

Evaluating the components of the expression sequentially:

a^3 = 8 \times 10^9

and

b^2 = 9 \times 10^{-8}

. Multiplying them yields

(8 \times 9) \times 10^{9-8} = 72 \times 10^1 = 720

. Dividing by the decimal

1.8

results in

720 / 1.8 = 400

. The square root of

400

yields

20

.

Step-by-Step Solution

1

Calculate the value of

a^3

8 \times 10^9

Apply the power of a product rule,

(xy)^n = x^n y^n

, and the power of a power rule,

(10^p)^q = 10^{p \cdot q}

.

2

Calculate the value of

b^2

9 \times 10^{-8}

Apply the power of a product rule and power of a power rule to

(3 \times 10^{-4})^2

.

3

Multiply

a^3

and

b^2

720

Multiply the coefficients (

8 \times 9 = 72

) and add the exponents of the base 10 (

10^9 \times 10^{-8} = 10^1

), resulting in

72 \times 10 = 720

.

4

Divide the product by

1.8

400

Substitute the values to get

720 / 1.8

, which simplifies to

7200 / 18 = 400

.

5

Take the square root of the quotient

20

Find the non-negative square root of

400

, which is

20

.

Key Concept

Simplifying numerical expressions containing combinations of exponent rules, roots, and scientific notation.

Estimated Time:1m 30s

Question 131Question →

A scientist estimates that the population of a certain bacteria culture doubles every 4 hours. If the culture starts with $3.0 \times 10^5$ bacteria, the population after 24 hours can be written in scientific notation as $a \times 10^7$ . What is the value of $a$ ?

Show answer & explanation

Answer: 1.92

Answer

The value of the coefficient

a

is

1.92

.

The correct answer is

1.92

. In 24 hours, a population that doubles every 4 hours undergoes 6 doubling cycles. The growth factor is

2^6 = 64

. Multiplying the initial population of

3.0 \times 10^5

by 64 yields

192 \times 10^5

. Converting this value into standard scientific notation gives

1.92 \times 10^7

. Comparing this expression to the format

a \times 10^7

shows that the coefficient

a

equals

1.92

.

Step-by-Step Solution

1

Determine the number of doubling periods.

6 doubling periods

Since the population doubles every 4 hours, over a span of 24 hours it will double

24 \div 4 = 6

times.

2

Calculate the growth multiplier.

64

Doubling 6 times is represented by the exponential expression

2^6

, which evaluates to 64.

3

Multiply the initial population by the growth factor.

192 \times 10^5

Multiply the coefficient of the initial population by the growth multiplier:

3.0 \times 64 = 192

, yielding a total population of

192 \times 10^5

.

4

Convert the resulting value to the required scientific notation format.

1.92 \times 10^7

To express

192 \times 10^5

in the standard scientific notation format

a \times 10^7

, divide 192 by 100 to get 1.92, and multiply the power of 10 by

10^2

(raising the exponent from 5 to 7).

Key Concept

Evaluating exponential growth and converting numbers to standard scientific notation format

Estimated Time:1m 30s

Question 132Question →

A researcher records the following list of seven temperatures (in degrees Fahrenheit) during a week, where $x$ and $y$ are unknown values:

88, 92, 75, 85, x, y, 90

The mean temperature for the seven days is $85.0^\circ\text{F}$ , and the range of the temperatures is $22.0^\circ\text{F}$ . If the highest temperature of the week is $y$ and the lowest temperature of the week is $x$ , what is the median temperature of the week?

82.5^\circ\text{F}

85.0^\circ\text{F}

86.5^\circ\text{F}

88.0^\circ\text{F}

90.0^\circ\text{F}

Show answer & explanation

Answer: $88.0^\circ\text{F}$

Answer

88.0^\circ\text{F}

To find the median temperature, we must first find the values of

x

and

y

. Since the mean of the seven temperatures is

85.0^\circ\text{F}

, their sum is

7 \times 85.0 = 595.0

. The sum of the five known temperatures is

88 + 92 + 75 + 85 + 90 = 430.0

. Therefore, the sum of the two unknown temperatures is

x + y = 595.0 - 430.0 = 165.0

. We are also given that the range of the temperatures is

22.0^\circ\text{F}

, and that

y

is the highest and

x

is the lowest temperature, which means

y - x = 22.0

. Solving this system of equations (

y + x = 165.0

and

y - x = 22.0

) yields

x = 71.5

and

y = 93.5

. Placing the seven temperatures in ascending order gives

71.5, 75, 85, 88, 90, 92, 93.5

. The median is the middle value (the 4th value) in this sorted list, which is

88.0^\circ\text{F}

.

Step-by-Step Solution

1

Determine the sum of the temperatures using the given mean of

85.0^\circ\text{F}

.

Total sum =

595.0

Since the mean of 7 temperatures is

85.0

, the sum of all temperatures must be

7 \times 85.0 = 595.0

.

2

Set up an equation for the sum of the unknown temperatures

x

and

y

.

x + y = 165.0

The sum of the five known temperatures is

88 + 92 + 75 + 85 + 90 = 430.0

. Thus,

x + y = 595.0 - 430.0 = 165.0

.

3

Set up and solve the system of equations with the range constraint to find

x

and

y

.

x = 71.5

and

y = 93.5

We are given that

y

is the maximum and

x

is the minimum, so the range is

y - x = 22.0

. Solving the system

y + x = 165.0

and

y - x = 22.0

by adding the equations gives

2y = 187.0 \implies y = 93.5

. Substituting back gives

x = 71.5

.

4

Sort the seven temperatures in ascending order and identify the median value.

Sorted list:

71.5, 75, 85, 88, 90, 92, 93.5

; Median =

88.0

For an odd number of data points (7), the median is the 4th value when the list is sorted. The 4th value in the sorted list is

88.0

.

Key Concept

Calculating the median of a dataset containing unknown values by deriving and solving a system of linear equations based on the mean and the range.

Alternative Method

Instead of setting up and solving the system of equations algebraically, one can test the median by using the fact that the sum of the deviations from the mean (

85

) must equal

0

. The deviations of the known numbers from

85

are:

(88-85) + (92-85) + (75-85) + (85-85) + (90-85) = 3 + 7 - 10 + 0 + 5 = 5

. Therefore, the sum of the deviations of

x

and

y

from

85

must be

-5

:

(x-85) + (y-85) = -5

, which simplifies to

x + y - 170 = -5

, or

x + y = 165

. Since

y - x = 22

, we can quickly find

x = 71.5

and

y = 93.5

, then sort the list to find the median.

Estimated Time:3m 0s

Question 133Question →

A local community center surveyed a group of $20$ students about the number of hours they volunteered last month. The table below displays the results of the survey, where $a$ and $b$ represent the number of students in their respective categories.

Hours Volunteered	Number of Students
$2$	$a$
$5$	$b$
$8$	$4$
$12$	$3$
$15$	$2$

If the mean number of hours volunteered per student for this group is $6.6$ , what is the median number of hours volunteered per student for these $20$ students?

2.0

3.5

5.0

6.6

8.0

Show answer & explanation

Answer: 5.0

Answer

The median number of hours volunteered per student is 5.0.

The correct answer is 5.0. To find this, we first establish two equations from the given information: the total count of students (

a + b + 9 = 20 \implies a + b = 11

) and the mean of the data (

\frac{2a + 5b + 32 + 36 + 30}{20} = 6.6 \implies 2a + 5b = 34

). Solving this system of equations gives

a = 7

and

b = 4

. To find the median of the 20 values, we look at the 10th and 11th sorted values. Since there are seven 2s (positions 1–7) and four 5s (positions 8–11), both the 10th and 11th values are 5, making the median 5.0.

Step-by-Step Solution

1

Set up an equation for the total number of students.

a + b = 11

The total number of students in the survey is 20. Summing the frequencies gives

a + b + 4 + 3 + 2 = 20

, which simplifies to

a + b = 11

.

2

Set up an equation for the mean of the dataset.

2a + 5b = 34

The mean of the data is 6.6. Using the formula for the weighted mean:

\frac{2a + 5b + 8(4) + 12(3) + 15(2)}{20} = 6.6

. Multiplying both sides by 20 gives

2a + 5b + 98 = 132

, which simplifies to

2a + 5b = 34

.

3

Solve the system of linear equations for

a

and

b

.

a = 7

and

b = 4

Multiply the first equation by 2 to get

2a + 2b = 22

. Subtract this from

2a + 5b = 34

to find

3b = 12 \implies b = 4

. Substitute

b = 4

back into the first equation to find

a = 7

.

4

Find the median of the 20 sorted values.

5.0

With 20 data points, the median is the average of the 10th and 11th data values in ascending order. Since there are seven 2s followed by four 5s, the 10th and 11th values are both 5. The average of 5 and 5 is 5.0.

Key Concept

Determining the median of a grouped frequency distribution by solving for missing frequencies using the total count and the mean.

Question 134Question →

A dataset consists of $7$ positive integers sorted in non-decreasing order. The mean of the dataset is $20$ , the median is $18$ , and the unique mode is $15$ . If the range of the dataset is $22$ , what is the minimum possible value of the largest number in the dataset?

27

29

30

35

37

Show answer & explanation

Answer: 30

Answer

30

The correct answer is 30 because minimizing the largest value requires minimizing the smallest value under the sum and uniqueness constraints. When the smallest value is 8, the largest value is 30, which allows for a valid, sorted sequence of positive integers where 15 is the unique mode: {8, 15, 15, 18, 26, 28, 30}.

Step-by-Step Solution

1

Set up the variables for the sorted dataset.

Let the seven positive integers in non-decreasing order be

x_1, x_2, x_3, x_4, x_5, x_6, x_7

. Since the median is the 4th value,

x_4 = 18

. The mean is

20

, so the sum of all elements is

7 \times 20 = 140

.

Establishing standard notation and utilizing the definitions of median and mean.

2

Incorporate the range into the sum equation.

x_1 + x_2 + x_3 + 18 + x_5 + x_6 + x_7 = 140

. Since the range is

22

, the largest number is

x_7 = x_1 + 22

. Substituting this into the sum gives

2x_1 + x_2 + x_3 + x_5 + x_6 = 100

.

Simplifying the system of equations by expressing the largest element in terms of the smallest element.

3

Apply the mode constraint.

The unique mode is

15

. Since

x_4 = 18

and the dataset is sorted,

15

must be in the lower half of the dataset and must repeat. Thus,

x_2 = x_3 = 15

. The equation simplifies to

2x_1 + 30 + x_5 + x_6 = 100 \implies 2x_1 + x_5 + x_6 = 70

.

Using the properties of the mode and sorted list to lock the values of the second and third elements.

4

Set up inequalities to minimize the largest value.

To minimize

x_7 = x_1 + 22

, we must minimize

x_1

, which is equivalent to maximizing

x_5

and

x_6

. Since

15

is the unique mode with a frequency of

2

, no other value can repeat. Thus,

x_5, x_6,

and

x_7

must be distinct from each other and larger than the median

18

. This means

x_5 \ge 19

. The upper bounds are

x_6 \le x_7 - 1 = x_1 + 21

and

x_5 \le x_7 - 2 = x_1 + 20

.

Applying the uniqueness constraint of the mode to restrict the values of the upper half of the dataset.

5

Solve for the minimum integer value of the smallest element.

Substitute the maximum bounds of

x_5

and

x_6

into the equation:

2x_1 + (x_1 + 20) + (x_1 + 21) \ge 70 \implies 4x_1 + 41 \ge 70 \implies 4x_1 \ge 29 \implies x_1 \ge 7.25

. Since

x_1

must be an integer, the minimum possible value is

x_1 = 8

. This yields a minimum largest value of

x_7 = 8 + 22 = 30

.

Solving the inequality system to find the smallest valid integer boundaries.

Key Concept

Descriptive Statistics and Optimization

Question 135Question →

A list of $15$ positive integers is sorted in non-decreasing order. The minimum value in the list is $10$ , the maximum value is $50$ , and the median is $30$ . If the unique mode of the list is $20$ , what is the maximum possible sum of the $15$ integers in the list?

488

503

506

508

518

Show answer & explanation

Answer: 508

Answer

508

The maximum sum is achieved by setting the frequency of the unique mode

20

to its maximum limit of

6

, occupying positions

2

through

7

(since the first element is

10

and the median at position

8

is

30

). This allows any other value to appear at most

5

times. To maximize the sum, we assign the maximum value

50

to

5

positions (positions

11

to

15

), and the next highest value

49

to the remaining

2

positions in the upper half (positions

9

and

10

). Summing these values gives

10 + 6(20) + 30 + 2(49) + 5(50) = 508

.

Step-by-Step Solution

1

Identify fixed elements based on the median, minimum, and maximum constraints.

For

15

sorted integers, the median is the 8th term:

x_8 = 30

. The minimum is

x_1 = 10

, and the maximum is

x_{15} = 50

.

Establishing these boundary points helps constrain the values of the other elements in the sorted list.

2

Determine the maximum possible frequency of the unique mode

20

.

Since

x_8 = 30

and the list is sorted,

20

can only occupy positions

2

through

7

(a maximum of

6

occurrences).

To maximize the overall sum, we want to maximize the frequency of the mode so that other large values (like

50

) can also appear as many times as possible.

3

Determine the maximum allowed frequency of other values under the unique mode constraint.

With the frequency of

20

set to

6

, any other number can appear at most

5

times.

Since

20

is the unique mode, no other value in the list can have a frequency equal to or greater than

6

.

4

Assign the largest possible values to the remaining positions in the upper half (

x_9

to

x_{14}

).

Set

x_{11} = x_{12} = x_{13} = x_{14} = x_{15} = 50

(frequency of

5

). The remaining two positions,

x_9

and

x_{10}

, are set to the next largest integer,

49

.

This maximizes the sum of the upper half of the list without violating the maximum frequency limit of

5

for any non-mode value.

5

Sum the elements of the optimal list.

Sum =

10 + 6(20) + 30 + 2(49) + 5(50) = 10 + 120 + 30 + 98 + 250 = 508

.

Adding all fifteen optimal values together gives the absolute maximum possible sum.

Key Concept

Using central tendency constraints (median, mode) and range limits to optimize a dataset sum.

Question 136Question →

An art gallery owner wants to display 5 paintings in a row on a wall. The owner selects these 5 paintings from a collection of 4 different landscape paintings and 4 different portrait paintings. The display must meet the following guidelines:

1. No two landscape paintings can be placed next to each other.
2. At least 2 landscape paintings must be displayed.

How many different arrangements of 5 paintings are possible?

Show answer & explanation

Answer: 2016

Answer

2016

The correct answer is 2016. By breaking down the problem into two mutually exclusive cases based on the number of landscape paintings (either 3 landscapes and 2 portraits, or 2 landscapes and 3 portraits), we can find the valid arrangements for each. For 3 landscapes, the only valid layout is

L-P-L-P-L

, yielding

P(4,3) \times P(4,2) = 288

arrangements. For 2 landscapes, there are

\binom{4}{2} = 6

layout configurations, each yielding

P(4,2) \times P(4,3) = 288

arrangements, for a total of 1728. Adding these two cases gives

288 + 1728 = 2016

.

Step-by-Step Solution

1

Determine the possible number of landscape paintings (

k

) that can be displayed.

k = 2

or

k = 3

Since at least 2 landscapes must be displayed,

k \ge 2

. Since no two landscapes can be adjacent in a 5-painting row, we cannot have 4 landscapes (as that would require at least 3 portraits to separate them, making the total count at least 7 paintings). Thus,

k

can only be 2 or 3.

2

Calculate the arrangements for the case with 3 landscape paintings and 2 portrait paintings.

288 arrangements

For 3 landscapes (

L

) and 2 portraits (

P

) to have no adjacent landscapes, the only possible pattern of positions is

L-P-L-P-L

. The number of ways to choose and arrange 3 landscapes from 4 is

P(4, 3) = 4 \times 3 \times 2 = 24

. The number of ways to choose and arrange 2 portraits from 4 is

P(4, 2) = 4 \times 3 = 12

. The total arrangements for this case is

24 \times 12 = 288

.

3

Calculate the arrangements for the case with 2 landscape paintings and 3 portrait paintings.

1728 arrangements

For 2 landscapes (

L

) and 3 portraits (

P

) to have no adjacent landscapes, we place the 3 portraits first:

\_ P \_ P \_ P \_

. We choose 2 of the 4 available spaces for the landscapes in

\binom{4}{2} = 6

ways. For each pattern, the number of ways to choose and arrange 2 landscapes from 4 is

P(4, 2) = 12

, and the number of ways to choose and arrange 3 portraits from 4 is

P(4, 3) = 24

. The total arrangements for this case is

6 \times 12 \times 24 = 1728

.

4

Sum the arrangements from both cases.

2016 arrangements

Since the two cases are mutually exclusive, we add their individual counts:

288 + 1728 = 2016

.

Key Concept

Permutations and Combinations with Constraints

Question 137Question →

A set of $10$ distinct positive integers has a median of $25$ and a range of $30$ . What is the greatest possible value of the mean of these $10$ integers?

25

29

31

33

37

Show answer & explanation

Answer: 33

Answer

The greatest possible value of the mean of the 10 integers is 33.

The correct answer is 33. To find the greatest possible mean of the 10 distinct positive integers, we must maximize their sum. Let the sorted integers be

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10}

. The median is the average of the 5th and 6th terms, so

\frac{x_5 + x_6}{2} = 25

, or

x_5 + x_6 = 50

. Since the integers are distinct, the maximum value for

x_5

is 24, which forces

x_6 = 26

. To maximize the sum of the first five terms, they should be consecutive integers ending at 24:

x_1 = 20, x_2 = 21, x_3 = 22, x_4 = 23, x_5 = 24

. Since the range is 30, the maximum value is

x_{10} = x_1 + 30 = 20 + 30 = 50

. To maximize the remaining terms in the upper half, we choose the largest possible distinct integers less than 50:

x_7 = 47, x_8 = 48, x_9 = 49

. The maximum sum is

20 + 21 + 22 + 23 + 24 + 26 + 47 + 48 + 49 + 50 = 330

, yielding a maximum mean of

330 / 10 = 33

.

Step-by-Step Solution

1

Define the variables and apply the median constraint.

Let the 10 sorted distinct positive integers be

x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10}

. The median is the average of the 5th and 6th terms:

\frac{x_5 + x_6}{2} = 25

, which means

x_5 + x_6 = 50

.

Since the number of terms is even, the median is the average of the two middle terms.

2

Maximize the first five integers to find the maximum value of the first term.

Since the integers are distinct, we must have

x_5 < x_6

. With

x_5 + x_6 = 50

, the maximum possible integer value for

x_5

is 24 (which makes

x_6 = 26

). To maximize the sum, we make the preceding terms as large as possible:

x_4 = 23

,

x_3 = 22

,

x_2 = 21

, and

x_1 = 20

.

To maximize the mean, we must maximize the sum of all terms, which requires making each term as large as possible within the distinct integer constraints.

3

Apply the range constraint to find the maximum value of the last term.

The range is 30, so

x_{10} - x_1 = 30

. Since the maximum value of

x_1

is 20, the maximum possible value for

x_{10}

is

20 + 30 = 50

.

The range of a dataset is the difference between the maximum and minimum values.

4

Maximize the remaining terms in the upper half of the dataset.

We have

x_6 = 26

. The remaining terms must satisfy

26 < x_7 < x_8 < x_9 < x_{10} = 50

. To maximize the sum, we choose the largest possible distinct integers for these slots:

x_9 = 49

,

x_8 = 48

, and

x_7 = 47

.

This maximizes the sum of the upper half of the dataset under the constraint that the maximum value is 50.

5

Calculate the maximum sum and the resulting maximum mean.

Sum =

20 + 21 + 22 + 23 + 24 + 26 + 47 + 48 + 49 + 50 = 330

. Mean =

\frac{330}{10} = 33

.

The mean is calculated by dividing the sum of the elements by the number of elements.

Key Concept

Maximizing the mean of a bounded dataset using median, range, and distinctness constraints.

Question 138Question →

A box contains only red, blue, and green marbles. The ratio of red marbles to blue marbles is $2:3$ , and the ratio of blue marbles to green marbles is $3:5$ . If one marble is drawn at random from the box, the probability of drawing a red marble is $P$ . If $6$ green marbles are added to the box and no other marbles are removed, the probability of drawing a red marble becomes $Q$ . Given that $P - Q = \frac{3}{115}$ , how many blue marbles were originally in the box?

3

8

12

15

20

Show answer & explanation

Answer: 12

Answer

12

The correct answer is 12. By expressing the original counts of red, blue, and green marbles as

2k

,

3k

, and

5k

, the total number of marbles is

10k

. The initial probability of drawing a red marble is

P = \frac{2k}{10k} = \frac{1}{5}

. When 6 green marbles are added, the new total becomes

10k + 6

, yielding a new probability

Q = \frac{2k}{10k + 6} = \frac{k}{5k + 3}

. Setting their difference to

\frac{3}{115}

results in

\frac{1}{5} - \frac{k}{5k + 3} = \frac{3}{115}

. Simplifying this rational equation gives

25k + 15 = 115

, which solves to

k = 4

. Therefore, the original number of blue marbles is

3k = 3(4) = 12

.

Step-by-Step Solution

1

Define the number of red, blue, and green marbles in terms of a variable

k

.

Let red marbles =

2k

, blue marbles =

3k

, and green marbles =

5k

. The initial total number of marbles is

2k + 3k + 5k = 10k

.

The given ratios are red:blue = 2:3 and blue:green = 3:5, which combine to a continuous ratio of red:blue:green = 2:3:5.

2

Express the initial probability

P

of drawing a red marble.

P = \frac{2k}{10k} = \frac{1}{5}

.

Probability is the number of favorable outcomes (red marbles) divided by the total number of outcomes.

3

Express the new probability

Q

of drawing a red marble after adding 6 green marbles.

The number of green marbles becomes

5k + 6

, so the new total is

10k + 6

. Thus,

Q = \frac{2k}{10k + 6}

.

Adding 6 green marbles increases the total marble count by 6 while the number of red marbles remains

2k

.

4

Set up the equation using the given difference

P - Q = \frac{3}{115}

and solve for

k

.

\frac{1}{5} - \frac{2k}{10k + 6} = \frac{3}{115}

\frac{1}{5} - \frac{k}{5k + 3} = \frac{3}{115}

\frac{(5k + 3) - 5k}{5(5k + 3)} = \frac{3}{115}

\frac{3}{25k + 15} = \frac{3}{115}

25k + 15 = 115 \implies 25k = 100 \implies k = 4

.

Solving the rational equation gives the multiplier

k

that determines the actual number of marbles.

5

Calculate the original number of blue marbles.

Blue marbles =

3k = 3(4) = 12

.

The question asks for the original number of blue marbles, which is represented by

3k

.

Key Concept

Basic Probability and Counting Methods

Estimated Time:3m 0s

Question 139Question →

A security passcode consists of a sequence of 4 digits chosen from the digits 0 through 9. To be valid, a passcode must contain at least one repeated digit, but no digit can appear more than twice. Additionally, the passcode cannot end with an odd digit. How many different valid security passcodes can be created?

Show answer & explanation

Answer: 2295

Answer

There are 2,295 different valid security passcodes.

The correct answer of 2,295 is found by subtracting all invalid passcodes from the total possible passcodes. The total number of 4-digit passcodes ending in an even digit is 5,000. The invalid passcodes are those with all distinct digits (2,520), those with a digit repeated three times (180), and those with all four digits identical (5). Subtracting these gives 5,000 - 2,520 - 180 - 5 = 2,295.

Step-by-Step Solution

1

Calculate the total number of 4-digit passcodes ending in an even digit.

5,000

The last digit must be even (0, 2, 4, 6, or 8) to not be odd, giving 5 choices. The first three digits can be any of the 10 digits from 0 through 9. By the Fundamental Counting Principle, the total number of passcodes is 10 * 10 * 10 * 5 = 5,000.

2

Calculate the number of invalid passcodes where all 4 digits are distinct.

2,520

For all 4 digits to be distinct, the last digit must be chosen from the 5 even digits. The remaining 3 positions must be filled with 3 distinct digits chosen from the remaining 9 digits. There are 9 * 8 * 7 = 504 ways to choose these. This gives 504 * 5 = 2,520 passcodes.

3

Calculate the number of invalid passcodes where a single digit is repeated three times.

180

If the repeated digit is the last digit, there are 5 choices for that digit, and the single distinct digit (9 choices) can be placed in any of the 3 remaining positions, giving 5 * 9 * 3 = 135 passcodes. If the repeated digit is not the last digit, the three identical digits occupy the first three positions, and the last digit (5 choices) is distinct from them (9 choices), giving 5 * 9 = 45 passcodes. In total, 135 + 45 = 180 passcodes.

4

Calculate the number of invalid passcodes where all four digits are identical.

5

Since the last digit must be even, all four digits must be the same even digit (0000, 2222, 4444, 6666, or 8888), which gives 5 passcodes.

5

Subtract the invalid passcodes from the total number of passcodes.

2,295

Subtracting the passcodes with all distinct digits (2,520) and those with a digit repeated three or four times (180 + 5 = 185) from the total of 5,000 gives 5,000 - 2,520 - 185 = 2,295.

Key Concept

Complementary counting using permutations and partition analysis

Alternative Method

Instead of complementary counting, count the valid cases directly: Case A where the digit frequencies are [2, 1, 1] (which yields 2,160 codes) and Case B where the digit frequencies are [2, 2] (which yields 135 codes). Adding these yields 2,160 + 135 = 2,295 codes.

Estimated Time:3m 0s

Question 140Question →

A high school debate club consists of 15 members: 6 freshmen, 5 sophomores, and 4 juniors. If a committee of 3 members is to be selected at random from the club, what is the probability that the committee will contain at least 1 freshman and at least 1 sophomore?

\frac{24}{91}

\frac{247}{455}

\frac{51}{91}

\frac{251}{455}

\frac{6}{7}

Show answer & explanation

Answer: $\frac{51}{91}$

Answer

The correct probability is

\frac{51}{91}

.

To find the probability that a randomly chosen 3-member committee from a club of 15 members (6 freshmen, 5 sophomores, 4 juniors) has at least 1 freshman and at least 1 sophomore, we can find the total number of possible committees and subtract the number of unfavorable committees. The total number of committees is

\binom{15}{3} = 455

. The unfavorable committees are those with no freshmen (chosen from the 9 sophomores and juniors:

\binom{9}{3} = 84

) or no sophomores (chosen from the 10 freshmen and juniors:

\binom{10}{3} = 120

). Because these two groups overlap when only juniors are chosen (

\binom{4}{3} = 4

), the total number of unfavorable committees is

84 + 120 - 4 = 200

by the Principle of Inclusion-Exclusion. Thus, there are

455 - 200 = 255

favorable committees, yielding a probability of

\frac{255}{455} = \frac{51}{91}

.

Step-by-Step Solution

1

Calculate the total number of possible 3-member committees that can be formed from the 15 club members.

\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455

total committees.

To find the probability, we first need to determine the size of the entire sample space (the total number of possible outcomes).

2

Determine the number of committees that do not meet the requirement of containing at least one freshman and at least one sophomore. This occurs if a committee has no freshmen, no sophomores, or neither.

Let

A

be the event of choosing a committee with no freshmen (only sophomores and juniors):

\binom{9}{3} = 84

ways. Let

B

be the event of choosing a committee with no sophomores (only freshmen and juniors):

\binom{10}{3} = 120

ways. Let

A \cap B

be the event of choosing a committee with neither freshmen nor sophomores (only juniors):

\binom{4}{3} = 4

ways.

It is easier to count the complement (the unfavorable outcomes) and subtract it from the total.

3

Use the Principle of Inclusion-Exclusion to find the total number of unfavorable committees (no freshmen or no sophomores).

|A \cup B| = |A| + |B| - |A \cap B| = 84 + 120 - 4 = 200

unfavorable committees.

Since the events of having no freshmen and having no sophomores overlap when only juniors are selected, we must subtract the intersection to avoid double-counting.

4

Subtract the number of unfavorable committees from the total number of committees to find the number of favorable committees, and then compute the probability.

Favorable committees:

455 - 200 = 255

ways. Probability:

\frac{255}{455} = \frac{51}{91}

.

Subtracting the complement from the total gives the number of valid outcomes, and dividing this by the total outcomes yields the desired probability.

Key Concept

Using combinations and the Principle of Inclusion-Exclusion to calculate probabilities of compound events.

Estimated Time:3m 0s

Pre-Algebra

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution