Number and Numeration Practice Questions — JAMB UTME — Page 2

Question 21Question →

A trader invested $\text{₦}50,000$ in a savings scheme that pays compound interest at a rate of $10\%$ per annum compounded annually. What is the total compound interest earned at the end of $2$ years?

\text{₦}10,500

\text{₦}60,500

\text{₦}10,000

\text{₦}5,000

Show answer & explanation

Answer: $\text{₦}10,500$

Answer

The total compound interest earned at the end of

2

years is

\text{₦}10,500

.

The compound interest is obtained by subtracting the principal from the total accumulated amount. Using

A = P(1 + r)^n

, the total amount is

\text{₦}60,500

. Subtracting the principal of

\text{₦}50,000

yields

\text{₦}10,500

.

Step-by-Step Solution

1

Calculate the total accumulated amount using the compound interest formula

A = P(1 + r)^n

A = 50,000 \times (1 + 0.10)^2 = 50,000 \times 1.21 = \text{₦}60,500

Determines the total value of the investment at the end of the duration.

2

Calculate the compound interest earned using

CI = A - P

CI = 60,500 - 50,000 = \text{₦}10,500

Subtracts the original principal from the total accumulated amount to find the interest portion.

Key Concept

Compound interest calculation

Question 22Question →

Given the universal set $U = \{x \in \mathbb{Z} : 1 \le x \le 20\}$ . Let $P = \{x \in U : x \text{ is a prime number}\}$ , $Q = \{x \in U : x \text{ is an odd integer}\}$ , and $R = \{x \in U : x \text{ is a multiple of } 3\}$ . What is the value of $n((P \cup Q) \cap R')$ ?

7

8

11

14

Show answer & explanation

Answer: 8

Answer

8

The correct value is 8. The union

P \cup Q

yields

\{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}

. Intersecting this set with

R'

means removing any element that is a multiple of 3. The multiples of 3 in

P \cup Q

are 3, 9, and 15. Removing these 3 elements from the 11 elements of

P \cup Q

leaves exactly 8 elements.

Step-by-Step Solution

1

Identify the elements of sets P, Q, and R within the universal set U.

U = \{1, 2, 3, \dots, 20\}

,

P = \{2, 3, 5, 7, 11, 13, 17, 19\}

,

Q = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19\}

, and

R = \{3, 6, 9, 12, 15, 18\}

.

Listing elements helps accurately compute set unions and intersections.

2

Find the union of sets P and Q, denoted as P ∪ Q.

P \cup Q = \{1, 2, 3, 5, 7, 9, 11, 13, 15, 17, 19\}

, so

n(P \cup Q) = 11

.

Combining elements of both P and Q without repetition gives their union.

3

Find the complement of set R with respect to U, denoted as R'.

R' = \{1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20\}

.

The complement set R' contains all elements in U that are not multiples of 3.

4

Determine the intersection of (P ∪ Q) and R'.

(P \cup Q) \cap R' = \{1, 2, 5, 7, 11, 13, 17, 19\}

. The number of elements is 8.

This removes the multiples of 3 (namely 3, 9, and 15) from the set

P \cup Q

.

Key Concept

Set operations including union, intersection, and set complementation.

Estimated Time:1m 30s

Question 23Question →

Find the real value of

x

that satisfies the exponential equation

\frac{8^{x + 2}}{4^{x - 1}} = 16^{x - 1}

Show answer & explanation

Answer: 4

Answer

The value of

x

is

4

.

Rewriting the terms in base 2 gives

2^{3(x+2)} / 2^{2(x-1)} = 2^{4(x-1)}

. Applying the quotient rule gives an exponent of

(3x + 6) - (2x - 2) = x + 8

on the left. Equating the exponents yields

x + 8 = 4x - 4

, which solves cleanly to

x = 4

.

Step-by-Step Solution

1

Express all terms using a common prime base of 2

The equation becomes

\frac{(2^3)^{x + 2}}{(2^2)^{x - 1}} = (2^4)^{x - 1}

.

Converting to a common base enables the use of index laws to simplify the equation.

2

Apply the power-of-a-power and quotient laws of indices

The left side simplifies to

2^{3(x+2) - 2(x-1)} = 2^{x + 8}

and the right side is

2^{4x - 4}

.

When dividing powers of the same base, exponents are subtracted:

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

.

3

Equate exponents and solve the linear equation

x + 8 = 4x - 4 \implies 3x = 12 \implies x = 4

.

Because the bases on both sides are identical, their respective exponents must be equal.

Key Concept

Solving exponential equations using common base conversion and laws of indices

Question 24Question →

A factory uses $8$ identical machines to produce $480$ items in $6$ hours. If $3$ of the machines break down, how many hours will it take the remaining machines to produce $600$ items at the same rate?

Show answer & explanation

Answer: 12

Answer

The remaining machines will take 12 hours to produce 600 items.

Each machine produces

\frac{480}{8 \times 6} = 10

items per hour. With

5

active machines, the combined rate is

50

items per hour. To reach

600

items, the time required is

\frac{600}{50} = 12

hours.

Step-by-Step Solution

1

Calculate the output rate per machine per hour

10 items per machine per hour

The combined rate of 8 machines is

480 \div 6 = 80

items per hour. Dividing by 8 machines gives

10

items/hour per machine.

2

Determine the new combined production rate

50 items per hour

With 3 machines out of service,

8 - 3 = 5

machines remain, working at a total rate of

5 \times 10 = 50

items per hour.

3

Calculate total hours required for the new target quantity

12 hours

Dividing the target quantity of

600

items by the rate of

50

items/hour yields

\frac{600}{50} = 12

hours.

Key Concept

Compound Proportion and Work Rate

Estimated Time:1m 30s

Question 25Question →

What is the result of the subtraction $41_5 - 23_5$ in base 5?

18_5

13_5

22_5

23_5

Show answer & explanation

Answer: $13_5$

Answer

13_5

The answer

13_5

is correct because borrowing 1 from the fives place contributes 5 to the units place, making the calculation

(1 + 5) - 3 = 3

for the units place and

3 - 2 = 1

for the fives place.

Step-by-Step Solution

1

Set up column subtraction for

41_5 - 23_5

in base 5.

In the units position, we evaluate

1 - 3

.

Since 1 is smaller than 3, a borrow operation from the fives column is required.

2

Borrow 1 from the fives column (reducing 4 to 3) and add the base value 5 to the units digit 1.

Units column value becomes

1 + 5 = 6

. Fives column value becomes 3.

When borrowing in base

n

, the borrowed amount is equal to

n

, which is 5 for base 5.

3

Subtract digits in each column.

Units column:

6 - 3 = 3

. Fives column:

3 - 2 = 1

. The result is

13_5

.

Completing the column subtraction yields the final digits in base 5.

Key Concept

Subtraction in Non-Decimal Number Bases

Estimated Time:45s

Question 26Question →

Convert the fractional binary number $0.1101_2$ to its equivalent base 10 (decimal) value. What is the decimal value?

Show answer & explanation

Answer: 0.8125

Answer

The base 10 value of

0.1101_2

is

0.8125

.

To convert a fractional binary number to decimal, expand each digit after the radix point using decreasing negative powers of 2 (

2^{-1}, 2^{-2}, 2^{-3}, 2^{-4}

). Evaluating

1(0.5) + 1(0.25) + 0(0.125) + 1(0.0625)

yields

0.8125

.

Step-by-Step Solution

1

Write the given binary fraction in place-value expansion using powers of 2

0.1101_2 = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 0 \cdot 2^{-3} + 1 \cdot 2^{-4}

Positions after the binary point represent negative powers of 2 starting from

2^{-1}

.

2

Evaluate each fractional component

2^{-1} = 0.5

,

2^{-2} = 0.25

,

2^{-3} = 0.125

,

2^{-4} = 0.0625

Calculating standard decimal values for binary fractional places.

3

Add the non-zero fractional terms together

0.5 + 0.25 + 0.0625 = 0.8125

Summing the decimal values gives the complete converted decimal representation.

Key Concept

Conversion of fractional numbers from base 2 to base 10

Question 27Question →

A student measured the length of a room as $4.00\text{ m}$ instead of the actual length of $5.00\text{ m}$ . What is the percentage error in the measurement?

20\%

25\%

80\%

2\%

Show answer & explanation

Answer: $20\%$

Answer

20\%

The absolute error is the difference between the measured value (

4.00\text{ m}

) and the actual value (

5.00\text{ m}

), which is

1.00\text{ m}

. Dividing the error by the actual value gives $\frac{1.00}{5.00} = 0.20$, which equals

20\%

when expressed as a percentage.

Step-by-Step Solution

1

Identify the actual value and the measured value.

Actual length =

5.00\text{ m}

, Measured length =

4.00\text{ m}

.

Percentage error calculations require establishing the true baseline quantity.

2

Calculate the absolute error.

Absolute Error =

|5.00 - 4.00| = 1.00\text{ m}

.

Error is defined as the magnitude of the difference between the true value and measured value.

3

Compute the percentage error.

Percentage Error = $\frac{1.00}{5.00} \times 100\% = 20\%$.

Percentage error is found by dividing the absolute error by the actual value and multiplying by

100\%

.

Key Concept

Percentage Error Calculation

Question 28Question →

What is the result of the subtraction $523_8 - 267_8$ in base 8?

234_8

256_8

244_8

344_8

Show answer & explanation

Answer: $234_8$

Answer

234_8

Performing place-by-place subtraction in base 8 requires borrowing 8 whenever a top digit is smaller than the bottom digit. Borrowing 1 from the tens place adds 8 to the units place (

3 + 8 - 7 = 4

). The tens place becomes 1, borrowing 1 from the hundreds place adds 8 (

1 + 8 - 6 = 3

). The hundreds place becomes 4, giving

4 - 2 = 2

. Thus, the answer is

234_8

.

Step-by-Step Solution

1

Subtract the units column (

3 - 7

in base 8)

Borrow

1

from the middle column (which represents

8

). The units position becomes

3 + 8 = 11

. Then

11 - 7 = 4

.

Since

3 < 7

, borrowing from the next higher position (base 8) is required.

2

Subtract the middle column (

1 - 6

in base 8)

After borrowing, the middle digit

2

becomes

1

. Borrow

1

from the hundreds column (representing

8

). The middle position becomes

1 + 8 = 9

. Then

9 - 6 = 3

.

The middle digit was reduced by

1

due to the previous borrow, requiring another borrow from the left.

3

Subtract the hundreds column (

4 - 2

in base 8)

The left digit

5

was reduced to

4

. Then

4 - 2 = 2

.

Complete the subtraction for the leading column.

4

Combine the resulting digits

234_8

Concatenating the results from left to right gives the final answer in base 8.

Key Concept

Subtraction in non-decimal number bases

Question 29Question →

A student evaluated the expression $\frac{0.0256 \times 1.5}{0.0064}$ by first rounding each number in the expression to $1$ significant figure before completing the computation. What is the percentage error in the student's result compared to the exact value?

66.67\%

40.00\%

166.67\%

33.33\%

Show answer & explanation

Answer: $66.67\%$

Answer

The percentage error in the student's result is

66.67\%

.

The exact evaluation yields

\frac{0.0384}{0.0064} = 6

. Rounding each quantity to

1

significant figure gives

0.03

,

2

, and

0.006

, which evaluates to

\frac{0.06}{0.006} = 10

. The absolute error is

|10 - 6| = 4

, and dividing by the true value

6

yields

\frac{4}{6} \times 100\% = 66.67\%

.

Step-by-Step Solution

1

Calculate the exact value of the expression

Numerator:

0.0256 \times 1.5 = 0.0384

. Division:

\frac{0.0384}{0.0064} = 6

.

Establishing the true reference value is required to calculate percentage error.

2

Round each number in the expression to

1

significant figure

0.0256 \rightarrow 0.03

,

1.5 \rightarrow 2

,

0.0064 \rightarrow 0.006

.

Following the approximation directive in the problem statement.

3

Calculate the estimated value using the rounded numbers

Estimated value =

\frac{0.03 \times 2}{0.006} = \frac{0.06}{0.006} = 10

.

Obtaining the student's evaluated result.

4

Determine the percentage error

Absolute Error =

|10 - 6| = 4

. Percentage Error =

\frac{4}{6} \times 100\% = 66.67\%

.

Percentage error is defined as

\frac{|\text{Estimated} - \text{True}|}{\text{True}} \times 100\%

.

Key Concept

Percentage error calculation with significant figure approximations

Estimated Time:2m 0s

Question 30Question →

A trader estimated the mass of a bag of rice to be $25\text{ kg}$ , but the actual mass of the bag was $20\text{ kg}$ . Calculate the percentage error in the trader's estimate.

Show answer & explanation

Answer: 25

Answer

The percentage error in the trader's estimate is

25\%

.

The absolute error is

25\text{ kg} - 20\text{ kg} = 5\text{ kg}

. Evaluating the error relative to the actual value gives

\frac{5\text{ kg}}{20\text{ kg}} = 0.25

, which equals

25\%

.

Step-by-Step Solution

1

Calculate the absolute error in measurement

Error =

|25 - 20| = 5\text{ kg}

Absolute error is the absolute difference between the estimated value and the true value.

2

Calculate the percentage error relative to the actual value

\text{Percentage Error} = \frac{5}{20} \times 100\% = 25\%

Percentage error must always be calculated using the actual (true) value as the denominator.

Key Concept

Percentage Error Calculation

Estimated Time:45s

Question 31Question →

What is the value of $(-38) \pmod{7}$ expressed in standard non-negative remainder form?

4

-3

3

5

Show answer & explanation

Answer: 4

Answer

4

The value

4

is correct because

-38 = 7 \times (-6) + 4

. The remainder

4

lies in the standard non-negative range

0 \le r < 7

. Alternatively, adding multiples of

7

to

-38

gives

-38 + 35 = -3

, and adding

7

once more gives

-3 + 7 = 4

.

Step-by-Step Solution

1

Find the largest multiple of the modulus

7

that is less than or equal to

-38

.

The multiple is

7 \times (-6) = -42

.

Modular arithmetic requires the remainder

r

to satisfy

0 \le r < 7

in standard form.

2

Calculate the remainder by subtracting the multiple from the dividend.

-38 - (-42) = -38 + 42 = 4

.

The remainder is the non-negative difference between the number and the multiple of the modulus.

Key Concept

Modular Arithmetic with Negative Numbers

Estimated Time:1m 0s

Question 32Question →

What is the larger real value of $x$ that satisfies the logarithmic equation $\log_x 8 + \log_4 x = \frac{5}{2}$ ?

8

4

6

16

Show answer & explanation

Answer: 8

Answer

The larger value of

x

that satisfies the equation is 8.

Using the change of base rule

\log_a b = \frac{\log_c b}{\log_c a}

, we express both logarithmic terms in base 2:

\log_x 8 = \frac{3}{\log_2 x}

and

\log_4 x = \frac{\log_2 x}{2}

. Setting

y = \log_2 x

gives

\frac{3}{y} + \frac{y}{2} = \frac{5}{2}

. Multiplying by

2y

yields

y^2 - 5y + 6 = 0

, which factors as

(y-2)(y-3) = 0

. Thus,

y = 2

or

y = 3

, giving solutions

x = 2^2 = 4

and

x = 2^3 = 8

. The larger solution is 8.

Step-by-Step Solution

1

Apply the change of base formula to express all logarithmic terms in base 2.

\log_x 8 = \frac{\log_2 8}{\log_2 x} = \frac{3}{\log_2 x} \quad \text{and} \quad \log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{1}{2}\log_2 x

Converting all terms to a common base (base 2) allows algebraic simplification.

2

Substitute

y = \log_2 x

into the given equation.

\frac{3}{y} + \frac{y}{2} = \frac{5}{2}

Using substitution converts the logarithmic expression into a rational algebraic equation.

3

Multiply the entire equation by

2y

to clear denominators and form a quadratic equation.

6 + y^2 = 5y \implies y^2 - 5y + 6 = 0

Clearing denominators transforms the relation into standard quadratic form.

4

Factor the quadratic equation to find the values of

y

.

(y - 2)(y - 3) = 0 \implies y = 2 \text{ or } y = 3

Factoring determines the possible powers of 2 for

x

.

5

Solve for

x

using

y = \log_2 x

and select the larger value.

x = 2^2 = 4 \quad \text{or} \quad x = 2^3 = 8. \text{ The larger value is } 8.

Converting back from

y

to

x

yields the final solutions for

x

.

Key Concept

Change of Base Theorem for Logarithms and Reduction to Quadratic Equations

Question 33Question →

What is the simplified form of the expression $\frac{a^3 \times a^4}{a^2}$ ?

a^5

a^{10}

a^6

a^9

Show answer & explanation

Answer: $a^5$

Answer

The simplified expression is

a^5

.

Combining terms with the same base using the product law gives

a^{3+4} = a^7

in the numerator. Next, dividing by

a^2

using the quotient law yields

a^{7-2} = a^5

.

Step-by-Step Solution

1

Simplify the numerator using the product law of indices

a^3 \times a^4 = a^{3+4} = a^7

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

2

Divide by the denominator using the quotient law of indices

\frac{a^7}{a^2} = a^{7-2} = a^5

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

Key Concept

Laws of Indices (Product and Quotient Rules)

Estimated Time:45s

Question 34Question →

Evaluate the value of the logarithmic expression $\log_3 7 \times \log_7 81$ .

Show answer & explanation

Answer: 4

Answer

The value of the expression is 4.

By applying the change of base formula

\log_7 81 = \frac{\log_3 81}{\log_3 7}

, the expression becomes

\log_3 7 \times \frac{\log_3 81}{\log_3 7} = \log_3 81

. Since

3^4 = 81

, the result is 4.

Step-by-Step Solution

1

Apply the change of base chain rule

\log_a b \cdot \log_b c = \log_a c

\log_3 7 \times \log_7 81 = \log_3 81

By change of base,

\log_7 81 = \frac{\log_3 81}{\log_3 7}

, so multiplying by

\log_3 7

cancels out the common factor.

2

Evaluate

\log_3 81

4

Since

3^4 = 81

, the logarithm base 3 of 81 is equal to 4.

Key Concept

Change of Base Property of Logarithms

Question 35Question →

A student spent $\frac{2}{5}$ of his monthly allowance on books and $\frac{1}{3}$ of the remaining amount on food. If he was left with $\text{N}4,000$ , what was his total monthly allowance in Naira ( $\text{N}$ )?

Show answer & explanation

Answer: 10000

Answer

The total monthly allowance was 10,000 Naira.

After spending

\frac{2}{5}

on books,

\frac{3}{5}

of the allowance remains. Spending

\frac{1}{3}

of this remainder on food accounts for

\frac{1}{5}

of the original allowance. Subtracting

\frac{1}{5}

from

\frac{3}{5}

leaves

\frac{2}{5}

of the total allowance, which is equal to

\text{N}4,000

. Solving

\frac{2}{5} \times \text{Total} = 4,000

yields

10,000

Naira.

Step-by-Step Solution

1

Determine the remaining fraction after the first expenditure

Fraction left =

\frac{3}{5}

The student spent

\frac{2}{5}

on books, leaving

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

of the total allowance.

2

Calculate the fraction of the total allowance spent on food

Fraction spent on food =

\frac{1}{5}

He spent

\frac{1}{3}

of the remaining

\frac{3}{5}

, which equals

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

of the whole allowance.

3

Calculate the final remaining fraction of the allowance

Final fraction left =

\frac{2}{5}

The remaining fraction is

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

.

4

Solve for the total allowance

Total allowance = 10,000 Naira

Since

\frac{2}{5}

of the total allowance equals

\text{N}4,000

, the total allowance is

\frac{4,000 \times 5}{2} = 10,000

Naira.

Key Concept

Sequential Fraction of Remainder Problems

Question 36Question →

In a survey of $100$ agricultural market traders in Lagos, $52$ sell cassava, $45$ sell yam, and $60$ sell plantain. Furthermore, $25$ sell both cassava and yam, $22$ sell both yam and plantain, and $28$ sell both cassava and plantain. If the number of traders who sell none of these three crops is twice the number of traders who sell all three crops, how many traders sell exactly two of these crops?

Show answer & explanation

Answer: 57

Answer

57 traders sell exactly two of these crops.

Using inclusion-exclusion, the total number of traders selling at least one crop is

82 + x

. Adding the

2x

traders selling none gives

82 + 3x = 100

, so

x = 6

. The number of traders selling exactly two crops is

(25 - 6) + (22 - 6) + (28 - 6) = 19 + 16 + 22 = 57

.

Step-by-Step Solution

1

Apply the Principle of Inclusion-Exclusion for three set unions.

n(C \cup Y \cup P) = 82 + x

, where

x = n(C \cap Y \cap P)

.

Summing single set cardinalities, subtracting pairwise intersections, and adding back the triple intersection accounts for all region overlaps.

2

Set up and solve the universal set cardinality equation.

x = 6

Since total traders

|U| = 100

and non-sellers equal

2x

, the equation

100 = (82 + x) + 2x

simplifies to

3x = 18

, giving

x = 6

.

3

Compute the sum of elements in regions representing exactly two sets.

57

Subtracting

x = 6

from each pairwise intersection isolates traders who sell only cassava & yam (19), only yam & plantain (16), and only cassava & plantain (22). Summing these gives

19 + 16 + 22 = 57

.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Partitioning Venn Diagram Regions

Question 37Question →

What is the simplified form of the expression $\frac{\sqrt{5}}{\sqrt{5} - \sqrt{3}}$ ?

\frac{5 + \sqrt{15}}{2}

\sqrt{5}

5 + \sqrt{15}

\frac{5 - \sqrt{15}}{2}

Show answer & explanation

Answer: $\frac{5 + \sqrt{15}}{2}$

Answer

\frac{5 + \sqrt{15}}{2}

To rationalize the denominator of

\frac{\sqrt{5}}{\sqrt{5} - \sqrt{3}}

, multiply both numerator and denominator by the conjugate

\sqrt{5} + \sqrt{3}

. The numerator becomes

\sqrt{5}(\sqrt{5} + \sqrt{3}) = 5 + \sqrt{15}

and the denominator becomes

(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2

, yielding

\frac{5 + \sqrt{15}}{2}

.

Step-by-Step Solution

1

Identify the conjugate of the denominator

The conjugate of

\sqrt{5} - \sqrt{3}

is

\sqrt{5} + \sqrt{3}

.

Multiplying a binomial surd by its conjugate eliminates the radical terms in the denominator using the difference of two squares identity

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

.

2

Multiply both the numerator and the denominator by the conjugate

\frac{\sqrt{5}(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}

Multiplying the fraction by

\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}

is equivalent to multiplying by

1

, preserving the value of the expression.

3

Expand the numerator and simplify the denominator

Numerator:

\sqrt{5} \times \sqrt{5} + \sqrt{5} \times \sqrt{3} = 5 + \sqrt{15}

. Denominator:

(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2

.

Applying the distributive law to the numerator and difference of squares to the denominator simplifies both parts.

4

Combine the simplified numerator and denominator

\frac{5 + \sqrt{15}}{2}

This is the simplified rationalized form.

Key Concept

Rationalization of Denominators

Question 38Question →

If $\frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}}$ is expressed in the simplified form $a + b\sqrt{6}$ , where $a$ and $b$ are rational numbers, what is the value of $a + b$ ?

5

7

9

13

Show answer & explanation

Answer: 7

Answer

The correct value of

a + b

is 7.

Multiplying by the conjugate

(3\sqrt{2} + 2\sqrt{3})

reduces the denominator to

18 - 12 = 6

and expands the numerator to

30 + 12\sqrt{6}

. Dividing by 6 yields

5 + 2\sqrt{6}

, giving

a = 5

and

b = 2

, which sums to 7.

Step-by-Step Solution

1

Multiply the numerator and the denominator by the conjugate of the denominator,

(3\sqrt{2} + 2\sqrt{3})

.

\frac{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}

Rationalizing eliminates the surd terms from the denominator using the difference of two squares.

2

Expand both numerator and denominator.

Denominator:

(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6

. Numerator:

(3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 + 12\sqrt{6} + 12 = 30 + 12\sqrt{6}

.

Apply algebraic identities

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

and

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

.

3

Divide the numerator by the denominator to simplify the expression.

\frac{30 + 12\sqrt{6}}{6} = 5 + 2\sqrt{6}

Both integer and radical coefficients are divisible by 6.

4

Equate

5 + 2\sqrt{6}

to

a + b\sqrt{6}

and calculate

a + b

.

a = 5, b = 2, so a + b = 5 + 2 = 7.

Matching corresponding rational and irrational components.

Key Concept

Rationalization of Binomial Denominators with Surds

Question 39Question →

In a class of $40$ students, $25$ study Mathematics and $18$ study Physics. If $3$ students study neither of the two subjects, how many students study both Mathematics and Physics?

Show answer & explanation

Answer: 6

Answer

6 students study both Mathematics and Physics.

Subtracting the

3

students who study neither subject from the class total of

40

leaves

37

students studying at least one subject. Adding those studying Mathematics (

25

) and Physics (

18

) totals

43

. The excess of

43

over

37

represents the

6

students who study both subjects.

Step-by-Step Solution

1

Subtract the number of students studying neither subject from the total number of students in the class.

n(M \cup P) = 40 - 3 = 37

This gives the number of students who belong to at least one of the two sets.

2

Set up the inclusion-exclusion formula

n(M \cup P) = n(M) + n(P) - n(M \cap P)

.

37 = 25 + 18 - n(M \cap P)

Summing

n(M)

and

n(P)

double-counts the students who study both subjects.

3

Solve for the intersection

n(M \cap P)

.

n(M \cap P) = 43 - 37 = 6

Subtracting the union from the sum of the individual sets isolates the intersection value.

Key Concept

Principle of Inclusion-Exclusion for Two Sets

Question 40Question →

If $5x \equiv 2 \pmod{11}$ , what is the value of $(x^3 - 4x) \pmod{11}$ expressed in standard non-negative remainder form?

7

-4

4

8

Show answer & explanation

Answer: 7

Answer

7

Solving the linear congruence

5x \equiv 2 \pmod{11}

gives

x \equiv 7 \pmod{11}

. Substituting

x = 7

into the expression

(x^3 - 4x)

yields

7^3 - 4(7) = 343 - 28 = 315

. Dividing 315 by 11 gives a quotient of 28 and a remainder of 7. Alternatively, working entirely modulo 11 gives

2 - 6 = -4 \equiv 7 \pmod{11}

. Thus, the canonical non-negative remainder is 7.

Step-by-Step Solution

1

Find the modular inverse of 5 modulo 11

The inverse is 9, since

5 \times 9 = 45 \equiv 1 \pmod{11}

.

To isolate

x

in the linear congruence

5x \equiv 2 \pmod{11}

, multiply both sides by the modular inverse of 5.

2

Solve for x modulo 11

x \equiv 2 \times 9 = 18 \equiv 7 \pmod{11}

.

Simplifying 18 modulo 11 gives the canonical value of

x

.

3

Evaluate

x^3 - 4x

modulo 11

x^3 \equiv 7^3 = 343 \equiv 2 \pmod{11}

and

4x \equiv 4(7) = 28 \equiv 6 \pmod{11}

, so

x^3 - 4x \equiv 2 - 6 = -4 \pmod{11}

.

Substitute

x \equiv 7

into the polynomial expression and reduce each term modulo 11.

4

Convert negative remainder to canonical non-negative form

-4 + 11 = 7 \pmod{11}

.

Modular remainders must be expressed within the standard range

[0, 10]

.

Key Concept

Linear modular congruences and negative remainder canonical reduction

Estimated Time:2m 0s

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