Number and Numeration Practice Questions — JAMB UTME — Page 3

Question 41Question →

Let the universal set be $U = \{x \in \mathbb{Z} : 1 \le x \le 10\}$ . If $P = \{x : x \text{ is a prime number within } U\}$ and $Q = \{x : x \text{ is an even number within } U\}$ , which of the following represents the set $(P \cup Q)'$ ?

\{2\}

\{3, 5, 7\}

\{1, 9\}

\{1, 3, 5, 7, 9\}

Show answer & explanation

Answer: $\{1, 9\}$

Answer

\{1, 9\}

The universal set

U

contains all integers from

1

to

10

. Prime numbers in this range form

P = \{2, 3, 5, 7\}

, and even numbers form

Q = \{2, 4, 6, 8, 10\}

. The union

P \cup Q

combines all elements from both sets:

\{2, 3, 4, 5, 6, 7, 8, 10\}

. Taking the complement of this union relative to

U

leaves the elements

\{1, 9\}

, making the option stating

\{1, 9\}

correct.

Step-by-Step Solution

1

List all elements of the universal set

U

and the subsets

P

and

Q

U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

,

P = \{2, 3, 5, 7\}

, and

Q = \{2, 4, 6, 8, 10\}

Explicit listing helps identify union and complement elements accurately.

2

Find the union of sets

P

and

Q

, denoted as

P \cup Q

P \cup Q = \{2, 3, 4, 5, 6, 7, 8, 10\}

The union includes all elements that belong to set

P

, set

Q

, or both.

3

Determine the complement

(P \cup Q)'

by subtracting

P \cup Q

from the universal set

U

(P \cup Q)' = U \setminus (P \cup Q) = \{1, 9\}

The complement contains all elements of

U

that are not present in

P \cup Q

.

Key Concept

Set Complement and Union Operations

Question 42Question →

If $x$ , $y$ , and $z$ are non-zero real numbers satisfying the exponential equation $2^x = 5^y = 100^z$ , what is the numerical value of the expression $z \left( \frac{2}{x} + \frac{2}{y} \right)$ ?

Show answer & explanation

Answer: 1

Answer

The numerical value of

z \left( \frac{2}{x} + \frac{2}{y} \right)

is

1

.

By setting

2^x = 5^y = 100^z = k

, we can write

2 = k^{1/x}

,

5 = k^{1/y}

, and

100 = k^{1/z}

. Factoring

100 = 2^2 \times 5^2

gives

k^{1/z} = (k^{1/x})^2 \times (k^{1/y})^2 = k^{2/x + 2/y}

. Equating exponents gives

\frac{1}{z} = \frac{2}{x} + \frac{2}{y}

, which upon multiplying by

z

yields

1

.

Step-by-Step Solution

1

Equate the given exponential expressions to a common constant

k

.

2^x = 5^y = 100^z = k

Introducing a common variable allows isolating each base exponent combination.

2

Express bases

2

,

5

, and

100

in terms of

k

using fractional indices.

2 = k^{\frac{1}{x}}

,

5 = k^{\frac{1}{y}}

,

100 = k^{\frac{1}{z}}

Applying the power law

(a^m)^{\frac{1}{m}} = a

isolates each base.

3

Express

100

using prime factorization of the other bases.

100 = 2^2 \times 5^2

Establishing a numerical relationship between

100

,

2

, and

5

links the exponential variables.

4

Substitute the

k

-expressions into

100 = 2^2 \times 5^2

and apply index multiplication laws.

k^{\frac{1}{z}} = \left(k^{\frac{1}{x}}\right)^2 \times \left(k^{\frac{1}{y}}\right)^2 = k^{\frac{2}{x}} \times k^{\frac{2}{y}} = k^{\frac{2}{x} + \frac{2}{y}}

Multiplying powers with the same base requires adding the exponents:

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

.

5

Equate exponents of identical bases.

\frac{1}{z} = \frac{2}{x} + \frac{2}{y}

If

k^a = k^b

for

k > 1

, then

a = b

.

6

Multiply both sides of the equation by

z

.

z \left( \frac{2}{x} + \frac{2}{y} \right) = 1

Rearranging the equation yields the exact numerical value of the requested expression.

Key Concept

Equating Exponents of Common Bases and Fractional Indices

Question 43Question →

In a survey of $150$ subscribers of a digital media platform, $75$ prefer High-Definition Audio ( $H$ ), $70$ prefer Offline Downloads ( $D$ ), and $65$ prefer Ad-free Listening ( $A$ ). It is observed that $35$ subscribers prefer both $H$ and $D$ , $30$ prefer both $D$ and $A$ , and $25$ prefer both $H$ and $A$ . If $15$ subscribers prefer none of these three features, how many subscribers prefer exactly two of these features?

Show answer & explanation

Answer: 45

Answer

The number of subscribers who prefer exactly two of the features is 45.

To find the number of subscribers who prefer exactly two features, we first calculate the cardinality of the union of all three sets as

150 - 15 = 135

. Applying the 3-set inclusion-exclusion formula gives the number of subscribers preferring all three features as

15

. Subtracting

15

from each pairwise intersection gives the exclusive regions:

20

for

H

and

D

only,

15

for

D

and

A

only, and

10

for

H

and

A

only. Summing these three exclusive regions gives

20 + 15 + 10 = 45

.

Step-by-Step Solution

1

Determine the total number of subscribers who prefer at least one feature.

|H \cup D \cup A| = 135

Subtracting the number of subscribers who prefer none of the features (

15

) from the universal set size (

150

) gives

|H \cup D \cup A| = 150 - 15 = 135

.

2

Apply the Principle of Inclusion-Exclusion for three sets to find the number of subscribers who prefer all three features.

|H \cap D \cap A| = 15

Substitute the known cardinalities into

|H \cup D \cup A| = |H| + |D| + |A| - |H \cap D| - |D \cap A| - |H \cap A| + |H \cap D \cap A|

to get

135 = 75 + 70 + 65 - (35 + 30 + 25) + |H \cap D \cap A|

, which simplifies to

135 = 120 + |H \cap D \cap A|

, giving

|H \cap D \cap A| = 15

.

3

Calculate the number of subscribers preferring exactly two features by subtracting the triple intersection from each pairwise intersection.

45 subscribers

Subscribers preferring only

H

and

D = 35 - 15 = 20

, only

D

and

A = 30 - 15 = 15

, and only

H

and

A = 25 - 15 = 10

. Summing these exclusive regions yields

20 + 15 + 10 = 45

.

Key Concept

Cardinality of set operations and 3-set inclusion-exclusion principle

Question 44Question →

What is the smallest non-negative integer $k$ that satisfies the modular congruence $7^{99} + k \equiv -15 \pmod{11}$ ?

10

1

7

8

Show answer & explanation

Answer: 10

Answer

The correct answer is 10.

Using Fermat's Little Theorem,

7^{10} \equiv 1 \pmod{11}

, which simplifies

7^{99} \pmod{11}

to

7^9 \equiv 8 \pmod{11}

. Reducing the right-hand side gives

-15 \equiv 7 \pmod{11}

. The modular equation

8 + k \equiv 7 \pmod{11}

yields

k \equiv -1 \pmod{11}

. Adding the modulus

11

gives the canonical positive remainder

10

.

Step-by-Step Solution

1

Simplify

7^{99} \pmod{11}

using Fermat's Little Theorem.

7^{10} \equiv 1 \pmod{11}

, so

7^{99} = (7^{10})^9 \times 7^9 \equiv 1^9 \times 7^9 \equiv 7^9 \pmod{11}

.

Since

11

is prime and

\gcd(7, 11) = 1

, Fermat's Little Theorem allows exponent reduction modulo

10

.

2

Compute

7^9 \pmod{11}

.

7^1 \equiv 7

,

7^2 = 49 \equiv 5

,

7^4 \equiv 5^2 = 25 \equiv 3

,

7^5 \equiv 7 \times 3 = 21 \equiv 10

,

7^9 = 7^5 \times 7^4 \equiv 10 \times 3 = 30 \equiv 8 \pmod{11}

.

Repeated squaring and modular multiplication efficiently reduces

7^9

modulo

11

.

3

Reduce the right-hand side

-15 \pmod{11}

.

-15 = -2(11) + 7 \equiv 7 \pmod{11}

.

Converting negative numbers into the standard non-negative remainder range

[0, 10]

.

4

Substitute remainders into the congruence and solve for

k

.

8 + k \equiv 7 \implies k \equiv 7 - 8 = -1 \pmod{11}

.

Linear algebraic rearrangement in modular arithmetic.

5

Convert the negative remainder

-1

to canonical non-negative form.

k = -1 + 11 = 10

.

The standard remainder must satisfy

0 \leq k < 11

.

Key Concept

Modular Exponentiation & Negative Remainder Reduction

Question 45Question →

Three business partners share a total profit of ₦60,000 in the ratio $1 : 2 : 3$ . What is the share of the partner who receives the largest portion, in naira?

Show answer & explanation

Answer: 30000

Answer

The share of the partner receiving the largest portion is ₦30,000.

Sum the parts of the ratio (

1 + 2 + 3 = 6

). The largest share corresponds to 3 parts out of 6. Calculating

\frac{3}{6} \times 60,000

yields 30,000 naira.

Step-by-Step Solution

1

Calculate the sum of all ratio parts

1 + 2 + 3 = 6 parts

The total amount is divided into equal parts represented by the sum of the ratio numbers.

2

Determine the monetary value of a single part

₦60,000 / 6 = ₦10,000 per part

Dividing the total sum by the total number of parts gives the value of one unit part.

3

Calculate the largest share corresponding to 3 parts

3 × ₦10,000 = ₦30,000

The largest portion of the ratio is 3 parts.

Key Concept

Direct Ratio Sharing

Question 46Question →

Find the smallest positive integer $n$ that simultaneously satisfies the linear modular congruences $3n \equiv 5 \pmod{13}$ and $4n \equiv 2 \pmod{9}$ .

Show answer & explanation

Answer: 32

Answer

The smallest positive integer satisfying both congruences is 32.

Solving the first congruence

3n \equiv 5 \pmod{13}

gives

n \equiv 6 \pmod{13}

(since

3 \times 9 \equiv 1 \pmod{13}

and

5 \times 9 = 45 \equiv 6 \pmod{13}

), which means

n

can be written as

13k + 6

. Substituting this into the second congruence

4n \equiv 2 \pmod{9}

yields

4(13k + 6) \equiv 2 \pmod{9} \implies 52k + 24 \equiv 2 \pmod{9}

. Reducing the coefficients modulo 9 gives

7k + 6 \equiv 2 \pmod{9} \implies 7k \equiv -4 \equiv 5 \pmod{9}

. Multiplying by 4 (the modular inverse of 7 modulo 9) yields

k \equiv 20 \equiv 2 \pmod{9}

. Setting

k = 2

yields the smallest positive integer

n = 13(2) + 6 = 32

.

Step-by-Step Solution

1

Solve the first modular congruence

3n \equiv 5 \pmod{13}

for

n

n \equiv 6 \pmod{13}

, which implies

n = 13k + 6

Multiplying both sides by the modular inverse of 3 modulo 13 (which is 9) isolates

n

.

2

Solve the second modular congruence

4n \equiv 2 \pmod{9}

for

n

n \equiv 5 \pmod{9}

Multiplying both sides by the modular inverse of 4 modulo 9 (which is 7) isolates

n

.

3

Substitute

n = 13k + 6

into

n \equiv 5 \pmod{9}

and simplify modulo 9

4k \equiv 8 \pmod{9}

Reducing 13 modulo 9 yields

4k

, and subtracting 6 from 5 yields

-1 \equiv 8 \pmod{9}

.

4

Solve for

k

and calculate the smallest positive integer

n

k \equiv 2 \pmod{9}

, giving

n = 13(2) + 6 = 32

Setting the integer parameter

k

to its minimum non-negative value

2

provides the smallest positive integer solution.

Key Concept

System of Linear Modular Congruences and Modular Inverses

Question 47Question →

What is the $10^{\text{th}}$ term of the arithmetic progression $3, 7, 11, 15, \dots$ ?

Show answer & explanation

Answer: 39

Answer

The

10^{\text{th}}

term of the arithmetic progression is

39

.

By applying the

n^{\text{th}}

term formula for an arithmetic progression

T_n = a + (n - 1)d

with first term

a = 3

, common difference

d = 4

, and term index

n = 10

, the calculation yields

T_{10} = 3 + (10 - 1) \times 4 = 3 + 36 = 39

.

Step-by-Step Solution

1

Identify the key parameters of the arithmetic progression from the given sequence.

First term

a = 3

, common difference

d = 7 - 3 = 4

, and number of terms

n = 10

.

These parameters are required to use the

n^{\text{th}}

term formula of an A.P.

2

Substitute the values into the formula

T_n = a + (n - 1)d

.

T_{10} = 3 + (10 - 1) \times 4

The formula relates the

n^{\text{th}}

term to the first term, common difference, and term position.

3

Evaluate the mathematical expression.

T_{10} = 3 + 9 \times 4 = 3 + 36 = 39

Perform multiplication before addition according to standard order of operations.

Key Concept

nth term of an Arithmetic Progression

Question 48Question →

What is the value of

x

that satisfies the exponential equation

\frac{4^{2x + 1} \times 8^{x - 1}}{16^x} = 32

?

2

1

\frac{5}{3}

11

Show answer & explanation

Answer: $2$

Answer

The value of

x

is

2

.

Converting all terms to base

2

gives

\frac{2^{4x+2} \times 2^{3x-3}}{2^{4x}} = 2^5

. Combining the powers on the left side yields

2^{3x-1} = 2^5

. Equating exponents gives

3x - 1 = 5

, which solves to

x = 2

.

Step-by-Step Solution

1

Express all base numbers in terms of a common prime base (base 2).

4 = 2^2

,

8 = 2^3

,

16 = 2^4

, and

32 = 2^5

.

Converting all terms to powers of 2 enables the application of index laws.

2

Substitute these prime base powers into the original equation and expand exponents.

\frac{(2^2)^{2x + 1} \times (2^3)^{x - 1}}{(2^4)^x} = 2^5 \implies \frac{2^{4x + 2} \times 2^{3x - 3}}{2^{4x}} = 2^5

Applying

(a^m)^n = a^{m \cdot n}

requires multiplying the outer exponent by every term in the inner exponent.

3

Apply index laws for multiplication (

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

) and division (

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

) on the left-hand side.

2^{(4x + 2) + (3x - 3) - 4x} = 2^5 \implies 2^{3x - 1} = 2^5

Powers with the same base are combined by adding numerator exponents and subtracting denominator exponents.

4

Equate the exponents since the bases are identical, and solve for

x

.

3x - 1 = 5 \implies 3x = 6 \implies x = 2

If

a^x = a^y

for

a > 0

and

a \neq 1

, then

x = y

.

Key Concept

Solving Exponential Equations using Laws of Indices

Estimated Time:1m 30s

Question 49Question →

Find the real value of

x

that satisfies the exponential equation

5^{2x - 1} \times 25^{x + 1} = 125^{x + 2}

Show answer & explanation

Answer: 5

Answer

The value of

x

is 5.

By converting all terms to base 5 (

25 = 5^2

and

125 = 5^3

), the equation becomes

5^{2x-1} \times 5^{2(x+1)} = 5^{3(x+2)}

. Simplifying exponents gives

5^{2x-1} \times 5^{2x+2} = 5^{3x+6}

. Adding the left-hand powers results in

5^{4x+1} = 5^{3x+6}

. Equating the exponents yields

4x + 1 = 3x + 6

, leading directly to

x = 5

.

Step-by-Step Solution

1

Express all terms with a common base of 5

5^{2x-1} \times 5^{2x+2} = 5^{3x+6}

Since

25 = 5^2

and

125 = 5^3

, using index laws

(a^m)^n = a^{mn}

allows all expressions to share base 5.

2

Apply the product rule of indices on the left side

5^{4x+1} = 5^{3x+6}

According to the product law

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

, the powers are added:

(2x-1) + (2x+2) = 4x+1

.

3

Equate powers of equal bases to solve for x

x = 5

Since bases are equal, exponents must be equal:

4x + 1 = 3x + 6 \implies 4x - 3x = 6 - 1 \implies x = 5

.

Key Concept

Indices and Laws of Indices

Question 50Question →

Pump A can fill a water reservoir in $6$ hours, while a drain pipe can empty the full reservoir in $15$ hours. Pump A is switched on to fill an empty reservoir while the drain pipe is accidentally left open. After $3$ hours, an identical pump, Pump B, is also switched on to assist Pump A while the drain pipe remains open. How many total hours will it take for the reservoir to become completely full?

Show answer & explanation

Answer: 5.625

Answer

The total time required to fill the reservoir completely is

5.625

hours.

To solve multi-stage work and rate problems involving opposing forces (filling vs. draining), calculate the net rate of change per unit of time for each stage. In stage one, Pump A adds

\frac{1}{6}

while the drain removes

\frac{1}{15}

, giving a net rate of

\frac{1}{10}

per hour. In 3 hours,

\frac{3}{10}

of the reservoir is filled, leaving

\frac{7}{10}

. In stage two, adding identical Pump B increases the filling rate to

2 \times \frac{1}{6} = \frac{1}{3}

. Subtracting the drain rate

\frac{1}{15}

yields a net rate of

\frac{4}{15}

per hour. Dividing the remaining

\frac{7}{10}

by

\frac{4}{15}

gives

2.625

hours. Adding the initial 3 hours yields a total of

5.625

hours.

Step-by-Step Solution

1

Determine individual hourly rates

Pump A rate =

+\frac{1}{6}

reservoir/hr, Drain rate =

-\frac{1}{15}

reservoir/hr

Rate is the reciprocal of the time required to complete the full job.

2

Calculate net rate and progress for the first 3 hours

Net rate =

\frac{1}{6} - \frac{1}{15} = \frac{1}{10}

reservoir/hr. Progress in 3 hours =

3 \times \frac{1}{10} = \frac{3}{10}

of the reservoir.

Only Pump A and the drain pipe are active during the initial 3-hour period.

3

Calculate remaining fraction of reservoir to be filled

Remaining portion =

1 - \frac{3}{10} = \frac{7}{10}

The total capacity of the reservoir is represented by

1

whole unit.

4

Calculate the combined rate after Pump B is added

New net rate =

\frac{1}{6} + \frac{1}{6} - \frac{1}{15} = \frac{1}{3} - \frac{1}{15} = \frac{4}{15}

reservoir/hr

Pump B is identical to Pump A, so its rate is also

\frac{1}{6}

reservoir/hr.

5

Find additional time needed and total time elapsed

Additional time =

\frac{7/10}{4/15} = \frac{7}{10} \times \frac{15}{4} = \frac{21}{8} = 2.625

hours. Total time =

3 + 2.625 = 5.625

hours.

Time equals remaining work divided by combined net rate, then added to elapsed time.

Key Concept

Work-Rate and Simultaneous Operations (Combined Filling and Emptying Rates)

Estimated Time:2m 30s

Question 51Question →

The first term of a geometric progression (GP) is $2$ and its common ratio is $3$ . What is the $4^{\text{th}}$ term of the progression?

54

162

11

24

Show answer & explanation

Answer: $54$

Answer

The

4^{\text{th}}

term of the geometric progression is

54

.

For a geometric progression with first term

a

and common ratio

r

, the

n^{\text{th}}

term is given by

T_n = a r^{n-1}

. Substituting

a = 2

,

r = 3

, and

n = 4

yields

T_4 = 2 \times 3^3 = 2 \times 27 = 54

.

Step-by-Step Solution

1

Identify the given parameters of the geometric progression.

First term

a = 2

, common ratio

r = 3

, and term position

n = 4

.

These values are directly provided in the question statement.

2

Apply the general formula for the

n^{\text{th}}

term of a geometric progression,

T_n = a r^{n-1}

.

T_4 = 2 \times 3^{4-1} = 2 \times 3^3

.

The exponent of the common ratio is always one less than the term index

n

.

3

Evaluate the exponent and multiply by the first term.

3^3 = 27

, so

T_4 = 2 \times 27 = 54

.

Performing standard arithmetic yields the exact term value.

Key Concept

Formula for the nth term of a Geometric Progression:

T_n = a r^{n-1}

Estimated Time:45s

Question 52Question →

A company purchased a processing machine for $\text{₦}200,000$ . The value of the machine depreciates at a compound rate of $10\%$ per annum. At the end of $2$ years, the machine was sold at a profit of $15\%$ based on its depreciated value. What was the selling price of the machine?

\text{₦}184,000

\text{₦}186,300

\text{₦}192,000

\text{₦}207,000

Show answer & explanation

Answer: $\text{₦}186,300$

Answer

The selling price of the machine was

\text{₦}186,300

.

The depreciated value after 2 years at a compound rate of

10\%

per annum is calculated as

V = 200,000 \times (0.90)^2 = \text{₦}162,000

. Selling the machine at a

15\%

profit on this depreciated value gives a selling price of

162,000 \times 1.15 = \text{₦}186,300

.

Step-by-Step Solution

1

Calculate the depreciated value of the machine after 2 years using the compound depreciation formula

V = P(1 - r)^n

.

V = 200,000 \times (1 - 0.10)^2 = 200,000 \times (0.90)^2 = 200,000 \times 0.81 = \text{₦}162,000

.

Compound depreciation reduces the asset's remaining book value by

10\%

each year.

2

Calculate the

15\%

profit based on the depreciated value.

\text{Profit} = 15\% \text{ of } \text{₦}162,000 = 0.15 \times 162,000 = \text{₦}24,300$.

The problem specifies that profit is made on the depreciated value.

3

Calculate the selling price by adding the profit to the depreciated value.

\text{Selling Price} = \text{₦}162,000 + \text{₦}24,300 = \text{₦}186,300$.

Selling price is equal to book value plus profit earned.

Key Concept

Compound Depreciation and Percentage Profit on Book Value

Estimated Time:2m 0s

Question 53Question →

What is the determinant of the $2 \times 2$ matrix $P = \begin{pmatrix} 5 & 2 \\ 3 & 4 \end{pmatrix}$ ?

Show answer & explanation

Answer: 14

Answer

The determinant of matrix

P

is

14

.

For any

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, the determinant is given by

ad - bc

. Substituting

a=5

,

b=2

,

c=3

, and

d=4

yields

(5 \times 4) - (2 \times 3) = 20 - 6 = 14

.

Step-by-Step Solution

1

Apply the determinant formula for a

2 \times 2

matrix

\begin{pmatrix} a & b \\ c & d \end{pmatrix}

, which is

\det(P) = ad - bc

.

\det(P) = (5)(4) - (2)(3)

The determinant of a

2 \times 2

matrix is defined as the product of the main diagonal elements minus the product of the off-diagonal elements.

2

Evaluate the arithmetic expression.

20 - 6 = 14

Perform multiplication followed by subtraction to get the final determinant value.

Key Concept

Determinant of a 2x2 Matrix

Estimated Time:45s

Question 54Question →

The $3^{\text{rd}}$ , $6^{\text{th}}$ , and $11^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the first term of the AP is $15$ , what is the $4^{\text{th}}$ term of the geometric progression?

Show answer & explanation

Answer: 125

Answer

The 4th term of the geometric progression is 125.

By writing the 3rd, 6th, and 11th terms of the AP as

15+2d

,

15+5d

, and

15+10d

, we utilize the geometric mean property

(15+5d)^2 = (15+2d)(15+10d)

to find

d=6

. This yields the GP terms

27, 45, 75

, giving a common ratio of

5/3

. Multiplying the third term

75

by

5/3

gives the 4th GP term as

125

.

Step-by-Step Solution

1

Write down the AP term expressions

T_3 = 15 + 2d

,

T_6 = 15 + 5d

,

T_{11} = 15 + 10d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

with initial term

a = 15

.

2

Apply the geometric progression condition

(15 + 5d)^2 = (15 + 2d)(15 + 10d)

If three terms

A, B, C

are in GP, then

B^2 = A \cdot C

.

3

Expand and solve the quadratic equation for the common difference

d

d = 6

Expanding gives

225 + 150d + 25d^2 = 225 + 180d + 20d^2 \implies 5d^2 = 30d \implies d = 6

because

d \neq 0

.

4

Find the terms and common ratio of the GP

G_1 = 27

,

G_2 = 45

,

G_3 = 75

, and common ratio

r = \frac{5}{3}

Substituting

d = 6

gives the GP terms, and dividing consecutive terms gives

r = \frac{45}{27} = \frac{5}{3}

.

5

Calculate the 4th term of the GP

G_4 = 125

Multiplying the 3rd term by the common ratio yields

75 \times \frac{5}{3} = 125

.

Key Concept

Combining Arithmetic Progression nth-term formulas with Geometric Progression consecutive-term properties

Estimated Time:2m 30s

Question 55Question →

The sum of the first $n$ terms of an arithmetic progression (AP) is given by $S_n = 2n^2 + 3n$ . The $3^{\text{rd}}$ term of this AP is equal to the $2^{\text{nd}}$ term of a geometric progression (GP). If the common ratio of the GP is $2$ , what is the sum of the first $4$ terms of the GP?

97.5

127.5

48.75

45.5

Show answer & explanation

Answer: $97.5$

Answer

97.5

Evaluating

S_3 - S_2

gives the 3rd AP term as

27 - 14 = 13

. Setting the 2nd GP term

a(2) = 13

yields

a = 6.5

. The sum of the first 4 terms of the GP is

6.5 \times (2^4 - 1) = 6.5 \times 15 = 97.5

.

Step-by-Step Solution

1

Find the 3rd term (

T_3

) of the AP using the given sum formula

S_n = 2n^2 + 3n

T_3 = S_3 - S_2 = [2(3)^2 + 3(3)] - [2(2)^2 + 3(2)] = [18 + 9] - [8 + 6] = 27 - 14 = 13

The

n

-th term of a sequence is equal to

S_n - S_{n-1}

.

2

Determine the first term (

a

) of the GP

Since

G_2 = 13

and common ratio

r = 2

,

a \cdot r^{2-1} = 13 \implies 2a = 13 \implies a = 6.5

The

n

-th term of a GP is given by

G_n = a r^{n-1}

.

3

Calculate the sum of the first 4 terms of the GP

S_4 = \frac{a(r^4 - 1)}{r - 1} = \frac{6.5(2^4 - 1)}{2 - 1} = 6.5 \times 15 = 97.5

The sum of the first

n

terms of a GP with

r > 1

is

S_n = \frac{a(r^n - 1)}{r - 1}

.

Key Concept

Combining Arithmetic Progression sum formula with Geometric Progression term and sum formulas

Estimated Time:2m 0s

Question 56Question →

What is the value of the logarithmic expression $\frac{1}{\log_3 12} + \frac{1}{\log_4 12}$ ?

1

\log_{12} 7

\frac{7}{12}

12

Show answer & explanation

Answer: $1$

Answer

1

Using the change of base relationship

\frac{1}{\log_a b} = \log_b a

, the expression simplifies to

\log_{12} 3 + \log_{12} 4

. By the product rule of logarithms, this equals

\log_{12}(3 \times 4) = \log_{12} 12 = 1

.

Step-by-Step Solution

1

Apply the change of base rule

\frac{1}{\log_a b} = \log_b a

to each term.

\frac{1}{\log_3 12} = \log_{12} 3 \quad \text{and} \quad \frac{1}{\log_4 12} = \log_{12} 4

Expressing both terms with a common base of

12

enables the use of logarithmic laws.

2

Apply the product law of logarithms

\log_b M + \log_b N = \log_b (M \times N)

.

\log_{12} 3 + \log_{12} 4 = \log_{12} (3 \times 4) = \log_{12} 12

The sum of logarithms with identical bases equals the logarithm of the product of their arguments.

3

Simplify

\log_{12} 12

.

1

The logarithm of any base to itself is always

1

(

\log_a a = 1

).

Key Concept

Change of Base Property and Logarithm Addition Law

Question 57Question →

What is the numerical value of the simplified expression $\frac{4}{\sqrt{5} - 1} - \sqrt{5}$ ?

Show answer & explanation

Answer: 1

Answer

The numerical value of the expression is 1.

Multiplying the top and bottom of

\frac{4}{\sqrt{5} - 1}

by its conjugate

(\sqrt{5} + 1)

simplifies the fraction to

\frac{4(\sqrt{5} + 1)}{4} = \sqrt{5} + 1

. Subtracting

\sqrt{5}

from

\sqrt{5} + 1

leaves 1.

Step-by-Step Solution

1

Rationalize the denominator of the fractional term

\sqrt{5} + 1

Multiply both numerator and denominator by the conjugate

(\sqrt{5} + 1)

to apply the difference of two squares identity

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

in the denominator.

2

Subtract the remaining surd term

1

Subtract

\sqrt{5}

from

\sqrt{5} + 1

, leaving the integer 1.

Key Concept

Rationalization of Binomial Denominators

Question 58Question →

If the expression $\frac{3 - \sqrt{5}}{3 + \sqrt{5}}$ is simplified and written in the form $a + b\sqrt{5}$ , where $a$ and $b$ are rational numbers, what is the numerical value of $a + b$ ?

Show answer & explanation

Answer: 2

Answer

The numerical value of

a + b

is

2

.

To express

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

in the standard form

a + b\sqrt{5}

, multiply both the numerator and denominator by the conjugate of the denominator, which is

(3 - \sqrt{5})

. The numerator expands to

(3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}

. The denominator becomes

3^2 - (\sqrt{5})^2 = 9 - 5 = 4

. Dividing gives

\frac{14}{4} - \frac{6}{4}\sqrt{5} = \frac{7}{2} - \frac{3}{2}\sqrt{5}

. Hence,

a = \frac{7}{2}

and

b = -\frac{3}{2}

, making

a + b = \frac{7}{2} - \frac{3}{2} = \frac{4}{2} = 2

.

Step-by-Step Solution

1

Multiply numerator and denominator by the conjugate of the denominator

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

To eliminate the surd from the denominator.

2

Expand both the numerator and the denominator

\frac{14 - 6\sqrt{5}}{4}

Using

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

for the numerator and difference of two squares

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

for the denominator.

3

Separate into rational component and radical coefficient

\frac{7}{2} - \frac{3}{2}\sqrt{5}

Simplifying fractions by dividing numerator and denominator by their greatest common divisor.

4

Calculate the sum

a + b

\frac{7}{2} - \frac{3}{2} = 2

Comparing

\frac{7}{2} - \frac{3}{2}\sqrt{5}

with

a + b\sqrt{5}

yields

a = \frac{7}{2}

and

b = -\frac{3}{2}

.

Key Concept

Rationalization of Binomial Denominators using Conjugates

Estimated Time:1m 30s

Question 59Question →

Given the universal set $U = \{x \in \mathbb{Z} : 1 \le x \le 15\}$ , let $A = \{x \in U : x \text{ is a multiple of } 3\}$ and $B = \{x \in U : x \text{ is an even number}\}$ . What is the cardinality of $(A \cup B)'$ ?

5

10

3

8

Show answer & explanation

Answer: $5$

Answer

The cardinality of

(A \cup B)'

is

5

.

The universal set contains

15

elements. The set of multiples of

3

within

U

contains

5

elements, and the set of even numbers contains

7

elements. Two numbers (

6

and

12

) belong to both sets. Subtracting the overlapping count gives

10

unique elements in the union. Subtracting

10

from the universal set total of

15

yields

5

elements in the complement.

Step-by-Step Solution

1

List elements of sets

U

,

A

, and

B

, and determine their cardinalities

U = \{1, 2, 3, \dots, 15\}

, so

n(U) = 15

.

A = \{3, 6, 9, 12, 15\}

(

n(A) = 5

).

B = \{2, 4, 6, 8, 10, 12, 14\}

(

n(B) = 7

).

Establishing explicit set memberships allows accurate counting of set operations.

2

Find the intersection

A \cap B

and compute the union cardinality

n(A \cup B)

A \cap B = \{6, 12\}

, so

n(A \cap B) = 2

. Using inclusion-exclusion:

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 5 + 7 - 2 = 10

.

The union includes all elements that are either multiples of 3, even, or both, avoiding double counting.

3

Calculate the complement cardinality

n((A \cup B)')

relative to

U

n((A \cup B)') = n(U) - n(A \cup B) = 15 - 10 = 5

. Explicitly,

(A \cup B)' = \{1, 5, 7, 11, 13\}

.

The complement set

(A \cup B)'

consists of all elements in the universal set

U

that are not in

A \cup B

.

Key Concept

Complement of Set Union and Inclusion-Exclusion Principle

Estimated Time:1m 0s

Question 60Question →

In a sports academy of $85$ athletes, $52$ participate in track events, $43$ participate in field events, and $12$ participate in neither track nor field events. How many athletes participate in both track and field events?

Show answer & explanation

Answer: 22

Answer

The number of athletes participating in both track and field events is 22.

To find the number of athletes in both events, first calculate the total number of athletes who take part in at least one event by subtracting the 12 non-participants from 85, giving 73. Adding the 52 track athletes and 43 field athletes yields 95, which double-counts those who participate in both. The difference between 95 and 73 is 22, representing the athletes in the intersection.

Step-by-Step Solution

1

Determine the cardinality of the union of track and field athletes

N(T \cup F) = 85 - 12 = 73

Athletes participating in neither event are outside the union of track and field sets.

2

Formulate the two-set inclusion-exclusion equation

N(T \cup F) = N(T) + N(F) - N(T \cap F)

Adding individual set cardinalities double-counts the intersection.

3

Substitute values and solve for the intersection

N(T \cap F) = 52 + 43 - 73 = 22

Rearranging the equation yields the number of athletes in both sets.

Key Concept

Cardinality of Sets and Principle of Inclusion-Exclusion

Number and Numeration

Answer

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