Number and Numeration Practice Questions — JAMB UTME — Page 11

Question 201Question →

The first, third, and seventh 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 $4$ , what is the sum of the first $4$ terms of the geometric progression?

60

28

124

92

Show answer & explanation

Answer: 60

Answer

The sum of the first

4

terms of the geometric progression is

60

.

The first three terms of the GP are

T_1 = 4

,

T_3 = 4 + 2d

, and

T_7 = 4 + 6d

. Equating

(4 + 2d)^2 = 4(4 + 6d)

yields

4d^2 - 8d = 0

, giving

d = 2

. The first four terms of the GP are

4, 8, 16, 32

, which sum to

4 + 8 + 16 + 32 = 60

.

Step-by-Step Solution

1

Express the terms of the arithmetic progression in terms of first term

a

and common difference

d

.

First term

T_1 = 4

, third term

T_3 = 4 + 2d

, seventh term

T_7 = 4 + 6d

.

The

n^{\text{th}}

term of an AP is defined as

T_n = a + (n-1)d

.

2

Set up the geometric progression condition

(T_3)^2 = T_1 \cdot T_7

to solve for

d

.

(4 + 2d)^2 = 4(4 + 6d) \implies 16 + 16d + 4d^2 = 16 + 24d \implies 4d^2 - 8d = 0 \implies d = 2

(since the AP is non-constant,

d \neq 0

).

Three terms

x, y, z

form a GP if and only if

y^2 = xz

.

3

Determine the terms and common ratio

r

of the GP.

First term

G_1 = 4

, second term

G_2 = 4 + 2(2) = 8

. Thus,

r = \frac{8}{4} = 2

.

The common ratio

r

is the quotient of consecutive terms of the GP.

4

Calculate the sum of the first

4

terms of the GP using

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

.

S_4 = \frac{4(2^4 - 1)}{2 - 1} = \frac{4(16 - 1)}{1} = 60

.

Formula for the sum of the first

n

terms of a geometric progression.

Key Concept

Arithmetic and Geometric Progression Inter-relationships

Question 202Question →

The $2^{\text{nd}}$ , $5^{\text{th}}$ , and $14^{\text{th}}$ terms of a non-constant arithmetic progression (AP) form the first three consecutive terms of a geometric progression (GP). If the sum of the first $6$ terms of the AP is $72$ , calculate the common difference of the AP.

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

Equating the square of the middle GP term

(a+4d)^2

to the product of the outer terms

(a+d)(a+13d)

yields

3d^2 = 6ad

, which simplifies to

d = 2a

. Substituting

2a = d

into the AP sum formula

S_6 = 3(2a+5d) = 72

gives

3(6d) = 72 \implies 18d = 72

, so

d = 4

.

Step-by-Step Solution

1

Express the

2^{\text{nd}}

,

5^{\text{th}}

, and

14^{\text{th}}

terms of the AP algebraically

T_2 = a + d

,

T_5 = a + 4d

,

T_{14} = a + 13d

The

n^{\text{th}}

term of an AP is given by

T_n = a + (n-1)d

.

2

Apply the consecutive terms property of a GP

(a + 4d)^2 = (a + d)(a + 13d)

For three consecutive terms of a GP, the square of the middle term equals the product of the first and third terms.

3

Simplify the quadratic equation to find the relationship between

a

and

d

a^2 + 8ad + 16d^2 = a^2 + 14ad + 13d^2 \implies 3d^2 = 6ad \implies d = 2a

Since the AP is non-constant,

d \neq 0

, allowing division by

3d

.

4

Formulate the sum of the first

6

terms of the AP

S_6 = 3(2a + 5d) = 72 \implies 2a + 5d = 24

The sum of the first

n

terms of an AP is

S_n = \frac{n}{2}[2a + (n-1)d]

.

5

Substitute

2a = d

into the linear equation and solve for

d

d + 5d = 24 \implies 6d = 24 \implies d = 4

Replacing

2a

with

d

reduces the equation to a single variable.

Key Concept

Relating non-consecutive terms of an Arithmetic Progression to form a Geometric Progression

Question 203Question →

Which of the following is the inverse of the matrix $M = \begin{pmatrix} 5 & 2 \\ 7 & 3 \end{pmatrix}$ ?

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

\begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix}

\begin{pmatrix} -5 & 2 \\ 7 & -3 \end{pmatrix}

\begin{pmatrix} 5 & 7 \\ 2 & 3 \end{pmatrix}

Show answer & explanation

Answer: $\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}$

Answer

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

For any non-singular 2x2 matrix

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

, the inverse is

\frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}

. Here,

ad - bc = (5)(3) - (2)(7) = 1

, so the inverse matrix is

\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

.

Step-by-Step Solution

1

Calculate the determinant of matrix M

\det(M) = (5)(3) - (2)(7) = 15 - 14 = 1

For a 2x2 matrix

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

, the determinant is given by

\det(M) = ad - bc

.

2

Find the adjugate matrix of M

\text{adj}(M) = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

The adjugate of a 2x2 matrix is obtained by swapping the main diagonal elements (

a

and

d

) and negating the off-diagonal elements (

b

and

c

).

3

Apply the matrix inverse formula

M^{-1} = \frac{1}{\det(M)} \text{adj}(M) = \frac{1}{1}\begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ -7 & 5 \end{pmatrix}

Multiplying the adjugate matrix by the reciprocal of the determinant gives the exact inverse matrix.

Key Concept

Inverse of a 2x2 Matrix

Question 204Question →

A trader deposited $\text{₦}40,000$ in a microfinance bank offering compound interest at the rate of $10\%$ per annum, compounded annually. What is the total interest earned by the trader at the end of $2$ years?

\text{₦}8,000

\text{₦}8,400

\text{₦}40,400

\text{₦}48,400

Show answer & explanation

Answer: $\text{₦}8,400$

Answer

The total interest earned by the trader at the end of

2

years is

\text{₦}8,400

.

Using the compound interest formula

A = P(1 + r)^n

, the total accumulated amount after 2 years is

40,000 \times (1.1)^2 = \text{₦}48,400

. Subtracting the original principal of

\text{₦}40,000

gives the compound interest earned:

48,400 - 40,000 = \text{₦}8,400

.

Step-by-Step Solution

1

Calculate the total accumulated amount (

A

) using the compound interest formula

A = P\left(1 + \frac{R}{100}\right)^n

.

A = 40,000\left(1 + \frac{10}{100}\right)^2 = 40,000 \times (1.1)^2 = 40,000 \times 1.21 = \text{₦}48,400

.

The compound interest formula yields the total value of the investment after 2 years.

2

Subtract the initial principal (

P

) from the total accumulated amount (

A

) to find the compound interest (

I

).

I = A - P = 48,400 - 40,000 = \text{₦}8,400

.

Interest earned is the difference between total final amount and initial principal.

Key Concept

Calculation of compound interest vs total accumulated amount

Question 205Question →

Arrange the following values related to arithmetic and geometric progressions in ascending order (from smallest to largest):

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from smallest to largest value is: the common ratio of the GP (4), the common difference of the AP (8), the 4th term of the AP (11), and the sum of the first 3 terms of the GP (13).

Evaluating each progression property gives numerical values of 4, 8, 11, and 13 respectively. Ordering these from least to greatest results in the order: common ratio of the GP (4), common difference of the AP (8), 4th term of the AP (11), and sum of the first 3 terms of the GP (13).

Step-by-Step Solution

1

Calculate the value for the first item (common ratio

r

)

Using

T_4 = T_2 \cdot r^2

, we have

80 = 5 r^2 \implies r^2 = 16 \implies r = 4

.

The terms of a GP follow

T_n = a r^{n-1}

.

2

Calculate the value for the second item (common difference

d

)

Using

T_5 = a + 4d

, we get

35 = 3 + 4d \implies 4d = 32 \implies d = 8

.

The nth term of an AP is given by

T_n = a + (n-1)d

.

3

Calculate the value for the third item (

4^{\text{th}}

term of AP)

T_4 = 2 + (4 - 1)(3) = 2 + 9 = 11

.

Direct application of the AP nth term formula.

4

Calculate the value for the fourth item (Sum of first

3

terms of GP)

S_3 = 1 + 3 + 9 = 13

(or using

S_n = \frac{a(r^n - 1)}{r - 1} = \frac{1(3^3 - 1)}{3 - 1} = 13

).

Sum of a finite geometric sequence.

5

Compare and arrange the calculated numerical values in ascending order

4 < 8 < 11 < 13

.

Ordering the numbers establishes the correct item sequence.

Key Concept

Evaluation of terms, differences, ratios, and sums in Arithmetic and Geometric Progressions.

Question 206Question →

What is the result of the subtraction $423_6 - 145_6$ expressed in base 6?

234_6

232_6

344_6

322_6

Show answer & explanation

Answer: $234_6$

Answer

234_6

Subtracting

145_6

from

423_6

column by column requires borrowing from higher place values. In the units column, borrowing 1 (worth 6) gives

3 + 6 - 5 = 4

. In the sixes column, borrowing 1 leaves 1 six, and

1 + 6 - 4 = 3

. In the thirty-sixes column,

3 - 1 = 2

. Thus, the correct result is

234_6

.

Step-by-Step Solution

1

Convert both base 6 numbers into base 10 (decimal).

423_6 = 4 \times 6^2 + 2 \times 6^1 + 3 \times 6^0 = 144 + 12 + 3 = 159_{10}

, and

145_6 = 1 \times 6^2 + 4 \times 6^1 + 5 \times 6^0 = 36 + 24 + 5 = 65_{10}

.

Converting both numbers to decimal provides a familiar base for subtraction.

2

Subtract the decimal numbers.

159_{10} - 65_{10} = 94_{10}

.

Perform standard decimal subtraction.

3

Convert the decimal result

94_{10}

back to base 6.

94 \div 6 = 15

R

4

;

15 \div 6 = 2

R

3

;

2 \div 6 = 0

R

2

. Arranging the remainders from bottom to top gives

234_6

.

Successive division by 6 extracts the base 6 digits.

Key Concept

Non-decimal base subtraction and conversions

Question 207Question →

What is the value of $(-43 \times 15) \pmod{8}$ expressed in standard non-negative remainder form?

3

-5

5

-3

Show answer & explanation

Answer: $3$

Answer

3

Reducing

-43

modulo

8

gives

5

, and

15

modulo

8

gives

7

. Multiplying these results yields

35 \equiv 3 \pmod{8}

, which is the canonical non-negative remainder.

Step-by-Step Solution

1

Reduce individual factors modulo 8

-43 \equiv 5 \pmod{8}

(since

-43 = -6 \times 8 + 5

) and

15 \equiv 7 \pmod{8}

(since

15 = 1 \times 8 + 7

)

Simplifying terms before multiplication makes arithmetic easier.

2

Multiply the reduced remainders

5 \times 7 = 35

Modular arithmetic preserves multiplication.

3

Reduce the product modulo 8 to obtain the canonical non-negative remainder

35 = 4 \times 8 + 3 \implies 35 \equiv 3 \pmod{8}

The final answer in modular arithmetic must fall within the range

[0, n-1]

.

Key Concept

Modular Arithmetic and Canonical Non-Negative Remainders

Question 208Question →

An investor deposited $\text{₦}60,000$ into a financial scheme for $2$ years at an annual interest rate of $r\%$ . If the difference between the compound interest (compounded annually) and the simple interest earned over the $2$ -year period is $\text{₦}384$ , calculate the value of $r$ .

Show answer & explanation

Answer: 8

Answer

The interest rate r is 8%

For a two-year investment period, the difference between compound interest (compounded annually) and simple interest equals the interest earned in the second year on the first year's interest, which is

P \left(\frac{r}{100}\right)^2

. Setting

60,000 \times \frac{r^2}{10,000} = 384

yields

6r^2 = 384

, giving

r^2 = 64

and

r = 8\%

.

Step-by-Step Solution

1

Express the simple interest for 2 years in terms of r

I_{\text{SI}} = 1,200r

Simple interest is calculated directly on the principal amount for the full term.

2

Express the compound interest for 2 years in terms of r

I_{\text{CI}} = 1,200r + 6r^2

Compound interest includes interest earned on the first year's interest.

3

Set up the equation for the difference between compound and simple interest

(1,200r + 6r^2) - 1,200r = 384 \implies 6r^2 = 384

The difference between CI and SI over 2 years isolates the interest-on-interest component.

4

Solve for r

r = 8

Dividing 384 by 6 gives 64, and taking the principal square root yields 8.

Key Concept

Difference between Compound Interest and Simple Interest for 2 years

Question 209Question →

In a survey of $90$ high school students regarding their participation in sports clubs, $40$ play Badminton ( $B$ ), $35$ play Volleyball ( $V$ ), and $42$ engage in Swimming ( $S$ ). It was found that $14$ play both Badminton and Volleyball, $12$ play both Volleyball and Swimming, and $15$ play both Badminton and Swimming. If $8$ students participate in none of these three sports, how many students participate in all three sports?

6

14

10

18

Show answer & explanation

Answer: $6$

Answer

The number of students participating in all three sports is

6

.

First, determine the number of students in the union of the three sports sets by subtracting the

8

non-participating students from the universal set of

90

, giving

82

. Then apply the Principle of Inclusion-Exclusion:

82 = 40 + 35 + 42 - (14 + 12 + 15) + x

, where

x

is the number of students participating in all three sports. Simplifying gives

82 = 76 + x

, which yields

x = 6

.

Step-by-Step Solution

1

Calculate the cardinality of the union of the three sets.

n(B \cup V \cup S) = n(U) - n((B \cup V \cup S)') = 90 - 8 = 82

Subtracting the students who do not participate in any sport from the total universal set gives the number of students participating in at least one sport.

2

Set up the Principle of Inclusion-Exclusion formula for three sets.

n(B \cup V \cup S) = n(B) + n(V) + n(S) - [n(B \cap V) + n(V \cap S) + n(B \cap S)] + n(B \cap V \cap S)

This fundamental relation accounts for overlapping subsets in a three-set system.

3

Substitute the known numerical values into the formula and solve for

n(B \cap V \cap S)

.

82 = 40 + 35 + 42 - (14 + 12 + 15) + n(B \cap V \cap S) \implies 82 = 117 - 41 + n(B \cap V \cap S) \implies 82 = 76 + n(B \cap V \cap S) \implies n(B \cap V \cap S) = 82 - 76 = 6

Simplifying the arithmetic expression directly isolates the unknown intersection value.

Key Concept

Principle of Inclusion-Exclusion for Three Sets

Estimated Time:1m 15s

Question 210Question →

A water storage tank is initially $\frac{1}{3}$ full. When $28\text{ litres}$ of water are added to the tank, it becomes $\frac{4}{5}$ full. What is the total capacity of the water tank?

35\text{ litres}

60\text{ litres}

84\text{ litres}

140\text{ litres}

Show answer & explanation

Answer: $60\text{ litres}$

Answer

The total capacity of the water tank is

60\text{ litres}

.

Subtracting the initial filled fraction (one-third) from the final filled fraction (four-fifths) yields seven-fifteenths. Since seven-fifteenths of the total capacity corresponds to 28 litres, dividing 28 by seven-fifteenths gives a total capacity of 60 litres.

Step-by-Step Solution

1

Find the fraction of the tank filled by adding the

28\text{ litres}

of water.

Fraction added =

\frac{4}{5} - \frac{1}{3} = \frac{12 - 5}{15} = \frac{7}{15}

.

Subtracting the initial fraction filled from the final fraction filled gives the fractional change.

2

Set up an equation equating the fractional change to the actual volume added.

\frac{7}{15} \times C = 28

, where

C

represents the total capacity of the tank.

The fractional change multiplied by total capacity equals the physical volume of water introduced.

3

Solve for the total capacity

C

.

C = 28 \times \frac{15}{7} = 4 \times 15 = 60\text{ litres}

.

Multiplying both sides by the reciprocal of

\frac{7}{15}

yields the total capacity.

Key Concept

Fractional Word Problems and Quantity Determination

Estimated Time:1m 30s

Question 211Question →

What is the simplified form of the expression $\frac{3\sqrt{2} + 2\sqrt{3}}{\sqrt{3} - \sqrt{2}}$ ?

12 + 5\sqrt{6}

5\sqrt{15} + 5\sqrt{10}

12 - 5\sqrt{6}

\sqrt{6}

Show answer & explanation

Answer: $12 + 5\sqrt{6}$

Answer

The simplified form is

12 + 5\sqrt{6}

.

Multiplying both numerator and denominator by the conjugate of the denominator,

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

, clears the square roots in the denominator (resulting in

3 - 2 = 1

). Expanding the numerator gives

3\sqrt{6} + 6 + 6 + 2\sqrt{6}

, which simplifies cleanly to

12 + 5\sqrt{6}

.

Step-by-Step Solution

1

Identify the conjugate of the denominator

The conjugate of

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

is

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

.

To rationalize a binomial denominator of the form

(\sqrt{a} - \sqrt{b})

, multiply numerator and denominator by

(\sqrt{a} + \sqrt{b})

.

2

Multiply the numerator and denominator by the conjugate

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

This removes the radical terms from the denominator using the difference of squares identity.

3

Expand the numerator and simplify the denominator

Denominator:

(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1

.
Numerator:

3\sqrt{2}\cdot\sqrt{3} + 3\sqrt{2}\cdot\sqrt{2} + 2\sqrt{3}\cdot\sqrt{3} + 2\sqrt{3}\cdot\sqrt{2} = 3\sqrt{6} + 6 + 6 + 2\sqrt{6}

.

Apply the distributive property and basic radical simplification rules.

4

Combine like terms in the numerator

(6 + 6) + (3\sqrt{6} + 2\sqrt{6}) = 12 + 5\sqrt{6}

.

Collect rational numbers together and like surd terms together.

Key Concept

Rationalization of Denominators with Binomial Surds

Question 212Question →

Given the universal set $U = \{x \in \mathbb{Z} : 1 \le x \le 20\}$ , with subsets $P = \{x \in U : x \text{ is a prime number}\}$ and $Q = \{x \in U : x \text{ is an odd integer}\}$ , what is the cardinal number of $(P \cup Q)'$ ?

9

11

10

8

Show answer & explanation

Answer: $9$

Answer

9

The universal set contains

20

elements. The union

P \cup Q

consists of all odd numbers and prime numbers between

1

and

20

, giving

11

unique elements:

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

. Subtracting these

11

elements from the total

20

elements in

U

gives

n((P \cup Q)') = 20 - 11 = 9

.

Step-by-Step Solution

1

Identify the elements of the universal set

U

and subsets

P

and

Q

.

U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}

, so

n(U) = 20

.

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

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

Explicitly listing the set elements helps correctly calculate the union.

2

Determine the union

P \cup Q

.

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

, which contains

11

elements.

The union combines all unique elements present in set

P

, set

Q

, or both.

3

Calculate the complement set

(P \cup Q)'

and its cardinality.

(P \cup Q)' = U \setminus (P \cup Q) = \{4, 6, 8, 10, 12, 14, 16, 18, 20\}

, so

n((P \cup Q)') = 20 - 11 = 9

.

The complement of a set contains all elements in the universal set

U

that are not in the given set.

Key Concept

Set Operations and Complement of Sets

Estimated Time:1m 0s

Question 213Question →

The sum of the first $4$ terms of a geometric progression (GP) with a common ratio of $2$ is $45$ . What is the $6^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 96

Answer

The

6^{\text{th}}

term of the geometric progression is

96

.

Using the sum formula

S_4 = \frac{a(2^4 - 1)}{2 - 1} = 45

gives

15a = 45

, so the first term

a

is

3

. Substituting

a = 3

and

r = 2

into the term formula

T_6 = a r^5

gives

3 \times 32 = 96

.

Step-by-Step Solution

1

Express the sum of the first 4 terms using the GP sum formula to find the first term

a

.

Setting up

45 = \frac{a(2^4 - 1)}{2 - 1}

yields

15a = 45

, so

a = 3

.

The sum formula

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

allows us to isolate the unknown initial term

a

when

S_4

and

r

are given.

2

Calculate the

6^{\text{th}}

term

T_6

using the

n^{\text{th}}

term formula

T_n = a r^{n-1}

.

T_6 = 3 \times 2^{6-1} = 3 \times 32 = 96

.

The exponent for the common ratio in the

n^{\text{th}}

term formula is

n - 1

, giving

5

as the exponent.

Key Concept

Sum and

n^{\text{th}}

term of a Geometric Progression

Question 214Question →

What is the value of $\frac{1}{\log_2 100} + \frac{1}{\log_5 100}$ ?

\frac{1}{2}

1

2

\frac{1}{7}

Show answer & explanation

Answer: $\frac{1}{2}$

Answer

\frac{1}{2}

Using the change of base identity

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

, the expression simplifies to

\log_{100} 2 + \log_{100} 5

. By the product rule of logarithms, this equals

\log_{100}(2 \times 5) = \log_{100} 10

. Since

100^{\frac{1}{2}} = 10

, the value is

\frac{1}{2}

.

Step-by-Step Solution

1

Apply the reciprocal change of base rule

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

.

\frac{1}{\log_2 100} = \log_{100} 2

and

\frac{1}{\log_5 100} = \log_{100} 5

.

Converting to a common base of

100

allows the use of logarithmic addition laws.

2

Combine the two logarithms using the product law

\log_c x + \log_c y = \log_c (xy)

.

\log_{100} 2 + \log_{100} 5 = \log_{100} (2 \times 5) = \log_{100} 10.

Adding logarithms with the same base is equivalent to taking the logarithm of the product of their arguments.

3

Evaluate

\log_{100} 10

.

Since

100^{\frac{1}{2}} = 10

,

\log_{100} 10 = \frac{1}{2}

.

The logarithm asks what power base

100

must be raised to in order to equal

10

.

Key Concept

Logarithms and Change of Base

Estimated Time:1m 30s

Question 215Question →

If $x$ is the smallest non-negative integer satisfying the modular congruence $7x + 4 \equiv 2 \pmod{13}$ , find the value of $(x^3 + 2x) \pmod{13}$ expressed as a canonical non-negative remainder.

Show answer & explanation

Answer: 6

Answer

The canonical non-negative remainder is 6.

Subtracting 4 from both sides of

7x + 4 \equiv 2 \pmod{13}

gives

7x \equiv -2 \equiv 11 \pmod{13}

. Multiplying by the modular inverse of 7 (which is 2, since

7 \times 2 = 14 \equiv 1 \pmod{13}

) yields

x \equiv 22 \equiv 9 \pmod{13}

. Evaluating

(9^3 + 2 \times 9) \pmod{13}

gives

(729 + 18) = 747

. Dividing 747 by 13 gives a quotient of 57 and a remainder of 6.

Step-by-Step Solution

1

Isolate the linear term in the congruence

7x \equiv -2 \equiv 11 \pmod{13}

Subtracting 4 from both sides simplifies the equation, and

-2 + 13 = 11

converts the negative remainder to positive form.

2

Solve for

x

by multiplying by the multiplicative inverse of 7 modulo 13

x \equiv 9 \pmod{13}

, so

x = 9

Since

7 \times 2 = 14 \equiv 1 \pmod{13}

, multiplying

7x \equiv 11 \pmod{13}

by 2 yields

x \equiv 22 \equiv 9 \pmod{13}

.

3

Evaluate

(x^3 + 2x) \pmod{13}

using modular reduction

6

9^3 = 729 = 56 \times 13 + 1 \equiv 1 \pmod{13}

and

2 \times 9 = 18 = 1 \times 13 + 5 \equiv 5 \pmod{13}

. Adding these gives

1 + 5 = 6

.

Key Concept

Solving linear modular congruences and modular polynomial evaluation

Estimated Time:1m 30s

Question 216Question →

The $4^{\text{th}}$ term of an arithmetic progression (AP) is $15$ and the $9^{\text{th}}$ term is $35$ . What is the sum of the first $12$ terms of the progression?

300

276

600

228

Show answer & explanation

Answer: $300$

Answer

The sum of the first

12

terms of the progression is

300

.

Subtracting the

4^{\text{th}}

term from the

9^{\text{th}}

term gives

5d = 35 - 15 = 20

, so the common difference

d = 4

. Substituting

d = 4

into the

4^{\text{th}}

term expression

a + 3d = 15

yields

a = 3

. Using the sum formula

S_n = \frac{n}{2}[2a + (n-1)d]

for

n = 12

,

a = 3

, and

d = 4

gives

S_{12} = 6 \times (6 + 44) = 300

.

Step-by-Step Solution

1

Set up the term equations using

T_n = a + (n-1)d

.

a + 3d = 15

and

a + 8d = 35

.

The

n^{\text{th}}

term formula for an AP is

T_n = a + (n-1)d

.

2

Solve the system of equations for

a

and

d

.

Subtracting the first equation from the second yields

5d = 20 \Rightarrow d = 4

. Substituting

d = 4

into

a + 3(4) = 15

gives

a = 3

.

Subtracting eliminates

a

to find the common difference

d

, which is then used to calculate the first term

a

.

3

Calculate the sum of the first

12

terms

S_{12}

.

S_{12} = \frac{12}{2}[2(3) + (12-1)4] = 6[6 + 44] = 6 \times 50 = 300

.

Applying the AP sum formula

S_n = \frac{n}{2}[2a + (n-1)d]

with

n = 12

,

a = 3

, and

d = 4

.

Key Concept

Arithmetic Progression

n^{\text{th}}

term and sum calculations

Question 217Question →

In a chemistry experiment, a student measured the mass of a substance as $12.8\text{ g}$ . If the actual mass of the substance is $12.5\text{ g}$ , calculate the percentage error in the student's measurement.

Show answer & explanation

Answer: 2.4

Answer

The percentage error in the student's measurement is

2.4\%

.

The percentage error is calculated by taking the absolute error (

0.3\text{ g}

), dividing it by the actual value (

12.5\text{ g}

), and multiplying the result by

100\%

, yielding

2.4\%

.

Step-by-Step Solution

1

Identify the actual value and the measured value

Actual value =

12.5\text{ g}

, Measured value =

12.8\text{ g}

Percentage error is computed relative to the true, actual value.

2

Calculate the absolute error

\text{Error} = |12.8 - 12.5| = 0.3\text{ g}

The error represents the difference between the measured value and the actual value.

3

Calculate the percentage error

\text{Percentage Error} = \frac{0.3}{12.5} \times 100\% = 2.4\%

Dividing the error by the actual value and multiplying by 100 converts the relative error into a percentage.

Key Concept

Percentage Error

Estimated Time:1m 30s

Question 218Question →

A team of $8$ identical excavators working together at a constant rate can clear a parcel of land in $15\text{ days}$ . All $8$ excavators work together for the first $3\text{ days}$ , after which $2$ excavators break down and are removed from the site. Assuming the remaining excavators continue working at the same constant rate, how many additional days will it take to complete the clearing of the land?

16\text{ days}

20\text{ days}

12\text{ days}

13\text{ days}

Show answer & explanation

Answer: $16\text{ days}$

Answer

The correct answer is

16\text{ days}

.

The total work required for the project is

120

excavator-days. During the first

3

days, the

8

excavators complete

24

excavator-days of work, leaving

96

excavator-days of work remaining. Since

2

excavators break down,

6

excavators remain. The additional time needed for these

6

excavators to clear the rest of the land is

96 \div 6 = 16

days.

Step-by-Step Solution

1

Calculate the total work required in excavator-days.

\text{Total Work} = 8 \text{ excavators} \times 15 \text{ days} = 120 \text{ excavator-days}

.

Work done is directly proportional to the product of rate (number of machines) and time.

2

Calculate the work completed in the first 3 days and the remaining work.

\text{Work Done} = 8 \times 3 = 24 \text{ excavator-days}

;

\text{Remaining Work} = 120 - 24 = 96 \text{ excavator-days}

.

Subtracting completed work from total work yields the portion left to be done.

3

Determine the number of remaining excavators and compute the additional days required.

\text{Remaining Excavators} = 8 - 2 = 6

;

\text{Additional Days} = \frac{96 \text{ excavator-days}}{6 \text{ excavators}} = 16 \text{ days}

.

Dividing the remaining work by the active rate gives the time needed to finish the project.

Key Concept

Inverse proportion and work-rate problem involving partial completion and workforce changes.

Question 219Question →

The $5^{\text{th}}$ term of an arithmetic progression (AP) is $18$ and its $11^{\text{th}}$ term is $42$ . Calculate the value of the $20^{\text{th}}$ term of this progression.

Show answer & explanation

Answer: 78

Answer

78

Using the AP general term formula

T_n = a + (n-1)d

, the equations

a + 4d = 18

and

a + 10d = 42

yield

d = 4

and

a = 2

. Evaluating

T_{20} = 2 + 19(4)

yields

78

.

Step-by-Step Solution

1

Set up simultaneous equations using the general term formula

T_n = a + (n-1)d

a + 4d = 18

and

a + 10d = 42

Relate given terms to the first term

a

and common difference

d

.

2

Solve for the common difference

d

6d = 24 \implies d = 4

Subtracting the 5th term equation from the 11th term equation eliminates

a

.

3

Solve for the first term

a

a + 16 = 18 \implies a = 2

Substitute the value of

d

back into the first equation.

4

Calculate the 20th term

T_{20} = 2 + 19(4) = 78

Apply the nth term formula for

n = 20

.

Key Concept

Determining terms of an Arithmetic Progression using simultaneous equations

Estimated Time:1m 30s

Question 220Question →

The sum of the first $n$ terms of an arithmetic progression (AP) is given by $S_n = 2n^2 + 3n$ . What is the $8^{\text{th}}$ term of the progression?

Show answer & explanation

Answer: 33

Answer

The

8^{\text{th}}

term of the arithmetic progression is

33

.

The

n^{\text{th}}

term of any sequence can be calculated using the identity

T_n = S_n - S_{n-1}

. Substituting

n = 8

gives

S_8 = 2(8)^2 + 3(8) = 152

, and substituting

n = 7

gives

S_7 = 2(7)^2 + 3(7) = 119

. Therefore,

T_8 = 152 - 119 = 33

.

Step-by-Step Solution

1

Calculate the sum of the first 8 terms (

S_8

) by substituting

n = 8

into

S_n = 2n^2 + 3n

.

S_8 = 2(8)^2 + 3(8) = 128 + 24 = 152

.

The sum formula provides the total sum of terms from

T_1

to

T_8

.

2

Calculate the sum of the first 7 terms (

S_7

) by substituting

n = 7

into

S_n = 2n^2 + 3n

.

S_7 = 2(7)^2 + 3(7) = 98 + 21 = 119

.

The sum formula provides the total sum of terms from

T_1

to

T_7

.

3

Subtract

S_7

from

S_8

to determine the value of the

8^{\text{th}}

term (

T_8

).

T_8 = S_8 - S_7 = 152 - 119 = 33

.

The difference between the sum of the first

n

terms and the sum of the first

(n-1)

terms yields the

n^{\text{th}}

term (

T_n = S_n - S_{n-1}

).

Key Concept

Finding the nth term of a sequence from the sum of the first n terms using

T_n = S_n - S_{n-1}

.

Estimated Time:1m 30s

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