Number and Numeration Practice Questions — JAMB UTME — Page 7

If $\log_2 x - 3\log_x 16 = 1$ , what is the product of all real values of $x$ that satisfy the equation?

2

16

-12

8

Show answer & explanation

Answer: $2$

Answer

The product of all real values of

x

satisfying the equation is

2

.

Using the change of base formula,

\log_x 16 = \frac{4}{\log_2 x}

. Substituting

u = \log_2 x

yields

u^2 - u - 12 = 0

, giving

u = 4

or

u = -3

. Converting back to

x

gives

x = 16

or

x = \frac{1}{8}

. Their product is

16 \times \frac{1}{8} = 2

.

Step-by-Step Solution

1

Apply the change of base formula to express

\log_x 16

in base 2.

\log_x 16 = \frac{\log_2 16}{\log_2 x} = \frac{4}{\log_2 x}

Standard logarithmic base conversion allows all terms to share base 2.

2

Substitute the change of base result into the original equation.

\log_2 x - 3 \left(\frac{4}{\log_2 x}\right) = 1 \implies \log_2 x - \frac{12}{\log_2 x} = 1

Simplifies the algebraic expression into a single logarithmic variable.

3

Substitute

u = \log_2 x

and transform into a quadratic equation.

u - \frac{12}{u} = 1 \implies u^2 - u - 12 = 0

Clears the fraction by multiplying through by

u

(where

u \neq 0

).

4

Factor the quadratic equation to find values of

u

.

(u - 4)(u + 3) = 0 \implies u = 4 \text{ or } u = -3

Determines the logarithmic roots.

5

Solve for

x

using the exponential form

x = 2^u

.

x_1 = 2^4 = 16

and

x_2 = 2^{-3} = \frac{1}{8}

Converts the logarithmic solution back to the original variable

x

.

6

Calculate the product of the roots

x_1 \cdot x_2

.

16 \times \frac{1}{8} = 2

Fulfills the specific evaluation requirement asked in the stem.

Key Concept

Logarithmic Change of Base and Quadratic Reducible Logarithmic Equations

Estimated Time:2m 0s

Question 122Question →

Given that $\frac{\sqrt{5} + 2}{\sqrt{5} - 2} = a + b\sqrt{5}$ , where $a$ and $b$ are rational numbers, what is the value of $a + b$ ?

13

11

9

7

Show answer & explanation

Answer: 13

Answer

The value of

a + b

is 13.

To simplify

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

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

\sqrt{5} + 2

. Expanding the numerator

(\sqrt{5} + 2)^2

gives

5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}

. The denominator simplifies to

(\sqrt{5})^2 - 2^2 = 5 - 4 = 1

. Thus, the expression becomes

9 + 4\sqrt{5}

. Comparing this to

a + b\sqrt{5}

gives

a = 9

and

b = 4

. Therefore,

a + b = 13

.

Step-by-Step Solution

1

Identify the conjugate of the denominator

The denominator is

\sqrt{5} - 2

, so its conjugate is

\sqrt{5} + 2

.

Multiplying by the conjugate rationalizes the binomial denominator using the difference of two squares.

2

Multiply the numerator and denominator by the conjugate

\frac{(\sqrt{5} + 2)(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \frac{(\sqrt{5} + 2)^2}{(\sqrt{5})^2 - 2^2}

This removes the radical from the denominator.

3

Expand both numerator and denominator

\frac{5 + 4\sqrt{5} + 4}{5 - 4} = \frac{9 + 4\sqrt{5}}{1} = 9 + 4\sqrt{5}

Using

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

for the numerator and

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

for the denominator.

4

Equate to

a + b\sqrt{5}

and solve for

a + b

a = 9

and

b = 4

, so

a + b = 9 + 4 = 13

.

Matching the rational part

a

and coefficient of the surd

b

gives the target sum.

Key Concept

Rationalization of Binomial Denominators

Estimated Time:1m 15s

Question 123Question →

If $\log_5 x + 2\log_x 5 = 3$ , what is the sum of all possible real values of $x$ ?

15

30

100

125

Show answer & explanation

Answer: 30

Answer

The sum of all possible real values of

x

is 30.

Applying the change of base identity

\log_x 5 = \frac{1}{\log_5 x}

converts the equation to

\log_5 x + \frac{2}{\log_5 x} = 3

. Setting

y = \log_5 x

yields

y^2 - 3y + 2 = 0

, giving solutions

y = 1

and

y = 2

. Converting back to

x

gives

x = 5^1 = 5

and

x = 5^2 = 25

. The sum of these values is

5 + 25 = 30

.

Step-by-Step Solution

1

Apply the change of base identity

\log_x 5 = \frac{1}{\log_5 x}

.

The equation becomes

\log_5 x + \frac{2}{\log_5 x} = 3

.

Converting all logarithmic terms to base 5 allows substitution and simplification.

2

Substitute

y = \log_5 x

and form a quadratic equation.

y + \frac{2}{y} = 3 \implies y^2 - 3y + 2 = 0

.

Multiplying through by

y

transforms the logarithmic equation into a standard quadratic form.

3

Solve the quadratic equation for

y

.

(y - 1)(y - 2) = 0 \implies y = 1 \text{ or } y = 2

.

Factoring determines the values of

y = \log_5 x

.

4

Convert back to

x

using exponentiation and calculate the sum.

For

y = 1

,

x = 5^1 = 5

. For

y = 2

,

x = 5^2 = 25

. Sum

= 5 + 25 = 30

.

Definition of logarithm

\log_a b = c \iff a^c = b

gives the values of

x

.

Key Concept

Logarithm Change of Base Formula and Quadratic Substitution

Estimated Time:1m 30s

Question 124Question →

What is the product of all real values of

x

that satisfy the exponential equation

4^{x+1} - 17 \times 2^x + 4 = 0

?

-4

1

-1

4

Show answer & explanation

Answer: $-4$

Answer

The product of all real values of

x

satisfying the equation is

-4

.

Expressing

4^{x+1}

as

4(2^x)^2

allows setting

y = 2^x

, giving

4y^2 - 17y + 4 = 0

. Solving this quadratic equation yields

y = \frac{1}{4}

and

y = 4

. Solving

2^x = \frac{1}{4} = 2^{-2}

gives

x = -2

, and

2^x = 4 = 2^2

gives

x = 2

. The product of these two real solutions is

(-2) \times 2 = -4

.

Step-by-Step Solution

1

Rewrite

4^{x+1}

using index laws

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

Converting all exponential terms to base

2

allows substitution into a quadratic form.

2

Substitute

y = 2^x

into the equation

4y^2 - 17y + 4 = 0

This transforms the exponential equation into a standard quadratic equation in terms of

y

.

3

Solve the quadratic equation for

y

(4y - 1)(y - 4) = 0 \implies y = \frac{1}{4} \text{ or } y = 4

Factoring

4y^2 - 16y - y + 4 = 0

yields the two valid values for

y

.

4

Solve for

x

using

y = 2^x

For

y = \frac{1}{4}

:

2^x = 2^{-2} \implies x_1 = -2

. For

y = 4

:

2^x = 2^2 \implies x_2 = 2

.

Applying the law of indices

a^m = a^n \implies m = n

determines the real solutions for

x

.

5

Calculate the product of the solutions

x_1 \times x_2 = (-2) \times 2 = -4

The question asks for the product of the real values of

x

.

Key Concept

Quadratic form exponential equations and index transformation rules

Question 125Question →

Determine the least non-negative integer congruent to $68 \pmod{9}$ .

Show answer & explanation

Answer: 5

Answer

The least non-negative integer congruent to

68 \pmod{9}

is 5.

Dividing 68 by 9 gives a quotient of 7 and a remainder of 5. Since

0 \le 5 < 9

, the value 5 is the standard non-negative remainder.

Step-by-Step Solution

1

Divide 68 by the modulus 9 using the division algorithm.

68 = 9 \times 7 + 5

Every integer

a

can be written uniquely as

a = nq + r

where

q

is the quotient and

0 \le r < n

.

2

Extract the non-negative remainder

r

.

r = 5

In modular arithmetic, the least non-negative integer congruent to

a \pmod{n}

is the remainder

r

when

a

is divided by

n

.

Key Concept

Modular Arithmetic Remainder

Question 126Question →

In a library, the ratio of science books to literature books was initially $4 : 5$ . After $60$ new science books were added to the collection and $20\%$ of the literature books were donated to a local school, the ratio of science books to literature books became $3 : 2$ . What was the total number of books (science and literature combined) originally in the library?

120

150

270

300

Show answer & explanation

Answer: $270$

Answer

The original total number of books in the library was

270

.

Let the original number of science books be

4x

and literature books be

5x

. After adding

60

science books, there are

4x + 60

science books. Donating

20\%

of literature books leaves

80\%

of

5x

, which equals

4x

literature books. Setting up the ratio

\frac{4x + 60}{4x} = \frac{3}{2}

yields

8x + 120 = 12x

, so

4x = 120

and

x = 30

. The original total number of books was

4x + 5x = 9x = 9(30) = 270

.

Step-by-Step Solution

1

Define initial quantities using a ratio multiplier

x

.

Initial science books =

4x

, initial literature books =

5x

, and total original books =

9x

.

The initial ratio given is

4 : 5

.

2

Calculate the updated quantities after books are added and donated.

New science books =

4x + 60

; new literature books =

5x \times (1 - 0.20) = 4x

.

60

science books were added and

20\%

of literature books were donated.

3

Set up the proportion for the new ratio of

3 : 2

.

\frac{4x + 60}{4x} = \frac{3}{2}

The new ratio of science to literature books is stated as

3 : 2

.

4

Solve the algebraic proportion for

x

.

2(4x + 60) = 3(4x) \implies 8x + 120 = 12x \implies 4x = 120 \implies x = 30.

Cross-multiplication simplifies the proportional equation to find the value of

x

.

5

Compute the original total number of books.

9x = 9 \times 30 = 270.

The original total collection consists of

9x

books.

Key Concept

Solving multi-step ratio changes and proportions with percentage alterations

Estimated Time:1m 30s

Question 127Question →

What is the simplified numerical value of $27^{\frac{2}{3}} \times 9^{-\frac{1}{2}}$ ?

Show answer & explanation

Answer: 3

Answer

The simplified numerical value of the expression is 3.

Evaluating

27^{\frac{2}{3}}

gives

(\sqrt[3]{27})^2 = 3^2 = 9

. Evaluating

9^{-\frac{1}{2}}

gives

\frac{1}{\sqrt{9}} = \frac{1}{3}

. Multiplying

9

by

\frac{1}{3}

yields

3

.

Step-by-Step Solution

1

Rewrite bases in terms of prime factors

27 = 3^3

and

9 = 3^2

Expressing bases in power-of-3 form allows direct application of index laws.

2

Simplify each indexed term using

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

and

a^{-n} = \frac{1}{a^n}

27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^2 = 9

and

9^{-\frac{1}{2}} = (3^2)^{-\frac{1}{2}} = 3^{-1} = \frac{1}{3}

Fractional indices denote roots and negative indices denote reciprocals.

3

Multiply the resulting numbers

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

Simplifying the final product yields the single numeric answer.

Key Concept

Fractional and Negative Index Laws

Question 128Question →

What is the smallest non-negative integer $x$ that satisfies the modular congruence $5x + 7 \equiv 2 \pmod{9}$ ?

8

4

1

7

Show answer & explanation

Answer: $8$

Answer

The smallest non-negative integer

x

is 8.

Subtracting

7

from both sides yields

5x \equiv -5 \pmod{9}

, which reduces to

5x \equiv 4 \pmod{9}

. Multiplying both sides by the modular multiplicative inverse of

5

modulo

9

(which is

2

, since

5 \times 2 = 10 \equiv 1 \pmod{9}

) gives

x \equiv 4 \times 2 = 8 \pmod{9}

. Thus, the smallest non-negative integer value is

8

.

Step-by-Step Solution

1

Isolate the term containing

x

by subtracting 7 from both sides of the congruence.

5x \equiv 2 - 7 \equiv -5 \pmod{9}

Standard algebraic balance operations apply under modular arithmetic.

2

Convert the negative remainder

-5

to its canonical non-negative representative in modulo 9.

-5 + 9 = 4 \implies 5x \equiv 4 \pmod{9}

Modular numbers are conventionally expressed in the non-negative range

[0, n-1]

.

3

Find the modular multiplicative inverse of 5 modulo 9 and multiply both sides by it.

Since

5 \times 2 = 10 \equiv 1 \pmod{9}

, the inverse is 2. Multiplying gives

x \equiv 4 \times 2 = 8 \pmod{9}

.

Multiplying by the modular inverse solves for

x

when

\gcd(5, 9) = 1

.

Key Concept

Linear Modular Congruences and Modular Inverses

Estimated Time:1m 30s

Question 129Question →

In a technology conference of $60$ software developers, $35$ write code in Python, $28$ write in Java, and $12$ write in both Python and Java. How many developers write in neither of these two languages?

Show answer & explanation

Answer: 9

Answer

The number of developers who write in neither Python nor Java is

9

.

The total number of developers writing in at least one language is given by

35 + 28 - 12 = 51

. Subtracting this from the universal set size of

60

gives

60 - 51 = 9

developers who write in neither language.

Step-by-Step Solution

1

Calculate the number of developers who write in at least one of the languages.

n(P \cup J) = 35 + 28 - 12 = 51

By the principle of inclusion-exclusion for two sets,

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

.

2

Subtract from the universal set total to get the complement.

n((P \cup J)') = 60 - 51 = 9

The number of elements outside the union is the universal set total minus the union cardinality.

Key Concept

Two-set inclusion-exclusion and complement cardinality

Estimated Time:1m 0s

Question 130Question →

Three farmers, Audu, Ngozi, and Tayo, shared a bag of fertilizer in the ratio $2 : 3 : 5$ . If Ngozi received $45\text{ kg}$ , what was the total mass of the fertilizer shared?

75\text{ kg}

135\text{ kg}

150\text{ kg}

225\text{ kg}

Show answer & explanation

Answer: $150\text{ kg}$

Answer

The total mass of the fertilizer shared was

150\text{ kg}

.

The ratio

2 : 3 : 5

indicates that the fertilizer is divided into

2 + 3 + 5 = 10

equal parts. Ngozi receives 3 of these parts, which equals

45\text{ kg}

. Dividing

45\text{ kg}

by 3 gives

15\text{ kg}

per part. Multiplying

15\text{ kg}

by the total of 10 parts gives

150\text{ kg}

.

Step-by-Step Solution

1

Find the total number of parts in the ratio.

Total parts =

2 + 3 + 5 = 10

parts.

The total quantity is divided into parts according to the given ratio

2 : 3 : 5

.

2

Determine the value of 1 part using Ngozi's share.

Value of 1 part =

\frac{45\text{ kg}}{3} = 15\text{ kg}

.

Ngozi corresponds to the middle term of the ratio, which is 3 parts out of 10.

3

Calculate the total mass by multiplying the value of 1 part by the total number of parts.

Total mass =

10 \times 15\text{ kg} = 150\text{ kg}

.

The total mass consists of all 10 parts combined.

Key Concept

Sharing quantities in a given ratio

Alternative Method

Use proportional fractions: Ngozi's fraction of the total is

\frac{3}{10}

. Let total mass be

M

. Then

\frac{3}{10} M = 45 \implies M = 45 \times \frac{10}{3} = 150\text{ kg}

.

Estimated Time:45s

Question 131Question →

A vessel contains a mixture of two liquids, $A$ and $B$ , in the ratio $5 : 3$ . If $16\text{ litres}$ of the mixture is drawn off and replaced with an equal volume of liquid $B$ , the ratio of liquid $A$ to liquid $B$ in the vessel becomes $1 : 1$ . What was the initial total volume of the mixture in the vessel, in litres?

Show answer & explanation

Answer: 80

Answer

The initial total volume of the mixture in the vessel was

80\text{ litres}

.

Let the initial volume of the mixture be

V

litres. Liquid

A

initially comprises

\frac{5}{8}V

litres and liquid

B

comprises

\frac{3}{8}V

litres. When

16\text{ litres}

of mixture is removed, the volume of liquid

A

removed is

\frac{5}{8} \times 16 = 10\text{ litres}

, and the volume of liquid

B

removed is

\frac{3}{8} \times 16 = 6\text{ litres}

. After adding

16\text{ litres}

of pure liquid

B

, the new volume of liquid

A

is

\frac{5}{8}V - 10

and the new volume of liquid

B

is

\frac{3}{8}V - 6 + 16 = \frac{3}{8}V + 10

. Since the new ratio is

1 : 1

, setting

\frac{5}{8}V - 10 = \frac{3}{8}V + 10

gives

\frac{2}{8}V = 20

, which simplifies to

V = 80\text{ litres}

.

Step-by-Step Solution

1

Define total initial volume as

V

and express the initial quantities of liquids

A

and

B

.

Liquid

A = \frac{5}{8}V

litres, Liquid

B = \frac{3}{8}V

litres.

The total ratio parts equal

5 + 3 = 8

.

2

Calculate the volume of each liquid removed in

16\text{ litres}

of mixture.

Volume of

A

removed

= 10

litres; Volume of

B

removed

= 6

litres.

The drawn-off mixture retains the original

5 : 3

ratio of liquids.

3

Formulate expressions for the quantities of

A

and

B

after adding

16\text{ litres}

of pure liquid

B

.

New amount of

A = \frac{5}{8}V - 10

; New amount of

B = \frac{3}{8}V + 10

.

16

litres of liquid

B

is added to the remaining quantity of

B

, which was

\frac{3}{8}V - 6

.

4

Equate the new quantities of

A

and

B

since the final ratio is

1 : 1

.

\frac{5}{8}V - 10 = \frac{3}{8}V + 10 \implies \frac{2}{8}V = 20 \implies V = 80

.

A final ratio of

1 : 1

implies equal quantities of both liquids.

Key Concept

Ratio modification through mixture removal and replacement

Estimated Time:2m 0s

Question 132Question →

If $2^{x-1} + 2^{x+1} = 160$ , what is the value of $3^{x-2}$ ?

27

81

243

12

Show answer & explanation

Answer: 81

Answer

81

Factoring

2^{x-1}

from the left-hand side of

2^{x-1} + 2^{x+1} = 160

yields

2^{x-1}(1 + 2^2) = 160

, which reduces to

5 \cdot 2^{x-1} = 160

. Dividing by

5

gives

2^{x-1} = 32 = 2^5

. Equating exponents gives

x - 1 = 5

, so

x = 6

. Substituting

x = 6

into

3^{x-2}

gives

3^{6-2} = 3^4 = 81

.

Step-by-Step Solution

1

Factor out the common term

2^{x-1}

from the expression

2^{x-1} + 2^{x+1}

.

2^{x-1}(1 + 2^2) = 160

Applying index law

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

allows factoring out

2^{x-1}

.

2

Simplify the bracketed terms and solve for

2^{x-1}

.

5 \cdot 2^{x-1} = 160 \implies 2^{x-1} = 32

Dividing both sides by

5

isolates the term with the unknown exponent.

3

Express

32

as a power of

2

and solve for

x

.

2^{x-1} = 2^5 \implies x - 1 = 5 \implies x = 6

Equating exponents since the bases are identical.

4

Substitute

x = 6

into the target expression

3^{x-2}

.

3^{6-2} = 3^4 = 81

Evaluating the power gives the final answer.

Key Concept

Factoring exponential expressions with common bases and applying laws of indices

Estimated Time:1m 30s

Question 133Question →

What is the value of $\frac{\log_4 81}{\log_4 3}$ ?

Show answer & explanation

Answer: 4

Answer

4

Applying the change of base property

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

, the ratio

\frac{\log_4 81}{\log_4 3}

reduces directly to

\log_3 81

. Since

3^4 = 81

, the value is 4.

Step-by-Step Solution

1

Apply the change of base formula to rewrite the ratio

\frac{\log_4 81}{\log_4 3} = \log_3 81

By the change of base identity,

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

.

2

Evaluate the logarithm

4

Since

3^4 = 81

,

\log_3 81 = 4

.

Key Concept

Change of Base Formula for Logarithms

Question 134Question →

A vessel initially contains $80\text{ litres}$ of water. After standing in the sun, $12\text{ litres}$ of water evaporates. What percentage of the original volume of water evaporated?

Show answer & explanation

Answer: 15

Answer

The percentage of the original volume of water that evaporated is

15\%

.

To find the percentage of evaporated water, divide the evaporated amount (

12\text{ litres}

) by the original total volume (

80\text{ litres}

) and multiply by

100\%

. This yields

\frac{12}{80} \times 100\% = 15\%

.

Step-by-Step Solution

1

Determine the fractional portion of the evaporated water relative to the initial total volume.

The fraction is

\frac{12}{80} = \frac{3}{20}

.

Percentage loss must be evaluated relative to the original starting amount.

2

Convert the resulting fraction into a percentage.

\frac{3}{20} \times 100\% = 15\%.

Multiplying a dimensionless ratio by 100 yields its percentage equivalent.

Key Concept

Calculating percentage change or loss relative to an initial amount

Estimated Time:45s

Question 135Question →

If $123_x = 38_{10}$ , what is the value of the base $x$ ?

Show answer & explanation

Answer: 5

Answer

The base

x

is 5.

Expanding

123_x

in terms of powers of

x

yields

1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3

. Setting this equal to the decimal value 38 produces the quadratic equation

x^2 + 2x + 3 = 38

, which simplifies to

x^2 + 2x - 35 = 0

. Factoring gives

(x + 7)(x - 5) = 0

, yielding solutions

x = -7

and

x = 5

. Since a number base must be a positive integer, the correct value for

x

is 5.

Step-by-Step Solution

1

Expand the base

x

number into decimal form using place values

123_x = 1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3

Each digit position in base

x

corresponds to a power of

x

, starting from

x^0

on the right.

2

Set up and rearrange the quadratic equation

x^2 + 2x + 3 = 38 \implies x^2 + 2x - 35 = 0

Subtracting 38 from both sides converts the equation into standard quadratic form

ax^2 + bx + c = 0

.

3

Solve the quadratic equation for

x

(x + 7)(x - 5) = 0 \implies x = -7 \text{ or } x = 5

Factoring gives the roots of the quadratic equation.

4

Select the valid positive base

x = 5

A base must be a positive integer greater than the largest digit appearing in the number (which is 3).

Key Concept

Place value expansion and base conversion to base 10

Question 136Question →

Three consecutive terms of an arithmetic progression are $x + 2$ , $3x - 1$ , and $4x + 1$ . What is the $10^{\text{th}}$ term of the progression?

63

77

70

35

Show answer & explanation

Answer: 70

Answer

The

10^{\text{th}}

term of the progression is 70.

Equating the differences between consecutive terms gives

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

, which simplifies to

2x - 3 = x + 2

, giving

x = 5

. The first term is

a = 5 + 2 = 7

and the common difference is

d = 14 - 7 = 7

. Substituting these into

T_n = a + (n - 1)d

for

n = 10

yields

T_{10} = 7 + 9(7) = 70

.

Step-by-Step Solution

1

Set up an equation using the common difference property of an arithmetic progression.

(3x - 1) - (x + 2) = (4x + 1) - (3x - 1)

In any arithmetic progression, the difference between consecutive terms is constant (

T_2 - T_1 = T_3 - T_2

).

2

Simplify and solve for

x

.

2x - 3 = x + 2 \implies x = 5

Subtract

x

from both sides and add 3 to both sides.

3

Find the first term

a

and the common difference

d

.

First term

a = 5 + 2 = 7

; second term

T_2 = 3(5) - 1 = 14

; common difference

d = 14 - 7 = 7

.

Substitute

x = 5

into the expressions for the terms.

4

Calculate the

10^{\text{th}}

term using

T_n = a + (n - 1)d

.

T_{10} = 7 + (10 - 1)(7) = 7 + 9(7) = 7 + 63 = 70

Apply

n = 10

,

a = 7

, and

d = 7

to the

n^{\text{th}}

term formula.

Key Concept

Arithmetic Progression - Consecutive terms and nth term evaluation

Question 137Question →

A rectangular lawn was measured to have a length of $15.0\text{ m}$ and a width of $8.0\text{ m}$ . If the actual length of the lawn is $16.0\text{ m}$ and the measured width is exact, what is the percentage error in the calculated area of the lawn?

Show answer & explanation

Answer: 6.25

Answer

The percentage error in the calculated area is

6.25\%

.

The actual area of the lawn is

16.0\text{ m} \times 8.0\text{ m} = 128\text{ m}^2

, while the measured area is

15.0\text{ m} \times 8.0\text{ m} = 120\text{ m}^2

. The error is

128 - 120 = 8\text{ m}^2

. Expressed as a percentage of the actual area,

\frac{8}{128} \times 100\% = 6.25\%

.

Step-by-Step Solution

1

Calculate the measured area of the lawn using the measured length and width

\text{Measured Area} = 15.0\text{ m} \times 8.0\text{ m} = 120\text{ m}^2

Area of a rectangle is length multiplied by width.

2

Calculate the true (actual) area of the lawn using the actual length and exact width

\text{Actual Area} = 16.0\text{ m} \times 8.0\text{ m} = 128\text{ m}^2

The actual dimensions determine the true surface area.

3

Determine the magnitude of the error in area

\text{Error} = |128\text{ m}^2 - 120\text{ m}^2| = 8\text{ m}^2

Error is defined as the absolute difference between actual and measured values.

4

Compute the percentage error relative to the actual area

\text{Percentage Error} = \frac{8}{128} \times 100\% = 6.25\%

Percentage error is always calculated as

\frac{\text{Error}}{\text{Actual Value}} \times 100\%

.

Key Concept

Percentage Error Calculation in Compound Quantities

Estimated Time:1m 30s

Question 138Question →

If $203_n - 114_n = 45_n$ , where $n$ represents a positive integer base, what is the value of $n$ ?

6

5

7

8

Show answer & explanation

Answer: $6$

Answer

The correct base is

6

.

Expanding the numbers in terms of powers of

n

gives

(2n^2 + 3) - (n^2 + n + 4) = 4n + 5

. Grouping like terms results in the quadratic equation

n^2 - 5n - 6 = 0

. Factoring yields

(n - 6)(n + 1) = 0

, giving

n = 6

as the only valid positive integer solution greater than

5

.

Step-by-Step Solution

1

Convert each term from base

n

to base 10 using positional expansion.

203_n = 2n^2 + 0n + 3 = 2n^2 + 3

,

114_n = 1n^2 + 1n + 4 = n^2 + n + 4

, and

45_n = 4n + 5

.

Converting all terms to a common decimal representation allows algebraic manipulation.

2

Substitute the expanded terms back into the original equation and simplify.

(2n^2 + 3) - (n^2 + n + 4) = 4n + 5 \implies n^2 - n - 1 = 4n + 5

.

Carefully distribute the negative sign across all terms of

(n^2 + n + 4)

.

3

Rearrange the expression into standard quadratic form

ax^2 + bx + c = 0

.

n^2 - 5n - 6 = 0

.

Subtracting

4n + 5

from both sides sets the quadratic equation to zero.

4

Factor the quadratic equation to find the valid base

n

.

(n - 6)(n + 1) = 0 \implies n = 6

or

n = -1

.

Since a base must be a positive integer strictly greater than any digit present in the equation (maximum digit is

5

),

n = 6

.

Key Concept

Solving unknown base equations using polynomial expansion in positional notation.

Question 139Question →

In a department of $70$ university lecturers, $40$ publish research in Journal $A$ , $30$ publish in Journal $B$ , and $25$ publish in Journal $C$ . It is known that $15$ publish in both Journals $A$ and $B$ , $12$ publish in both Journals $B$ and $C$ , $10$ publish in both Journals $A$ and $C$ , and $5$ publish in all three journals. How many lecturers do not publish in any of these three journals?

7

12

17

63

Show answer & explanation

Answer: $7$

Answer

The number of lecturers who do not publish in any of the three journals is

7

.

Using the inclusion-exclusion principle for three sets,

n(A \cup B \cup C) = 40 + 30 + 25 - 15 - 12 - 10 + 5 = 63

. The number of lecturers publishing in none of the journals is the complement of this union relative to the universal set of

70

, which is

70 - 63 = 7

.

Step-by-Step Solution

1

Apply the Principle of Inclusion-Exclusion for three sets to find the total number of lecturers who publish in at least one journal,

n(A \cup B \cup C)

.

n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(A \cap C)] + n(A \cap B \cap C)

Elements counted multiple times in pairwise intersections must be subtracted, and the central triple intersection must be added back.

2

Substitute the given numerical values into the formula.

n(A \cup B \cup C) = 40 + 30 + 25 - (15 + 12 + 10) + 5 = 95 - 37 + 5 = 63

To evaluate the total cardinality of the union.

3

Subtract

n(A \cup B \cup C)

from the universal set size

n(U)

.

n((A \cup B \cup C)') = n(U) - n(A \cup B \cup C) = 70 - 63 = 7

The number of lecturers publishing in none of the journals corresponds to the complement of the union of all three sets.

Key Concept

Principle of Inclusion-Exclusion for Three Sets and Set Complement

Question 140Question →

In a secondary school of $800$ students, $45\%$ of the students are girls. If $\frac{1}{6}$ of the girls and $\frac{1}{4}$ of the boys wear eyeglasses, what is the total number of students in the school who wear eyeglasses?

Show answer & explanation

Answer: 170

Answer

170 students

Calculating

45\%

of

800

gives

360

girls, leaving

800 - 360 = 440

boys. Taking

\frac{1}{6}

of

360

gives

60

girls with eyeglasses, and taking

\frac{1}{4}

of

440

gives

110

boys with eyeglasses. Adding these two quantities yields

60 + 110 = 170

students.

Step-by-Step Solution

1

Find the total number of girls

360 girls

45% of the total 800 students are girls

2

Find the total number of boys

440 boys

Subtracting the number of girls (360) from total students (800) gives the number of boys

3

Find the number of girls wearing eyeglasses

60 girls

One-sixth of the 360 girls wear eyeglasses

4

Find the number of boys wearing eyeglasses

110 boys

One-fourth of the 440 boys wear eyeglasses

5

Sum the girls and boys wearing eyeglasses

170 students

Adding 60 girls and 110 boys gives the total students wearing eyeglasses

Key Concept

Fractions and Percentages of Quantities

Number and Numeration

Answer

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