Algebra Practice Questions — JAMB UTME — Page 2

Find the set of real values of $x$ that satisfies both inequalities $x^2 - 3x - 10 < 0$ and $3 - 2x \le 1$ .

1 \le x < 5

-2 < x \le 1

x < -2 \text{ or } x > 5

-2 < x < 5

Show answer & explanation

Answer: $1 \le x < 5$

Answer

The set of real values satisfying both inequalities is

1 \le x < 5

.

Solving the quadratic inequality

x^2 - 3x - 10 < 0

gives the open interval

-2 < x < 5

. Solving the linear inequality

3 - 2x \le 1

gives

-2x \le -2

, which upon dividing by

-2

and reversing the inequality sign becomes

x \ge 1

. Finding the overlapping values that satisfy both inequalities gives

1 \le x < 5

.

Step-by-Step Solution

1

Solve the quadratic inequality

x^2 - 3x - 10 < 0

.

Factor into

(x - 5)(x + 2) < 0

. Critical values are

x = -2

and

x = 5

. Since the inequality is strictly less than zero, the solution region is

-2 < x < 5

.

The quadratic expression is negative between its real roots.

2

Solve the linear inequality

3 - 2x \le 1

.

Subtract 3 from both sides:

-2x \le -2

. Divide by

-2

and flip the inequality sign:

x \ge 1

.

Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.

3

Find the intersection of the two solution sets.

Combine

-2 < x < 5

and

x \ge 1

to get

1 \le x < 5

.

Values of

x

must satisfy both conditions simultaneously.

Key Concept

Solving simultaneous linear and quadratic inequalities

Question 22Question →

The second term of a geometric progression (G.P.) with positive terms is $6$ and the fifth term is $48$ . What is the sum of the first $6$ terms of the progression?

93

96

189

192

Show answer & explanation

Answer: $189$

Answer

The sum of the first

6

terms is

189

.

Using the

n

-th term formula

T_n = a r^{n-1}

, we form two equations:

a r = 6

and

a r^4 = 48

. Dividing the fifth term equation by the second term equation gives

r^3 = 8

, so the common ratio

r = 2

. Substituting

r = 2

back gives the first term

a = 3

. Finally, applying the sum formula

S_6 = \frac{3(2^6 - 1)}{2 - 1}

gives

3 \times 63 = 189

.

Step-by-Step Solution

1

Set up equations for the given terms using the

n

-th term formula

T_n = a r^{n-1}

.

T_2 = a r = 6

and

T_5 = a r^4 = 48

.

The

n

-th term of a G.P. is defined by

T_n = a r^{n-1}

.

2

Divide the equation for

T_5

by the equation for

T_2

to find the common ratio

r

.

\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2.

Dividing eliminates the first term

a

and allows solving for

r

directly.

3

Substitute

r = 2

into

T_2 = 6

to find the first term

a

.

a(2) = 6 \implies a = 3.

Knowing

r

allows calculating

a

from any known term.

4

Calculate the sum of the first

6

terms using

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

.

S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(64 - 1)}{1} = 3 \times 63 = 189.

Applying the G.P. sum formula for

n = 6

,

a = 3

, and

r = 2

gives the total sum.

Key Concept

Geometric Progression term formula (

T_n = a r^{n-1}

) and sum of

n

terms formula (

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

).

Estimated Time:1m 30s

Question 23Question →

What are the values of $y$ for the real solution pairs $(x, y)$ that satisfy the simultaneous equations $y - 3x = 2$ and $y = x^2 - x + 5$ ?

5

or

11

1

or

3

-1

or

-7

3

or

5

Show answer & explanation

Answer: $5$ or $11$

Answer

5

or

11

Substituting

y = 3x + 2

into the quadratic equation

y = x^2 - x + 5

gives

3x + 2 = x^2 - x + 5

, which simplifies to

x^2 - 4x + 3 = 0

. Solving this yields

x = 1

and

x = 3

. Substituting these into the linear equation gives

y = 3(1) + 2 = 5

and

y = 3(3) + 2 = 11

. Therefore, the possible values of

y

are

5

or

11

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation

y = 3x + 2

Linear equations can easily be substituted into quadratic equations.

2

Equate the linear expression for

y

to the quadratic equation

3x + 2 = x^2 - x + 5

Both expressions represent

y

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

x^2 - 4x + 3 = 0

Subtract

3x

and

2

from both sides.

4

Factorize and solve for

x

(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3

Finding the roots of the quadratic equation gives the

x

-coordinates of the solution pairs.

5

Substitute each

x

-value back into

y = 3x + 2

to find

y

When

x = 1

,

y = 3(1) + 2 = 5

. When

x = 3

,

y = 3(3) + 2 = 11

.

The question specifically asks for the values of

y

.

Key Concept

Simultaneous Linear and Quadratic Equations

Estimated Time:1m 30s

Question 24Question →

Which of the following represents the complete set of real values of $x$ that satisfy the inequality $\frac{x^2 - x - 6}{x - 1} \le 0$ ?

x \le -2

or

1 < x \le 3

x \le -2

or

1 \le x \le 3

-2 \le x < 1

or

x \ge 3

-2 \le x \le 3

with

x \neq 1

Show answer & explanation

Answer: $x \le -2$ or $1 < x \le 3$

Answer

The complete solution set is

x \le -2

or

1 < x \le 3

.

The expression

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

evaluates to a non-positive value (

\le 0

) when the numerator and denominator have opposite signs or when the numerator is zero. Evaluating across the critical boundaries

x = -2, 1, 3

while excluding

x = 1

yields the solution set

x \le -2

or

1 < x \le 3

.

Step-by-Step Solution

1

Factor the quadratic numerator and state the domain restriction.

The numerator factors as

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

, giving the rational inequality

\frac{(x - 3)(x + 2)}{x - 1} \le 0

with

x \neq 1

.

Factoring isolates the critical boundary points where the expression can change sign.

2

Identify all critical numbers.

The critical values are

x = -2

,

x = 1

, and

x = 3

.

These points partition the real number line into sub-intervals.

3

Test points in each interval to determine the sign of the rational function

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

.

For

x < -2

,

f(x) \le 0

; for

-2 < x < 1

,

f(x) > 0

; for

1 < x < 3

,

f(x) \le 0

; for

x > 3

,

f(x) > 0

.

Determining where the function is negative or zero identifies the regions satisfying

\le 0

.

4

Combine intervals and include non-undefined endpoints.

Endpoints

x = -2

and

x = 3

make the numerator zero (included), while

x = 1

makes the denominator zero (excluded). Thus,

x \le -2

or

1 < x \le 3

.

Division by zero must be excluded from the solution set.

Key Concept

Solving rational inequalities by finding critical values, using test intervals, and respecting domain restrictions.

Question 25Question →

The variable $y$ varies inversely as $x$ . If $y = 6$ when $x = 4$ , what is the value of $y$ when $x = 8$ ?

3

12

16

48

Show answer & explanation

Answer: 3

Answer

The value of

y

when

x = 8

is 3.

Since

y

varies inversely as

x

, the formula connecting them is

y = \frac{k}{x}

. Substituting

y = 6

when

x = 4

gives

6 = \frac{k}{4}

, which yields

k = 24

. Using this constant, when

x = 8

,

y = \frac{24}{8} = 3

.

Step-by-Step Solution

1

Write the relationship for inverse variation.

y = \frac{k}{x}

Inverse variation means

y

is inversely proportional to

x

, where

k

is the constant of variation.

2

Substitute given values

y = 6

and

x = 4

to solve for

k

.

6 = \frac{k}{4} \implies k = 6 \times 4 = 24

To complete the equation of variation, the constant

k

must be determined.

3

Substitute

k = 24

and

x = 8

into the variation equation to find

y

.

y = \frac{24}{8} = 3

Evaluating the relationship at

x = 8

gives the requested value.

Key Concept

Inverse Variation

Question 26Question →

A quantity $y$ is partly constant and partly varies directly as $x$ . When $x = 2$ , $y = 10$ , and when $x = 5$ , $y = 19$ . What is the value of $y$ when $x = 8$ ?

28

35

40

25

Show answer & explanation

Answer: 28

Answer

The value of

y

when

x = 8

is 28.

By representing partial variation as

y = c + kx

, substituting the given conditions gives two simultaneous linear equations:

10 = c + 2k

and

19 = c + 5k

. Subtracting the first equation from the second yields

3k = 9

, so

k = 3

. Substituting

k = 3

back into the first equation yields

c = 4

. The general equation is

y = 4 + 3x

. Evaluating at

x = 8

gives

y = 4 + 3(8) = 28

.

Step-by-Step Solution

1

Set up the partial variation equation

y = c + kx

, where

c

is the constant part and

k

is the constant of variation.

Partial variation consists of a fixed term plus a variable term.

2

Substitute given values to form simultaneous equations

Equation 1:

10 = c + 2k

Equation 2:

19 = c + 5k

Two pairs of

(x, y)

values are provided to solve for the two unknown constants

c

and

k

.

3

Solve for

k

and

c

Subtract Equation 1 from Equation 2:

9 = 3k \implies k = 3

.
Substitute

k = 3

into Equation 1:

10 = c + 2(3) \implies c = 4

.
Thus,

y = 4 + 3x

.

Eliminating

c

yields the value of

k

, which is then used to find

c

.

4

Calculate

y

for

x = 8

y = 4 + 3(8) = 4 + 24 = 28

.

Substitute the required value of

x

into the established formula.

Key Concept

Partial Variation and Simultaneous Equations

Question 27Question →

Given that $P$ varies directly as the square of $r$ , and $P = 48$ when $r = 4$ , calculate the value of $P$ when $r = 6$ .

Show answer & explanation

Answer: 108

Answer

The value of

P

when

r = 6

is

108

.

Since

P

varies directly as

r^2

, the relationship is expressed as

P = k r^2

. Substituting the given values

P = 48

and

r = 4

yields

48 = 16k

, so

k = 3

. Substituting

k = 3

and

r = 6

into the equation gives

P = 3 \times 6^2 = 3 \times 36 = 108

.

Step-by-Step Solution

1

Set up the variation equation using constant of variation

k

P = k r^2

Direct variation with the square of a variable means

P

is directly proportional to

r^2

.

2

Substitute the initial values

P = 48

and

r = 4

to determine

k

48 = k \times 4^2 \implies 48 = 16k \implies k = 3

Finding the variation constant

k

allows us to establish a specific relationship between

P

and

r

.

3

Calculate

P

for

r = 6

using the specific equation

P = 3 r^2

P = 3 \times 6^2 = 3 \times 36 = 108

Evaluating the formula with the new input

r = 6

yields the required value of

P

.

Key Concept

Direct variation involving square powers

Question 28Question →

Given that $z$ varies directly as $x^2$ and inversely as $\sqrt{y}$ , and $z = 12$ when $x = 2$ and $y = 9$ , what is the value of $z$ when $x = 3$ and $y = 16$ ?

\frac{81}{4}

\frac{27}{2}

\frac{243}{16}

\frac{64}{9}

Show answer & explanation

Answer: $\frac{81}{4}$

Answer

\frac{81}{4}

The joint variation formula is

z = \frac{k x^2}{\sqrt{y}}

. Substituting the given values

x = 2, y = 9, z = 12

gives

12 = \frac{4k}{3}

, which yields

k = 9

. Evaluating

z

for

x = 3

and

y = 16

gives

z = \frac{9 \times 3^2}{\sqrt{16}} = \frac{81}{4}

.

Step-by-Step Solution

1

Set up the joint variation equation

z = \frac{k x^2}{\sqrt{y}}

Direct variation means

x^2

is in the numerator, and inverse variation means

\sqrt{y}

is in the denominator.

2

Substitute the initial values to solve for the constant of variation

k

12 = \frac{k (2)^2}{\sqrt{9}} \implies 12 = \frac{4k}{3} \implies 4k = 36 \implies k = 9

Using

x = 2

,

y = 9

, and

z = 12

allows us to find the constant

k

.

3

Calculate the new value of

z

using

x = 3

and

y = 16

z = \frac{9 (3)^2}{\sqrt{16}} = \frac{9 \times 9}{4} = \frac{81}{4}

Substitute

k = 9

,

x = 3

, and

y = 16

into the variation formula.

Key Concept

Joint Variation involving powers and roots

Estimated Time:1m 30s

Question 29Question →

The electrical resistance $R$ of a wire varies directly as its length $L$ and inversely as the square of its diameter $d$ . If a wire of length $36\text{ m}$ and diameter $3\text{ mm}$ has a resistance of $16\ \Omega$ , what is the resistance, in ohms, of a wire of the same material with a length of $45\text{ m}$ and a diameter of $5\text{ mm}$ ?

Show answer & explanation

Answer: 7.2

Answer

The resistance of the wire is 7.2 ohms.

The equation governing the relation is

R = \frac{kL}{d^2}

. Substituting the initial parameters

R=16\ \Omega

,

L=36\text{ m}

, and

d=3\text{ mm}

gives

16 = \frac{36k}{9} = 4k

, which yields

k = 4

. Using

k = 4

with the new dimensions

L=45\text{ m}

and

d=5\text{ mm}

gives

R = \frac{4 \times 45}{5^2} = \frac{180}{25} = 7.2\ \Omega

.

Step-by-Step Solution

1

Formulate the joint variation equation

R = \frac{kL}{d^2}

Direct variation places length

L

in the numerator and inverse variation of the square of diameter

d

places

d^2

in the denominator.

2

Determine the variation constant

k

k = 4

Substituting

R = 16

,

L = 36

, and

d = 3

gives

16 = \frac{36k}{9} \implies 16 = 4k \implies k = 4

.

3

Calculate the new resistance

R = 7.2\ \Omega

Substituting

k = 4

,

L = 45

, and

d = 5

into

R = \frac{kL}{d^2}

yields

R = \frac{4 \times 45}{25} = 7.2

.

Key Concept

Joint Variation involving direct proportionality and inverse square law

Question 30Question →

The hourly operational cost, $C$ Naira, of an industrial water pump is partly constant and partly varies jointly as the flow rate, $r$ in litres per second, and the square of the pressure head, $h$ in metres. When $r = 10\text{ L/s}$ and $h = 4\text{ m}$ , the operational cost is $\text{N}620$ . When $r = 15\text{ L/s}$ and $h = 2\text{ m}$ , the operational cost is $\text{N}380$ . What is the operational cost in Naira when $r = 20\text{ L/s}$ and $h = 3\text{ m}$ ?

Show answer & explanation

Answer: 668

Answer

The operational cost when

r = 20

and

h = 3

is 668 Naira.

The partial and joint variation relationship is defined by

C = k_1 + k_2 r h^2

. Substituting the two given states gives the simultaneous equations

620 = k_1 + 160k_2

and

380 = k_1 + 60k_2

. Subtracting these equations gives

100k_2 = 240

, so

k_2 = 2.4

. Substituting

k_2 = 2.4

into the second equation yields

k_1 = 236

. Finally, evaluating

C

for

r = 20

and

h = 3

gives

C = 236 + 2.4(20)(3^2) = 236 + 432 = 668

.

Step-by-Step Solution

1

Set up the variation equation

C = k_1 + k_2 r h^2

, where

k_1

is the constant part and

k_2

is the constant of joint variation.

The problem states that

C

is partly constant (

k_1

) and partly varies jointly as

r

and

h^2

(

k_2 r h^2

).

2

Form simultaneous linear equations using the given data points

(1)

620 = k_1 + 160k_2

and (2)

380 = k_1 + 60k_2

Substituting

r = 10, h = 4, C = 620

gives

10 \times 4^2 = 160

. Substituting

r = 15, h = 2, C = 380

gives

15 \times 2^2 = 60

.

3

Solve for the constants

k_1

and

k_2

k_2 = 2.4

and

k_1 = 236

Subtracting equation (2) from (1) eliminates

k_1

, giving

100k_2 = 240 \implies k_2 = 2.4

. Substituting back into equation (2) gives

k_1 = 380 - 60(2.4) = 236

.

4

Calculate the operational cost for the target parameters

C = 236 + 2.4 \times 20 \times 3^2 = 668

Substitute

k_1 = 236

,

k_2 = 2.4

,

r = 20

, and

h = 3

into the variation formula.

Key Concept

Partial and Joint Variation

Question 31Question →

What is the range of values of $x$ that satisfies the inequality $3 - 2x > 9$ ?

x < -3

x > -3

x < 3

x > 3

Show answer & explanation

Answer: $x < -3$

Answer

x < -3

Subtracting 3 from both sides yields

-2x > 6

. Dividing both sides by

-2

and reversing the inequality symbol gives

x < -3

.

Step-by-Step Solution

1

Subtract 3 from both sides of the inequality.

-2x > 6

To isolate the variable term

-2x

on the left-hand side.

2

Divide both sides by

-2

and reverse the inequality sign.

x < -3

Dividing an inequality by a negative number requires reversing the direction of the inequality sign.

Key Concept

Reversing inequality signs upon multiplication or division by negative numbers

Estimated Time:45s

Question 32Question →

Find the set of real values of $x$ that satisfies the inequality $\frac{3 - x}{x + 2} \geq 1$ .

-2 < x \leq \frac{1}{2}

-2 \leq x \leq \frac{1}{2}

x < -2 \text{ or } x \geq \frac{1}{2}

x \geq \frac{1}{2}

Show answer & explanation

Answer: $-2 < x \leq \frac{1}{2}$

Answer

-2 < x \leq \frac{1}{2}

Subtracting 1 from both sides yields

\frac{1 - 2x}{x + 2} \geq 0

. The critical points are

x = \frac{1}{2}

and

x = -2

. Testing values shows that the fraction is positive for

-2 < x < \frac{1}{2}

and equal to zero at

x = \frac{1}{2}

. Since

x = -2

causes division by zero, it is excluded from the interval, giving

-2 < x \leq \frac{1}{2}

.

Step-by-Step Solution

1

Subtract 1 from both sides of the inequality to set one side to zero.

\frac{3 - x}{x + 2} - 1 \geq 0

Direct cross-multiplication is invalid because the sign of

(x + 2)

depends on

x

.

2

Combine the terms over a common denominator.

\frac{(3 - x) - (x + 2)}{x + 2} \geq 0 \implies \frac{1 - 2x}{x + 2} \geq 0

Simplifying the numerator yields a clear rational inequality expression.

3

Identify the critical points and domain restrictions.

Numerator critical point:

x = \frac{1}{2}

; Denominator restriction:

x \neq -2

.

The quotient changes sign around

x = \frac{1}{2}

and

x = -2

, and division by zero is undefined.

4

Test the intervals

(-\infty, -2)

,

(-2, \frac{1}{2}]

, and

(\frac{1}{2}, \infty)

.

For

x \in (-2, \frac{1}{2}]

, the expression

\frac{1 - 2x}{x + 2}

is non-negative.

When

x = 0

,

\frac{1}{2} > 0

(positive). Outside this interval, the ratio is negative.

Key Concept

Solving Rational and Linear/Quadratic Inequalities

Estimated Time:2m 0s

Question 33Question →

Find the number of integers that satisfy the compound inequality $-3 < 2x + 1 \le 9$ .

Show answer & explanation

Answer: 6

Answer

The number of integer solutions satisfying the inequality is 6.

Subtracting 1 across

-3 < 2x + 1 \le 9

gives

-4 < 2x \le 8

. Dividing by 2 yields

-2 < x \le 4

. The integers in this interval are

-1, 0, 1, 2, 3,

and

4

, giving a total of 6 integer solutions.

Step-by-Step Solution

1

Subtract 1 from all parts of the compound inequality.

-4 < 2x \le 8

Isolate the variable term

2x

in the middle.

2

Divide all parts by 2.

-2 < x \le 4

Solve for

x

by undoing the coefficient of 2.

3

Identify the set of integer solutions within the interval

(-2, 4]

.

x \in \{-1, 0, 1, 2, 3, 4\}

The endpoint

-2

is excluded due to the strict inequality (

<

), while the endpoint

4

is included due to the inclusive inequality (

\\le

).

4

Count the elements in the solution set.

6

There are 6 distinct integer values in the set.

Key Concept

Solving compound linear inequalities and identifying integer solution sets.

Question 34Question →

A variable $y$ is partly constant and partly varies directly as $\sqrt{x}$ . Given that $y = 26$ when $x = 16$ , and $y = 38$ when $x = 49$ , what is the value of $y$ when $x = 64$ ?

Show answer & explanation

Answer: 42

Answer

The value of

y

when

x = 64

is

42

.

The relationship is given by the partial variation equation

y = a + b\sqrt{x}

. Substituting the pairs

(16, 26)

and

(49, 38)

yields the linear system

a + 4b = 26

and

a + 7b = 38

. Solving this system gives the constants

a = 10

and

b = 4

. Substituting

x = 64

into

y = 10 + 4\sqrt{64}

results in

y = 10 + 4(8) = 42

.

Step-by-Step Solution

1

Formulate the partial variation equation.

y = a + b\sqrt{x}

, where

a

and

b

are constants of variation.

Partial variation implies

y

is the sum of a constant term

a

and a term directly proportional to

\sqrt{x}

.

2

Set up simultaneous equations using the given pairs of

(x, y)

.

Equation 1:

a + 4b = 26

Equation 2:

a + 7b = 38

Evaluating

\sqrt{16} = 4

and

\sqrt{49} = 7

simplifies the relationship into two linear equations in two unknowns.

3

Solve for constants

a

and

b

.

b = 4

and

a = 10

Subtracting Equation 1 from Equation 2 yields

3b = 12 \implies b = 4

, and substituting

b = 4

back into Equation 1 gives

a = 10

.

4

Calculate

y

when

x = 64

.

y = 10 + 4(8) = 42

Using the specific formula

y = 10 + 4\sqrt{x}

for

x = 64

gives

y = 10 + 32 = 42

.

Key Concept

Partial Variation with Simultaneous Equations

Question 35Question →

The second, fourth, and eighth terms of an arithmetic progression (AP) with a non-zero common difference form three consecutive terms of a geometric progression (GP). If the sum of the first $5$ terms of the AP is $45$ , what is the first term of the AP?

3

5

9

15

Show answer & explanation

Answer: 3

Answer

The first term of the AP is 3.

By expressing the 2nd, 4th, and 8th terms as

a+d

,

a+3d

, and

a+7d

, the geometric mean condition

(a+3d)^2 = (a+d)(a+7d)

reduces to

d = a

. Substituting

d = a

into the sum formula

S_5 = 5(a+2d) = 45

gives

15a = 45

, yielding a first term of 3.

Step-by-Step Solution

1

Express the 2nd, 4th, and 8th terms of the AP in terms of first term

a

and common difference

d

T_2 = a + d

,

T_4 = a + 3d

, and

T_8 = a + 7d

The

n

-th term of an AP is given by

T_n = a + (n-1)d

.

2

Apply the condition for consecutive terms of a GP

(a + 3d)^2 = (a + d)(a + 7d) \implies a^2 + 6ad + 9d^2 = a^2 + 8ad + 7d^2 \implies 2d^2 = 2ad \implies d = a

If three terms

x, y, z

form a GP, then

y^2 = xz

. Since

d \neq 0

, dividing by

2d

gives

d = a

.

3

Use the sum of the first 5 terms of the AP to set up an equation for

a

and

d

S_5 = \frac{5}{2}[2a + 4d] = 45 \implies 5(a + 2d) = 45 \implies a + 2d = 9

The sum of the first

n

terms of an AP is

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

.

4

Substitute

d = a

into the sum equation to solve for

a

a + 2(a) = 9 \implies 3a = 9 \implies a = 3

Substituting

d = a

simplifies the linear equation to solve directly for

a

.

Key Concept

Combining Arithmetic and Geometric Progression properties to set up and solve simultaneous equations.

Question 36Question →

An arithmetic progression (A.P.) has a first term of $5$ and a common difference of $3$ . What is the $8^{\text{th}}$ term of this progression?

26

29

32

23

Show answer & explanation

Answer: $26$

Answer

The

8^{\text{th}}

term of the arithmetic progression is

26

.

The

8^{\text{th}}

term is calculated using the standard formula

T_n = a + (n - 1)d

. Substituting

a = 5

,

d = 3

, and

n = 8

gives

T_8 = 5 + 7(3) = 26

.

Step-by-Step Solution

1

Identify the given parameters from the problem

First term

a = 5

, common difference

d = 3

, and term position

n = 8

.

These are the values required for substitution into the

n^{\text{th}}

term formula of an A.P.

2

Write down the general formula for the

n^{\text{th}}

term of an arithmetic progression

T_n = a + (n - 1)d

This formula defines any term in an arithmetic progression based on its position.

3

Substitute the values into the formula and simplify

T_8 = 5 + (8 - 1) \times 3 = 5 + 7 \times 3 = 5 + 21 = 26

Performing the multiplication before addition yields the value of the

8^{\text{th}}

term.

Key Concept

Arithmetic Progression (A.P.)

n^{\text{th}}

term formula

Estimated Time:45s

Question 37Question →

The 3rd term of an arithmetic progression (AP) is $14$ and its 7th term is $34$ . If the $n$ -th term of this AP is equal to the 4th term of a geometric progression (GP) whose first term is $2$ and common ratio is $3$ , what is the value of $n$ ?

10

11

12

9

Show answer & explanation

Answer: 11

Answer

The value of

n

is

11

.

The AP has first term

a = 4

and common difference

d = 5

, giving

T_n = 4 + 5(n-1) = 5n - 1

. The 4th term of the GP is

2 \times 3^{3} = 54

. Setting

5n - 1 = 54

gives

5n = 55

, so

n = 11

.

Step-by-Step Solution

1

Find the first term

a

and common difference

d

of the arithmetic progression.

a = 4

and

d = 5

The

n

-th term of an AP is given by

T_n = a + (n-1)d

. Using given terms:

T_3 = a + 2d = 14

and

T_7 = a + 6d = 34

. Subtracting the first equation from the second yields

4d = 20 \Rightarrow d = 5

. Substituting

d = 5

into

a + 2(5) = 14

gives

a = 4

.

2

Calculate the 4th term of the geometric progression (

G_4

).

G_4 = 54

The

m

-th term of a GP is given by

G_m = a_{gp} \cdot r^{m-1}

. With first term

a_{gp} = 2

and ratio

r = 3

,

G_4 = 2 \cdot 3^{4-1} = 2 \cdot 3^3 = 2 \cdot 27 = 54

.

3

Equate

T_n

to

G_4

and solve for

n

.

n = 11

Set

T_n = G_4 \Rightarrow 4 + (n-1)5 = 54

. Simplifying gives

(n-1)5 = 50 \Rightarrow n-1 = 10 \Rightarrow n = 11

.

Key Concept

Solving simultaneous AP/GP equations using the

n

-th term formulas

T_n = a + (n-1)d

and

G_n = a r^{n-1}

.

Estimated Time:2m 0s

Question 38Question →

Which of the following ranges of $x$ satisfies the quadratic inequality $2x^2 - 7x + 3 < 0$ ?

\frac{1}{2} < x < 3

x < \frac{1}{2} \text{ or } x > 3

-3 < x < -\frac{1}{2}

x < -3 \text{ or } x > -\frac{1}{2}

Show answer & explanation

Answer: $\frac{1}{2} < x < 3$

Answer

\frac{1}{2} < x < 3

Factorizing

2x^2 - 7x + 3 < 0

gives

(2x - 1)(x - 3) < 0

. The roots of the quadratic equation

(2x - 1)(x - 3) = 0

are

x = \frac{1}{2}

and

x = 3

. Because the coefficient of

x^2

is positive, the quadratic curve opens upwards and is strictly negative between the two roots. Therefore, the inequality is satisfied when

\frac{1}{2} < x < 3

.

Step-by-Step Solution

1

Factorize the quadratic expression

2x^2 - 7x + 3 = (2x - 1)(x - 3)

Factorization allows us to find the critical values of the inequality.

2

Determine the critical values by setting the factored expression to zero

x = \frac{1}{2}

and

x = 3

These points mark the boundary values where the quadratic expression changes sign.

3

Test intervals or apply sign analysis for strict inequality

< 0

The product

(2x - 1)(x - 3)

is negative for

\frac{1}{2} < x < 3

Since the leading coefficient is positive (

2 > 0

), the parabola opens upwards and takes negative values strictly between its real roots.

Key Concept

Solving Quadratic Inequalities by Factorization and Interval Sign Analysis

Estimated Time:1m 15s

Question 39Question →

The fourth term of an arithmetic progression (A.P.) is $15$ and the ninth term is $35$ . What is the common difference of the progression?

Show answer & explanation

Answer: 4

Answer

The common difference of the arithmetic progression is

4

.

Using the A.P. term formula

T_n = a + (n-1)d

, the fourth term gives

a + 3d = 15

and the ninth term gives

a + 8d = 35

. Subtracting the two equations yields

5d = 20

, which simplifies directly to

d = 4

.

Step-by-Step Solution

1

Express the given terms using the n-th term formula

T_n = a + (n-1)d

a + 3d = 15

and

a + 8d = 35

The

n^{\text{th}}

term formula relates any term to the first term (

a

) and common difference (

d

).

2

Subtract the equation for the fourth term from the ninth term

5d = 20

Subtracting eliminates the first term

a

and leaves a simple equation in terms of

d

.

3

Divide by

5

to solve for

d

d = 4

Dividing both sides of

5d = 20

by

5

gives the common difference.

Key Concept

Finding the common difference of an Arithmetic Progression given two non-consecutive terms

Question 40Question →

Find the number of non-negative integer values of $x$ that satisfy the quadratic inequality $x^2 - 5x - 14 \le 0$ .

Show answer & explanation

Answer: 8

Answer

8

Factoring the quadratic expression gives

(x - 7)(x + 2) \le 0

, which evaluates to the solution interval

-2 \le x \le 7

. Restricting this interval to non-negative integers (

x \ge 0

) yields the set

\{0, 1, 2, 3, 4, 5, 6, 7\}

, which contains exactly 8 values.

Step-by-Step Solution

1

Factor the quadratic equation

x^2 - 5x - 14 = 0

(x - 7)(x + 2) = 0

, yielding critical roots at

x = 7

and

x = -2

Finding the roots determines the boundary points for the quadratic inequality.

2

Determine the solution set for the inequality

x^2 - 5x - 14 \le 0

-2 \le x \le 7

The quadratic expression is negative or zero between its two real roots.

3

Identify and count the non-negative integers in the interval

[-2, 7]

The non-negative integers are

0, 1, 2, 3, 4, 5, 6, 7

, giving a total of 8 values.

Non-negative integers consist of zero and all positive whole numbers within the solution range.

Key Concept

Solving quadratic inequalities and identifying discrete non-negative integer solution sets

Algebra

Answer

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