Chapter 4: Factorization and Algebraic Manipulation

Mathematics

Class 92025–26 SNC27 Questions SolvedEnglish Medium

Review Exercise 4 — Class 9 Mathematics Solutions

All 27 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Review Exercise 4

5 Questions (22 Sub-parts)

Quick Solution Step by Step

Just the working. Switch to Step by Step for the full explanation.

Quick solutions are ready for 14 of 15 questions. The rest show the full step-by-step solution.

Question 1

EasyApproved by Miss Aniya • Sep 9, 2026

Four options are given against each statement. Encircle the correct option.

Solution

This question contains 10 multiple-choice questions covering factorization techniques, highest common factor (HCF), least common multiple (LCM), polynomial relations, square roots, and degree of polynomials from Chapter 4.

Step by Step

Q1Part i

EasyApproved by Miss Aniya • Sep 9, 2026

The factorization of $12 x + 36$ is:

Multiple Choice

$12 (x + 3)$

Answer

$12 (3 x)$

$12 (3 x + 1)$

$x (12 + 36 x)$

Hint

Factor out the greatest common factor $(GCF)$ of $12$ and $36$ , which is $12$ .

Solution

Identify the common factor.

Find the greatest common factor $(GCF)$ of the terms $12 x$ and $36$ :
$\begin{matrix} 12 x & = 12 \times x \\ 36 & = 12 \times 3 \end{matrix}$
The common factor is $12$ .
Factor out $12$ .
$\begin{matrix} 12 x + 36 & = 12 (x) + 12 (3) \\ = 12 (x + 3) \end{matrix}$
Analyze distractors.
- Option (b): $12 (3 x) = 36 x \neq 12 x + 36$
- Option (c): $12 (3 x + 1) = 36 x + 12 \neq 12 x + 36$
- Option (d): $x (12 + 36 x) = 12 x + 36 x^{2} \neq 12 x + 36$
Thus, the correct choice is option (a).

Answer

$Option (a): 12 (x + 3)$

Step by Step

Q1Part ii

EasyUnder Review

The factors of $4 x^{2} - 12 y + 9$ are:

Multiple Choice

$(2 x + 3)^{2}$

$(2 x - 3)^{2}$

Answer

$(2 x - 3) (2 x + 3)$

$(2 + 3 x) (2 - 3 x)^{2}$

Hint

The $12 y$ in the book is a misprint for $12 x$ . Recognize $4 x^{2} - 12 x + 9$ as a perfect square trinomial $a^{2} - 2 a b + b^{2} = (a - b)^{2}$ , where $a = 2 x$ and $b = 3$ .

Solution

Read the expression correctly.

The book prints the middle term as $- 12 y$ , but every option contains only $x$ , and no product of these options can give a $y$ -term. The middle term is meant to be $- 12 x$ , so we factorize $4 x^{2} - 12 x + 9$ .
Recognize the algebraic identity.

Recall the perfect square expansion formula:
$\begin{matrix} (a - b)^{2} & = a^{2} - 2 a b + b^{2} \end{matrix}$
Rewrite the terms in square form.
$\begin{matrix} 4 x^{2} - 12 x + 9 & = (2 x)^{2} - 2 (2 x) (3) + 3^{2} \\ = (2 x - 3)^{2} \end{matrix}$
Analyze distractors.
- Option (a): $(2 x + 3)^{2} = 4 x^{2} + 12 x + 9$ (middle term has $+ 12 x$ instead of $- 12 x$ )
- Option (c): $(2 x - 3) (2 x + 3) = 4 x^{2} - 9$ (difference of two squares)
- Option (d): a cubic expression, so it cannot equal a quadratic
Thus, the correct choice is option (b).

Answer

$Option (b): (2 x - 3)^{2}$

Step by Step

Q1Part iii

EasyApproved by Miss Aniya • Sep 9, 2026

The HCF of $a^{3} b^{3}$ and $a b^{2}$ is:

Multiple Choice

$a^{3} b^{3}$

$a b^{2}$

Answer

$a^{4} b^{5}$

$a^{2} b$

Hint

The $HCF$ of monomials is obtained by taking the lowest power of each common variable: $a^{\min (3, 1)} = a$ and $b^{\min (3, 2)} = b^{2}$ .

Solution

Express each monomial in prime power factors.

The given expressions are:
$\begin{matrix} First term & = a^{3} b^{3} \\ Second term & = a^{1} b^{2} \end{matrix}$
Take the lowest exponent of each common variable.

To find the Highest Common Factor $(HCF)$ :
- For base $a$ : $\min (3, 1) = 1 ⟹ a^{1} = a$
- For base $b$ : $\min (3, 2) = 2 ⟹ b^{2}$
HCF=a1b2=ab2
Analyze distractors.
- Option (a): $a^{3} b^{3}$ is the LCM of the variables.
- Option (c): $a^{4} b^{5}$ is the product $a^{3} b^{3} \times a b^{2}$ .
- Option (d): $a^{2} b$ does not divide either term completely.
Thus, the correct choice is option (b).

Answer

$Option (b): a b^{2}$

Step by Step

Q1Part iv

EasyApproved by Miss Aniya • Sep 9, 2026

The LCM of $16 x^{2}, 4 x$ and $30 x y$ is:

Multiple Choice

$480 x^{3} y$

$240 x y$

$240 x^{2} y$

Answer

$120 x^{4} y$

Hint

Find the $LCM$ of the numerical coefficients $16$ , $4$ , and $30$ (which is $240$ ), then take the highest power of each variable: $x^{2}$ and $y$ .

Solution

Find the LCM of the numerical coefficients.

Prime factorization of the numbers $16$ , $4$ , and $30$ :
$\begin{matrix} 16 & = 2^{4} \\ 4 & = 2^{2} \\ 30 & = 2 \times 3 \times 5 \end{matrix}$
Taking the highest power of each prime factor:
$\begin{matrix} LCM (16, 4, 30) & = 2^{4} \times 3^{1} \times 5^{1} \\ = 16 \times 15 \\ = 240 \end{matrix}$
Find the LCM of the variable parts.
- For base $x$ : highest power among $x^{2}, x^{1}, x^{1}$ is $x^{2}$ .
- For base $y$ : highest power among $y^{0}, y^{0}, y^{1}$ is $y$ .
Combining numerical and algebraic factors:
LCM=240×x2×y=240x2y
Analyze distractors.
- Option (a): $480 x^{3} y$ unnecessarily doubles the coefficient and cubes $x$ .
- Option (b): $240 x y$ misses the required power $x^{2}$ .
- Option (d): $120 x^{4} y$ has an incorrect numerical multiple and power of $x$ .
Thus, the correct choice is option (c).

Answer

$Option (c): 240 x^{2} y$

Step by Step

Q1Part v

EasyApproved by Miss Aniya • Sep 9, 2026

Product of LCM and HCF = __________ of two polynomials.

Multiple Choice

sum

difference

product

Answer

quotient

Hint

Recall the fundamental relationship between two polynomials $P (x)$ and $Q (x)$ : $LCM \times HCF = P (x) \times Q (x)$ .

Solution

State the fundamental relationship.

For any two algebraic polynomials $P (x)$ and $Q (x)$ , their relationship with their Least Common Multiple (LCM) and Highest Common Factor (HCF) is given by:
$\begin{matrix} LCM \times HCF & = P (x) \times Q (x) \end{matrix}$
Interpret the identity in words.

The left-hand side is the product of LCM and HCF. The right-hand side is the product of the two polynomials. Therefore:
$\begin{matrix} Product of LCM and HCF & = Product of the two polynomials \end{matrix}$
Thus, the correct choice is option (c).

Answer

$Option (c): product$

Step by Step

Q1Part vi

EasyApproved by Miss Aniya • Sep 9, 2026

The square root of $x^{2} - 6 x + 9$ is:

Multiple Choice

$\pm (x - 3)$

Answer

$\pm (x + 3)$

$x - 3$

$x + 3$

Hint

Express the quadratic trinomial as a perfect square $(x - 3)^{2}$ and take its algebraic square root $\pm (x - 3)$ .

Solution

Factorize the quadratic expression as a perfect square.

Apply the identity $a^{2} - 2 a b + b^{2} = (a - b)^{2}$ :
$\begin{matrix} x^{2} - 6 x + 9 & = (x)^{2} - 2 (x) (3) + 3^{2} \\ = (x - 3)^{2} \end{matrix}$
Take the square root.

In algebra, taking the square root of an expression yields both positive and negative branches:
$\begin{matrix} \sqrt{x^{2} - 6 x + 9} & = \sqrt{(x - 3)^{2}} \\ = \pm (x - 3) \end{matrix}$
Analyze distractors.
- Option (b): Has $+ 3$ inside the bracket, which corresponds to $x^{2} + 6 x + 9$ .
- Options (c) and (d): Omit the essential $\pm$ sign required for complete algebraic square roots.
Thus, the correct choice is option (a).

Answer

$Option (a): \pm (x - 3)$

Step by Step

Q1Part vii

EasyApproved by Miss Aniya • Sep 9, 2026

The LCM of $(a - b)^{2}$ and $(a - b)^{4}$ is:

Multiple Choice

$(a - b)^{2}$

$(a - b)^{3}$

$(a - b)^{4}$

Answer

$(a - b)^{6}$

Hint

For powers of the same polynomial base, the $LCM$ is the factor with the highest exponent, $\max (2, 4) = 4$ .

Solution

Understand LCM of identical expressions with different powers.

When two polynomial expressions share the same common base $(a - b)$ :
- Their HCF is the base raised to the smallest exponent.
- Their LCM is the base raised to the greatest exponent.
Compare the exponents.

The exponents are $2$ and $4$ :
$\begin{matrix} LCM & = (a - b)^{\max (2, 4)} \\ = (a - b)^{4} \end{matrix}$
Analyze distractors.
- Option (a): $(a - b)^{2}$ is the HCF, not the LCM.
- Option (b): $(a - b)^{3}$ is not an exact common multiple.
- Option (d): $(a - b)^{6}$ is the product $(a - b)^{2} \times (a - b)^{4}$ , not the least common multiple.
Thus, the correct choice is option (c).

Answer

$Option (c): (a - b)^{4}$

Step by Step

Q1Part viii

EasyApproved by Miss Aniya • Sep 9, 2026

Factorization of $x^{3} + 3 x^{2} + 3 x + 1$ is:

Multiple Choice

$(x + 1)^{3}$

Answer

$(x - 1)^{3}$

$(x + 1) (x^{2} + x + 1)$

$(x - 1) (x^{2} - x + 1)$

Hint

Apply the cube of a binomial expansion $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ with $a = x$ and $b = 1$ .

Solution

Recall the cubic binomial expansion identity.
$\begin{matrix} (a + b)^{3} & = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} \end{matrix}$
Match the given polynomial with the identity.

Let $a = x$ and $b = 1$ :
$\begin{matrix} x^{3} + 3 x^{2} + 3 x + 1 & = (x)^{3} + 3 (x)^{2} (1) + 3 (x) (1)^{2} + (1)^{3} \\ = (x + 1)^{3} \end{matrix}$
Analyze distractors.
- Option (b): $(x - 1)^{3} = x^{3} - 3 x^{2} + 3 x - 1$ (alternating signs)
- Option (c): $(x + 1) (x^{2} + x + 1) = x^{3} + 2 x^{2} + 2 x + 1 \neq x^{3} + 3 x^{2} + 3 x + 1$
- Option (d): Expands to $x^{3} - 2 x^{2} + 2 x - 1$
Thus, the correct choice is option (a).

Answer

$Option (a): (x + 1)^{3}$

Step by Step

Q1Part ix

EasyApproved by Miss Aniya • Sep 9, 2026

Cubic polynomial has degree:

Multiple Choice

Answer

Hint

The degree of a polynomial is determined by the highest exponent of the variable. For a cubic polynomial, the highest power is $3$ .

Solution

Define polynomial classifications by degree.

In algebra, polynomials in a single variable are classified by their highest exponent (degree):
- Degree $1$ : Linear polynomial (e.g., $a x + b$ )
- Degree $2$ : Quadratic polynomial (e.g., $a x^{2} + b x + c$ )
- Degree $3$ : Cubic polynomial (e.g., $a x^{3} + b x^{2} + c x + d$ )
- Degree $4$ : Biquadratic or quartic polynomial (e.g., $a x^{4} + \dots$ )
Conclude the degree of a cubic polynomial.
$\begin{matrix} Degree of cubic polynomial & = 3 \end{matrix}$
Thus, the correct choice is option (c).

Answer

$Option (c): 3$

Step by Step

Q1Part x

EasyApproved by Miss Aniya • Sep 9, 2026

One of the factors of $x^{3} - 27$ is:

Multiple Choice

$x - 3$

Answer

$x + 3$

$x^{2} - 3 x + 9$

Both $a$ and $c$

Hint

Factor using the difference of cubes identity $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ . Note that the quadratic factor has a $+ 3 x$ middle term, making $(x - 3)$ the only valid option.

Solution

Recall the difference of cubes algebraic identity.
$\begin{matrix} a^{3} - b^{3} & = (a - b) (a^{2} + a b + b^{2}) \end{matrix}$
Factorize $x^{3} - 27$ .

Write $27$ as $3^{3}$ :
$\begin{matrix} x^{3} - 27 & = (x)^{3} - (3)^{3} \\ = (x - 3) ((x)^{2} + (x) (3) + 3^{2}) \\ = (x - 3) (x^{2} + 3 x + 9) \end{matrix}$
The two true factors of $x^{3} - 27$ are $(x - 3)$ and $(x^{2} + 3 x + 9)$ .
Scrutinize the given options carefully.
- Option (a): $x - 3$ is a correct factor.
- Option (b): $x + 3$ is not a factor.
- Option (c): $x^{2} - 3 x + 9$ has a negative middle term $- 3 x$ , whereas the true factor has $+ 3 x$ . Therefore, $x^{2} - 3 x + 9$ is not a factor!
- Option (d): "Both a and c" is a distractor trap based on misreading the sign in option (c).
Because only $(x - 3)$ is a valid factor among the options, option (a) is the only correct answer.

Answer

$Option (a): x - 3$

Step by Step

Question 2

MediumApproved by Miss Aniya • Sep 9, 2026

Factorize the following expressions:

Solution

This question consists of 9 sub-parts covering key factorization techniques: taking out common monomial factors, sum and difference of cubes, quadratic middle-term splitting, completing the square for fourth-degree polynomials, and quadratic substitution for products of linear factors.

Step by Step

Q2Part (i)

EasyApproved by Miss Aniya • Sep 9, 2026

$4 x^{3} + 18 x^{2} - 12 x$

Quick Solution

Under Review

\begin{matrix} 4 x^{3} + 18 x^{2} - 12 x & = 2 x (2 x^{2}) + 2 x (9 x) - 2 x (6) \\ = 2 x (2 x^{2} + 9 x - 6) \end{matrix}

Step by Step

Q2Part (ii)

EasyApproved by Miss Aniya • Sep 9, 2026

$x^{3} + 64 y^{3}$

Quick Solution

Under Review

\begin{matrix} x^{3} + 64 y^{3} & = (x)^{3} + (4 y)^{3} \\ = (x + 4 y) [x^{2} - x (4 y) + (4 y)^{2}] \\ = (x + 4 y) (x^{2} - 4 x y + 16 y^{2}) \end{matrix}

Step by Step

Q2Part (iii)

EasyApproved by Miss Aniya • Sep 9, 2026

$x^{3} y^{3} - 8$

Quick Solution

Under Review

\begin{matrix} x^{3} y^{3} - 8 & = (x y)^{3} - (2)^{3} \\ = (x y - 2) [(x y)^{2} + 2 x y + 2^{2}] \\ = (x y - 2) (x^{2} y^{2} + 2 x y + 4) \end{matrix}

Step by Step

Q2Part (iv)

MediumApproved by Miss Aniya • Sep 9, 2026

$- x^{2} - 23 x - 60$

Quick Solution

Under Review

\begin{matrix} - x^{2} - 23 x - 60 & = - (x^{2} + 23 x + 60) \\ = - (x^{2} + 20 x + 3 x + 60) \\ = - [x (x + 20) + 3 (x + 20)] \\ = - (x + 20) (x + 3) \end{matrix}

Step by Step

Q2Part (v)

EasyApproved by Miss Aniya • Sep 9, 2026

$2 x^{2} + 7 x + 3$

Quick Solution

Under Review

\begin{matrix} 2 x^{2} + 7 x + 3 & = 2 x^{2} + 6 x + x + 3 \\ = 2 x (x + 3) + 1 (x + 3) \\ = (x + 3) (2 x + 1) \end{matrix}

Step by Step

Q2Part (vi)

MediumApproved by Miss Aniya • Sep 9, 2026

$x^{4} + 64$

Quick Solution

Under Review

\begin{matrix} x^{4} + 64 & = (x^{2})^{2} + 2 (x^{2}) (8) + 8^{2} - 16 x^{2} \\ = (x^{2} + 8)^{2} - (4 x)^{2} \\ = (x^{2} + 8 + 4 x) (x^{2} + 8 - 4 x) \\ = (x^{2} + 4 x + 8) (x^{2} - 4 x + 8) \end{matrix}

Step by Step

Q2Part (vii)

MediumApproved by Miss Aniya • Sep 9, 2026

$x^{4} + 2 x^{2} + 9$

Quick Solution

Under Review

\begin{matrix} x^{4} + 2 x^{2} + 9 & = (x^{2})^{2} + 2 (x^{2}) (3) + 3^{2} - 6 x^{2} + 2 x^{2} \\ = (x^{2} + 3)^{2} - 4 x^{2} \\ = (x^{2} + 3)^{2} - (2 x)^{2} \\ = (x^{2} + 3 + 2 x) (x^{2} + 3 - 2 x) \\ = (x^{2} + 2 x + 3) (x^{2} - 2 x + 3) \end{matrix}

Step by Step

Q2Part (viii)

HardApproved by Miss Aniya • Sep 9, 2026

$(x + 3) (x + 4) (x + 5) (x + 6) - 360$

Quick Solution

Under Review

\begin{matrix} Expression & = [(x + 3) (x + 6)] [(x + 4) (x + 5)] - 360 \\ = (x^{2} + 9 x + 18) (x^{2} + 9 x + 20) - 360 \\ Let y & = x^{2} + 9 x \\ Expression & = (y + 18) (y + 20) - 360 \\ = y^{2} + 38 y + 360 - 360 \\ = y (y + 38) \\ = (x^{2} + 9 x) (x^{2} + 9 x + 38) \\ = x (x + 9) (x^{2} + 9 x + 38) \end{matrix}

Step by Step

Q2Part (ix)

HardApproved by Miss Aniya • Sep 9, 2026

$(x^{2} + 6 x + 3) (x^{2} + 6 x - 9) + 36$

Quick Solution

Under Review

\begin{matrix} Let y & = x^{2} + 6 x \\ Expression & = (y + 3) (y - 9) + 36 \\ = y^{2} - 6 y - 27 + 36 \\ = y^{2} - 6 y + 9 \\ = (y - 3)^{2} \\ = (x^{2} + 6 x - 3)^{2} \end{matrix}

Step by Step

Q3Part (i)

MediumApproved by Miss Aniya • Sep 9, 2026

$4 x^{3} + 12 x^{2}, 8 x^{2} + 16 x$

Quick Solution

Under Review

\begin{matrix} 4 x^{3} + 12 x^{2} & = 2^{2} \cdot x^{2} (x + 3) \\ 8 x^{2} + 16 x & = 2^{3} \cdot x (x + 2) \\ HCF & = 4 x \\ LCM & = 2^{3} \cdot x^{2} (x + 3) (x + 2) \\ = 8 x^{2} (x + 3) (x + 2) \end{matrix}

Step by Step

Q3Part (ii)

Medium

$x^{3} + 3 x^{2} - 4 x, x^{2} - x - 6$

Quick Solution

Under Review

\begin{matrix} x^{3} + 3 x^{2} - 4 x & = x (x^{2} + 4 x - x - 4) \\ = x (x + 4) (x - 1) \\ x^{2} - x - 6 & = x^{2} - 3 x + 2 x - 6 \\ = (x - 3) (x + 2) \\ HCF & = 1 \\ LCM & = x (x + 4) (x - 1) (x - 3) (x + 2) \end{matrix}

Step by Step

Q3Part (iii)

MediumApproved by Miss Aniya • Sep 9, 2026

$x^{2} + 8 x + 16, x^{2} - 16$

Quick Solution

Under Review

\begin{matrix} x^{2} + 8 x + 16 & = (x + 4)^{2} \\ x^{2} - 16 & = (x - 4) (x + 4) \\ HCF & = x + 4 \\ LCM & = (x + 4)^{2} (x - 4) \end{matrix}

Step by Step

Q3Part (iv)

Medium

$x^{3} - 9 x, x^{2} - 4 x + 3$

Quick Solution

Under Review

\begin{matrix} x^{3} - 9 x & = x (x^{2} - 9) \\ = x (x - 3) (x + 3) \\ x^{2} - 4 x + 3 & = x^{2} - 3 x - x + 3 \\ = (x - 3) (x - 1) \\ HCF & = x - 3 \\ LCM & = x (x - 3) (x + 3) (x - 1) \end{matrix}

Step by Step

Question 4

Medium

Find square root by factorization and division method of the expression $16 x^{4} + 8 x^{2} + 1$ .

Quick Solution

Under Review

\begin{matrix} 16 x^{4} + 8 x^{2} + 1 & = (4 x^{2})^{2} + 2 (4 x^{2}) (1) + 1^{2} \\ = (4 x^{2} + 1)^{2} \\ \sqrt{16 x^{4} + 8 x^{2} + 1} & = \pm (4 x^{2} + 1) \\ Division: (16 x^{4} + 8 x^{2} + 1) - (4 x^{2})^{2} & = 8 x^{2} + 1 \\ (8 x^{2} + 1) - (8 x^{2} + 1) (1) & = 0 \\ Root & = \pm (4 x^{2} + 1) \end{matrix}

Step by Step

Question 5

MediumApproved by Miss Aniya • Sep 9, 2026

Huria is analyzing the total cost of her loan, modeled by the expression $C (x) = x^{2} - 8 x + 15$ , where $x$ represents the number of years. What is the optimal repayment period for Huria's loan?

Hint

Set the cost function to zero: $C (x) = 0$ . Factor the quadratic trinomial $x^{2} - 8 x + 15 = (x - 3) (x - 5) = 0$ . This gives roots $x = 3$ and $x = 5$ . The earlier break-even point $x = 3 years$ is the optimal time to clear the loan cost.

Quick solution not ready yet. Showing the full step-by-step solution.

Solution

Formulate the objective from the mathematical model.

The loan cost function is modeled by:
$\begin{matrix} C (x) & = x^{2} - 8 x + 15 \end{matrix}$
where $x \geq 0$ represents the number of years. In financial analysis, the optimal repayment period corresponds to clearing the net loan cost or balance to zero, which means finding the values of $x$ such that $C (x) = 0$ .
Factorize the quadratic cost function.

Factor the quadratic expression by splitting the middle term. Find two numbers whose product is $1 \times 15 = 15$ and whose sum is $- 8$ . These numbers are $- 3$ and $- 5$ :
$\begin{matrix} C (x) & = x^{2} - 8 x + 15 \\ = x^{2} - 5 x - 3 x + 15 \\ = x (x - 5) - 3 (x - 5) \\ = (x - 5) (x - 3) \end{matrix}$
Solve for the zero-cost repayment periods.

Set $C (x) = 0$ :
$\begin{matrix} (x - 5) (x - 3) & = 0 \end{matrix}$
By the zero-product principle:
$\begin{matrix} x - 3 = 0 & ⟹ x = 3 \\ x - 5 = 0 & ⟹ x = 5 \end{matrix}$
Pedagogical & Financial Interpretation.
1. Zero Cost / Break-even Years: The loan cost clears to zero at $x = 3 years$ and $x = 5 years$ .
2. Optimal Repayment Decision:
  - Clearing the loan at the earliest zero-cost opportunity ( $x = 3 years$ ) avoids holding debt over additional years and is the most practical optimal repayment period.
  - The interval between $3$ and $5$ years forms the cost-effective repayment window where the cost function remains non-positive ( $C (x) \leq 0$ ).
  - Specifically, the mathematical minimum of the quadratic occurs at the vertex:
  xmin=−b2a=−−82(1)=4 years At x=4 years, C(4)=(4)2−8(4)+15=−1 (the maximum net cost reduction).
3. Conclusion: The critical zero-cost repayment periods are 3 years and 5 years, with 3 years being the optimal earliest time to eliminate loan cost.

Answer

$x = 3 years (or critical repayment periods: 3 years and 5 years)$

Step by Step

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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