Maharashtra Board Class 9 Maths Chapter 3 Set 3 Algebra Standard Part 1 Polynomials Solutions

Question 1. Write the correct alternative answer for each of the following questions.
 

i. Which of the following is a polynomial?
(a) \( \frac{x}{y} \)
(b) \( x^{\sqrt{2}} - 3x \)
(c) \( x^{-2} + 7 \)
(d) \( \sqrt{2} x^2 + \frac{1}{2} \)
Answer: (d) \( \sqrt{2} x^2 + \frac{1}{2} \)
In simple words: A polynomial can only have variables with non-negative whole number exponents. In option (d), the exponent of x is 2, which is a whole number, making it a valid polynomial.

🎯 Exam Tip: Always check the powers of the variables; they must be non-negative integers (0, 1, 2, ...). Coefficients can be any real numbers, including roots and fractions.

Question 1. Choose the correct alternative answer for each of the following questions:

(i) Which of the following is a polynomial?
(a) \( \frac{x}{y} \)
(b) \( x^{\sqrt{2}} - 3x \)
(c) \( x^{-2} + 7 \)
(d) \( \sqrt{2}x^2 + \frac{1}{2} \)

(ii) What is the degree of the polynomial \( \sqrt{7} \)?
(a) \( \frac{1}{2} \)
(b) 5
(c) 2
(d) 0

(iii) What is the degree of the zero polynomial?
(a) 0
(b) 1
(c) undefined
(d) any real number

(iv) What is the degree of the polynomial \( 2x^2 + 5x^3 + 7 \)?
(a) 3
(b) 2
(c) 5
(d) 7

(v) What is the coefficient form of \( x^3 - 1 \)?
(a) (1, -1)
(b) (3, -1)
(c) (1, 0, 0, -1)
(d) (1, 3, -1)

(vi) If \( p(x) = x^2 - 7\sqrt{7}x + 3 \), then \( p(7\sqrt{7}) = \)?
(a) 3
(b) \( 7\sqrt{7} \)
(c) \( 42\sqrt{7} + 3 \)
(d) \( 49\sqrt{7} \)
Answer:
(i) (d) \( \sqrt{2}x^2 + \frac{1}{2} \)
(ii) (d) 0
(iii) (c) undefined
(iv) (a) 3
(v) (c) (1, 0, 0, -1)
(vi) (a) 3
In simple words: For (i), a polynomial must have non-negative integer exponents, which only option (d) satisfies. For (ii), any non-zero constant has a degree of 0. For (iii), the degree of the zero polynomial is not defined. For (iv), the highest exponent of the variable is 3. For (v), we include the missing \( x^2 \) and \( x \) terms with 0 coefficients to get (1, 0, 0, -1). For (vi), substituting the value makes the variable terms cancel out, leaving only the constant 3.

🎯 Exam Tip: Remember that the degree of a non-zero constant polynomial is always 0, whereas the degree of a zero polynomial is undefined. Double-check that all missing powers of the variable are represented with a coefficient of 0 when writing the coefficient form.

Question vii. When \( x = -1 \), what is the value of the polynomial \( 2x^3 + 2x \)?
(a) 4
(b) 2
(c) -2
(d) -4
Answer: (a) 4
In simple words: To find the value, we substitute \( x = -1 \) into the polynomial expression and calculate the final numerical result.

🎯 Exam Tip: Always pay close attention to negative signs when raising a negative number to an odd power like 3.

Question viii. If \( x - 1 \) is a factor of the polynomial \( 3x^2 + mx \), then find the value of \( m \).
(a) 2
(b) -2
(c) -3
(d) 3
Answer: (c) -3
In simple words: Since \( x - 1 \) is a factor, substituting \( x = 1 \) into the polynomial must make it equal to 0. Solving \( 3(1)^2 + m(1) = 0 \) gives us \( m = -3 \).

🎯 Exam Tip: Use the Factor Theorem: if \( x - a \) is a factor of a polynomial \( P(x) \), then \( P(a) = 0 \). This is a highly scoring and frequently asked question type.

Question ix. Multiply \( (x^2 - 3)(2x - 7x^3 + 4) \) and write the degree of the product.
(a) 5
(b) 3
(c) 2
(d) 0
Answer: (a) 5
In simple words: The degree of a product of two polynomials is found by adding the highest powers of both parts, which is \( 2 + 3 = 5 \).

🎯 Exam Tip: Save time in the exam by adding the highest exponents of the individual terms instead of multiplying the entire expression out.

Question x. Which of the following is a linear polynomial?
(a) \( x + 5 \)
(b) \( x^2 + 5 \)
(c) \( x^3 + 5 \)
(d) \( x^4 + 5 \)
Answer: (a) \( x + 5 \)
In simple words: A linear polynomial is an expression where the highest power of the variable \( x \) is exactly 1.

🎯 Exam Tip: Remember the classification of polynomials by degree: degree 1 is linear, degree 2 is quadratic, and degree 3 is cubic.

Hints:
v. \( x^3 - 1 = x^3 + 0x^2 + 0x - 1 \)
vi. \( p(7\sqrt{7}) = (7\sqrt{7})^2 - (7\sqrt{7})(7\sqrt{7}) + 3 = 3 \)
vii. \( p(-1) = 2(-1)^3 + 2(-1) = -2 - 2 = -4 \)
viii. \( p(1) = 0 \)
\( \implies 3(1)^2 + m(1) = 0 \)
\( \implies 3 + m = 0 \)
\( \implies m = -3 \)
ix. Here, degree of first polynomial = 2 and degree of second polynomial = 3
\( \implies \) Degree of polynomial obtained by multiplication = 2 + 3 = 5

Question 2. Write the degree of the polynomial for each of the following.
(i) \( 5 + 3x^4 \)
(ii) \( 7 \)
(iii) \( ax^7 + bx^9 \) (a, b are constants)
Answer:
(i) \( 5 + 3x^4 \)
Here, the highest power of \( x \) is 4.
\( \implies \) Degree of the polynomial = 4

(ii) \( 7 = 7x^0 \)
\( \implies \) Degree of the polynomial = 0

(iii) \( ax^7 + bx^9 \)
Here, the highest power of \( x \) is 9.
\( \implies \) Degree of the polynomial = 9
In simple words: The degree of a polynomial is the highest exponent of the variable in that expression. For any non-zero constant number like 7, the degree is always 0 because there is no variable term.

🎯 Exam Tip: Always look for the term with the highest exponent to find the degree, and remember that non-zero constant numbers always have a degree of 0.

Question 3. Write the following polynomials in standard form. [1 Mark each]
(i) \( 4x^2 + 7x^4 - x^3 - x + 9 \)
Answer:
(i) \( 7x^4 - x^3 + 4x^2 - x + 9 \)
In simple words: To write a polynomial in standard form, you just need to rearrange the terms so that the powers of x go from the highest to the lowest.

🎯 Exam Tip: Don't forget to keep the correct positive or negative sign with each term when you rearrange them into descending order.

Question 4. Write the following polynomial in coefficient form.
(i) \( x^4 + 16 \)
(ii) \( m^5 + 2m^2 + 3m + 15 \)
Answer:
(i) \( x^4 + 16 \)
Index form = \( x^4 + 0x^3 + 0x^2 + 0x + 16 \)
\( \therefore \) Coefficient form of the polynomial = \( (1, 0, 0, 0, 16) \)

(ii) \( m^5 + 2m^2 + 3m + 15 \)
Index form = \( m^5 + 0m^4 + 0m^3 + 2m^2 + 3m + 15 \)
\( \dots \) Coefficient form of the polynomial = \( (1, 0, 0, 2, 3, 15) \). This systematic representation captures all coefficients in order of descending powers.
In simple words: To write a polynomial in coefficient form, first fill in any missing powers of the variable with zero coefficients, and then list only the numbers inside brackets.

🎯 Exam Tip: Do not forget to include 0 for any missing powers of the variable when writing the index form, otherwise your coefficient list will be incorrect.

Question 5. Write the index form of the polynomial using variable x from its coefficient form.
(i) \( (3, -2, 0, 7, 18) \)
(ii) \( (6, 1, 0, 7) \)
(iii) \( (4, 5, -3, 0) \)
Answer:
(i) Number of coefficients = 5
\( \therefore \) Degree = \( 5 - 1 = 4 \)
\( \therefore \) Index form = \( 3x^4 - 2x^3 + 0x^2 + 7x + 18 \)

(ii) Number of coefficients = 4
\( \therefore \) Degree = \( 4 - 1 = 3 \)
\( \therefore \) Index form = \( 6x^3 + x^2 + 0x + 7 \)

(iii) Number of coefficients = 4
\( \therefore \) Degree = \( 4 - 1 = 3 \)
\( \therefore \) Index form = \( 4x^3 + 5x^2 - 3x + 0 \). The degree of the polynomial is always one less than the total number of coefficients in the given list.
In simple words: Count the number of coefficients, subtract 1 to find the highest power of x, and then attach the powers of x to each number in order.

🎯 Exam Tip: Always check that the number of terms in your final index form matches the number of coefficients given in the question.

Question 6. Add the following polynomials.
(i) \( 7x^4 - 2x^3 + x + 10 \); \( 3x^4 + 15x^3 + 9x^2 - 8x + 2 \)
(ii) \( 3p^3q + 2p^2q + 7 \); \( 2p^2q + 4pq - 2p^3q \)
Answer:
(i) \( (7x^4 - 2x^3 + x + 10) + (3x^4 + 15x^3 + 9x^2 - 8x + 2) \)
= \( 7x^4 - 2x^3 + x + 10 + 3x^4 + 15x^3 + 9x^2 - 8x + 2 \)
= \( 7x^4 + 3x^4 - 2x^3 + 15x^3 + 9x^2 + x - 8x + 10 + 2 \)
= \( 10x^4 + 13x^3 + 9x^2 - 7x + 12 \)

(ii) \( (3p^3q + 2p^2q + 7) + (2p^2q + 4pq - 2p^3q) \)
= \( 3p^3q + 2p^2q + 7 + 2p^2q + 4pq - 2p^3q \)
= \( 3p^3q - 2p^3q + 2p^2q + 2p^2q + 4pq + 7 \)
= \( p^3q + 4p^2q + 4pq + 7 \)
In simple words: To add polynomials, we group the like terms (terms with the same variables and exponents) together and then add their coefficients.

🎯 Exam Tip: Always group like terms carefully and double-check the signs of each term before adding them to avoid simple calculation errors.

Question 7. Subtract the second polynomial from the first.
(i) \( 5x^2 - 2y + 9 \); \( 3x^2 + 5y - 7 \)
(ii) \( 2x^2 + 3x + 5 \); \( x^2 - 2x + 3 \)
Answer:
(i) \( (5x^2 - 2y + 9) - (3x^2 + 5y - 7) \)
= \( 5x^2 - 2y + 9 - 3x^2 - 5y + 7 \)
= \( 5x^2 - 3x^2 - 2y - 5y + 9 + 7 \)
= \( 2x^2 - 7y + 16 \)

(ii) \( (2x^2 + 3x + 5) - (x^2 - 2x + 3) \)
= \( 2x^2 + 3x + 5 - x^2 + 2x - 3 \)
= \( 2x^2 - x^2 + 3x + 2x + 5 - 3 \)
= \( x^2 + 5x + 2 \)
In simple words: When subtracting a polynomial, change the sign of every term inside the second polynomial and then combine the like terms.

🎯 Exam Tip: Remember that subtracting a negative term makes it positive. Be extremely careful when distributing the negative sign across the entire second polynomial.

Question 8. Multiply the following polynomials.
(i) \( (m^3 - 2m + 3)(m^4 - 2m^2 + 3m + 2) \)
(ii) \( (5m^3 - 2)(m^2 - m + 3) \)
Answer:
(i) \( (m^3 - 2m + 3)(m^4 - 2m^2 + 3m + 2) \)
= \( m^3(m^4 - 2m^2 + 3m + 2) - 2m(m^4 - 2m^2 + 3m + 2) + 3(m^4 - 2m^2 + 3m + 2) \)
= \( m^7 - 2m^5 + 3m^4 + 2m^3 - 2m^5 + 4m^3 - 6m^2 - 4m + 3m^4 - 6m^2 + 9m + 6 \)
= \( m^7 - 4m^5 + 6m^4 + 6m^3 - 12m^2 + 5m + 6 \)

(ii) \( (5m^3 - 2)(m^2 - m + 3) \)
= \( 5m^3(m^2 - m + 3) - 2(m^2 - m + 3) \)
= \( 5m^5 - 5m^4 + 15m^3 - 2m^2 + 2m - 6 \)
In simple words: To multiply polynomials, multiply each term of the first polynomial by every term of the second polynomial, and then add the resulting terms together.

🎯 Exam Tip: When multiplying terms, remember to add their exponents (e.g., \( m^3 \times m^4 = m^7 \)) and carefully combine any like terms at the end.

Question 8. (ii) Multiply the given polynomials: \( (5m^3 - 2)(m^2 - m + 3) \)
Answer:
\( (5m^3 - 2)(m^2 - m + 3) \)
\( = 5m^3(m^2 - m + 3) - 2(m^2 - m + 3) \)
\( = 5m^5 - 5m^4 + 15m^3 - 2m^2 + 2m - 6 \)
In simple words: To multiply these two expressions, we multiply each term in the first bracket by every term in the second bracket and then combine any terms that are alike.

🎯 Exam Tip: Always double-check the signs when multiplying terms, especially when dealing with negative coefficients like \( -2 \).

Question 9. Divide polynomial \( 3x^3 - 8x^2 + x + 7 \) by \( x - 3 \) using synthetic method and write the quotient and remainder.
Answer:
Dividend = \( 3x^3 - 8x^2 + x + 7 \)
\( \therefore \) Coefficient form of dividend = \( (3, -8, 1, 7) \)
Divisor = \( x - 3 \)
\( \dots \) Opposite of \( -3 \) is \( 3 \)

Coefficient form of quotient = \( (3, 1, 4) \)
\( \therefore \) Quotient = \( 3x^2 + x + 4 \) and
Remainder = \( 19 \)
In simple words: Synthetic division is a quick way to divide a polynomial by a linear term like \( x - 3 \). We use only the coefficients to perform simple multiplication and addition to find the final quotient and remainder.

🎯 Exam Tip: Remember to write the final quotient in polynomial form by decreasing the degree by one from the original dividend.

Question 10. For which value of m, \( x + 3 \) is the factor of the polynomial \( x^3 - 2mx + 21 \)?
Answer:
Here, \( p(x) = x^3 - 2mx + 21 \)
\( (x + 3) \) is a factor of \( x^3 - 2mx + 21 \).
\( \therefore \) By factor theorem,
Remainder = \( 0 \)
\( \therefore p(-3) = 0 \)
\( p(x) = x^3 - 2mx + 21 \)
\( \therefore p(-3) = (-3)^3 - 2(m)(-3) + 21 \)
\( \therefore 0 = -27 + 6m + 21 \)
\( \therefore -6 + 6m = 0 \)
\( \therefore 6m = 6 \)
\( \therefore m = 1 \)
\( \therefore x + 3 \) is the factor of \( x^3 - 2mx + 21 \) for \( m = 1 \).
In simple words: Since \( x + 3 \) is a factor, substituting \( x = -3 \) into the polynomial must make the entire expression equal to zero. Solving this simple equation gives us the value of \( m \) as 1.

🎯 Exam Tip: Always substitute the value with the opposite sign of the constant in the factor (i.e., use \( -3 \) for \( x + 3 \)) to correctly apply the factor theorem.

Question 11. At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is \(5x^2 - 3y^2\), \(7y^2 + 2xy\) and \(9x^2 + 4xy\) respectively. At the beginning of the year 2017, \(x^2 + xy - y^2\), \(5xy\) and \(3x^2 + xy\) persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages?
Answer:
Total population of villages at the end of 2016 = \((5x^2 - 3y^2) + (7y^2 + 2xy) + (9x^2 + 4xy)\)
\(= 5x^2 + 9x^2 - 3y^2 + 7y^2 + 2xy + 4xy\)
\(= 14x^2 + 4y^2 + 6xy\) ......(i)

Total number of persons who went to other village at the beginning of 2017 = \((x^2 + xy - y^2) + (5xy) + (3x^2 + xy)\)
\(= x^2 + 3x^2 - y^2 + xy + 5xy + xy\)
\(= 4x^2 - y^2 + 7xy\) ... (ii)

Remaining total population of villages = Total population at the end of 2016 \(-\) total number of persons who went to other village at the beginning of 2017
\(= 14x^2 + 4y^2 + 6xy - (4x^2 - y^2 + 7xy)\) ... [From (i) and (ii)]
\(= 14x^2 + 4y^2 + 6xy - 4x^2 + y^2 - 7xy\)
\(= 14x^2 - 4x^2 + 4y^2 + y^2 + 6xy - 7xy\)
\(= 10x^2 + 5y^2 - xy\)
\(\therefore\) The remaining total population of the three villages is \(10x^2 + 5y^2 - xy\).
In simple words: To find the remaining population, we first add up the starting populations of all three villages, then add up all the people who left, and finally subtract the group that left from the starting total.

🎯 Exam Tip: Be very careful with the negative sign when subtracting polynomials; it changes the sign of every term inside the bracket during expansion.

Question 12. Polynomials \(bx^2 + x + 5\) and \(bx^3 - 2x + 5\) are divided by polynomial \(x - 3\) and the remainders are \(m\) and \(n\) respectively. If \(m - n = 0\), then find the value of \(b\).
Answer:
When polynomial \(bx^2 + x + 5\) is divided by \((x - 3)\), the remainder is \(m\).
\(\therefore\) By remainder theorem,
Remainder = \(p(3) = m\)
\(p(x) = bx^2 + x + 5\)
\(\dots p(3) = b(3)^2 + 3 + 5\)
\(\therefore m = b(9) + 8\)
\(m = 9b + 8\) ...(i)

When polynomial \(bx^3 - 2x + 5\) is divided by \(x - 3\) the remainder is \(n\).
\(\therefore\) Remainder = \(q(3) = n\)
\(q(x) = bx^3 - 2x + 5\)
\(\therefore q(3) = b(3)^3 - 2(3) + 5\)
\(\therefore n = 27b - 6 + 5\)
\(n = 27b - 1\) ...(ii)

Given that \(m - n = 0\)
\( \implies m = n \)
Substituting the values from (i) and (ii):
\( \implies 9b + 8 = 27b - 1 \)
\( \implies 8 + 1 = 27b - 9b \)
\( \implies 9 = 18b \)
\( \implies b = \frac{9}{18} \)
\( \implies b = \frac{1}{2} \)
\(\therefore\) The value of \(b\) is \(\frac{1}{2}\).
In simple words: We use the remainder theorem by putting \(x = 3\) into both expressions to find the remainders \(m\) and \(n\). Since \(m - n = 0\), we set the two remainders equal to each other and solve for \(b\).

🎯 Exam Tip: Remember that the remainder theorem states that when a polynomial \(p(x)\) is divided by \((x - a)\), the remainder is \(p(a)\). Always show the substitution step clearly to secure full marks.

Question 13. Simplify.
\( (8m^2 + 3m - 6) - (9m - 7) + (3m^2 - 2m + 4) \)
Answer:
\( (8m^2 + 3m - 6) - (9m - 7) + (3m^2 - 2m + 4) \)
\( = 8m^2 + 3m - 6 - 9m + 7 + 3m^2 - 2m + 4 \)
\( = 8m^2 + 3m^2 + 3m - 9m - 2m - 6 + 7 + 4 \)
\( = 11m^2 - 8m + 5 \)
This simplified expression represents the sum and difference of the given quadratic polynomials.
In simple words: To simplify this, we first remove the parentheses while being careful with the minus sign, then group the like terms together, and finally add or subtract them.

🎯 Exam Tip: Be extremely careful with the negative sign outside the parentheses, as it changes the sign of every term inside when you expand it.

Question 14. Which polynomial is to be subtracted from \( x^2 + 13x + 7 \) to get the polynomial \( 3x^2 + 5x - 4 \)?
Answer:
Let the required polynomial be A.
\( \therefore (x^2 + 13x + 7) - A = 3x^2 + 5x - 4 \)
\( \therefore A = (x^2 + 13x + 7) - (3x^2 + 5x - 4) \)
\( = x^2 + 13x + 7 - 3x^2 - 5x + 4 \)
\( = x^2 - 3x^2 + 13x - 5x + 7 + 4 \)
\( = -2x^2 + 8x + 11 \)
\( \therefore -2x^2 + 8x + 11 \) must be subtracted from \( x^2 + 13x + 7 \) to get \( 3x^2 + 5x - 4 \). This algebraic method helps us easily find an unknown polynomial by rearranging the equation.
In simple words: Think of this like a simple number puzzle: 'What do you subtract from 10 to get 3?' You do 10 minus 3 to get 7. Similarly, we subtract the second polynomial from the first one to find our answer.

🎯 Exam Tip: Always write down the equation with an unknown variable like A first, then rearrange it carefully to avoid sign errors.

Question 15. Which polynomial is to be added to \( 4m + 2n + 3 \) to get the polynomial \( 6m + 3n + 10 \)?
Answer:
Let the required polynomial be A.
\( \therefore (4m + 2n + 3) + A = 6m + 3n + 10 \)
\( \therefore A = (6m + 3n + 10) - (4m + 2n + 3) \)
\( = 6m + 3n + 10 - 4m - 2n - 3 \)
\( = 6m - 4m + 3n - 2n + 10 - 3 \)
\( = 2m + n + 7 \)
\( \therefore 2m + n + 7 \) must be added to \( 4m + 2n + 3 \) to get \( 6m + 3n + 10 \). By performing this subtraction, we determine the exact difference needed to reach the target polynomial.
In simple words: To find what to add to a first number to get a second number, you just subtract the first number from the second number.

🎯 Exam Tip: Remember to distribute the negative sign to all terms inside the parentheses when subtracting the polynomial.

Question 1. Read the following passage, write the appropriate amount in the boxes and discuss.
Govind, who is a dry land farmer from Shiralas has a 5 acre field. His family includes his wife, two children and his old mother. He borrowed one lakh twenty five thousand rupees from the bank for one year as agricultural loan at 10 p.c.p.a. He cultivated soyabean in \( x \) acres and cotton and tur in \( y \) acres. The expenditure he incurred was as follows:
He spent Rs. 10,000 on seeds. The expenses for fertilizers and pesticides for the soyabean crop was Rs. \( 2000x \) and Rs. \( 4000x^2 \) were spent on wages and cultivation of land. He spent Rs. \( 8000y \) on fertilizers and pesticides and Rs. \( 9000y^2 \) for wages and cultivation of land for the cotton and tur crops.
Let us write the total expenditure on all the crops by using variables \( x \) and \( y \).
Rs. \( 10000 + 2000x + 4000x^2 + 8000y + 9000y^2 \)
He harvested \( 5x^2 \) quintals soyabean and sold it at Rs. 2800 per quintal. The cotton crop yield was \( \frac{5}{3} y^2 \) quintals which fetched Rs. 5000 per quintal. The tur crop yield was \( 4y \) quintals and was sold at Rs. 4000 per quintal. Write the total income in rupees that was obtained by selling the entire farm produce, with the help of an expression using variables \( x \) and \( y \). (Textbook pg. no. 44)
Answer:
Total income = Income from soyabean crop + Income from cotton crop + Income from tur crop
= Rs. \( (5x^2 \times 2800) + \left(\frac{5}{3} y^2 \times 5000\right) + (4y \times 4000) \)
= Rs. \( \left(14000x^2 + \frac{25000}{3}y^2 + 16000y\right) \). This algebraic expression represents the total revenue Govind generated from his entire seasonal harvest.
In simple words: To find the total money earned, we multiply the quantity of each crop harvested by its selling price per quintal and then add them all together.

🎯 Exam Tip: When forming algebraic expressions for total income, always multiply the quantity by the rate for each individual item before adding them together.

Question 2. We have seen the example of expenditure and income (in terms of polynomials) of Govind who is a dry land farmer. He has borrowed rupees one lakh twenty-five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a. He had spent Rs. 10,000 on seeds. The expenses on soyabean crop was Rs. \( 2000x \) for fertilizers and pesticides and Rs. \( 4000x^2 \) was spent on wages and cultivation. He spent Rs. \( 8000y \) on fertilizers and pesticides and Rs. \( 9000y^2 \) on cultivation and wages for cotton and tur crop. Find the total amount repaid to the bank and write the polynomial expression for his total expenditure.
Answer:
1. Amount repaid to the bank:
Loan Amount = Rs. \( 1,25,000 \)
Rate of Interest = \( 10\% \) p.c.p.a. for 1 year
Interest = \( 1,25,000 \times \frac{10}{100} = \text{Rs. } 12,500 \)
Total Amount Repaid = Loan Amount + Interest = \( 1,25,000 + 12,500 = \text{Rs. } 1,37,500 \)

2. Total expenditure expression:
Total Expenditure = Rs. \( (10000 + 2000x + 4000x^2 + 8000y + 9000y^2) \). This polynomial represents the sum of all fixed and variable costs incurred during cultivation.
In simple words: The total money repaid to the bank includes the original loan plus the 10% interest, and the total expenditure is the sum of all money spent on seeds, fertilizers, and labor.

🎯 Exam Tip: Remember to calculate simple interest using the formula \( I = \frac{P \times R \times T}{100} \) and add it to the principal to find the total amount repaid.

Question. His total income was Rs. \( (14000x^2 + \frac{25000}{3}y^2 + 16000y) \). By taking \( x = 2 \), \( y = 3 \) write the income expenditure account of Govind’s farming.
Answer:
By substituting the values \( x = 2 \) and \( y = 3 \) into the given algebraic expressions, we can calculate the individual income components for Govind's farming. This helps us systematically organize his financial records into a standard credit and debit format.
\( \text{Income from soyabean} = 14000x^2 = 14000(2)^2 = 14000 \times 4 = \text{Rs. } 56,000 \)
\( \text{Income from cotton} = \frac{25000}{3}y^2 = \frac{25000}{3}(3)^2 = \frac{25000}{3} \times 9 = 25000 \times 3 = \text{Rs. } 75,000 \)
\( \text{Income from tur} = 16000y = 16000(3) = \text{Rs. } 48,000 \)

Credit (Income)

Debit (Expenses)

In simple words: We find the actual income values by substituting x = 2 and y = 3 into the formulas. Then, we list all the earnings under Credit and all the spending under Debit to show the complete farming balance sheet.

🎯 Exam Tip: Always show the step-by-step substitution of variables before drawing the final tables to ensure you get full marks for calculation steps.