Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra More Ques

I. Multiple Choice Questions

 

Question 1. Which of the following is a monomial?
(a) \( 4x^2 \)
(b) \( a + b \)
(c) \( a + b + c \)
(d) \( a + b + c + d \)
Answer: (a) \( 4x^2 \)
In simple words: A monomial is an algebraic expression with only one term. Terms are separated by plus or minus signs. \( 4x^2 \) has only one term, so it is a monomial.

🎯 Exam Tip: Remember that terms in an expression are separated by addition (+) or subtraction (-) signs. Monomials have just one term, binomials have two, and trinomials have three.

 

Question 2. Which of the following is trinomial?
(a) \( -7z \)
(b) \( z^2 - 4y^2 \)
(c) \( x^2y - xy^2 + y \)
(d) \( 12a - 9ab + 5b - 3 \)
Answer: (c) \( x^2y - xy^2 + y \)
In simple words: A trinomial is an algebraic expression that has exactly three terms. In the option \( x^2y - xy^2 + y \), there are three terms: \( x^2y \), \( -xy^2 \), and \( y \). This makes it a trinomial.

🎯 Exam Tip: Count the number of terms carefully. A term is a single number, variable, or product of numbers and variables. Terms are separated by plus or minus signs.

 

Question 3. The sum of \( 5x^2 \); \( -7x^2 \); \( 8x^2 \); \( 11x^2 \) and \( -9x^2 \) is .........
(a) \( 2x^2 \)
(b) \( 4x^2 \)
(c) \( 6x^2 \)
(d) \( 8x^2 \)
Answer: (d) \( 8x^2 \)
In simple words: To find the sum of these terms, just add their number parts because they all have the same variable part, \( x^2 \). When you add \( 5 - 7 + 8 + 11 - 9 \), you get \( 8 \). So the sum is \( 8x^2 \). Adding like terms is like combining similar items.

🎯 Exam Tip: Remember you can only add or subtract like terms. Like terms have the same variables raised to the same powers. Just add or subtract their coefficients.

 

Question 4. The area of a rectangle with length \( 2l^2m \) and breadth \( 3lm^2 \) is .........
(a) \( 6l^3m^3 \)
(b) \( l^3m^3 \)
(c) \( 2l^3m^3 \)
(d) \( 4l^3m^3 \)
Answer: (a) \( 6l^3m^3 \)
In simple words: To find the area of a rectangle, you multiply its length by its breadth. Here, you multiply the numbers (2 x 3 = 6) and then add the powers of the same letters (\( l^2 \times l = l^3 \) and \( m \times m^2 = m^3 \)). This gives you \( 6l^3m^3 \).

🎯 Exam Tip: When multiplying terms with exponents, remember the rule \( a^m \times a^n = a^{m+n} \). Multiply the coefficients and add the exponents of the same bases.

 

Question 5. The coefficient of \( x^2 \) and \( x \) in \( 2x^3 - 5x^2 + 6x - 3 \) are respectively .........
(a) 2, -5
(b) 2, 6
(c) -5, 6
(d) -5, -3
Answer: (c) -5, 6
In simple words: The coefficient is the number multiplied by a variable. For \( x^2 \), the number in front is \( -5 \). For \( x \), the number in front is \( 6 \). We must also include the sign. So, the coefficients are \( -5 \) and \( 6 \).

🎯 Exam Tip: Always pay attention to the sign in front of the term when identifying coefficients. A minus sign means the coefficient is negative.

 

Question 6. In the system \( 6x - 2y = 3 \); \( kx - y = 2 \) has a unique solution then .........
(a) \( k = 3 \)
(b) \( k \neq 3 \)
(c) \( k = 4 \)
(d) \( k \neq 4 \)
Answer: (b) \( k \neq 3 \)
In simple words: For a system of two linear equations to have a unique solution, the ratio of the coefficients of \( x \) must not be equal to the ratio of the coefficients of \( y \). This means \( \frac{6}{k} \neq \frac{-2}{-1} \). Solving this gives \( k \neq 3 \). This condition ensures the lines intersect at one point.

🎯 Exam Tip: For a unique solution in linear equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), the condition is \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).

 

Question 7. A system of two linear equations in two variables is inconsistent. If their graphs .........
(a) coincide
(b) intersect only at a point
(c) do not intersect at any point
(d) cut the x-axis
Answer: (c) do not intersect at any point
In simple words: An inconsistent system of equations means there is no solution that works for both equations at the same time. Graphically, this happens when the lines are parallel and never cross each other. So, they do not intersect at any point.

🎯 Exam Tip: Inconsistent systems have no solutions and their graphs are parallel lines. Consistent systems have at least one solution (intersecting or coinciding lines).

 

Question 8. The system of equation \( x - 4y = 8 \); \( 3x - 12y = 24 \) ..........
(a) has infinitely many solutions
(b) has no solution
(c) has a unique solution
(d) may or may not have a solution
Answer: (a) has infinitely many solutions
In simple words: If you simplify the second equation \( (3x - 12y = 24) \) by dividing everything by 3, you get \( x - 4y = 8 \), which is exactly the same as the first equation. This means both equations represent the same line, so every point on that line is a solution. When two equations are identical, they have infinite solutions.

🎯 Exam Tip: For equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \), if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the system has infinitely many solutions (the lines coincide).

 

Question 9. The solution set of \( x - ay = 4 \) and \( x + y = 0 \) is \( (1, -1) \). The value of \( a \) is .........
(a) -1
(b) 1
(c) -3
(d) 3
Answer: (d) 3
In simple words: Since \( (1, -1) \) is the solution, it means when \( x = 1 \) and \( y = -1 \), both equations are true. Substitute these values into the first equation \( x - ay = 4 \). You get \( 1 - a(-1) = 4 \), which simplifies to \( 1 + a = 4 \). Solving for \( a \) gives \( a = 3 \).

🎯 Exam Tip: When a solution set \( (x, y) \) is given, always substitute the \( x \) and \( y \) values into the equation(s) to find unknown constants.

 

Question 10. The solution set of \( x + y = 7 \); \( x - y = 3 \) is .........
(a) (-5, -2)
(b) (-5, 2)
(c) (5, 2)
(d) (2, 5)
Answer: (c) (5, 2)
In simple words: You can solve this system of equations by adding the two equations together. \( (x + y) + (x - y) = 7 + 3 \) gives \( 2x = 10 \), so \( x = 5 \). Then substitute \( x = 5 \) into \( x + y = 7 \) to get \( 5 + y = 7 \), which means \( y = 2 \). Thus, the solution is \( (5, 2) \).

🎯 Exam Tip: Systems of linear equations can often be solved quickly using elimination or substitution methods. In this case, adding the equations eliminates \( y \).

II. Answer Following Questions

 

Question 1. What must be added to \( x^4 - 3x^2 + 2x + 6 \) to get \( x^4 - 2x^3 - x + 8 \)?
Answer: To find the expression that needs to be added, we subtract the first polynomial \( (x^4 - 3x^2 + 2x + 6) \) from the target polynomial \( (x^4 - 2x^3 - x + 8) \).
Let the required expression be \( A \).
So, \( (x^4 - 3x^2 + 2x + 6) + A = x^4 - 2x^3 - x + 8 \)
Now, we isolate \( A \):
\( A = (x^4 - 2x^3 - x + 8) - (x^4 - 3x^2 + 2x + 6) \)
Distribute the negative sign to all terms inside the second parenthesis:
\( A = x^4 - 2x^3 - x + 8 - x^4 + 3x^2 - 2x - 6 \)
Combine like terms:
\( A = (x^4 - x^4) - 2x^3 + 3x^2 + (-x - 2x) + (8 - 6) \)
\( A = 0 - 2x^3 + 3x^2 - 3x + 2 \)
\( A = -2x^3 + 3x^2 - 3x + 2 \)
So, the expression \( -2x^3 + 3x^2 - 3x + 2 \) must be added. This method is similar to finding how much money you need if you have some and want to reach a target amount.
In simple words: To find what you need to add, simply subtract the first expression from the second one. Remember to change all signs when you subtract. Then, group similar terms and add them up.

🎯 Exam Tip: When subtracting polynomials, be very careful with the signs. The negative sign outside the parenthesis applies to every term inside it. Combine like terms accurately for the final answer.

 

Question 2. What must be subtracted from \( y^4 + 2y^3 - 3y^2 + 8 \) to get \( y^4 - 2y^3 + 6 \)?
Answer: Let \( A \) be the expression that must be subtracted.
So, \( (y^4 + 2y^3 - 3y^2 + 8) - A = y^4 - 2y^3 + 6 \)
To find \( A \), we can rearrange the equation:
\( A = (y^4 + 2y^3 - 3y^2 + 8) - (y^4 - 2y^3 + 6) \)
Now, we distribute the negative sign to the terms in the second parenthesis:
\( A = y^4 + 2y^3 - 3y^2 + 8 - y^4 + 2y^3 - 6 \)
Combine the like terms:
\( A = (y^4 - y^4) + (2y^3 + 2y^3) - 3y^2 + (8 - 6) \)
\( A = 0 + 4y^3 - 3y^2 + 2 \)
\( A = 4y^3 - 3y^2 + 2 \)
Therefore, the expression \( 4y^3 - 3y^2 + 2 \) must be subtracted. This is similar to finding how much you spent if you know what you started with and what you have left.
In simple words: To find what needs to be subtracted, take the first expression and subtract the target expression from it. Make sure to change the signs of all terms you are subtracting. Then, combine similar terms.

🎯 Exam Tip: When setting up problems asking "what must be subtracted to get...", remember that the unknown expression (what you subtract) is found by subtracting the final result from the initial expression.

 

Question 3. The area of a rectangle is \( x^4 + 9x^2 + 20 \) square units and its length is \( x^2 + 4 \) units. Find its breadth in terms of \( x \).
Answer: We know that the area of a rectangle is given by the formula: Area = Length \( \times \) Breadth.
Let the breadth of the rectangle be \( b \).
Given: Area \( = x^4 + 9x^2 + 20 \)
Given: Length \( = x^2 + 4 \)
So, we have the equation:
\( x^4 + 9x^2 + 20 = (x^2 + 4) \times b \)
To find \( b \), we divide the area by the length:
\( b = \frac{x^4 + 9x^2 + 20}{x^2 + 4} \)
We perform polynomial long division:

The result of the division is \( x^2 + 5 \). This tells us how many times the length fits into the area.
Therefore, the breadth of the rectangle is \( x^2 + 5 \) units.
In simple words: To find the breadth of a rectangle, you divide its area by its length. We use polynomial long division to divide the given area expression by the given length expression. The answer we get from this division is the breadth.

🎯 Exam Tip: When dividing polynomials, ensure that all powers of the variable (even those with a zero coefficient, like \( 0x^3 \) or \( 0x \)) are included in the dividend for correct alignment during long division. This helps prevent errors.

 

Question 4. Solve \( 3x + 4y = 24 \); \( 20x - 11y = 47 \) using the cross multiplication method.
Answer: First, write the equations in the standard form \( ax + by + c = 0 \):
1) \( 3x + 4y - 24 = 0 \)
2) \( 20x - 11y - 47 = 0 \)

Now, apply the cross multiplication formula. For \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), the solution is given by:
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)

Using the coefficients from our equations:
\( a_1 = 3, b_1 = 4, c_1 = -24 \)
\( a_2 = 20, b_2 = -11, c_2 = -47 \)

Substitute these values into the formula:
\( \frac{x}{(4)(-47) - (-11)(-24)} = \frac{y}{(-24)(20) - (-47)(3)} = \frac{1}{(3)(-11) - (20)(4)} \)

Calculate the denominators:
For \( x \): \( (4)(-47) - (-11)(-24) = -188 - 264 = -452 \)
For \( y \): \( (-24)(20) - (-47)(3) = -480 - (-141) = -480 + 141 = -339 \)
For \( 1 \): \( (3)(-11) - (20)(4) = -33 - 80 = -113 \)

So the equation becomes:
\( \frac{x}{-452} = \frac{y}{-339} = \frac{1}{-113} \)

To find \( x \), set the first and third parts equal:
\( \frac{x}{-452} = \frac{1}{-113} \)
\( x = \frac{-452}{-113} \)
\( x = 4 \)

To find \( y \), set the second and third parts equal:
\( \frac{y}{-339} = \frac{1}{-113} \)
\( y = \frac{-339}{-113} \)
\( y = 3 \)

The solution set for the system of equations is \( (4, 3) \). This method provides a structured way to find the point where two lines cross.
In simple words: First, rewrite both equations so they equal zero. Then, use the cross-multiplication rule where you set up three fractions: \( x \) over one calculation, \( y \) over another, and \( 1 \) over a third. Calculate each part, then solve for \( x \) and \( y \).

🎯 Exam Tip: When using the cross multiplication method, carefully write down the coefficients \( a_1, b_1, c_1, a_2, b_2, c_2 \) with their correct signs. A small sign error can lead to a completely wrong answer.

 

Question 5. A fraction such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get \( \frac{18}{11} \), but if the numerator is increased by 8 and the denominator is doubled, we get \( \frac{2}{5} \). Find the fraction.
Answer: Let the original fraction be \( \frac{x}{y} \), where \( x \) is the numerator and \( y \) is the denominator.

**Condition 1:** If the numerator is multiplied by 3 and the denominator is reduced by 3, the new fraction is \( \frac{18}{11} \).
\( \frac{3x}{y-3} = \frac{18}{11} \)
Cross-multiply:
\( 11(3x) = 18(y-3) \)
\( 33x = 18y - 54 \)
Rearrange into standard form \( ax + by + c = 0 \):
\( 33x - 18y + 54 = 0 \)
Divide by 3 to simplify:
\( 11x - 6y + 18 = 0 \quad \ldots(1) \)

**Condition 2:** If the numerator is increased by 8 and the denominator is doubled, the new fraction is \( \frac{2}{5} \).
\( \frac{x+8}{2y} = \frac{2}{5} \)
Cross-multiply:
\( 5(x+8) = 2(2y) \)
\( 5x + 40 = 4y \)
Rearrange into standard form \( ax + by + c = 0 \):
\( 5x - 4y + 40 = 0 \quad \ldots(2) \)

Now, we use the cross-multiplication method for equations (1) and (2).
For \( 11x - 6y + 18 = 0 \), we have \( a_1 = 11, b_1 = -6, c_1 = 18 \).
For \( 5x - 4y + 40 = 0 \), we have \( a_2 = 5, b_2 = -4, c_2 = 40 \).

The cross-multiplication formula is:
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)

Substituting the values, we get:
\( \frac{x}{(-6)(40) - (-4)(18)} = \frac{y}{(18)(5) - (40)(11)} = \frac{1}{(11)(-4) - (5)(-6)} \)

Calculate the denominators:
Denominator for \( x \): \( (-6)(40) - (-4)(18) = -240 - (-72) = -240 + 72 = -168 \)
Denominator for \( y \): \( (18)(5) - (40)(11) = 90 - 440 = -350 \)
Denominator for \( 1 \): \( (11)(-4) - (5)(-6) = -44 - (-30) = -44 + 30 = -14 \)

So the equation becomes:
\( \frac{x}{-168} = \frac{y}{-350} = \frac{1}{-14} \)

To find \( x \):
\( \frac{x}{-168} = \frac{1}{-14} \)
\( x = \frac{-168}{-14} \)
\( x = 12 \)

To find \( y \):
\( \frac{y}{-350} = \frac{1}{-14} \)
\( y = \frac{-350}{-14} \)
\( y = 25 \)

Thus, the numerator is 12 and the denominator is 25. Therefore, the required fraction is \( \frac{12}{25} \). This type of problem often involves setting up algebraic equations based on word descriptions.
In simple words: First, write down the original fraction as \( x/y \). Then, turn each given condition into an equation with \( x \) and \( y \). Solve these two equations using the cross-multiplication method to find the values of \( x \) and \( y \), which will give you the fraction.

🎯 Exam Tip: Always convert word problems into algebraic equations carefully, ensuring each condition translates correctly. Double-check your calculations, especially with negative signs, during the cross-multiplication steps.

 

Question 6. One number is greater than thrice the other number by 2. If 4 times the smaller number exceeds the greater by 5, find the numbers.
Answer: Let the greater number be \( x \) and the smaller number be \( y \).

**From the first condition:** "One number is greater than thrice the other number by 2."
This means the greater number \( x \) is 2 more than three times the smaller number \( y \).
\( x = 3y + 2 \)
Rearranging this equation gives:
\( x - 3y = 2 \quad \ldots(1) \)

**From the second condition:** "4 times the smaller number exceeds the greater by 5."
This means four times the smaller number \( (4y) \) is 5 more than the greater number \( (x) \).
\( 4y = x + 5 \)
Rearranging this equation gives:
\( -x + 4y = 5 \quad \ldots(2) \)

Now, we have a system of two linear equations:
1) \( x - 3y = 2 \)
2) \( -x + 4y = 5 \)

We can solve this system by adding equation (1) and equation (2). Notice that the \( x \) terms will cancel out:
\( (x - 3y) + (-x + 4y) = 2 + 5 \)
\( x - 3y - x + 4y = 7 \)
\( y = 7 \)

Now that we have the value of \( y \), substitute \( y = 7 \) into equation (1) to find \( x \):
\( x - 3(7) = 2 \)
\( x - 21 = 2 \)
\( x = 2 + 21 \)
\( x = 23 \)

So, the greater number is 23 and the smaller number is 7. These types of problems help in developing analytical thinking by converting verbal statements into mathematical expressions.
In simple words: Set up two equations based on the two statements given in the problem. Use \( x \) for the greater number and \( y \) for the smaller. Then, solve these equations together, for example, by adding them to cancel out one variable, to find the values of \( x \) and \( y \).

🎯 Exam Tip: When translating word problems, ensure that "exceeds by" or "is greater than by" are correctly represented. For example, "A exceeds B by 5" means \( A = B + 5 \).

 

Question 7. The cost of 11 pencils and 3 erasers is Rs 50 and the cost of 8 pencils and 3 erasers is Rs 38. Find the cost of 5 pencils and 5 erasers.
Answer: Let the cost of one pencil be Rs \( x \) and the cost of one eraser be Rs \( y \).

**From the first condition:** "The cost of 11 pencils and 3 erasers is Rs 50."
This can be written as an equation:
\( 11x + 3y = 50 \quad \ldots(1) \)

**From the second condition:** "The cost of 8 pencils and 3 erasers is Rs 38."
This can be written as:
\( 8x + 3y = 38 \quad \ldots(2) \)

Now we have a system of two linear equations. We can solve it by subtracting equation (2) from equation (1) to eliminate \( y \):
\( (11x + 3y) - (8x + 3y) = 50 - 38 \)
\( 11x + 3y - 8x - 3y = 12 \)
\( 3x = 12 \)
To find \( x \), divide both sides by 3:
\( x = \frac{12}{3} \)
\( x = 4 \)
So, the cost of one pencil is Rs 4.

Next, substitute the value of \( x = 4 \) into equation (1) to find \( y \):
\( 11(4) + 3y = 50 \)
\( 44 + 3y = 50 \)
Subtract 44 from both sides:
\( 3y = 50 - 44 \)
\( 3y = 6 \)
To find \( y \), divide both sides by 3:
\( y = \frac{6}{3} \)
\( y = 2 \)
So, the cost of one eraser is Rs 2.

Finally, we need to find the cost of 5 pencils and 5 erasers:
Cost \( = 5 \times (\text{cost of one pencil}) + 5 \times (\text{cost of one eraser}) \)
Cost \( = 5(4) + 5(2) \)
Cost \( = 20 + 10 \)
Cost \( = 30 \)
The total required cost is Rs 30. Solving such problems helps in daily life calculations involving purchases.
In simple words: First, create two equations from the given information about pencils and erasers. Solve these equations to find the cost of one pencil \( (x) \) and one eraser \( (y) \). Then, use these single costs to find the total cost of 5 pencils and 5 erasers.

🎯 Exam Tip: Look for opportunities to use elimination when solving systems of equations, especially if coefficients of one variable are the same (like \( 3y \) in both equations here). This simplifies the calculation significantly.