Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Unit Exercise 3

Question 1. Solve \( \frac { 1 }{ 3 } (x + y - 5) = y - z = 2x - 11 = 9 - (x + 2z) \)
Answer:
Let's solve the given system of equations step by step to find the values of x, y, and z.
We have: \( \frac { 1 }{ 3 } (x + y - 5) = y - z = 2x - 11 = 9 - (x + 2z) \)

First, take the first two parts:
\( \frac { 1 }{ 3 } (x + y - 5) = y - z \)
Multiply both sides by 3 to clear the fraction:
\( x + y - 5 = 3(y - z) \)
\( x + y - 5 = 3y - 3z \)
Rearrange the terms to form a linear equation:
\( x + y - 3y + 3z = 5 \)
\( x - 2y + 3z = 5 \)....(1)

Next, take the second and third parts:
\( y - z = 2x - 11 \)
Rearrange the terms:
\( -2x + y - z = -11 \)
Multiply by -1 to make the x coefficient positive:
\( 2x - y + z = 11 \)....(2)

Then, take the third and fourth parts:
\( 2x - 11 = 9 - (x + 2z) \)
\( 2x - 11 = 9 - x - 2z \)
Move x and z terms to the left side and constants to the right side:
\( 2x + x + 2z = 9 + 11 \)
\( 3x + 2z = 20 \)....(3)

Now, we have a system of three linear equations:
1) \( x - 2y + 3z = 5 \)
2) \( 2x - y + z = 11 \)
3) \( 3x + 2z = 20 \)

We can eliminate 'y' from equations (1) and (2). Multiply equation (2) by 2:
(1) \( \times 1 \implies x - 2y + 3z = 5 \)....(1)
(2) \( \times 2 \implies 4x - 2y + 2z = 22 \)....(4)

Subtract equation (4) from equation (1):
\( (x - 2y + 3z) - (4x - 2y + 2z) = 5 - 22 \)
\( x - 2y + 3z - 4x + 2y - 2z = -17 \)
\( -3x + 0y + z = -17 \)
\( -3x + z = -17 \)
Multiply by -1:
\( 3x - z = 17 \)....(5)

Now we have two equations with x and z:
(3) \( 3x + 2z = 20 \)
(5) \( 3x - z = 17 \)

Subtract equation (5) from equation (3):
\( (3x + 2z) - (3x - z) = 20 - 17 \)
\( 3x + 2z - 3x + z = 3 \)
\( 3z = 3 \)
\( z = \frac{3}{3} \)
\( z = 1 \)

Substitute the value of \( z = 1 \) into equation (3):
\( 3x + 2(1) = 20 \)
\( 3x + 2 = 20 \)
\( 3x = 20 - 2 \)
\( 3x = 18 \)
\( x = \frac{18}{3} \)
\( x = 6 \)

Substitute the values of \( x = 6 \) and \( z = 1 \) into equation (2):
\( 2(6) - y + 1 = 11 \)
\( 12 - y + 1 = 11 \)
\( 13 - y = 11 \)
\( -y = 11 - 13 \)
\( -y = -2 \)
\( y = 2 \)

The values are \( x = 6 \), \( y = 2 \), and \( z = 1 \). This means we have found the unique solution for the system of linear equations.
In simple words: We solved a set of math puzzles, called equations, that were linked together. By carefully changing and combining them, we found out what numbers x, y, and z stand for.

🎯 Exam Tip: When solving a system of equations, systematically eliminate one variable at a time using addition or subtraction. Always double-check your final answers by plugging them back into the original equations.

Question 2. One hundred and fifty students are admitted to a school. They are distrbuted over three sections A, B and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section C exceeds the number of students of section A by the number of students in section B, find the number of students in the three sections.
Answer:
Let's find the number of students in each section by setting up equations.
Let the number of students in section A be "x".
Let the number of students in section B be "y".
Let the number of students in section C be "z".

From the first condition, the total number of students is 150:
\( x + y + z = 150 \)....(1)

From the second condition, if 6 students are moved from A to C, all sections have an equal number of students:
Students in A become \( x - 6 \).
Students in C become \( z + 6 \).
So, \( x - 6 = y = z + 6 \).
From \( x - 6 = z + 6 \):
\( x - z = 6 + 6 \)
\( x - z = 12 \)....(2)

From the third condition, 4 times the students in section C exceeds students in A by students in B:
\( 4z = x + y \)
Rearrange this equation:
\( x + y - 4z = 0 \)....(3)

Now we have the system of equations:
1) \( x + y + z = 150 \)
2) \( x - z = 12 \)
3) \( x + y - 4z = 0 \)

Subtract equation (3) from equation (1):
\( (x + y + z) - (x + y - 4z) = 150 - 0 \)
\( x + y + z - x - y + 4z = 150 \)
\( 0x + 0y + 5z = 150 \)
\( 5z = 150 \)
\( z = \frac{150}{5} \)
\( z = 30 \)

Substitute the value of \( z = 30 \) into equation (2):
\( x - 30 = 12 \)
\( x = 12 + 30 \)
\( x = 42 \)

Substitute the values of \( x = 42 \) and \( z = 30 \) into equation (1):
\( 42 + y + 30 = 150 \)
\( y + 72 = 150 \)
\( y = 150 - 72 \)
\( y = 78 \)

So, the number of students in section A is 42, in section B is 78, and in section C is 30. We made sure to follow all the conditions given in the problem statement for student distribution.
In simple words: We used clues about how 150 students were split into three groups to find out exactly how many students were in each group. We found 42 students in A, 78 in B, and 30 in C.

🎯 Exam Tip: Break down word problems into individual conditions and translate each condition into a mathematical equation. This systematic approach simplifies complex problems into manageable steps.

Question 3. In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens the tens digit exceeds the unit digit. Find the original number.
Answer:
Let's determine the original three-digit number using the given clues.
Let the hundreds digit be \( x \).
Let the tens digit be \( y \).
Let the unit digit be \( z \).
The original number can be written as \( 100x + 10y + z \).

By the first condition: When the tens and hundreds digits are interchanged, the new number is \( 100y + 10x + z \). This new number is 54 more than three times the original number.
\( 100y + 10x + z = 54 + 3(100x + 10y + z) \)
\( 100y + 10x + z = 54 + 300x + 30y + 3z \)
Rearrange the terms:
\( 10x - 300x + 100y - 30y + z - 3z = 54 \)
\( -290x + 70y - 2z = 54 \)
Divide the entire equation by -2 to simplify:
\( 145x - 35y + z = -27 \)....(1)

By the second condition: If 198 is added to the number, the digits are reversed. The reversed number is \( 100z + 10y + x \).
\( 198 + (100x + 10y + z) = 100z + 10y + x \)
Rearrange the terms:
\( 100x - x + 10y - 10y + z - 100z = -198 \)
\( 99x - 99z = -198 \)
Divide by 99:
\( x - z = -2 \)....(2)

By the third condition: The tens digit exceeds the hundreds digit by twice the tens digit exceeds the unit digit. (This sentence seems to have a small repetition "twice as that of the tens the tens digit exceeds the unit digit". Interpreting it as: "the tens digit exceeds the hundreds digit by twice the difference between the tens digit and the unit digit.")
So, \( y - x = 2(y - z) \)
\( y - x = 2y - 2z \)
Rearrange the terms:
\( -x + y - 2y + 2z = 0 \)
\( -x - y + 2z = 0 \)
Multiply by -1 to make \( x \) positive:
\( x + y - 2z = 0 \)....(3)

Now we have a system of three linear equations:
1) \( 145x - 35y + z = -27 \)
2) \( x - z = -2 \)
3) \( x + y - 2z = 0 \)

From (2), we can express \( z \) in terms of \( x \):
\( z = x + 2 \)

Substitute \( z = x + 2 \) into equation (3):
\( x + y - 2(x + 2) = 0 \)
\( x + y - 2x - 4 = 0 \)
\( -x + y - 4 = 0 \)
\( -x + y = 4 \)....(5)

Substitute \( z = x + 2 \) into equation (1):
\( 145x - 35y + (x + 2) = -27 \)
\( 146x - 35y + 2 = -27 \)
\( 146x - 35y = -27 - 2 \)
\( 146x - 35y = -29 \)....(4)

Now we have a system with two variables (x and y) from equations (4) and (5):
(4) \( 146x - 35y = -29 \)
(5) \( -x + y = 4 \)

From equation (5), \( y = x + 4 \). Substitute this into equation (4):
\( 146x - 35(x + 4) = -29 \)
\( 146x - 35x - 140 = -29 \)
\( 111x = -29 + 140 \)
\( 111x = 111 \)
\( x = \frac{111}{111} \)
\( x = 1 \)

Now find \( y \) using \( y = x + 4 \):
\( y = 1 + 4 \)
\( y = 5 \)

Now find \( z \) using \( z = x + 2 \):
\( z = 1 + 2 \)
\( z = 3 \)

The digits are \( x = 1 \), \( y = 5 \), and \( z = 3 \).
Therefore, the original number is \( 100(1) + 10(5) + 3 = 100 + 50 + 3 = 153 \). This three-digit number meets all the given conditions.
In simple words: We used math clues about a mystery three-digit number. By setting up equations for each clue and solving them step by step, we found the digits and the original number, which is 153.

🎯 Exam Tip: When dealing with digit problems, always represent the number using place values (100x + 10y + z). Carefully translate each condition into an equation, as misinterpretation is a common error.

Question 4. Find the least common multiple of \( xy(k^2 +1) + k(x^2 + y^2) \) and \( xy(k^2 - 1) + k(x^2 - y^2) \).
Answer:
Let's find the LCM of the two given expressions by factoring them.
First expression: \( xy(k^2 + 1) + k(x^2 + y^2) \)
\( = xyk^2 + xy + kx^2 + ky^2 \)
Rearrange and group terms:
\( = xyk^2 + kx^2 + xy + ky^2 \)
\( = kx(yk + x) + y(x + ky) \)
\( = (xk + y)(yk + x) \) or \( (xk + y)(x + yk) \)

Second expression: \( xy(k^2 - 1) + k(x^2 - y^2) \)
\( = xyk^2 - xy + kx^2 - ky^2 \)
Rearrange and group terms:
\( = xyk^2 + kx^2 - xy - ky^2 \)
\( = kx(yk + x) - y(x + ky) \)
This factoring looks incorrect based on the result in the OCR. Let's re-evaluate the grouping to match the provided solution's factors.

Let's try grouping differently for the first expression:
\( xyk^2 + xy + kx^2 + ky^2 \)
\( = kx^2 + xyk^2 + ky^2 + xy \)
\( = kx(x + yk) + y(ky + x) \)
\( = (kx + y)(x + yk) \)
This matches the OCR's \( (xk + y) (x + yk) \).

Now for the second expression:
\( xyk^2 - xy + kx^2 - ky^2 \)
\( = kx^2 + xyk^2 - ky^2 - xy \)
\( = kx(x + yk) - y(ky + x) \)
\( = (kx - y)(x + yk) \)
This matches the OCR's \( (x + yk)(xk - y) \).

So, we have:
First expression: \( (xk + y)(x + yk) \)
Second expression: \( (x + yk)(xk - y) \)

To find the Least Common Multiple (LCM), we take all unique factors with their highest power.
The common factor is \( (x + yk) \).
The unique factors are \( (xk + y) \) and \( (xk - y) \).

So, the L.C.M. is \( (x + yk)(xk + y)(xk - y) \).
We know that \( (A+B)(A-B) = A^2 - B^2 \). Here, \( (xk + y)(xk - y) = (xk)^2 - y^2 = x^2k^2 - y^2 \).
Therefore, the L.C.M. is \( (x + yk)(x^2k^2 - y^2) \). This result combines the unique parts after factoring, similar to finding the LCM of numbers.
In simple words: We took two long math expressions and broke them down into smaller parts that multiply together. Then, we found the smallest expression that both of the original expressions can divide into evenly.

🎯 Exam Tip: For LCM of polynomials, factorize each polynomial completely first. The LCM is the product of all unique factors, each raised to the highest power it appears in any factorization.

Question 5. Find the GCD of the following by division algorithm \( 2x^4 + 13x^3 + 27x^2 + 23x + 7 \), \( x^3 + 3x^2 + 3x + 1 \), \( x^2 + 2x + 1 \)
(i) Find the G.C.D. of p(x) and g(x)
Answer:
Let's find the Greatest Common Divisor (GCD) using the division algorithm for the given polynomials.
Let \( p(x) = 2x^4 + 13x^3 + 27x^2 + 23x + 7 \)
Let \( g(x) = x^3 + 3x^2 + 3x + 1 \)
Let \( r(x) = x^2 + 2x + 1 \)

(i) To find the G.C.D. of \( p(x) \) and \( g(x) \), we perform polynomial long division.
Divide \( p(x) \) by \( g(x) \):

2x + 7
________________
x³+3x²+3x+1 | 2x⁴+13x³+27x²+23x+7
- (2x⁴+ 6x³+ 6x²+ 2x)
________________
7x³+21x²+21x+7
- (7x³+21x²+21x+7)
________________
0

Since the remainder is 0, \( g(x) \) is the G.C.D. of \( p(x) \) and \( g(x) \).
So, G.C.D. \( (p(x), g(x)) = x^3 + 3x^2 + 3x + 1 \). This means that \( x^3 + 3x^2 + 3x + 1 \) can perfectly divide both \( p(x) \) and \( g(x) \).
In simple words: We divided the first long math expression by the second one, just like doing long division with numbers. Since there was no remainder, the second expression is the biggest common factor for both of them.

🎯 Exam Tip: Remember to arrange polynomials in descending order of powers before performing long division. If the remainder is zero, the divisor is the GCD; otherwise, continue the process with the divisor and the remainder.

Question 5. (ii) Find the G.C.D. of r(x) and the G.C.D. of p(x) and g(x)
Answer:
From part (i), we found G.C.D. \( (p(x), g(x)) = x^3 + 3x^2 + 3x + 1 \).
We are also given \( r(x) = x^2 + 2x + 1 \).
Now we need to find the G.C.D. of \( (x^3 + 3x^2 + 3x + 1) \) and \( (x^2 + 2x + 1) \).
Let's perform polynomial long division of \( x^3 + 3x^2 + 3x + 1 \) by \( x^2 + 2x + 1 \):

x + 1
_____________
x²+2x+1 | x³+3x²+3x+1
- (x³+2x²+ x)
_____________
x²+2x+1
- (x²+2x+1)
_____________
0

Since the remainder is 0, \( x^2 + 2x + 1 \) is the G.C.D.
Therefore, G.C.D. of the three polynomials \( p(x), g(x), r(x) \) is \( x^2 + 2x + 1 \). This means \( x^2 + 2x + 1 \) is the largest polynomial that divides all three original polynomials without a remainder.
In simple words: We found the biggest common factor of the first two math expressions. Then, we used that answer and the third math expression to find the biggest common factor for all three. The final common factor is \( x^2 + 2x + 1 \).

🎯 Exam Tip: When finding the GCD of three or more polynomials, find the GCD of any two first, and then find the GCD of that result with the third polynomial. This breaks down the problem effectively.

Question 6. Reduce the given Rational expressions to its lowest form
(i) \( \frac{x^{3 a}-8}{x^{2 a}+2 x^{a}+4} \)
Answer:
To reduce the rational expression, we need to factorize the numerator and the denominator.
The numerator is \( x^{3a} - 8 \). This can be written as \( (x^a)^3 - 2^3 \).
This is in the form of \( a^3 - b^3 \), which factors as \( (a - b)(a^2 + ab + b^2) \).
Here, \( a = x^a \) and \( b = 2 \).
So, \( x^{3a} - 8 = (x^a - 2)((x^a)^2 + x^a \cdot 2 + 2^2) \)
\( = (x^a - 2)(x^{2a} + 2x^a + 4) \)

Now, substitute this back into the original expression:
\( \frac{x^{3 a}-8}{x^{2 a}+2 x^{a}+4} = \frac{(x^a - 2)(x^{2a} + 2x^a + 4)}{x^{2 a}+2 x^{a}+4} \)

We can cancel out the common term \( (x^{2a} + 2x^a + 4) \) from both the numerator and the denominator, as long as it's not zero. This simplifies the expression to its simplest form.
\( = x^a - 2 \)
This simplified form is the reduced version of the original complex fraction. It's a key step in simplifying algebraic fractions.
In simple words: We took the top part of the fraction and broke it down using a special math rule. Then, we saw that a part of the top was the same as the bottom, so we canceled them out. This left us with a much simpler answer.

🎯 Exam Tip: Look for common algebraic identities like \( a^3 - b^3 \) or \( a^2 - b^2 \) to factorize expressions. This is often the first step in simplifying rational expressions.

Question 6. (ii) \( \frac{10 x^{3}-25 x^{2}+4 x-10}{-4-10 x^{2}} \)
Answer:
To reduce this rational expression, we factorize the numerator and the denominator.
Numerator: \( 10x^3 - 25x^2 + 4x - 10 \)
Group the terms:
\( = (10x^3 - 25x^2) + (4x - 10) \)
Factor out common terms from each group:
\( = 5x^2(2x - 5) + 2(2x - 5) \)
Factor out the common binomial \( (2x - 5) \):
\( = (2x - 5)(5x^2 + 2) \)

Denominator: \( -4 - 10x^2 \)
Factor out a common term, -2:
\( = -2(2 + 5x^2) \)
\( = -2(5x^2 + 2) \)

Now, substitute the factored forms back into the original expression:
\( \frac{10 x^{3}-25 x^{2}+4 x-10}{-4-10 x^{2}} = \frac{(2x - 5)(5x^2 + 2)}{-2(5x^2 + 2)} \)

We can cancel out the common term \( (5x^2 + 2) \) from both the numerator and the denominator, provided it is not zero.
\( = \frac{2x - 5}{-2} \)
We can rewrite this by splitting the terms or factoring the negative sign:
\( = - \frac{2x - 5}{2} = \frac{-(2x - 5)}{2} = \frac{-2x + 5}{2} = \frac{5 - 2x}{2} \)
Alternatively, from \( \frac{2x - 5}{-2} \), we can write \( \frac{2x}{-2} - \frac{5}{-2} = -x + \frac{5}{2} \). Both forms are correct. This step demonstrates that complex fractions can often be simplified to much simpler forms.
In simple words: We broke down the top and bottom parts of the fraction into their multiplying pieces. We found a part that was common to both and canceled it out. This made the fraction much easier to read and understand.

🎯 Exam Tip: When factorizing polynomials with four terms, try grouping. For the denominator, always look for common factors, including negative ones, to simplify the expression for easier cancellation.

Question 7. Simplify \( \frac{1}{p} + \frac{1}{q+r} \left[ 1 + \frac{q^2+r^2-p^2}{2qr} \right] \div \left[ \frac{1}{p} - \frac{1}{q+r} \right] \)
Answer:
Let's simplify the given expression step by step. This problem combines fraction arithmetic and algebraic identities.

First, simplify the term inside the square brackets: \( 1 + \frac{q^2+r^2-p^2}{2qr} \)
Combine with 1 by finding a common denominator:
\( = \frac{2qr}{2qr} + \frac{q^2+r^2-p^2}{2qr} \)
\( = \frac{2qr + q^2+r^2-p^2}{2qr} \)
Notice that \( q^2 + 2qr + r^2 \) is \( (q+r)^2 \).
So, \( = \frac{(q+r)^2 - p^2}{2qr} \)
This is in the form \( A^2 - B^2 = (A - B)(A + B) \), where \( A = (q+r) \) and \( B = p \).
\( = \frac{(q+r-p)(q+r+p)}{2qr} \)

Now let's work on the division part. The expression is \( \left[ \frac{1}{p} + \frac{1}{q+r} \right] \div \left[ \frac{1}{p} - \frac{1}{q+r} \right] \cdot \left[ 1 + \frac{q^2+r^2-p^2}{2qr} \right] \)
(The OCR shows \( \frac{1}{p} + \frac{1}{q+r} \) at the start, which suggests a modification to the original question. I will follow the OCR's provided solution steps to match the derived answer).

Let's re-evaluate the full expression as given in the OCR, which appears to be:
\( \left( \frac{1}{p} + \frac{1}{q+r} \right) \times \left( 1 + \frac{q^2+r^2-p^2}{2qr} \right) \div \left( \frac{1}{p} - \frac{1}{q+r} \right) \)

Let's simplify the first part: \( \frac{1}{p} + \frac{1}{q+r} \)
Common denominator is \( p(q+r) \):
\( = \frac{q+r}{p(q+r)} + \frac{p}{p(q+r)} \)
\( = \frac{p+q+r}{p(q+r)} \)

Now simplify the third part: \( \frac{1}{p} - \frac{1}{q+r} \)
Common denominator is \( p(q+r) \):
\( = \frac{q+r}{p(q+r)} - \frac{p}{p(q+r)} \)
\( = \frac{q+r-p}{p(q+r)} \)

Now, put all the simplified parts back into the expression:
\( \frac{p+q+r}{p(q+r)} \times \frac{(q+r-p)(q+r+p)}{2qr} \div \frac{q+r-p}{p(q+r)} \)

When dividing by a fraction, we multiply by its reciprocal:
\( \frac{p+q+r}{p(q+r)} \times \frac{(q+r-p)(q+r+p)}{2qr} \times \frac{p(q+r)}{q+r-p} \)

Now, cancel out common terms:
The \( p(q+r) \) in the first term's denominator cancels with the \( p(q+r) \) in the last term's numerator.
The \( (q+r-p) \) in the second term's numerator cancels with the \( (q+r-p) \) in the last term's denominator.

We are left with:
\( (p+q+r) \times \frac{(q+r+p)}{2qr} \)
Since \( (p+q+r) \) is the same as \( (q+r+p) \), we can write:
\( = \frac{(p+q+r)^2}{2qr} \). This is the fully simplified form of the expression. It required careful expansion and factoring.
In simple words: We had a big math problem with fractions and multiplications. We broke it into smaller pieces, solved each piece, and then put them back together. We also flipped some fractions for division. In the end, we got a much simpler answer.

🎯 Exam Tip: When simplifying complex algebraic expressions, always work inside parentheses and brackets first. Remember that division by a fraction is equivalent to multiplication by its reciprocal, and look for common factors to cancel.

Question 8. Arul, Ravi and Ram working together can clean a store in 6 hours. Working alone, Ravi takes twice as long to clean the store as Arul does. Ram needs three times as long as Arul does. How long would it take each if they are working alone?
Answer:
Let's figure out how long each person takes to clean the store alone. This is a work-rate problem.
Let the time taken by Arul to clean the store alone be \( x \) hours.
Let the time taken by Ravi to clean the store alone be \( y \) hours.
Let the time taken by Ram to clean the store alone be \( z \) hours.

The amount of work done in 1 hour by Arul is \( \frac{1}{x} \).
The amount of work done in 1 hour by Ravi is \( \frac{1}{y} \).
The amount of work done in 1 hour by Ram is \( \frac{1}{z} \).

From the first condition, Arul, Ravi, and Ram working together clean the store in 6 hours:
\( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{6} \)....(1)

From the second condition, Ravi takes twice as long as Arul:
\( y = 2x \)
This means \( \frac{1}{y} = \frac{1}{2x} \). Or, if we rearrange, \( \frac{1}{x} = \frac{2}{y} \).
So, \( \frac{1}{x} - \frac{2}{y} = 0 \)....(2)

From the third condition, Ram needs three times as long as Arul:
\( z = 3x \)
This means \( \frac{1}{z} = \frac{1}{3x} \). Or, if we rearrange, \( \frac{1}{x} = \frac{3}{z} \).
So, \( \frac{1}{x} - \frac{3}{z} = 0 \)....(3)

To simplify, let \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), and \( c = \frac{1}{z} \).
The system of equations becomes:
1) \( a + b + c = \frac{1}{6} \)
2) \( a - 2b = 0 \)
3) \( a - 3c = 0 \)

From (2), \( a = 2b \).
From (3), \( a = 3c \).
So, \( 2b = 3c \implies b = \frac{3}{2}c \)

Substitute \( a = 3c \) and \( b = \frac{3}{2}c \) into equation (1):
\( 3c + \frac{3}{2}c + c = \frac{1}{6} \)
To combine, find a common denominator (2):
\( \frac{6c}{2} + \frac{3c}{2} + \frac{2c}{2} = \frac{1}{6} \)
\( \frac{11c}{2} = \frac{1}{6} \)
\( c = \frac{1}{6} \times \frac{2}{11} \)
\( c = \frac{2}{66} = \frac{1}{33} \)

Now find \( a \) and \( b \):
\( a = 3c = 3 \times \frac{1}{33} = \frac{3}{33} = \frac{1}{11} \)
\( b = \frac{3}{2}c = \frac{3}{2} \times \frac{1}{33} = \frac{3}{66} = \frac{1}{22} \)

Recall that \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), \( c = \frac{1}{z} \).
\( \frac{1}{x} = \frac{1}{11} \implies x = 11 \) hours.
\( \frac{1}{y} = \frac{1}{22} \implies y = 22 \) hours.
\( \frac{1}{z} = \frac{1}{33} \implies z = 33 \) hours.

So, Arul takes 11 hours, Ravi takes 22 hours, and Ram takes 33 hours to clean the store alone. This problem shows how individual work rates combine to form a group rate.
In simple words: We figured out how long it takes each person to clean a store by themselves. Arul takes 11 hours, Ravi takes 22 hours (which is twice Arul's time), and Ram takes 33 hours (which is three times Arul's time).

🎯 Exam Tip: In work-rate problems, always express each person's contribution as a fraction of work done per unit time (e.g., \( \frac{1}{x} \) for x hours). This makes it easy to add their contributions when working together.

Question 9. Find the square root of \( 289x^4 - 612x^3 + 970x^2 - 684x + 361 \).
Answer:
To find the square root of a polynomial, we use the long division method, similar to finding the square root of a number.

Given polynomial: \( 289x^4 - 612x^3 + 970x^2 - 684x + 361 \)

Step 1: Find the square root of the first term, \( \sqrt{289x^4} = 17x^2 \). Write this as the first term of the quotient and also as the divisor.
Multiply \( 17x^2 \) by \( 17x^2 \) to get \( 289x^4 \). Subtract this from the polynomial.


17x² - 18x + 19
_________________
/ 289x⁴ - 612x³ + 970x² - 684x + 361
17x² | 289x⁴
- (289x⁴)
_________________
0 - 612x³ + 970x²


Step 2: Bring down the next two terms (\( -612x^3 + 970x^2 \)). Double the quotient obtained so far (\( 2 \times 17x^2 = 34x^2 \)). This is the new partial divisor. Find a term to append to \( 34x^2 \) such that when multiplied by this term, it matches the first term of the new dividend (\( -612x^3 \)).
\( -612x^3 \div 34x^2 = -18x \). So, append \( -18x \) to the divisor and the quotient.
New divisor: \( 34x^2 - 18x \). Multiply \( (34x^2 - 18x) \) by \( (-18x) \).
\( (34x^2 - 18x)(-18x) = -612x^3 + 324x^2 \). Subtract this.


17x² - 18x + 19
_________________
/ 289x⁴ - 612x³ + 970x² - 684x + 361
17x² | 289x⁴
- (289x⁴)
_________________
0 - 612x³ + 970x²
(34x²-18x) | -(-612x³ + 324x²)
_________________
0 + 646x² - 684x + 361


Step 3: Bring down the last two terms (\( -684x + 361 \)). Double the quotient obtained so far (\( 2 \times (17x^2 - 18x) = 34x^2 - 36x \)). This is the new partial divisor. Find a term to append such that when multiplied by this term, it matches the first term of the new dividend (\( 646x^2 \)).
\( 646x^2 \div 34x^2 = 19 \). So, append \( +19 \) to the divisor and the quotient.
New divisor: \( 34x^2 - 36x + 19 \). Multiply \( (34x^2 - 36x + 19) \) by \( (19) \).
\( (34x^2 - 36x + 19)(19) = 646x^2 - 684x + 361 \). Subtract this.


17x² - 18x + 19
_________________
/ 289x⁴ - 612x³ + 970x² - 684x + 361
17x² | 289x⁴
- (289x⁴)
_________________
0 - 612x³ + 970x²
(34x²-18x) | -(-612x³ + 324x²)
_________________
0 + 646x² - 684x + 361
(34x²-36x+19)| -(646x² - 684x + 361)
_________________
0

Since the remainder is 0, the square root of the polynomial is \( |17x^2 - 18x + 19| \). The absolute value is used because a square root can be positive or negative. Polynomial long division is a powerful tool for this.
In simple words: We used a special long division method, like how you find the square root of a regular number, but with a long math expression instead. We kept dividing until nothing was left, and the answer we got on top is the square root.

🎯 Exam Tip: When finding the square root of a polynomial by long division, ensure the polynomial is written in descending powers of the variable. Remember to double the 'quotient' part when forming the next divisor, and pay close attention to signs during subtraction.

Question 10. Solve \( \sqrt { y+1 } + \sqrt { 2y-5 } = 3 \)
Answer:
Let's solve this equation involving square roots. This requires isolating the square roots and squaring both sides.
Given equation: \( \sqrt { y+1 } + \sqrt { 2y-5 } = 3 \)

To remove the square roots, we square both sides. First, it's often easier to isolate one square root.
Move one square root term to the other side:
\( \sqrt { y+1 } = 3 - \sqrt { 2y-5 } \)

Now, square both sides:
\( (\sqrt { y+1 })^2 = (3 - \sqrt { 2y-5 })^2 \)
\( y+1 = 3^2 - 2 \cdot 3 \cdot \sqrt { 2y-5 } + (\sqrt { 2y-5 })^2 \)
\( y+1 = 9 - 6\sqrt { 2y-5 } + (2y-5) \)
\( y+1 = 9 - 5 + 2y - 6\sqrt { 2y-5 } \)
\( y+1 = 4 + 2y - 6\sqrt { 2y-5 } \)

Now, isolate the remaining square root term:
\( 6\sqrt { 2y-5 } = 4 + 2y - y - 1 \)
\( 6\sqrt { 2y-5 } = y + 3 \)

Square both sides again to eliminate the last square root:
\( (6\sqrt { 2y-5 })^2 = (y + 3)^2 \)
\( 36(2y-5) = y^2 + 2(3)y + 3^2 \)
\( 72y - 180 = y^2 + 6y + 9 \)

Move all terms to one side to form a quadratic equation:
\( 0 = y^2 + 6y - 72y + 9 + 180 \)
\( 0 = y^2 - 66y + 189 \)

Now, solve the quadratic equation \( y^2 - 66y + 189 = 0 \). We can factor this equation.
We need two numbers that multiply to 189 and add up to -66.
Let's list factors of 189: (1, 189), (3, 63), (7, 27), (9, 21).
Notice that \( -3 \) and \( -63 \) multiply to 189 and add to \( -66 \).
So, \( (y - 3)(y - 63) = 0 \)

This gives two possible solutions for \( y \):
\( y - 3 = 0 \implies y = 3 \)
\( y - 63 = 0 \implies y = 63 \)

We must check these solutions in the original equation to ensure they are valid (to avoid extraneous solutions that sometimes arise from squaring both sides).

Check \( y = 3 \):
\( \sqrt { 3+1 } + \sqrt { 2(3)-5 } = \sqrt{4} + \sqrt{6-5} = \sqrt{4} + \sqrt{1} = 2 + 1 = 3 \)
This is true, so \( y = 3 \) is a valid solution.

Check \( y = 63 \):
\( \sqrt { 63+1 } + \sqrt { 2(63)-5 } = \sqrt{64} + \sqrt{126-5} = \sqrt{64} + \sqrt{121} = 8 + 11 = 19 \)
This is not equal to 3, so \( y = 63 \) is an extraneous solution and is not valid.

Thus, the only valid solution is \( y = 3 \). Solving radical equations often requires this final verification step.
In simple words: We had a math puzzle with square roots. We squared both sides of the equation two times to get rid of the square roots. This gave us a normal "y squared" problem. We solved that and found two possible answers, but only one of them worked when we put it back into the very first puzzle.

🎯 Exam Tip: Always check your solutions for radical equations in the original equation. Squaring both sides can introduce extraneous solutions that do not satisfy the original problem.

Question 11. A boat takes 1.6 hours longer to go 36 kins up a river than down the river. If the speed of the water current is 4 km per hr, what is the speed of the boat in still water?
Answer:
Let's find the speed of the boat in still water using the given information about its travel time upstream and downstream.
Let the speed of the boat in still water be \( x \) km/hr.
The speed of the water current is 4 km/hr.

When the boat goes downstream (with the current), its effective speed is \( (x + 4) \) km/hr.
When the boat goes upstream (against the current), its effective speed is \( (x - 4) \) km/hr. (Note: \( x \) must be greater than 4 for the boat to move upstream).

Distance covered is 36 km for both upstream and downstream travel.
Time taken = Distance / Speed.

Time taken to go 36 km downstream: \( T_{down} = \frac{36}{x+4} \) hours.
Time taken to go 36 km upstream: \( T_{up} = \frac{36}{x-4} \) hours.

The problem states that the boat takes 1.6 hours longer to go upstream than downstream.
So, \( T_{up} - T_{down} = 1.6 \)
\( \frac{36}{x-4} - \frac{36}{x+4} = 1.6 \)

To solve this equation, find a common denominator for the fractions on the left side:
\( \frac{36(x+4) - 36(x-4)}{(x-4)(x+4)} = 1.6 \)
\( \frac{36x + 144 - 36x + 144}{x^2 - 16} = 1.6 \)
\( \frac{288}{x^2 - 16} = 1.6 \)

To simplify 1.6, we can write it as \( \frac{16}{10} = \frac{8}{5} \).
So, \( \frac{288}{x^2 - 16} = \frac{8}{5} \)

Cross-multiply:
\( 288 \times 5 = 8 \times (x^2 - 16) \)
\( 1440 = 8x^2 - 128 \)

Add 128 to both sides:
\( 1440 + 128 = 8x^2 \)
\( 1568 = 8x^2 \)

Divide by 8:
\( x^2 = \frac{1568}{8} \)
\( x^2 = 196 \)

Take the square root of both sides:
\( x = \pm \sqrt{196} \)
\( x = \pm 14 \)

Since speed cannot be negative, we take the positive value.
\( x = 14 \)

The speed of the boat in still water is 14 km/hr. This speed is greater than the current's speed, so the boat can indeed travel upstream.
In simple words: We used how long a boat took to go up and down a river to find its speed without the river's push. By setting up an equation with the boat's speed (x) and the river's speed, we solved for x and found it was 14 km per hour.

🎯 Exam Tip: For boat and stream problems, remember that downstream speed is (boat speed + stream speed), and upstream speed is (boat speed - stream speed). Always ensure that the boat's speed is greater than the stream's speed for upstream travel to be possible.

Question 12. Is it possible to design a rectangular park of perimeter 320 m and area 4800 m2? If so find its length and breadth.
Answer:
Let's check if such a park can be designed and, if so, find its dimensions. This involves using the formulas for perimeter and area of a rectangle.
Let the length of the rectangular park be \( l \) meters.
Let the breadth of the rectangular park be \( b \) meters.

The perimeter of the park is given as 320 m.
Formula for perimeter: \( 2(l + b) = 320 \)
Divide by 2:
\( l + b = 160 \)....(1)

The area of the park is given as 4800 m\(^2\).
Formula for area: \( l \times b = 4800 \)....(2)

From equation (1), we can express \( l \) in terms of \( b \):
\( l = 160 - b \)

Substitute this expression for \( l \) into equation (2):
\( (160 - b) \times b = 4800 \)
\( 160b - b^2 = 4800 \)

Rearrange the terms to form a quadratic equation:
\( b^2 - 160b + 4800 = 0 \)

To determine if real solutions exist for \( b \) (and thus if such a park is possible), we can use the discriminant (\( \Delta = B^2 - 4AC \)) of the quadratic formula. For real solutions, \( \Delta \ge 0 \).
Here, \( A = 1 \), \( B = -160 \), \( C = 4800 \).
\( \Delta = (-160)^2 - 4(1)(4800) \)
\( \Delta = 25600 - 19200 \)
\( \Delta = 6400 \)

Since \( \Delta = 6400 \) which is \( > 0 \), there are two distinct real solutions for \( b \). This means it is indeed possible to design such a park.

Now, let's find the values of \( b \) by factoring the quadratic equation:
We need two numbers that multiply to 4800 and add up to -160.
Consider factors of 4800. We can try numbers related to 160. Half of 160 is 80. Numbers around 80 that multiply to 4800 are 120 and 40 (since \( 120 \times 40 = 4800 \) and \( 120 + 40 = 160 \)).
So, the factors are -120 and -40.
\( (b - 120)(b - 40) = 0 \)

This gives two possible values for \( b \):
\( b - 120 = 0 \implies b = 120 \)
\( b - 40 = 0 \implies b = 40 \)

If \( b = 120 \) m, then from \( l = 160 - b \):
\( l = 160 - 120 = 40 \) m.

If \( b = 40 \) m, then from \( l = 160 - b \):
\( l = 160 - 40 = 120 \) m.

Both sets of dimensions represent the same park, just with length and breadth swapped. Typically, length is considered greater than or equal to breadth.
So, the length of the park is 120 m and the breadth is 40 m. This confirms the park's feasibility and dimensions.
In simple words: We used the given perimeter and area of a park to see if it could exist. By writing down equations for its length and width, we found specific numbers that worked, meaning the park can be designed with a length of 120 meters and a breadth of 40 meters.

🎯 Exam Tip: For geometry problems involving quadratic equations, first set up the equations for perimeter and area. Then, use the discriminant (\( B^2 - 4AC \)) to quickly check if real solutions exist before proceeding to find the exact dimensions.

Question 13. At t minutes past 2 pm, the time needed to 3 pm is 3 minutes less than \( \frac{t^{2}}{4} \) Find t.
Answer: First, let's find the time remaining until 3 pm. From 2 pm to 3 pm is 60 minutes. So, 't' minutes past 2 pm means there are \( (60 - t) \) minutes left until 3 pm. According to the problem, this time remaining \( (60 - t) \) is 3 minutes less than \( \frac{t^{2}}{4} \). So we can write the equation: \( 60 - t = \frac{t^{2}}{4} - 3 \) Now, we need to solve this equation for 't'. Multiply the entire equation by 4 to remove the fraction: \( 4(60 - t) = t^{2} - 12 \) \( 240 - 4t = t^{2} - 12 \) Rearrange the terms to form a quadratic equation (move everything to one side): \( t^{2} + 4t - 12 - 240 = 0 \) \( t^{2} + 4t - 252 = 0 \) We need to find two numbers that multiply to -252 and add up to 4. These numbers are 18 and -14. So, we can factor the quadratic equation: \( (t + 18)(t - 14) = 0 \) This gives two possible values for 't': \( t + 18 = 0 \implies t = -18 \) \( t - 14 = 0 \implies t = 14 \) Since time cannot be negative, we take the positive value. Therefore, \( t = 14 \) minutes. This means 14 minutes past 2 pm, or 2:14 pm.In simple words: We set up an equation that says the time left until 3 pm (which is 60 minus 't') is 3 minutes less than \( t^2 \) divided by 4. Solving this equation gives us two answers, but because time cannot be negative, we pick the positive one. So, 't' is 14 minutes.

🎯 Exam Tip: Remember to always check if your calculated values make sense in the real world, especially for quantities like time, distance, or speed which cannot be negative.

Question 14. The number of seats in a row is equal to the total number of rows in a hall. The total number of seats in the hall will increase by 375 if the number of rows is doubled and the number of seats in each row is reduced by 5. Find the number of rows in the hall at the beginning.
Answer: Let 'x' be the original number of rows in the hall. Since the number of seats in a row is equal to the number of rows, the original number of seats in each row is also 'x'. The total number of seats in the hall at the beginning is \( \text{rows} \times \text{seats per row} = x \times x = x^2 \). Now, let's look at the new situation: The number of rows is doubled, so new rows \( = 2x \). The number of seats in each row is reduced by 5, so new seats per row \( = x - 5 \). The new total number of seats in the hall is \( (2x)(x - 5) \). The problem states that the total number of seats will increase by 375. So, New total seats \( = \) Original total seats \( + 375 \). \( (2x)(x - 5) = x^2 + 375 \) Now, we need to solve this equation for 'x': \( 2x^2 - 10x = x^2 + 375 \) Move all terms to one side to form a quadratic equation: \( 2x^2 - x^2 - 10x - 375 = 0 \) \( x^2 - 10x - 375 = 0 \) We need to find two numbers that multiply to -375 and add up to -10. These numbers are -25 and 15. So, we can factor the quadratic equation: \( (x - 25)(x + 15) = 0 \) This gives two possible values for 'x': \( x - 25 = 0 \implies x = 25 \) \( x + 15 = 0 \implies x = -15 \) Since the number of rows cannot be negative, we choose the positive value. Therefore, the number of rows in the hall at the beginning was 25. Each row also had 25 seats.In simple words: We start with 'x' rows and 'x' seats per row, making \( x^2 \) total seats. When rows double and seats per row drop by 5, the new total seats are \( (2x)(x-5) \). This new total is 375 more than the old total. We solve this math puzzle and find that the number of rows must be 25, as you cannot have negative rows.

🎯 Exam Tip: When setting up algebraic equations for word problems, carefully define your variables and translate each piece of information into a mathematical expression or equation. Always discard negative solutions for real-world quantities like counts or measurements.

Question 15. If \( \alpha \) and \( \beta \) are the roots of the polynomial \( f(x) - x^2 - 2x + 3 \), find the polynomial whose roots are
(i) \( \alpha + 2, \beta + 2 \)
Answer: We are given that \( \alpha \) and \( \beta \) are the roots of the polynomial \( x^2 - 2x + 3 = 0 \). For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is \( -\frac{b}{a} \) and the product of the roots is \( \frac{c}{a} \). In our case, \( a=1, b=-2, c=3 \). So, Sum of the roots: \( \alpha + \beta = -(\frac{-2}{1}) = 2 \) Product of the roots: \( \alpha\beta = \frac{3}{1} = 3 \) (i) We need to find a new polynomial whose roots are \( \alpha + 2 \) and \( \beta + 2 \). Let the new roots be \( \alpha' = \alpha + 2 \) and \( \beta' = \beta + 2 \). First, find the sum of these new roots: Sum of new roots \( = \alpha' + \beta' = (\alpha + 2) + (\beta + 2) \) \( = \alpha + \beta + 4 \) We already know \( \alpha + \beta = 2 \), so: Sum of new roots \( = 2 + 4 = 6 \) Next, find the product of these new roots: Product of new roots \( = \alpha' \beta' = (\alpha + 2)(\beta + 2) \) \( = \alpha\beta + 2\alpha + 2\beta + 4 \) \( = \alpha\beta + 2(\alpha + \beta) + 4 \) We know \( \alpha\beta = 3 \) and \( \alpha + \beta = 2 \), so: Product of new roots \( = 3 + 2(2) + 4 \) \( = 3 + 4 + 4 = 11 \) A quadratic polynomial with roots \( \alpha' \) and \( \beta' \) is given by the formula: \( x^2 - (\text{Sum of new roots})x + (\text{Product of new roots}) = 0 \) Substitute the values we found: \( x^2 - (6)x + 11 = 0 \) So, the new polynomial is \( x^2 - 6x + 11 = 0 \). This method helps to derive new polynomials from modified roots without solving for the original roots.In simple words: First, we found the sum and product of the original roots from the given equation. Then, we used these to find the sum and product of the new roots. Finally, we put these new sum and product values into the general formula for a quadratic equation to get the new polynomial.

🎯 Exam Tip: Remember the relationships between the roots and coefficients of a quadratic equation. This shortcut (sum and product of roots formulas) is vital for solving problems involving transformations of roots without needing to find the actual roots.

Question 15. (ii) \( \frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1} \)
Answer: From part (i), we know for \( x^2 - 2x + 3 = 0 \): \( \alpha + \beta = 2 \) \( \alpha\beta = 3 \) Let the new roots be \( \alpha'' = \frac{\alpha-1}{\alpha+1} \) and \( \beta'' = \frac{\beta-1}{\beta+1} \). First, find the sum of these new roots: Sum of new roots \( = \frac{\alpha-1}{\alpha+1} + \frac{\beta-1}{\beta+1} \) To add these fractions, we find a common denominator, which is \( (\alpha+1)(\beta+1) \): \( = \frac{(\alpha-1)(\beta+1) + (\beta-1)(\alpha+1)}{(\alpha+1)(\beta+1)} \) Expand the numerator: \( = \frac{(\alpha\beta + \alpha - \beta - 1) + (\alpha\beta + \beta - \alpha - 1)}{\alpha\beta + \alpha + \beta + 1} \) Combine like terms in the numerator: \( \alpha \) and \( -\alpha \) cancel out, \( -\beta \) and \( \beta \) cancel out. \( = \frac{2\alpha\beta - 2}{\alpha\beta + \alpha + \beta + 1} \) Substitute the values \( \alpha\beta = 3 \) and \( \alpha + \beta = 2 \): \( = \frac{2(3) - 2}{3 + 2 + 1} \) \( = \frac{6 - 2}{6} = \frac{4}{6} = \frac{2}{3} \) So, the sum of the new roots is \( \frac{2}{3} \). Next, find the product of these new roots: Product of new roots \( = (\frac{\alpha-1}{\alpha+1})(\frac{\beta-1}{\beta+1}) \) \( = \frac{(\alpha-1)(\beta-1)}{(\alpha+1)(\beta+1)} \) Expand the numerator and denominator: \( = \frac{\alpha\beta - \alpha - \beta + 1}{\alpha\beta + \alpha + \beta + 1} \) \( = \frac{\alpha\beta - (\alpha + \beta) + 1}{\alpha\beta + (\alpha + \beta) + 1} \) Substitute the values \( \alpha\beta = 3 \) and \( \alpha + \beta = 2 \): \( = \frac{3 - (2) + 1}{3 + (2) + 1} \) \( = \frac{3 - 2 + 1}{3 + 2 + 1} = \frac{2}{6} = \frac{1}{3} \) So, the product of the new roots is \( \frac{1}{3} \). Finally, form the new quadratic polynomial using the formula: \( x^2 - (\text{Sum of new roots})x + (\text{Product of new roots}) = 0 \) \( x^2 - (\frac{2}{3})x + (\frac{1}{3}) = 0 \) To remove fractions, multiply the entire equation by 3: \( 3x^2 - 2x + 1 = 0 \) This is the required polynomial. It is important to carefully manage algebraic fractions when combining terms for sums and products.In simple words: We used the sum and product of the original roots to find the sum and product of the new, more complex roots by doing fraction addition and multiplication. Then, we plugged these new sum and product values into the standard quadratic equation formula and cleared the fractions to get the final polynomial.

🎯 Exam Tip: Be very careful with algebraic manipulations, especially when dealing with fractions. Make sure to combine like terms correctly in the numerator and denominator before substituting the values of \( \alpha+\beta \) and \( \alpha\beta \).

Question 16. If -4 is a root of the equation \( x^2 + px - 4 = 0 \) and if the equation \( x^2 + px + q = 0 \) has equal roots, find the values of p and q.
Answer: We are given two conditions to find the values of 'p' and 'q'. Condition 1: -4 is a root of the equation \( x^2 + px - 4 = 0 \). If -4 is a root, it means when we substitute \( x = -4 \) into the equation, the equation holds true. \( (-4)^2 + p(-4) - 4 = 0 \) \( 16 - 4p - 4 = 0 \) \( 12 - 4p = 0 \) \( 12 = 4p \) \( p = \frac{12}{4} \) \( p = 3 \) So, we have found the value of \( p \). Condition 2: The equation \( x^2 + px + q = 0 \) has equal roots. For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, its discriminant \( \Delta \) must be zero. The discriminant is given by \( b^2 - 4ac \). In this equation, \( a=1, b=p, c=q \). So, \( b^2 - 4ac = 0 \) \( p^2 - 4(1)(q) = 0 \) \( p^2 - 4q = 0 \) We already found \( p = 3 \). Substitute this value into the equation: \( (3)^2 - 4q = 0 \) \( 9 - 4q = 0 \) \( 9 = 4q \) \( q = \frac{9}{4} \) Thus, we have found the value of \( q \). This problem shows how properties of roots are used to solve for unknown coefficients in polynomial equations.In simple words: First, we used the fact that -4 is a root of the first equation. We put -4 in place of 'x' to find 'p'. Then, we used the value of 'p' in the second equation. Since the second equation has "equal roots," we set its special number (called the discriminant) to zero, which helped us find 'q'.

🎯 Exam Tip: Remember two key concepts: if a value is a root, it satisfies the equation; and for equal roots, the discriminant \( b^2 - 4ac \) must always be zero.

Question 17. Two farmers Senthil and Ravi cultivate three varieties of grains namely rice, wheat and ragi. If the sale (in Rs) of three varieties of grains by both the farmers in the month of April is given by the matrix.
April sale in Rs
rice wheat ragi
\( A = \begin{bmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{bmatrix} \) Senthil
Ravi
and the May month sale (in Rs) is exactly twice as that of the April month sale for each variety.
(i) What is the average sales of the months April and May.
Answer: Let matrix A represent the sales in April: \( A = \begin{bmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{bmatrix} \) The problem states that the May month sale is exactly twice that of the April month sale for each variety. Let matrix B represent the sales in May. So, \( B = 2A \) \( B = 2 \begin{bmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{bmatrix} \) \( B = \begin{bmatrix} 2 \times 500 & 2 \times 1000 & 2 \times 1500 \\ 2 \times 2500 & 2 \times 1500 & 2 \times 500 \end{bmatrix} \) \( B = \begin{bmatrix} 1000 & 2000 & 3000 \\ 5000 & 3000 & 1000 \end{bmatrix} \) Now, we need to find the average sales of the months April and May. Average sales \( = \frac{1}{2} (A + B) \) First, calculate \( A + B \): \( A + B = \begin{bmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{bmatrix} + \begin{bmatrix} 1000 & 2000 & 3000 \\ 5000 & 3000 & 1000 \end{bmatrix} \) To add matrices, we add the corresponding elements: \( A + B = \begin{bmatrix} 500+1000 & 1000+2000 & 1500+3000 \\ 2500+5000 & 1500+3000 & 500+1000 \end{bmatrix} \) \( A + B = \begin{bmatrix} 1500 & 3000 & 4500 \\ 7500 & 4500 & 1500 \end{bmatrix} \) Now, divide by 2 to find the average: Average sales \( = \frac{1}{2} \begin{bmatrix} 1500 & 3000 & 4500 \\ 7500 & 4500 & 1500 \end{bmatrix} \) \( = \begin{bmatrix} \frac{1500}{2} & \frac{3000}{2} & \frac{4500}{2} \\ \frac{7500}{2} & \frac{4500}{2} & \frac{1500}{2} \end{bmatrix} \) \( = \begin{bmatrix} 750 & 1500 & 2250 \\ 3750 & 2250 & 750 \end{bmatrix} \) This matrix shows the average sales for each variety by each farmer over April and May. Understanding matrix operations simplifies calculations for large datasets.In simple words: First, we figured out the May sales by doubling the April sales for each item. Then, we added the April and May sales matrices together. Finally, we divided the total by two to get the average sales matrix.

🎯 Exam Tip: Remember that when performing scalar multiplication or addition/subtraction on matrices, the operation is applied to each individual element of the matrix. Always ensure matrices have compatible dimensions for addition or scalar multiplication.

Question 17. (ii) If the sales continue to increase in the same way in the successive months, what will be sales in the month of August?
Answer: We are given that May sales are 2 times April sales. This means the increase factor is 2 each month. Let's list the increase factor for each month compared to April sales: - April sale: \( 1 \times (\text{April sale}) \) - May sale: \( 2 \times (\text{April sale}) \) (as given) - June sale: \( 2 \times (\text{May sale}) = 2 \times (2 \times \text{April sale}) = 4 \times (\text{April sale}) \) - July sale: \( 2 \times (\text{June sale}) = 2 \times (4 \times \text{April sale}) = 8 \times (\text{April sale}) \) - August sale: \( 2 \times (\text{July sale}) = 2 \times (8 \times \text{April sale}) = 16 \times (\text{April sale}) \) So, the sales in the month of August will be 16 times the sales in April. Let A be the matrix for April sales: \( A = \begin{bmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{bmatrix} \) Sales in the month of August \( = 16 \times A \) \( = 16 \begin{bmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{bmatrix} \) Multiply each element of the matrix by 16: \( = \begin{bmatrix} 16 \times 500 & 16 \times 1000 & 16 \times 1500 \\ 16 \times 2500 & 16 \times 1500 & 16 \times 500 \end{bmatrix} \) \( = \begin{bmatrix} 8000 & 16000 & 24000 \\ 40000 & 24000 & 8000 \end{bmatrix} \) This shows how exponential growth applies to sales figures over time.In simple words: We know sales double each month. So, August sales will be 16 times the April sales (because May is 2x, June 4x, July 8x, and August 16x). We just multiply every number in the April sales matrix by 16 to get the August sales.

🎯 Exam Tip: Pay attention to keywords like "increase in the same way" which often imply a consistent multiplicative factor or exponential growth. Carefully track the multiplier for each successive period.

Question 18. If \( \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \cos\theta & -\sin\theta \\ x & \cos\theta \end{bmatrix} = I_2 \), find x.
Answer: We are given the matrix equation: \( \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \begin{bmatrix} \cos\theta & -\sin\theta \\ x & \cos\theta \end{bmatrix} = I_2 \) Here, \( I_2 \) is the 2x2 identity matrix, which is \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \). Let's perform the matrix multiplication on the left side. To multiply two matrices, we multiply rows by columns. The element in the first row, first column of the product matrix will be: \( (\cos\theta)(\cos\theta) + (\sin\theta)(x) = \cos^2\theta + x\sin\theta \) The element in the first row, second column of the product matrix will be: \( (\cos\theta)(-\sin\theta) + (\sin\theta)(\cos\theta) = -\cos\theta\sin\theta + \cos\theta\sin\theta = 0 \) The element in the second row, first column of the product matrix will be: \( (-\sin\theta)(\cos\theta) + (\cos\theta)(x) = -\sin\theta\cos\theta + x\cos\theta \) The element in the second row, second column of the product matrix will be: \( (-\sin\theta)(-\sin\theta) + (\cos\theta)(\cos\theta) = \sin^2\theta + \cos^2\theta = 1 \) So, the product matrix is: \( \begin{bmatrix} \cos^2\theta + x\sin\theta & 0 \\ -\sin\theta\cos\theta + x\cos\theta & 1 \end{bmatrix} \) Now, we set this equal to the identity matrix \( I_2 \): \( \begin{bmatrix} \cos^2\theta + x\sin\theta & 0 \\ -\sin\theta\cos\theta + x\cos\theta & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) For two matrices to be equal, their corresponding elements must be equal. From the first row, first column: \( \cos^2\theta + x\sin\theta = 1 \) We know the trigonometric identity \( \sin^2\theta + \cos^2\theta = 1 \). So, \( 1 - \sin^2\theta + x\sin\theta = 1 \) Subtract 1 from both sides: \( -\sin^2\theta + x\sin\theta = 0 \) Factor out \( \sin\theta \): \( \sin\theta(x - \sin\theta) = 0 \) This implies either \( \sin\theta = 0 \) or \( x - \sin\theta = 0 \). If \( \sin\theta = 0 \), then the value of x can be anything, which is not a unique solution. However, we also have the second row, first column element. From the second row, first column: \( -\sin\theta\cos\theta + x\cos\theta = 0 \) Factor out \( \cos\theta \): \( \cos\theta(x - \sin\theta) = 0 \) This implies either \( \cos\theta = 0 \) or \( x - \sin\theta = 0 \). For a unique value of x, we must consider the case where \( \sin\theta \neq 0 \) and \( \cos\theta \neq 0 \) (i.e., \( \theta \) is not a multiple of \( \frac{\pi}{2} \)). In this general case, for both equations to hold, we must have \( x - \sin\theta = 0 \).
\( \implies x = \sin\theta \) This value for x satisfies both conditions simultaneously, which is the most general solution.In simple words: We multiplied the two matrices on the left side of the equation. Then, we made this new matrix equal to the identity matrix (a special matrix with 1s on the diagonal and 0s elsewhere). By comparing the elements in the first row, first column, and second row, first column of both matrices, we found that 'x' must be equal to \( \sin\theta \) for all cases where sine and cosine are not both zero.

🎯 Exam Tip: When solving matrix equations, ensure you perform matrix multiplication correctly (row by column). For equations involving trigonometric functions, remember fundamental identities like \( \sin^2\theta + \cos^2\theta = 1 \) and consider all conditions implied by the matrix equality to find a consistent solution for the unknown variable.

Question 19. Given \( A=\begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix}, B=\begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix}, C=\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \) and if \( BA = C^2 \), find p and q.
Answer: We are given three matrices A, B, C, and a condition \( BA = C^2 \). We need to find the values of 'p' and 'q'. First, let's calculate \( BA \): \( BA = \begin{bmatrix} 0 & -q \\ 1 & 0 \end{bmatrix} \begin{bmatrix} p & 0 \\ 0 & 2 \end{bmatrix} \) Multiply rows of B by columns of A: \( BA = \begin{bmatrix} (0)(p) + (-q)(0) & (0)(0) + (-q)(2) \\ (1)(p) + (0)(0) & (1)(0) + (0)(2) \end{bmatrix} \) \( BA = \begin{bmatrix} 0 + 0 & 0 - 2q \\ p + 0 & 0 + 0 \end{bmatrix} \) \( BA = \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix} \) Next, let's calculate \( C^2 \), which is \( C \times C \): \( C^2 = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \) Multiply rows of C by columns of C: \( C^2 = \begin{bmatrix} (2)(2) + (-2)(2) & (2)(-2) + (-2)(2) \\ (2)(2) + (2)(2) & (2)(-2) + (2)(2) \end{bmatrix} \) \( C^2 = \begin{bmatrix} 4 - 4 & -4 - 4 \\ 4 + 4 & -4 + 4 \end{bmatrix} \) \( C^2 = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix} \) Now, we use the given condition \( BA = C^2 \): \( \begin{bmatrix} 0 & -2q \\ p & 0 \end{bmatrix} = \begin{bmatrix} 0 & -8 \\ 8 & 0 \end{bmatrix} \) For these two matrices to be equal, their corresponding elements must be equal. Comparing the element in the first row, second column: \( -2q = -8 \) Divide both sides by -2: \( q = \frac{-8}{-2} \) \( q = 4 \) Comparing the element in the second row, first column: \( p = 8 \) So, the values are \( p = 8 \) and \( q = 4 \). This example shows how matrix algebra can be used to solve for unknown elements within matrices.In simple words: First, we multiplied matrices B and A together. Then, we calculated C multiplied by itself (C squared). We set these two resulting matrices equal to each other. By matching up the numbers in the same positions in both matrices, we were able to find the values for 'p' and 'q'.

🎯 Exam Tip: Always perform matrix multiplication and squaring operations carefully, ensuring that each element is calculated correctly by summing the products of the corresponding row and column entries. Then, equate corresponding elements of the resulting matrices to solve for unknowns.

Question 20. Given \( A=\begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix}, B=\begin{bmatrix} 6 & 3 \\ 8 & 5 \end{bmatrix}, C=\begin{bmatrix} 3 & 6 \\ 1 & 1 \end{bmatrix} \) find the matrix D, such that \( CD - AB = 0 \)
Answer: We are given three matrices A, B, C, and an equation \( CD - AB = 0 \). We need to find matrix D. From the equation \( CD - AB = 0 \), we can rearrange it to \( CD = AB \). First, let's calculate the product \( AB \): \( AB = \begin{bmatrix} 3 & 0 \\ 4 & 5 \end{bmatrix} \begin{bmatrix} 6 & 3 \\ 8 & 5 \end{bmatrix} \) Multiply rows of A by columns of B: \( AB = \begin{bmatrix} (3)(6) + (0)(8) & (3)(3) + (0)(5) \\ (4)(6) + (5)(8) & (4)(3) + (5)(5) \end{bmatrix} \) \( AB = \begin{bmatrix} 18 + 0 & 9 + 0 \\ 24 + 40 & 12 + 25 \end{bmatrix} \) \( AB = \begin{bmatrix} 18 & 9 \\ 64 & 37 \end{bmatrix} \) Now we have the equation \( CD = AB \), which is: \( \begin{bmatrix} 3 & 6 \\ 1 & 1 \end{bmatrix} D = \begin{bmatrix} 18 & 9 \\ 64 & 37 \end{bmatrix} \) Let matrix D be a 2x2 matrix with elements \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \). So, \( \begin{bmatrix} 3 & 6 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 18 & 9 \\ 64 & 37 \end{bmatrix} \) Perform the matrix multiplication on the left side: \( \begin{bmatrix} 3a + 6c & 3b + 6d \\ 1a + 1c & 1b + 1d \end{bmatrix} = \begin{bmatrix} 18 & 9 \\ 64 & 37 \end{bmatrix} \) This gives us a system of four linear equations by equating corresponding elements: 1. \( 3a + 6c = 18 \) (Equation 1) 2. \( a + c = 64 \) (Equation 2) 3. \( 3b + 6d = 9 \) (Equation 3) 4. \( b + d = 37 \) (Equation 4) Let's solve for 'a' and 'c' using Equations 1 and 2. From Equation 2, \( a = 64 - c \). Substitute this into Equation 1: \( 3(64 - c) + 6c = 18 \) \( 192 - 3c + 6c = 18 \) \( 192 + 3c = 18 \) \( 3c = 18 - 192 \) \( 3c = -174 \) \( c = \frac{-174}{3} \) \( c = -58 \) Now, substitute \( c = -58 \) back into \( a = 64 - c \): \( a = 64 - (-58) \) \( a = 64 + 58 \) \( a = 122 \) Next, let's solve for 'b' and 'd' using Equations 3 and 4. From Equation 4, \( b = 37 - d \). Substitute this into Equation 3: \( 3(37 - d) + 6d = 9 \) \( 111 - 3d + 6d = 9 \) \( 111 + 3d = 9 \) \( 3d = 9 - 111 \) \( 3d = -102 \) \( d = \frac{-102}{3} \) \( d = -34 \) Now, substitute \( d = -34 \) back into \( b = 37 - d \): \( b = 37 - (-34) \) \( b = 37 + 34 \) \( b = 71 \) So, the elements of matrix D are \( a = 122, b = 71, c = -58, d = -34 \). Therefore, the matrix D is: \( D = \begin{bmatrix} 122 & 71 \\ -58 & -34 \end{bmatrix} \) This problem illustrates how matrix equations can be broken down into systems of linear equations to find unknown matrices.In simple words: First, we rewrote the main equation to be \( CD = AB \). We then calculated the product of matrices A and B. Next, we assumed matrix D had unknown letters for its numbers. We multiplied C by this unknown matrix D and set it equal to the AB product. This gave us four mini-equations that we solved to find all the numbers for matrix D.

🎯 Exam Tip: When solving for an unknown matrix in a matrix equation like \( CD = AB \), first calculate the known products (like AB). Then, represent the unknown matrix (D) with variables and perform the remaining matrix multiplication. Finally, equate corresponding elements to form and solve a system of linear equations.