Exercise 1 Algebra
Solve \(\frac{1}{3}(x + y - 5) = y - z = 2x - 11 = 9 - (x + 2z)\)
Show Solution
Step 1: Set each part equal to a common variable \( k \):
1. \(\frac{1}{3}(x + y - 5) = k\)
Step 2: From equation 3:
\(2x - 11 = k \Rightarrow x = \frac{k + 11}{2}\)
Step 3: From equation 1:
\(x + y - 5 = 3k \Rightarrow y = 3k - x + 5\)
Step 4: From equation 2:
\(z = y - k = (3k - x + 5) - k = 2k - x + 5\)
Step 5: Substitute \(x\) and \(z\) into equation 4:
\(9 - \left(\frac{k + 11}{2} + 2(2k - \frac{k + 11}{2} + 5)\right) = k\)
Step 6: Solve for \(k\):
After simplifying: \(k = 3\)
Step 7: Find \(x, y, z\):
\(x = \frac{3 + 11}{2} = 7\)
Final Answer: \(x = 7\), \(y = 7\), \(z = 4\)
Exercise 2 Word Problem
One hundred and fifty students are admitted to a school. They are distributed 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.
Show Solution
Step 1: Let:
A = Number of students in section A
Step 2: Total students equation:
\(A + B + C = 150\)
Step 3: After shifting 6 from A to C:
\(A - 6 = B = C + 6\)
So \(B = A - 6\) and \(C = A - 12\)
Step 4: Substitute into total:
\(A + (A - 6) + (A - 12) = 150\)
Step 5: Find B and C:
\(B = 56 - 6 = 50\)
Step 6: Verify second condition:
\(4C - A = 4(44) - 56 = 176 - 56 = 120\)
Step 7: Alternative approach:
Second condition: \(4C = A + B\)
Final Answer: Please verify the problem statement as there appears to be inconsistency in the given conditions.
Exercise 3 Number Theory
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 digit exceeds the unit digit. Find the original number.
Show Solution
Step 1: Let the number be \(100a + 10b + c\) where:
a = hundreds digit
Step 2: First condition (interchanging hundreds and tens):
\(100b + 10a + c = 3(100a + 10b + c) + 54\)
Step 3: Second condition (adding 198 reverses digits):
\(100a + 10b + c + 198 = 100c + 10b + a\)
Step 4: Third condition (relationship between digits):
\(b - a = 2(b - c)\)
Step 5: Substitute \(c = a + 2\) into \(b = 2c - a\):
\(b = 2(a + 2) - a = a + 4\)
Step 6: Substitute \(b\) and \(c\) into first condition:
\(70(a + 4) - 290a - 2(a + 2) = 54\)
Step 7: Find \(b\) and \(c\):
\(b = 1 + 4 = 5\)
Final Answer: The original number is 153
Exercise 4 Algebra
Find the least common multiple of \( xy(k^2 + 1) + k(x^2 + y^2) \) and \( xy(k^2 - 1) + k(x^2 - y^2) \)
Show Solution
Step 1: Let's denote the expressions:
\( P = xy(k^2 + 1) + k(x^2 + y^2) \)
Step 2: Factorize P:
\( P = xyk^2 + xy + kx^2 + ky^2 \)
Step 3: Factorize Q:
\( Q = xyk^2 - xy + kx^2 - ky^2 \)
Step 4: Identify common factors:
Both P and Q have the common factor \( (x + yk) \)
Step 5: Find LCM:
LCM = \( (x + yk) \times (kx + y)(kx - y) \)
Final Answer: The LCM is \( (x + yk)(k^2x^2 - y^2) \)
Exercise 5 Polynomials
Find the GCD of the following by division algorithm
\[ 2x^4 + 13x^3 + 27x^2 + 23x + 7, \, x^3 + 3x^2 + 3x + 1, \, x^2 + 2x + 1 \]
Show Solution
Step 1: First, find GCD of the last two polynomials:
\( f(x) = x^3 + 3x^2 + 3x + 1 \)
Step 2: Factorize \( g(x) \):
\( x^2 + 2x + 1 = (x + 1)^2 \)
Step 3: Check if \( (x + 1) \) divides \( f(x) \):
\( f(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0 \)
Step 4: Perform polynomial division or factorize \( f(x) \):
\( f(x) = (x + 1)(x^2 + 2x + 1) = (x + 1)^3 \)
Step 5: GCD of \( f(x) \) and \( g(x) \) is \( (x + 1)^2 \)
Step 6: Now find GCD with the first polynomial:
\( h(x) = 2x^4 + 13x^3 + 27x^2 + 23x + 7 \)
Step 7: Check if \( (x + 1) \) divides \( h(x) \):
\( h(-1) = 2 - 13 + 27 - 23 + 7 = 0 \)
Step 8: Perform polynomial division:
\( h(x) = (x + 1)(2x^3 + 11x^2 + 16x + 7) \)
Step 9: Check again for \( (x + 1) \):
\( 2(-1)^3 + 11(-1)^2 + 16(-1) + 7 = -2 + 11 - 16 + 7 = 0 \)
Step 10: Final factorization:
\( h(x) = (x + 1)^2(2x^2 + 9x + 7) \)
Step 11: The common factor with previous GCD is \( (x + 1)^2 \)
Final Answer: The GCD is \( (x + 1)^2 \)
Exercise 6 Rational Expressions
Reduce the given Rational expressions to its lowest form
\[\begin{array}{ccc}
\text{(i)} & \frac{x^{3a} - 8}{x^{2a} + 2x^a + 4} & \text{(ii)} \\
& \frac{10x^3 - 25x^2 + 4x - 10}{-4 - 10x^2} & \\
\end{array}\]
Show Solution
Part (i): Simplify \(\frac{x^{3a} - 8}{x^{2a} + 2x^a + 4}\)
Step 1: Recognize the numerator as difference of cubes:
\( x^{3a} - 8 = (x^a)^3 - 2^3 = (x^a - 2)(x^{2a} + 2x^a + 4) \)
Step 2: The denominator matches the second factor:
\( x^{2a} + 2x^a + 4 \) (same as denominator)
Step 3: Cancel common factors:
\( \frac{(x^a - 2)\cancel{(x^{2a} + 2x^a + 4)}}{\cancel{x^{2a} + 2x^a + 4}} = x^a - 2 \)
Part (ii): Simplify \(\frac{10x^3 - 25x^2 + 4x - 10}{-4 - 10x^2}\)
Step 1: Factor numerator by grouping:
\( (10x^3 - 25x^2) + (4x - 10) = 5x^2(2x - 5) + 2(2x - 5) = (5x^2 + 2)(2x - 5) \)
Step 2: Factor denominator:
\( -4 - 10x^2 = -2(5x^2 + 2) \)
Step 3: Cancel common factors:
\( \frac{(5x^2 + 2)(2x - 5)}{-2(5x^2 + 2)} = \frac{2x - 5}{-2} = \frac{5 - 2x}{2} \)
Final Answers:
Exercise 7 Algebra
Simplify
\[\frac{1}{p} + \frac{1}{q + r} \times \left( 1 + \frac{q^2 + r^2 - p^2}{2qr} \right)\]
\[p = q + r\]
Show Solution
Step 1: Substitute \( p = q + r \) into the expression:
\( \frac{1}{q + r} + \frac{1}{q + r} \times \left( 1 + \frac{q^2 + r^2 - (q + r)^2}{2qr} \right) \)
Step 2: Expand \( (q + r)^2 \) in the numerator:
\( q^2 + r^2 - (q^2 + 2qr + r^2) = -2qr \)
Step 3: Simplify the fraction:
\( \frac{-2qr}{2qr} = -1 \)
Step 4: Now the expression becomes:
\( \frac{1}{q + r} + \frac{1}{q + r} \times (1 - 1) = \frac{1}{q + r} + 0 \)
Final Answer: \( \frac{1}{q + r} \)
Exercise 8 Work Problems
Arul, Madan and Ram working together can clean a store in 6 hours. Working alone, Madan 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?
Show Solution
Step 1: Let Arul's time alone be \( x \) hours
Step 2: Then:
Madan's time = \( 2x \) hours
Step 3: Work rates (portion of job per hour):
Arul: \( \frac{1}{x} \)
Step 4: Combined work rate:
\( \frac{1}{x} + \frac{1}{2x} + \frac{1}{3x} = \frac{1}{6} \)
Step 5: Find common denominator (6x):
\( \frac{6}{6x} + \frac{3}{6x} + \frac{2}{6x} = \frac{1}{6} \)
Step 6: Solve for \( x \):
\( 66 = 6x \)
Step 7: Find others' times:
Madan: \( 2 \times 11 = 22 \) hours
Final Answer:
Exercise 9 Algebra
Find the square root of
\[289x^4 - 612x^3 + 970x^2 - 684x + 361.\]
Show Solution
Step 1: Assume the square root is of the form:
\( \sqrt{289x^4 - 612x^3 + 970x^2 - 684x + 361} = ax^2 + bx + c \)
Step 2: Square both sides:
\( (ax^2 + bx + c)^2 = a^2x^4 + 2abx^3 + (2ac + b^2)x^2 + 2bcx + c^2 \)
Step 3: Compare coefficients:
1. \( a^2 = 289 \Rightarrow a = 17 \) (taking positive root)
Final Answer: \( 17x^2 - 18x + 19 \)
Exercise 10 Equations
Solve
\[\sqrt{y + 1} + \sqrt{2y - 5} = 3\]
Show Solution
Step 1: Determine domain:
\( y + 1 \geq 0 \Rightarrow y \geq -1 \)
Step 2: Let \( \sqrt{y + 1} = a \) and \( \sqrt{2y - 5} = b \), so \( a + b = 3 \)
Step 3: Square both sides:
\( a^2 + b^2 + 2ab = 9 \)
Step 4: Also \( a^2 - b^2 = (y + 1) - (2y - 5) = -y + 6 \)
Step 5: Factor difference of squares:
\( a^2 - b^2 = (a + b)(a - b) = 3(a - b) = -y + 6 \)
Step 6: Solve system:
\( a + b = 3 \)
Step 7: Substitute back and solve:
\( 2\sqrt{y + 1} = 3 + (6 - y)/3 \)
Step 8: Solve quadratic:
\( y = \frac{66 \pm \sqrt{4356 - 756}}{2} = \frac{66 \pm 60}{2} \)
Step 9: Check solutions:
For \( y = 63 \): \( \sqrt{64} + \sqrt{121} = 8 + 11 = 19 \neq 3 \) → invalid
Final Answer: \( y = 3 \)
Exercise 11 Motion Problems
A boat takes 1.6 hours longer to go 36 kms 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?
Show Solution
Step 1: Let \( x \) = speed of boat in still water (km/h)
Step 2: Downstream speed = \( x + 4 \) km/h
Step 3: Time difference equation:
\( \frac{36}{x - 4} - \frac{36}{x + 4} = 1.6 \)
Step 4: Multiply through by \( (x - 4)(x + 4) \):
\( 36(x + 4) - 36(x - 4) = 1.6(x^2 - 16) \)
Step 5: Solve for \( x^2 \):
\( 1.6x^2 = 313.6 \)
Final Answer: The boat's speed in still water is 14 km/h
Exercise 12 Geometry
Is it possible to design a rectangular park of perimeter 320 m and area 4800 \( m^2 \)? If so find its length and breadth.
Show Solution
Step 1: Let length = \( l \) m, breadth = \( b \) m
Step 2: Perimeter equation:
\( 2(l + b) = 320 \) ⇒ \( l + b = 160 \)
Step 3: Area equation:
\( l \times b = 4800 \)
Step 4: Form quadratic equation:
\( t^2 - (l + b)t + lb = 0 \) ⇒ \( t^2 - 160t + 4800 = 0 \)
Step 5: Solve quadratic:
\( t = \frac{160 \pm \sqrt{25600 - 19200}}{2} = \frac{160 \pm \sqrt{6400}}{2} = \frac{160 \pm 80}{2} \)
Step 6: Solutions:
\( t = 120 \) m or \( t = 40 \) m
Final Answer: Yes, possible with length = 120 m and breadth = 40 m
Exercise 13 Time Problems
At \( t \) minutes past 2 pm, the time needed to 3 pm is 3 minutes less than \(\frac{t^2}{4}\). Find \( t \).
Show Solution
Step 1: Time from 2 pm to 3 pm is 60 minutes
Step 2: Time remaining when it's \( t \) minutes past 2 pm:
\( 60 - t \)
Step 3: According to problem:
\( 60 - t = \frac{t^2}{4} - 3 \)
Step 4: Rearrange equation:
\( \frac{t^2}{4} + t - 63 = 0 \)
Step 5: Solve quadratic:
\( t = \frac{-4 \pm \sqrt{16 + 1008}}{2} = \frac{-4 \pm \sqrt{1024}}{2} = \frac{-4 \pm 32}{2} \)
Step 6: Solutions:
\( t = 14 \) or \( t = -18 \)
Final Answer: \( t = 14 \) minutes
Exercise 14 Word Problem
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.
Show Solution
Step 1: Let original number of rows = \( n \)
Step 2: Original total seats = \( n \times n = n^2 \)
Step 3: After changes:
New rows = \( 2n \)
Step 4: According to problem:
\( 2n(n - 5) = n^2 + 375 \)
Step 5: Expand and simplify:
\( 2n^2 - 10n = n^2 + 375 \)
Step 6: Solve quadratic:
\( n = \frac{10 \pm \sqrt{100 + 1500}}{2} = \frac{10 \pm \sqrt{1600}}{2} = \frac{10 \pm 40}{2} \)
Step 7: Solutions:
\( n = 25 \) or \( n = -15 \)
Final Answer: Original number of rows was 25
Exercise 15 Polynomials
If \(\alpha\) and \(\beta\) are the roots of the polynomial \(f(x) = x^2 - 2x + 3\), find the polynomial whose roots are (i) \(\alpha + 2, \beta + 2\) (ii) \(\frac{\alpha - 1}{\alpha + 1}, \frac{\beta - 1}{\beta + 1}\).
Show Solution
Part (i): Polynomial with roots \( \alpha + 2, \beta + 2 \)
Step 1: From given polynomial:
\( \alpha + \beta = 2 \) (sum of roots)
Step 2: For new roots \( \alpha + 2 \) and \( \beta + 2 \):
Sum: \( (\alpha + 2) + (\beta + 2) = \alpha + \beta + 4 = 2 + 4 = 6 \)
Step 3: New polynomial:
\( x^2 - (\text{sum})x + (\text{product}) = x^2 - 6x + 11 \)
Part (ii): Polynomial with roots \( \frac{\alpha - 1}{\alpha + 1}, \frac{\beta - 1}{\beta + 1} \)
Step 1: Let \( y = \frac{\alpha - 1}{\alpha + 1} \), solve for \( \alpha \):
\( y(\alpha + 1) = \alpha - 1 \)
Step 2: Since \( \alpha \) is root of original polynomial:
\( \left( \frac{y + 1}{1 - y} \right)^2 - 2\left( \frac{y + 1}{1 - y} \right) + 3 = 0 \)
Step 3: Multiply through by \( (1 - y)^2 \):
\( (y + 1)^2 - 2(y + 1)(1 - y) + 3(1 - y)^2 = 0 \)
Step 4: Expand and simplify:
\( y^2 + 2y + 1 - 2(1 - y^2) + 3(1 - 2y + y^2) = 0 \)
Final Answers:
Exercise 16 Quadratic Equations
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 \(q\).
Show Solution
Step 1: Since \(-4\) is root of first equation:
\( (-4)^2 + p(-4) - 4 = 0 \)
Step 2: Second equation \( x^2 + 3x + q = 0 \) has equal roots
Step 3: For equal roots, discriminant = 0:
\( b^2 - 4ac = 0 \)
Final Answer: \( p = 3 \), \( q = \frac{9}{4} \)
Exercise 17 Matrices
Two farmers Thilagan and Kausigan cultivates three varieties of grains namely rice, wheat and ragi. If the sale (in ₹) of three varieties of grains by both the farmers in the month of April is given by the matrix:
\[ A = \begin{pmatrix} 500 & 1000 & 1500 \\ 2500 & 1500 & 500 \end{pmatrix} \]
and the May month sale is exactly twice as that of the April month sale for each variety.
Show Solution
Part (i): Average sales of April and May
Step 1: May sales matrix (twice April):
\( B = 2A = \begin{pmatrix} 1000 & 2000 & 3000 \\ 5000 & 3000 & 1000 \end{pmatrix} \)
Step 2: Average sales:
\( \frac{A + B}{2} = \frac{A + 2A}{2} = \frac{3A}{2} = \begin{pmatrix} 750 & 1500 & 2250 \\ 3750 & 2250 & 750 \end{pmatrix} \)
Part (ii): Sales in August
Step 1: Each month sales double from previous month
Step 2: April to August is 4 months (May, June, July, August)
Step 3: August sales:
\( A \times 2^4 = 16A = \begin{pmatrix} 8000 & 16000 & 24000 \\ 40000 & 24000 & 8000 \end{pmatrix} \)
Final Answers:
Exercise 18 Matrices
If \( \cos \theta \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} + \sin \theta \begin{pmatrix} x & -\cos \theta \\ \cos \theta & x \end{pmatrix} = I_2 \), find \( x \).
Show Solution
Step 1: Perform the scalar multiplication:
\( \begin{pmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\cos \theta \sin \theta & \cos^2 \theta \end{pmatrix} + \begin{pmatrix} x \sin \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & x \sin \theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
Step 2: Add the matrices:
\( \begin{pmatrix} \cos^2 \theta + x \sin \theta & 0 \\ 0 & \cos^2 \theta + x \sin \theta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
Step 3: Equate corresponding elements:
\( \cos^2 \theta + x \sin \theta = 1 \)
Step 4: Solve for \( x \):
\( x \sin \theta = 1 - \cos^2 \theta \)
Final Answer: \( x = \sin \theta \)
Exercise 19 Matrices
Given \( A = \begin{pmatrix} p & 0 \\ 0 & 2 \end{pmatrix}, B = \begin{pmatrix} 0 & -q \\ 1 & 0 \end{pmatrix}, C = \begin{pmatrix} 2 & -2 \\ 2 & 2 \end{pmatrix} \) and if \( BA = C^2 \), find \( p \) and \( q \).
Show Solution
Step 1: Compute \( BA \):
\( BA = \begin{pmatrix} 0 & -q \\ 1 & 0 \end{pmatrix} \begin{pmatrix} p & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & -2q \\ p & 0 \end{pmatrix} \)
Step 2: Compute \( C^2 \):
\( C^2 = \begin{pmatrix} 2 & -2 \\ 2 & 2 \end{pmatrix} \begin{pmatrix} 2 & -2 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 4-4 & -4-4 \\ 4+4 & -4+4 \end{pmatrix} = \begin{pmatrix} 0 & -8 \\ 8 & 0 \end{pmatrix} \)
Step 3: Set \( BA = C^2 \):
\( \begin{pmatrix} 0 & -2q \\ p & 0 \end{pmatrix} = \begin{pmatrix} 0 & -8 \\ 8 & 0 \end{pmatrix} \)
Step 4: Equate corresponding elements:
\( -2q = -8 \) ⇒ \( q = 4 \)
Final Answer: \( p = 8 \), \( q = 4 \)
Exercise 20 Matrices
\( A = \begin{pmatrix} 3 & 0 \\ 4 & 5 \end{pmatrix}, B = \begin{pmatrix} 6 & 3 \\ 8 & 5 \end{pmatrix}, C = \begin{pmatrix} 3 & 6 \\ 1 & 1 \end{pmatrix} \) find the matrix \( D \), such that \( CD - AB = 0 \)
Show Solution
Step 1: Compute \( AB \):
\( AB = \begin{pmatrix} 3 & 0 \\ 4 & 5 \end{pmatrix} \begin{pmatrix} 6 & 3 \\ 8 & 5 \end{pmatrix} = \begin{pmatrix} 18+0 & 9+0 \\ 24+40 & 12+25 \end{pmatrix} = \begin{pmatrix} 18 & 9 \\ 64 & 37 \end{pmatrix} \)
Step 2: The equation \( CD - AB = 0 \) implies \( CD = AB \)
Step 3: To find \( D \), we need \( D = C^{-1} AB \)
Step 4: Find inverse of \( C \):
\( C = \begin{pmatrix} 3 & 6 \\ 1 & 1 \end{pmatrix} \)
Step 5: Multiply \( C^{-1} AB \):
\( D = \begin{pmatrix} -1/3 & 2 \\ 1/3 & -1 \end{pmatrix} \begin{pmatrix} 18 & 9 \\ 64 & 37 \end{pmatrix} \)
Final Answer: \( D = \begin{pmatrix} 122 & 71 \\ -58 & -34 \end{pmatrix} \)