CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 09

Question 1. If \( \frac{1}{2} \) is a root of the equation \( x^2 + kx - \frac{5}{4} = 0 \), then the value of k is
(a) 2
(b) -2
(c) \( \frac{1}{4} \)
(d) \( \frac{1}{2} \)
Answer: (a) 2
In simple words: Substitute \( x = \frac{1}{2} \) into the given equation to get \( \frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0 \). Solving this simple equation gives \( k = 2 \).

Exam Tip: Always plug the given root directly into the quadratic equation to find any unknown coefficient quickly.

Question 2. The non-zero root of \( 3x - 5x^2 = 0 \) is
(a) \( \frac{5}{3} \)
(b) \( \frac{3}{5} \)
(c) \( \frac{9}{5} \)
(d) \( \frac{5}{9} \)
Answer: (b) \( \frac{3}{5} \)
In simple words: Factor the equation as \( x(3 - 5x) = 0 \). Since we need the non-zero root, set \( 3 - 5x = 0 \), which gives \( x = \frac{3}{5} \).

Exam Tip: Avoid dividing by \( x \) directly as you might lose the zero root, although here only the non-zero root is asked.

Question 3. Which constant should be added and subtracted to solve the quadratic equation \( 4x^2 - \sqrt{3}x - 5 = 0 \) by the method of completing the square ?
(a) \( \frac{9}{16} \)
(b) \( \frac{3}{16} \)
(c) \( \frac{3}{4} \)
(d) \( \frac{\sqrt{3}}{4} \)
Answer: (b) \( \frac{3}{16} \)
In simple words: Write the expression as \( (2x)^2 - 2(2x)\left(\frac{\sqrt{3}}{4}\right) - 5 \). To complete the square, we need to add and subtract the square of \( \frac{\sqrt{3}}{4} \), which is \( \frac{3}{16} \).

Exam Tip: When the leading term is a perfect square like \( 4x^2 \), express it as \( (2x)^2 \) before determining the term to complete the square.

Question 4. The product of two successive positive integral multiples of 5 is 300. Then the numbers are
(a) 25, 30
(b) 15, 20
(c) 10, 15
(d) 30, 35
Answer: (b) 15, 20
In simple words: Among the options, only 15 and 20 are consecutive positive multiples of 5 whose product is exactly 300.

Exam Tip: Checking options is often much faster for multiple-choice questions than solving the quadratic equation from scratch.

Question 5. Value of ‘k’ for which the quadratic equation \( 2x^2 - kx + k = 0 \) has equal roots is
(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8
Answer: (d) 0, 8
In simple words: For equal roots, the discriminant must be 0. Thus, \( (-k)^2 - 4(2)(k) = 0 \implies k^2 - 8k = 0 \), which gives \( k = 0 \) or \( k = 8 \).

Exam Tip: Don't forget that \( k = 0 \) is also a mathematically valid root of the discriminant equation even if it changes the expression's behavior.

Question 6. If \( ax^2 + bx + c = 0 \) has equal roots, then c =
(a) \( \frac{-b}{2a} \)
(b) \( \frac{b}{2a} \)
(c) \( \frac{-b^2}{4a} \)
(d) \( \frac{b^2}{4a} \)
Answer: (d) \( \frac{b^2}{4a} \)
In simple words: Since the roots are equal, the discriminant \( b^2 - 4ac = 0 \). Solving for \( c \) gives \( c = \frac{b^2}{4a} \).

Exam Tip: Memorizing basic discriminant relationships helps to write direct relations for coefficients quickly during exams.

Question 7. If 2 is a root of the equation \( x^2 + bx + 12 = 0 \) and the equation \( x^2 + bx + q = 0 \) has equal roots, then q =
(a) 8
(b) -8
(c) 16
(d) -16
Answer: (c) 16
In simple words: Substitute \( x = 2 \) into the first equation to find \( b = -8 \). Then, substitute \( b = -8 \) into the second equation and set the discriminant to zero to get \( q = 16 \).

Exam Tip: Solve multi-step questions systematically by finishing one equation first before moving to the next.

Question 8. If \( x = 1 \) is a common root of the equation \( ax^2 + ax + 3 = 0 \) and \( x^2 + x + b = 0 \) then ab = _______
(a) 3
(b) 3.5
(c) 6
(d) -3
Answer: (a) 3
In simple words: Substitute \( x = 1 \) into both equations. This gives \( a = -1.5 \) and \( b = -2 \). Multiplying them yields \( ab = 3 \).

Exam Tip: Substituting a known root of 1 simplifies the equation to the sum of its coefficients, which makes solving very fast.

Question 9. The value of c for which the equation \( ax^2 + 2bx + c = 0 \) has equal roots is
(a) \( \frac{b^2}{a} \)
(b) \( \frac{b^2}{4a} \)
(c) \( \frac{a^2}{b} \)
(d) \( \frac{a^2}{4b} \)
Answer: (a) \( \frac{b^2}{a} \)
In simple words: Set the discriminant to zero: \( (2b)^2 - 4ac = 0 \implies 4b^2 - 4ac = 0 \). Solving for \( c \) gives \( c = \frac{b^2}{a} \).

Exam Tip: Be careful with the coefficient of \( x \); here it is \( 2b \), so its square is \( 4b^2 \), not \( b^2 \).

Question 10. If the sum and product of the roots of the equation \( kx^2 + 6x + 4k = 0 \) are equal then k =
(a) \( -\frac{3}{2} \)
(b) \( \frac{3}{2} \)
(c) \( \frac{2}{3} \)
(d) \( -\frac{2}{3} \)
Answer: (a) \( -\frac{3}{2} \)
In simple words: The sum of roots is \( -\frac{6}{k} \) and the product of roots is \( \frac{4k}{k} = 4 \). Equating them gives \( -\frac{6}{k} = 4 \implies k = -\frac{3}{2} \).

Exam Tip: The formulas for sum (\( -b/a \)) and product (\( c/a \)) of roots are highly useful shortcuts for such comparison problems.

Question 11. If one root of the equation \( 4x^2 - 2x + (\alpha - 4) = 0 \) be the reciprocal of the other then \( \alpha = \)
(a) 8
(b) -8
(c) 4
(d) -4
Answer: (a) 8
In simple words: If roots are reciprocals, their product is 1. Thus, \( \frac{\alpha - 4}{4} = 1 \implies \alpha - 4 = 4 \implies \alpha = 8 \).

Exam Tip: Remember that whenever roots are reciprocal to each other, the coefficient of \( x^2 \) must equal the constant term (\( a = c \)).

Question 12. Find two positive consecutive odd integers, sum of whose squares is 290
(a) 13, 14
(b) 11, 13
(c) -12, 11
(d) 7, 8
Answer: (b) 11, 13
In simple words: The numbers must be positive and odd. Checking option (b), \( 11^2 + 13^2 = 121 + 169 = 290 \), which fits perfectly.

Exam Tip: Eliminate options immediately if they do not satisfy simple conditions like being "odd" or "positive".

Question 13. If \( \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = 2\frac{1}{6} \) the value of x are
(a) \( \frac{9}{13}, \frac{4}{13} \)
(b) \( \frac{7}{13}, \frac{5}{13} \)
(c) \( \frac{9}{13}, \frac{7}{13} \)
(d) None of the options
Answer: (a) \( \frac{9}{13}, \frac{4}{13} \)
In simple words: Let \( y = \sqrt{\frac{x}{1-x}} \). The equation becomes \( y + \frac{1}{y} = \frac{13}{6} \). Solving this gives \( y = \frac{3}{2} \) or \( \frac{2}{3} \), which yields \( x = \frac{9}{13} \) or \( \frac{4}{13} \).

Exam Tip: Substituting a complex term with a temporary variable \( y \) simplifies the equation into a standard quadratic form.

Question 14. If \( \alpha, \beta \) are the roots of \( x^2 + px + 12 = 0 \) and \( \alpha - \beta = 1 \) then
(a) \( p = \pm 3 \)
(b) \( p = \pm 5 \)
(c) \( p = \pm 7 \)
(d) \( p = \pm 8 \)
Answer: (c) \( p = \pm 7 \)
In simple words: Use the identity \( (\alpha + \beta)^2 - (\alpha - \beta)^2 = 4\alpha\beta \). Substituting the values gives \( p^2 - 1 = 48 \implies p^2 = 49 \implies p = \pm 7 \).

Exam Tip: The algebraic identity linking sum, difference, and product of roots is extremely useful for such problems.

Question 15. The sum of a number and its square is 240. The number is
(a) 4
(b) 5
(c) 6
(d) 15
Answer: (d) 15
In simple words: Let the number be \( x \). The equation is \( x + x^2 = 240 \implies x^2 + x - 240 = 0 \). Factoring gives \( (x - 15)(x + 16) = 0 \). So the positive value is 15.

Exam Tip: You can quickly test options by adding each number to its square to see if it equals 240.

Question 16. The roots of the quadratic equation \( 3x^2 - 2\sqrt{6}x + 2 = 0 \) are
(a) \( \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}} \)
(b) \( \sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}} \)
(c) \( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \)
(d) \( \frac{4}{\sqrt{3}}, \frac{4}{\sqrt{3}} \)
Answer: (a) \( \sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}} \)
In simple words: Factor the expression as \( (\sqrt{3}x - \sqrt{2})^2 = 0 \). This yields two equal real roots: \( x = \frac{\sqrt{2}}{\sqrt{3}} = \sqrt{\frac{2}{3}} \).

Exam Tip: If the first and last terms are perfect squares, check if the middle term fits the perfect square trinomial formula \( (a-b)^2 \).

Question 17. The difference of two numbers is 5 and the difference of their reciprocals is \( \frac{1}{10} \), then numbers are
(a) \( \pm 5, \pm 10 \)
(b) 5, 10
(c) 10, 5
(d) -5, -10
Answer: (a) \( \pm 5, \pm 10 \)
In simple words: Let the numbers be \( x \) and \( y \). Since \( x - y = 5 \) and \( \frac{1}{y} - \frac{1}{x} = \frac{1}{10} \), solving gives the values as either (10, 5) or (-5, -10), which can be written together as \( \pm 5, \pm 10 \).

Exam Tip: Be mindful of signs when taking reciprocals, as the difference can be positive for both positive and negative pairs.

Question 18. The altitude of a right \( \Delta \) is 7 cm less than the base x and the hypotenuse is 13 cm. The quadratic representation of above equation is
(a) \( x(x - 7) = 13 \)
(b) \( x^2 + (x - 7)^2 = 13^2 \)
(c) \( x^2 + (x + 7)^2 = 13^2 \)
(d) None of the options
Answer: (b) \( x^2 + (x - 7)^2 = 13^2 \)
In simple words: Using the Pythagorean theorem, \( \text{base}^2 + \text{altitude}^2 = \text{hypotenuse}^2 \). Substituting base \( = x \) and altitude \( = x - 7 \) gives \( x^2 + (x - 7)^2 = 13^2 \).

Exam Tip: Always sketch a quick right-angled triangle to visualize base, altitude, and hypotenuse relations.

SA-1

Question 1. Does \( (x - 1)^2 + 2(x + 1) = 0 \) have a real root ? Justify your answer.
Answer: Simplifying the given equation:
\( x^2 - 2x + 1 + 2x + 2 = 0 \)
\( \implies x^2 + 3 = 0 \)
\( \implies x^2 = -3 \)
Since the square of any real number cannot be negative, there is no real value of \( x \) that satisfies this equation. Therefore, it does not have any real roots.
In simple words: Simplifying the equation gives \( x^2 = -3 \). No real number squared can equal a negative number, so there are no real roots.

Exam Tip: Alternatively, calculate the discriminant of \( x^2 + 0x + 3 = 0 \): \( D = 0^2 - 4(1)(3) = -12 < 0 \), confirming that the roots are not real.

Question 2. Is \( x(x + 1) + 8 = (x + 2)(x + 3) \) quadratic equation ? Justify it.
Answer: Let us simplify both sides:
LHS: \( x^2 + x + 8 \)
RHS: \( x^2 + 5x + 6 \)
Equating them:
\( x^2 + x + 8 = x^2 + 5x + 6 \)
\( \implies 4x - 2 = 0 \)
Since the highest power of \( x \) in this simplified equation is 1, it is a linear equation, not a quadratic equation.
In simple words: After simplification, the \( x^2 \) terms on both sides cancel each other out, so the equation is not quadratic.

Exam Tip: Do not conclude that an equation is quadratic just by looking at the unexpanded terms; always expand and simplify completely first.

Question 3. State whether the quadratic equation \( (x - 1)(x + 2) + 2 = 0 \) has two distinct real roots. Justify it.
Answer: First, write the equation in standard form:
\( x^2 + x - 2 + 2 = 0 \implies x^2 + x = 0 \)
Here, \( a = 1, b = 1, c = 0 \).
The discriminant is \( D = b^2 - 4ac = 1^2 - 4(1)(0) = 1 \).
Since \( D > 0 \), the quadratic equation has two distinct real roots.
In simple words: The equation simplifies to \( x(x + 1) = 0 \), which has two different real answers, 0 and -1.

Exam Tip: A positive discriminant \( D > 0 \) always guarantees two distinct and real roots.

Question 4. Is 0.2 a root of the equation \( x^2 - 0.4 = 0 \) ? Justify it.
Answer: To test if 0.2 is a root, plug \( x = 0.2 \) into the LHS of the equation:
LHS \( = (0.2)^2 - 0.4 = 0.04 - 0.4 = -0.36 \neq 0 \)
Since LHS is not equal to zero, 0.2 is not a root of the equation.
In simple words: Multiplying 0.2 by itself gives 0.04, not 0.4. Because \( 0.04 - 0.4 \) is not zero, 0.2 is not a solution.

Exam Tip: Be careful with decimal multiplications, as a common mistake is thinking \( 0.2 \times 0.2 \) equals \( 0.4 \).

Question 5. If one root of the quadratic equation \( 2x^2 - 3x + p = 0 \) is 3, find the other root of the quadratic equation. Also find the value of ‘p’.
Answer: Since 3 is a root, it must satisfy the equation:
\( 2(3)^2 - 3(3) + p = 0 \implies 18 - 9 + p = 0 \implies p = -9 \)
Now, the equation is \( 2x^2 - 3x - 9 = 0 \).
Let the second root be \( \beta \). The sum of roots is:
\( 3 + \beta = -\frac{b}{a} = \frac{3}{2} \implies \beta = \frac{3}{2} - 3 = -\frac{3}{2} \)
So, the other root is \( -\frac{3}{2} \) and \( p = -9 \).
In simple words: Plug 3 into the equation to find \( p = -9 \). Then, use the sum of roots formula to find that the other root is \( -1.5 \).

Exam Tip: Using root relationship formulas like sum (\( -b/a \)) or product (\( c/a \)) is faster than re-factoring the entire quadratic equation.

Question 6. Find the value of ‘c’ such that equation \( 4x^2 - 2(c + 1)x + (c + 4) = 0 \) has real and equal roots.
Answer: For equal roots, the discriminant \( D \) must be zero:
\( D = [-2(c + 1)]^2 - 4(4)(c + 4) = 0 \)
\( \implies 4(c + 1)^2 - 16(c + 4) = 0 \)
Divide by 4:
\( c^2 + 2c + 1 - 4c - 16 = 0 \implies c^2 - 2c - 15 = 0 \)
\( \implies (c - 5)(c + 3) = 0 \implies c = 5 \) or \( c = -3 \).
In simple words: Setting the discriminant to zero gives a quadratic equation in \( c \). Solving it gives the values of \( c \) as 5 and -3.

Exam Tip: Simplify the discriminant expression by dividing out any common constant factors before expanding it.

Question 7. Is it possible to design a rectangular park of perimeter 80 m and area 400 m\(^2\) ? If so find its length and breadth.
Answer: Let the length be \( L \) and breadth be \( B \).
Perimeter \( = 2(L + B) = 80 \implies B = 40 - L \)
Area \( = L \times B = 400 \implies L(40 - L) = 400 \implies L^2 - 40L + 400 = 0 \)
Checking the discriminant:
\( D = (-40)^2 - 4(1)(400) = 1600 - 1600 = 0 \)
Since \( D = 0 \), real roots exist, so it is possible.
Solving \( (L - 20)^2 = 0 \implies L = 20 \) m.
Then \( B = 40 - 20 = 20 \) m.
In simple words: Yes, it is possible. The park will have a length of 20 meters and a breadth of 20 meters, which makes it a square.

Exam Tip: A square is a special type of rectangle, so equal dimensions are perfectly acceptable here.

Question 8. If one root of quadratic equation with rational co-efficient is \( 2 + \sqrt{3} \). Find the other root.
Answer: Since irrational roots of quadratic equations with rational coefficients always occur in conjugate pairs, if one root is \( 2 + \sqrt{3} \), the other root must be its conjugate, which is \( 2 - \sqrt{3} \).
In simple words: Square-root roots always come in pairs with opposite signs. So, the partner root of \( 2 + \sqrt{3} \) is \( 2 - \sqrt{3} \).

Exam Tip: This conjugate pair rule only holds true when all coefficients of the quadratic equation are rational.

Question 9. The sum of a number and its reciprocal is \( 2\frac{1}{2} \). Find the number.
Answer: Let the number be \( x \).
\( x + \frac{1}{x} = \frac{5}{2} \implies 2(x^2 + 1) = 5x \implies 2x^2 - 5x + 2 = 0 \)
\( \implies (2x - 1)(x - 2) = 0 \implies x = 2 \) or \( x = \frac{1}{2} \).
The number is either 2 or \( \frac{1}{2} \).
In simple words: Solving the equation gives the number as 2, and its reciprocal is \( \frac{1}{2} \).

Exam Tip: The solutions to reciprocal-sum equations will always be reciprocals of one another.

Question 10. Find ‘k’ so that \( x^2 - 2x(1 + 3k) + 7(3 + 2k) = 0 \) have equal roots.
Answer: For equal roots, \( D = b^2 - 4ac = 0 \):
\( [-2(1 + 3k)]^2 - 4(1)(7(3 + 2k)) = 0 \)
\( \implies 4(1 + 3k)^2 - 28(3 + 2k) = 0 \)
Divide by 4:
\( 1 + 6k + 9k^2 - 21 - 14k = 0 \implies 9k^2 - 8k - 20 = 0 \)
\( \implies 9k^2 - 18k + 10k - 20 = 0 \implies (9k + 10)(k - 2) = 0 \)
Thus, \( k = 2 \) or \( k = -\frac{10}{9} \).
In simple words: Setting the discriminant to zero leads to a quadratic equation in \( k \). Solving it gives \( k = 2 \) or \( k = -\frac{10}{9} \).

Exam Tip: Be careful when expanding \( (1+3k)^2 \) to write all three terms, including the middle term \( 6k \).

Question 11. Form a quadratic equation whose roots are \( -\frac{1}{3} \) and \( \frac{5}{2} \).
Answer: Let the roots be \( \alpha = -\frac{1}{3} \) and \( \beta = \frac{5}{2} \).
Sum of roots \( = -\frac{1}{3} + \frac{5}{2} = \frac{13}{6} \)
Product of roots \( = \left(-\frac{1}{3}\right)\left(\frac{5}{2}\right) = -\frac{5}{6} \)
The quadratic equation is:
\( x^2 - (\text{sum})x + \text{product} = 0 \implies x^2 - \frac{13}{6}x - \frac{5}{6} = 0 \)
Multiply by 6:
\( 6x^2 - 13x - 5 = 0 \)
In simple words: Calculate the sum and product of the roots, substitute them into the formula, and clear the denominators.

Exam Tip: Expressing the final equation with integer coefficients by multiplying out denominators is standard practice.

Question 12. Find the nature of roots of \( 3\sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0 \)
Answer: Let us calculate the discriminant \( D \):
\( D = b^2 - 4ac = (-2\sqrt{2})^2 - 4(3\sqrt{3})(-2\sqrt{3}) \)
\( D = 8 + 72 = 80 \)
Since the discriminant \( D > 0 \), the equation has real and distinct roots.
In simple words: Since the discriminant is positive, the equation will have two different real roots.

Exam Tip: If \( D \) is positive but not a perfect square, the roots are real and distinct but irrational.

SA-II

Question 1. Find the roots of the quadratic equation \( 2x^2 - \sqrt{5}x - 2 = 0 \) using the quadratic formula.
Answer: Here, \( a = 2, b = -\sqrt{5}, c = -2 \).
Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-\sqrt{5}) \pm \sqrt{(-\sqrt{5})^2 - 4(2)(-2)}}{2(2)} \)
\( x = \frac{\sqrt{5} \pm \sqrt{5 + 16}}{4} = \frac{\sqrt{5} \pm \sqrt{21}}{4} \)
Thus, the roots are \( \frac{\sqrt{5} + \sqrt{21}}{4} \) and \( \frac{\sqrt{5} - \sqrt{21}}{4} \).
In simple words: Plug the coefficients into the quadratic formula and simplify the term under the square root to get the two roots.

Exam Tip: Leave the final answer in surd form if the term inside the square root is not a perfect square.

Question 2. The difference of square of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the number.
Answer: Let the larger number be \( x \) and smaller number be \( y \).
\( x^2 - y^2 = 45 \) and \( y^2 = 4x \)
Substitute \( y^2 \) in the first equation:
\( x^2 - 4x - 45 = 0 \implies (x - 9)(x + 5) = 0 \)
Since they are natural numbers, we choose \( x = 9 \).
Then \( y^2 = 4(9) = 36 \implies y = 6 \) (rejecting negative value).
Thus, the numbers are 9 and 6.
In simple words: Substituting the second relation into the first gives a quadratic equation in \( x \). Solving it gives the larger number as 9 and the smaller as 6.

Exam Tip: Since natural numbers are positive integers, always reject negative roots during the final selection.

Question 3. Solve : \( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \)
Answer: Factor by grouping terms:
\( b^2x(a^2x + 1) - 1(a^2x + 1) = 0 \)
\( \implies (b^2x - 1)(a^2x + 1) = 0 \)
This gives:
\( b^2x - 1 = 0 \implies x = \frac{1}{b^2} \)
\( a^2x + 1 = 0 \implies x = -\frac{1}{a^2} \)
So, the roots are \( \frac{1}{b^2} \) and \( -\frac{1}{a^2} \).
In simple words: Grouping terms allows us to factor out \( (a^2x + 1) \), which directly splits the equation into two simple linear parts.

Exam Tip: Factoring by grouping is much quicker than using the quadratic formula when dealing with variables as coefficients.

Question 4. If a student had walked 1 km/hr faster, he would have taken 15 minutes less to walk 3 km. Find the rate at which he was walking.
Answer: Let the original speed be \( x \) km/hr.
Original time \( = \frac{3}{x} \) hours.
New speed \( = x + 1 \) km/hr, and new time \( = \frac{3}{x+1} \) hours.
Given time difference is 15 minutes \( = \frac{15}{60} = \frac{1}{4} \) hour.
\( \frac{3}{x} - \frac{3}{x+1} = \frac{1}{4} \implies 3 \left(\frac{x+1-x}{x(x+1)}\right) = \frac{1}{4} \)
\( \implies \frac{3}{x^2+x} = \frac{1}{4} \implies x^2 + x - 12 = 0 \)
\( \implies (x + 4)(x - 3) = 0 \).
Since speed must be positive, we choose \( x = 3 \) km/hr.
In simple words: Setting up the relation for time difference leads to the equation \( x^2 + x - 12 = 0 \). Solving it gives the speed as 3 km/hr.

Exam Tip: Always convert time differences to hours if speeds are given in km/hr to keep units consistent.

Question 5. Determine the roots of \( 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \)
Answer: First, factor the constant term:
\( 2a^2 + 5ab + 2b^2 = (2a + b)(a + 2b) \)
Now write the quadratic equation as:
\( 9x^2 - 3[ (2a + b) + (a + 2b) ]x + (2a + b)(a + 2b) = 0 \)
This can be factored as:
\( [3x - (2a + b)][3x - (a + 2b)] = 0 \)
This yields:
\( x = \frac{2a + b}{3} \) and \( x = \frac{a + 2b}{3} \).
In simple words: Factor the last polynomial expression first, and then use those factors to split the middle term and solve.

Exam Tip: Splitting the middle term using factored constant components is a highly elegant way to solve complex algebraic quadratics.

Question 6. Solve for x : \( 10 \left( \frac{4x - 3}{2x + 1} \right) - 3 \left( \frac{2x + 1}{4x - 3} \right) = 3 \), \( x \neq -\frac{1}{2} \), \( x \neq \frac{3}{4} \)
Answer: Let \( y = \frac{4x - 3}{2x + 1} \).
The equation becomes \( 10y - \frac{3}{y} = 3 \implies 10y^2 - 3y - 3 = 0 \)
\( \implies 10y^2 - 6y + 5y - 3 = 0 \implies (2y + 1)(5y - 3) = 0 \)
Thus, \( y = -\frac{1}{2} \) or \( y = \frac{3}{5} \).

Case 1: \( \frac{4x - 3}{2x + 1} = -\frac{1}{2} \implies 8x - 6 = -2x - 1 \implies 10x = 5 \implies x = \frac{1}{2} \)

Case 2: \( \frac{4x - 3}{2x + 1} = \frac{3}{5} \implies 20x - 15 = 6x + 3 \implies 14x = 18 \implies x = \frac{9}{7} \)
The roots are \( x = \frac{1}{2} \) and \( x = \frac{9}{7} \).
In simple words: Replacing the fractional group with \( y \) simplifies the work. Solving the quadratic gives \( y \), which we then solve to find the real \( x \) values.

Exam Tip: Substitution reduces algebraic clutter and prevents arithmetic mistakes in complex rational equations.

Question 7. For what value of ‘m’, the equation \( 2mx^2 - 2(1 + 2m)x + (3 + 2m) = 0 \) have real but distinct roots ? When will the roots be equal ?
Answer: First, find the discriminant \( D \):
\( D = b^2 - 4ac = [-2(1 + 2m)]^2 - 4(2m)(3 + 2m) \)
\( D = 4(1 + 4m + 4m^2) - 8m(3 + 2m) = 4 + 16m + 16m^2 - 24m - 16m^2 = 4 - 8m \)

For real and distinct roots, we need \( D > 0 \):
\( 4 - 8m > 0 \implies m < \frac{1}{2} \) (where \( m \neq 0 \) to maintain a quadratic equation).

For equal roots, we need \( D = 0 \):
\( 4 - 8m = 0 \implies m = \frac{1}{2} \).
In simple words: Simplifying the discriminant gives \( D = 4 - 8m \). The roots are distinct if \( m < 1/2 \) (excluding 0) and equal if \( m = 1/2 \).

Exam Tip: Always state that \( m \neq 0 \) for the equation to remain quadratic, as the coefficient of \( x^2 \) must never be zero.

Question 8. Find the roots of the quadratic equation by using the quadratic formula in each of the following :
(i) \( -3x^2 + 5x + 12 = 0 \)
(ii) \( x^2 - 3\sqrt{5}x + 10 = 0 \)
(iii) \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \)
Answer:
(i) For \( -3x^2 + 5x + 12 = 0 \):
\( D = 5^2 - 4(-3)(12) = 25 + 144 = 169 \)
\( x = \frac{-5 \pm \sqrt{169}}{2(-3)} = \frac{-5 \pm 13}{-6} \implies x = -\frac{4}{3} \) or \( x = 3 \)

(ii) For \( x^2 - 3\sqrt{5}x + 10 = 0 \):
\( D = (-3\sqrt{5})^2 - 4(1)(10) = 45 - 40 = 5 \)
\( x = \frac{3\sqrt{5} \pm \sqrt{5}}{2} \implies x = 2\sqrt{5} \) or \( x = \sqrt{5} \)

(iii) For \( \frac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \), multiply by 2 to get \( x^2 - 2\sqrt{11}x + 2 = 0 \):
\( D = (-2\sqrt{11})^2 - 4(1)(2) = 44 - 8 = 36 \)
\( x = \frac{2\sqrt{11} \pm \sqrt{36}}{2} = \frac{2\sqrt{11} \pm 6}{2} = \sqrt{11} \pm 3 \)
The roots are \( \sqrt{11} + 3 \) and \( \sqrt{11} - 3 \).
In simple words: Apply the quadratic formula to each part. In part (iii), clearing the fraction first makes the calculations much cleaner.

Exam Tip: Be careful with signs when writing down the formula's terms, particularly when \( a \) or \( c \) is negative.

Question 9. Find the roots of the following quadratic equations by the factorization method.
(a) \( \frac{2}{5}x^2 - x - \frac{3}{5} = 0 \)
(b) \( 3\sqrt{2}x^2 - 5x - \sqrt{2} = 0 \)
Answer:
(a) Multiply by 5: \( 2x^2 - 5x - 3 = 0 \implies 2x^2 - 6x + x - 3 = 0 \)
\( \implies 2x(x - 3) + 1(x - 3) = 0 \implies (2x + 1)(x - 3) = 0 \)
Thus, the roots are \( x = 3 \) and \( x = -\frac{1}{2} \).

(b) For \( 3\sqrt{2}x^2 - 5x - \sqrt{2} = 0 \):
Product needed is \( 3\sqrt{2} \times (-\sqrt{2}) = -6 \) and sum is \( -5 \).
\( 3\sqrt{2}x^2 - 6x + x - \sqrt{2} = 0 \implies 3\sqrt{2}x(x - \sqrt{2}) + 1(x - \sqrt{2}) = 0 \)
\( \implies (3\sqrt{2}x + 1)(x - \sqrt{2}) = 0 \)
Thus, the roots are \( x = \sqrt{2} \) and \( x = -\frac{1}{3\sqrt{2}} = -\frac{\sqrt{2}}{6} \).
In simple words: Split the middle term based on the product of the first and last terms to group and factor.

Exam Tip: Always double check your split terms to ensure they sum to the middle coefficient and multiply to the correct product.

Question 9(c). Solve : \( 21x^2 - 2x + \frac{1}{21} = 0 \)
Answer: Multiply the entire equation by 21:
\( 441x^2 - 42x + 1 = 0 \)
This is a perfect square trinomial:
\( (21x - 1)^2 = 0 \implies 21x - 1 = 0 \implies x = \frac{1}{21} \)
Thus, the equation has equal roots: \( x = \frac{1}{21}, \frac{1}{21} \).
In simple words: Removing the fraction reveals that the equation is a simple perfect square, giving two identical roots at \( \frac{1}{21} \).

Exam Tip: Check for perfect squares (\( a^2 \pm 2ab + b^2 \)) to avoid using the quadratic formula needlessly.

Question 10. Find the discriminant of the equation \( 3x^2 - 2x + \frac{1}{3} = 0 \) and hence find the nature of its roots. Find them, if they are real.
Answer: Here, \( a = 3, b = -2, c = \frac{1}{3} \).
Discriminant \( D = b^2 - 4ac = (-2)^2 - 4(3)\left(\frac{1}{3}\right) = 4 - 4 = 0 \)
Since \( D = 0 \), the roots are real and equal.
The roots are given by:
\( x = -\frac{b}{2a} = \frac{2}{2(3)} = \frac{1}{3} \)
So, the roots are \( \frac{1}{3}, \frac{1}{3} \).
In simple words: The discriminant is exactly zero, meaning the equation has two identical real roots, which are both \( \frac{1}{3} \).

Exam Tip: Whenever \( D = 0 \), the root is always simplified to \( -b / 2a \).

Question 11. If -5 is a root of the quadratic equation \( 2x^2 + 2px - 15 = 0 \) and the quadratic equation \( p(x^2 + x) + k = 0 \) has equal roots, find the value of k.
Answer: Since -5 is a root of \( 2x^2 + 2px - 15 = 0 \):
\( 2(-5)^2 + 2p(-5) - 15 = 0 \implies 50 - 10p - 15 = 0 \implies 10p = 35 \implies p = \frac{7}{2} \)
Now substitute \( p = \frac{7}{2} \) into the second equation:
\( \frac{7}{2}(x^2 + x) + k = 0 \implies \frac{7}{2}x^2 + \frac{7}{2}x + k = 0 \)
Since it has equal roots, its discriminant must be 0:
\( \left(\frac{7}{2}\right)^2 - 4\left(\frac{7}{2}\right)k = 0 \implies \frac{49}{4} - 14k = 0 \implies 14k = \frac{49}{4} \implies k = \frac{7}{8} \)
Thus, the value of \( k \) is \( \frac{7}{8} \).
In simple words: Use the first equation to find \( p = \frac{7}{2} \). Put this into the second equation and make the discriminant zero to get \( k = \frac{7}{8} \).

Exam Tip: Be careful with fractional calculations when checking discriminant values to avoid arithmetic slips.

Question 12. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer: Let the base be \( x \) cm. Then the altitude is \( (x - 7) \) cm.
Using the Pythagorean theorem:
\( x^2 + (x - 7)^2 = 13^2 \implies x^2 + x^2 - 14x + 49 = 169 \)
\( \implies 2x^2 - 14x - 120 = 0 \implies x^2 - 7x - 60 = 0 \)
\( \implies (x - 12)(x + 5) = 0 \).
Since side length must be positive, we select \( x = 12 \) cm.
Base \( = 12 \) cm and altitude \( = 12 - 7 = 5 \) cm.
In simple words: Use the Pythagorean theorem to set up a quadratic equation. Solving it gives the base as 12 cm and the height as 5 cm.

Exam Tip: Recognizing standard Pythagorean triplets like (5, 12, 13) can immediately verify your result.

LONG ANSWER TYPE QUESTIONS

Question 1. Rs.6500 is divided equally among a certain number of persons. Had there been 15 more persons, each would have got Rs.30 less. Find the original number of persons.
Answer: Let the original number of persons be \( x \).
Original share of each person \( = \frac{6500}{x} \)
New share with 15 more persons \( = \frac{6500}{x + 15} \)
According to the problem:
\( \frac{6500}{x} - \frac{6500}{x + 15} = 30 \)
Divide by 30:
\( \frac{650}{3x} - \frac{650}{3(x + 15)} = 1 \implies 650 \left( \frac{15}{x^2 + 15x} \right) = 30 \implies \frac{3250}{x^2 + 15x} = 1 \)
\( \implies x^2 + 15x - 3250 = 0 \implies (x + 65)(x - 50) = 0 \)
Since the count of people must be positive, we choose \( x = 50 \).
The original number of people is 50.
In simple words: Solving the equation \( x^2 + 15x - 3250 = 0 \) gives the original number of people as 50.

Exam Tip: Simplify equations by dividing out large common constants early on to make factoring the quadratic easier.

Question 2. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed.
Answer: Let the original average speed be \( x \) km/hr.
Time taken for the first part \( = \frac{63}{x} \) hours.
Time taken for the second part \( = \frac{72}{x + 6} \) hours.
Total time:
\( \frac{63}{x} + \frac{72}{x + 6} = 3 \)
Divide by 3:
\( \frac{21}{x} + \frac{24}{x + 6} = 1 \implies 21(x + 6) + 24x = x(x + 6) \)
\( \implies 45x + 126 = x^2 + 6x \implies x^2 - 39x - 126 = 0 \)
\( \implies (x - 42)(x + 3) = 0 \).
Since speed must be positive, \( x = 42 \) km/hr.
The original average speed was 42 km/hr.
In simple words: Form an equation based on the total travel time of 3 hours. Solving it gives the original train speed as 42 km/hr.

Exam Tip: Dividing both sides by 3 simplifies the numerical coefficients and makes factoring much faster.

Question 3. A plane left 40 minutes late due to bad weather and in order to reach its destination 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Answer: Let the usual speed of the plane be \( x \) km/hr.
Usual time \( = \frac{1600}{x} \) hours, and new time \( = \frac{1600}{x + 400} \) hours.
Given time difference is 40 minutes \( = \frac{40}{60} = \frac{2}{3} \) hours.
\( \frac{1600}{x} - \frac{1600}{x + 400} = \frac{2}{3} \)
Divide by 2:
\( \frac{800}{x} - \frac{800}{x + 400} = \frac{1}{3} \implies 800 \left(\frac{400}{x(x + 400)}\right) = \frac{1}{3} \)
\( \implies x^2 + 400x - 960000 = 0 \implies (x + 1200)(x - 800) = 0 \)
Since speed must be positive, \( x = 800 \) km/hr.
The usual speed of the plane is 800 km/hr.
In simple words: Setting up the time difference equation leads to \( x^2 + 400x - 960000 = 0 \). Solving it gives the speed as 800 km/hr.

Exam Tip: With very large numbers like 960000, look at the non-zero digits (96 and 4) to find the factors (1200 and 800) easily.

Question 4. If the roots of the equation \( (a^2 + b^2)x^2 - 2(ac + bd)x + (c^2 + d^2) = 0 \) are equal, prove that \( \frac{a}{b} = \frac{c}{d} \)
Answer: Since roots are equal, the discriminant \( D \) must be zero:
\( D = [-2(ac + bd)]^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \)
\( \implies 4(ac + bd)^2 - 4(a^2 + b^2)(c^2 + d^2) = 0 \)
Divide by 4 and expand:
\( (a^2c^2 + 2abcd + b^2d^2) - (a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 0 \)
\( \implies 2abcd - a^2d^2 - b^2c^2 = 0 \)
Multiply by -1:
\( a^2d^2 - 2abcd + b^2c^2 = 0 \implies (ad - bc)^2 = 0 \)
\( \implies ad - bc = 0 \implies ad = bc \implies \frac{a}{b} = \frac{c}{d} \)
Hence Proved.
In simple words: Setting the discriminant to zero and expanding the terms simplifies to \( (ad - bc)^2 = 0 \). This shows that \( ad = bc \), which proves the required ratio.

Exam Tip: Be very organized when expanding long expressions with many variables to prevent sign errors.

Question 5. A natural number, when increased by 12, equal 160 times its reciprocal. Find the number.
Answer: Let the number be \( x \).
\( x + 12 = \frac{160}{x} \implies x^2 + 12x - 160 = 0 \)
\( \implies (x + 20)(x - 8) = 0 \)
Since \( x \) is a natural number, it must be positive. Thus, \( x = 8 \).
The number is 8.
In simple words: Form the equation \( x^2 + 12x - 160 = 0 \) and solve it to find the positive natural number, which is 8.

Exam Tip: Always verify your final answer with the original word problem: \( 8 + 12 = 20 \) and \( 160 / 8 = 20 \).

Question 6. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age to Nisha. Find the present ages of both Asha and Nisha.
Answer: Let Nisha's present age be \( x \) years.
Asha's present age \( = x^2 + 2 \) years.
Age difference \( = (x^2 + 2 - x) \) years.
When Nisha grows to her mother's age, Asha's age will be:
\( (x^2 + 2) + (x^2 - x + 2) = 2x^2 - x + 4 \) years.
According to the problem:
\( 2x^2 - x + 4 = 10x - 1 \implies 2x^2 - 11x + 5 = 0 \)
\( \implies 2x^2 - 10x - x + 5 = 0 \implies (2x - 1)(x - 5) = 0 \)
So \( x = 5 \) or \( x = \frac{1}{2} \).
Since a mother's age cannot be 2.25 when the daughter is 0.5, we choose \( x = 5 \) years.
Thus, Nisha is 5 years old and Asha is \( 5^2 + 2 = 27 \) years old.
In simple words: Setting up a quadratic for their ages gives Nisha's age as 5 and Asha's age as 27.

Exam Tip: Always check if any fractional age values make logical sense in parent-child age problems.

Question 7. A motor-boat takes 2 hours more to cover a distance of 30 km upstream than it takes to cover the same distance down stream. If the speed of the stream is 2 km/hr, find the speed of the boat in still water.
Answer: Let the speed of the boat in still water be \( x \) km/hr.
Upstream speed \( = x - 2 \) km/hr, and Downstream speed \( = x + 2 \) km/hr.
Time taken:
\( \frac{30}{x - 2} - \frac{30}{x + 2} = 2 \)
Divide by 2:
\( \frac{15}{x - 2} - \frac{15}{x + 2} = 1 \implies 15 \left( \frac{x + 2 - x + 2}{x^2 - 4} \right) = 1 \)
\( \implies \frac{60}{x^2 - 4} = 1 \implies x^2 - 4 = 60 \implies x^2 = 64 \implies x = 8 \) km/hr.
The speed of the boat in still water is 8 km/hr.
In simple words: Use upstream and downstream rate formulas to set up the time difference equation. Solving it gives the boat's speed as 8 km/hr.

Exam Tip: Remember that upstream speed is always boat speed minus stream speed, while downstream is boat speed plus stream speed.

Question 8. The perimeter of right angled triangle is five times the length of its shortest side. The numerical value of the area of the triangle is 15 times the numerical value of the length of the shortest side. Find the lengths of the three sides of the triangle.
Answer: Let the sides of the right-angled triangle be \( a, b, c \) where \( a \) is the shortest side and \( c \) is the hypotenuse.
Perimeter: \( a + b + c = 5a \implies b + c = 4a \implies c = 4a - b \)
By Pythagoras theorem:
\( a^2 + b^2 = c^2 \implies a^2 + b^2 = (4a - b)^2 = 16a^2 - 8ab + b^2 \)
\( \implies a^2 = 16a^2 - 8ab \implies 8b = 15a \implies b = \frac{15}{8}a \)
Area of the triangle:
\( \text{Area} = \frac{1}{2} a b = \frac{15}{16} a^2 \)
We are given:
\( \frac{15}{16} a^2 = 15a \implies a = 16 \) (since \( a \neq 0 \))
Then \( b = \frac{15}{8}(16) = 30 \), and \( c = \sqrt{16^2 + 30^2} = 34 \).
The lengths of the sides are 16, 30, and 34.
In simple words: Using perimeter and Pythagoras relations, we find \( b = \frac{15}{8}a \). Substituting this into the area condition gives the sides as 16, 30, and 34.

Exam Tip: Keep the variables related to each other so that you can solve for one single variable at the end.

Question 9. Check whether the equation \( 6x^2 - 7x + 2 = 0 \) has real roots and if it has find then by the method of completing the squares.
Answer: First, find the discriminant:
\( D = b^2 - 4ac = (-7)^2 - 4(6)(2) = 49 - 48 = 1 > 0 \), so real roots exist.
Now, solve by completing the square:
\( 6x^2 - 7x + 2 = 0 \implies x^2 - \frac{7}{6}x + \frac{1}{3} = 0 \)
Add and subtract \( \left(\frac{7}{12}\right)^2 = \frac{49}{144} \):
\( x^2 - \frac{7}{6}x + \frac{49}{144} - \frac{49}{144} + \frac{1}{3} = 0 \)
\( \implies \left(x - \frac{7}{12}\right)^2 = \frac{49}{144} - \frac{48}{144} = \frac{1}{144} \)
\( \implies x - \frac{7}{12} = \pm \frac{1}{12} \)
\( \implies x = \frac{7 \pm 1}{12} \implies x = \frac{2}{3} \) or \( x = \frac{1}{2} \).
In simple words: The discriminant is positive, so the roots are real. Completing the square gives the roots as \( \frac{2}{3} \) and \( \frac{1}{2} \).

Exam Tip: Divide the entire equation by the coefficient of \( x^2 \) as the very first step of completing the square.

Question 10. If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 sq cm. However, if twice the area of large square is added to three times the area of the smaller square, the result is 203 sq. cm. Find the sides of the two squares.
Answer: Let the side of the larger square be \( x \) cm, and side of the smaller square be \( y \) cm.
\( x^2 - 2y^2 = 14 \) and \( 2x^2 + 3y^2 = 203 \)
Multiply the first equation by 2:
\( 2x^2 - 4y^2 = 28 \)
Subtract this from the second equation:
\( 7y^2 = 175 \implies y^2 = 25 \implies y = 5 \) cm.
Now find \( x^2 \):
\( x^2 - 2(25) = 14 \implies x^2 = 64 \implies x = 8 \) cm.
The sides of the squares are 8 cm and 5 cm.
In simple words: Treat \( x^2 \) and \( y^2 \) as simple variables. Solving the system of equations gives the sides of the squares as 8 cm and 5 cm.

Exam Tip: When the variables only appear as squares, solve for the squares directly before taking roots to find the side lengths.

Question 11. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the original duration of the flight.
Answer: Let the original speed be \( x \) km/hr.
Original time \( = \frac{600}{x} \) hours, and new time \( = \frac{600}{x - 200} \) hours.
Time difference is 30 minutes \( = \frac{1}{2} \) hour.
\( \frac{600}{x - 200} - \frac{600}{x} = \frac{1}{2} \implies 600 \left( \frac{200}{x^2 - 200x} \right) = \frac{1}{2} \)
\( \implies x^2 - 200x - 240000 = 0 \implies (x - 600)(x + 400) = 0 \)
Since speed must be positive, \( x = 600 \) km/hr.
Original duration \( = \frac{600}{600} = 1 \) hour.
In simple words: Setting up the time difference equation gives the original speed as 600 km/hr, which means the flight originally took exactly 1 hour.

Exam Tip: Be sure to answer the specific question asked (original duration, i.e., time), not just stop after finding the speed.

Question 12. In a class test, the sum of the marks obtained by p in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.
Answer: Let the marks in Mathematics be \( x \). Then marks in Science \( = 28 - x \).
New conditions:
Mathematics \( = x + 3 \)
Science \( = (28 - x) - 4 = 24 - x \)
Product of marks:
\( (x + 3)(24 - x) = 180 \implies 24x - x^2 + 72 - 3x = 180 \)
\( \implies x^2 - 21x + 108 = 0 \implies (x - 12)(x - 9) = 0 \)
Thus, \( x = 12 \) or \( x = 9 \).

Case 1: If Mathematics \( = 12 \), then Science \( = 28 - 12 = 16 \).
Case 2: If Mathematics \( = 9 \), then Science \( = 28 - 9 = 19 \).
The marks are either (12, 16) or (9, 19).
In simple words: Setting up the product equation gives two possible scenarios: either 12 in Math and 16 in Science, or 9 in Math and 19 in Science.

Exam Tip: Since both cases yield positive integer marks, write both scenarios down to secure full credit.

Question 13. The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.
Answer: Let the shorter side be \( x \) m.
Diagonal \( = x + 60 \) m, and longer side \( = x + 30 \) m.
By Pythagoras theorem:
\( x^2 + (x + 30)^2 = (x + 60)^2 \implies x^2 + x^2 + 60x + 900 = x^2 + 120x + 3600 \)
\( \implies x^2 - 60x - 2700 = 0 \implies (x - 90)(x + 30) = 0 \)
Since length must be positive, \( x = 90 \) m.
Shorter side \( = 90 \) m, and longer side \( = 90 + 30 = 120 \) m.
In simple words: Pythagoras theorem gives a quadratic equation in \( x \). Solving it gives the shorter side as 90 m and the longer side as 120 m.

Exam Tip: Verify your dimensions (90, 120, 150) using the simple ratio 3:4:5 to confirm it is correct.

Question 14. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the largest number. Find the two numbers.
Answer: Let the larger number be \( x \) and the smaller number be \( y \).
\( x^2 - y^2 = 180 \) and \( y^2 = 8x \)
Substitute \( y^2 \) into the first equation:
\( x^2 - 8x - 180 = 0 \implies (x - 18)(x + 10) = 0 \)
Thus, \( x = 18 \) (rejecting \( x = -10 \) since \( y^2 = 8(-10) = -80 \), which is impossible for a real number \( y \)).
If \( x = 18 \), \( y^2 = 8(18) = 144 \implies y = \pm 12 \).
The numbers are 18 and 12, or 18 and -12.
In simple words: Substituting \( y^2 = 8x \) into the first equation gives \( x = 18 \), which then gives \( y = 12 \) or \( -12 \).

Exam Tip: Remember to write both positive and negative values for the smaller number since its square is used.

Question 15. Sum of the areas of two squares is 468 m\(^2\). If the difference of their perimeters is 24 m, find the sides of the two squares.
Answer: Let the sides of the squares be \( x \) m and \( y \) m, where \( x > y \).
\( x^2 + y^2 = 468 \)
Difference of perimeters:
\( 4x - 4y = 24 \implies x - y = 6 \implies x = y + 6 \)
Substitute this into the area equation:
\( (y + 6)^2 + y^2 = 468 \implies 2y^2 + 12y + 36 - 468 = 0 \)
\( \implies 2y^2 + 12y - 432 = 0 \implies y^2 + 6y - 216 = 0 \)
\( \implies (y + 18)(y - 12) = 0 \)
Since length must be positive, \( y = 12 \) m.
Then \( x = 12 + 6 = 18 \) m.
The sides of the squares are 18 m and 12 m.
In simple words: Setup the area and perimeter relations to get a quadratic equation. Solving it gives the sides as 18 m and 12 m.

Exam Tip: Dividing your quadratic by common factors reduces the size of the constants and simplifies factoring.

Question 16. Two water laps together can fill a tank in \( 9\frac{3}{8} \) hrs. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: Let the smaller tap take \( x \) hours. Then the larger tap takes \( (x - 10) \) hours.
Together they fill it in \( 9\frac{3}{8} = \frac{75}{8} \) hours.
In 1 hour:
\( \frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75} \implies \frac{2x - 10}{x^2 - 10x} = \frac{8}{75} \)
Divide by 2:
\( \frac{x - 5}{x^2 - 10x} = \frac{4}{75} \implies 75x - 375 = 4x^2 - 40x \)
\( \implies 4x^2 - 115x + 375 = 0 \implies 4x^2 - 100x - 15x + 375 = 0 \)
\( \implies 4x(x - 25) - 15(x - 25) = 0 \implies (4x - 15)(x - 25) = 0 \)
Since \( x \) must be greater than 10, we choose \( x = 25 \) hours.
The smaller tap takes 25 hours and the larger tap takes 15 hours.
In simple words: Formulate a work-rate equation. Solving the resulting quadratic gives 25 hours for the smaller tap and 15 hours for the larger one.

Exam Tip: Always verify if any of the positive mathematical roots must be rejected based on the word constraints.

Question 17. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Answer: Let the tens digit be \( x \) and units digit be \( y \).
The number is \( 10x + y \).
\( x \times y = 18 \)
When 63 is subtracted:
\( (10x + y) - 63 = 10y + x \implies 9x - 9y = 63 \implies x - y = 7 \)
Substitute \( y = \frac{18}{x} \):
\( x - \frac{18}{x} = 7 \implies x^2 - 7x - 18 = 0 \implies (x - 9)(x + 2) = 0 \)
Since digits must be positive, \( x = 9 \).
Then \( y = 2 \).
The number is 92.
In simple words: Using the digits relation, we find \( x - y = 7 \). Combining with the product \( xy = 18 \) shows that the original number is 92.

Exam Tip: Digits of a number can only be single-digit positive integers between 1 and 9.

Value Based Questions

Question 1. In a society, certain number of people worked for three campaigns. Some worked for “ Say no to plastic”. Some for “Say no to crackers” and the remaining for “Say no to child labour”. A person is selected at random from the society. The probability of getting a person from first campaign is 3/10 and that of 2nd campaign is 2/5. Find the probability of getting a person who worked for the third campaign. If the number of persons in 2nd campaign was 20, find the total number of persons involved in the three campaigns. Are such campaigns helpful in spreading awareness among the people? Give your views.
Answer: Let the probabilities of the three campaigns sum to 1:
\( P(C_1) = \frac{3}{10} \) and \( P(C_2) = \frac{2}{5} = \frac{4}{10} \)
\( P(C_3) = 1 - \left(\frac{3}{10} + \frac{4}{10}\right) = 1 - \frac{7}{10} = \frac{3}{10} \)
Given that the number of persons in the 2nd campaign is 20:
\( \frac{2}{5} \times \text{Total} = 20 \implies \text{Total} = \frac{20 \times 5}{2} = 50 \)
Yes, such campaigns are highly effective in educating the public on essential social and environmental concerns.
In simple words: The probability for the third campaign is \( 3/10 \), and the total number of people involved in the campaigns is 50.

Exam Tip: Express all probabilities with a common denominator first to simplify additions and subtractions.

Question 2. If the price of petrol is increased by Rs. 2 per litre, a person will have to buy 1 litre less petrol for Rs. 1740. Find the original price of petrol at a time.
Answer: Let the original price be Rs. \( x \) per litre.
Original quantity \( = \frac{1740}{x} \), and new quantity \( = \frac{1740}{x+2} \)
\( \frac{1740}{x} - \frac{1740}{x+2} = 1 \implies 1740 \left( \frac{2}{x^2 + 2x} \right) = 1 \)
\( \implies x^2 + 2x - 3480 = 0 \implies (x + 60)(x - 58) = 0 \)
Since price must be positive, \( x = 58 \).
The original price of petrol was Rs. 58 per litre.

a) Why do you think the price of petrol is increasing day by day?
The rising global demand, limited natural reserves, and geopolitical factors are primary reasons for the increase.
b) What should we do to save petrol?
We should adopt carpooling, use public transit, ride bicycles, and switch to electric vehicles.
In simple words: Set up the quantity difference equation to get \( x^2 + 2x - 3480 = 0 \). This shows that the original price was Rs. 58 per litre.

Exam Tip: For large constants like 3480, find factors close to \( \sqrt{3480} \approx 59 \).

Question 3. One fourth of a group of people claim they are creative, twice the square root of the group claim to be caring and the remaining 15 claim they are optimistic. Find the total number of people in the group.
Answer: Let the total number of people be \( x \).
\( \frac{x}{4} + 2\sqrt{x} + 15 = x \implies 3x - 8\sqrt{x} - 60 = 0 \)
Let \( y = \sqrt{x} \):
\( 3y^2 - 8y - 60 = 0 \implies (3y + 10)(y - 6) = 0 \)
Since \( y \) must be positive, \( y = 6 \).
Thus, \( \sqrt{x} = 6 \implies x = 36 \).
The total number of people is 36.

a) How many persons in the group are creative?
Creative persons \( = \frac{36}{4} = 9 \).
b) According to you, which one of the above three values is more important for development of a society?
Being caring is incredibly vital because it builds empathy and strong, supportive relationships in the community.
In simple words: Substituting \( \sqrt{x} = y \) yields a quadratic equation. Solving it shows there are 36 people in total, with 9 being creative.

Exam Tip: Substitution of \( \sqrt{x} \) with \( y \) is a great technique to solve equations with mixed powers of \( x \).

Question 4. In the centre of a rectangular plot of land of dimensions 120 m x 100 m a rectangular portion is to be covered with trees so that the area of the remaining part of the plot is 10500m\(^2\). Find the dimensions of the area to be planted.
Answer: Let the width of the remaining path around the planted plot be \( x \) meters.
Planted area dimensions: \( (120 - 2x) \) by \( (100 - 2x) \).
Total area \( = 120 \times 100 = 12000 \) m\(^2\).
Planted area \( = 12000 - 10500 = 1500 \) m\(^2\).
\( (120 - 2x)(100 - 2x) = 1500 \implies 4(60 - x)(50 - x) = 1500 \)
\( \implies x^2 - 110x + 2625 = 0 \implies (x - 75)(x - 35) = 0 \)
Since \( x \) cannot be larger than half the breadth (50 m), we choose \( x = 35 \) m.
Planted area dimensions are:
Length \( = 120 - 2(35) = 50 \) m.
Breadth \( = 100 - 2(35) = 30 \) m.
The dimensions are 50 m x 30 m.

a) Which social act is being discussed here? Give its advantages.
Afforestation (planting trees) is discussed, which increases oxygen production, reduces pollution, and preserves the environment.
In simple words: The planted region is 1500 m\(^2\). Solving the quadratic gives its dimensions as 50 m by 30 m.

Exam Tip: Verify if your chosen value of \( x \) fits within the real-world limits of the plot dimensions.

Question 5. Mr. Ahuja has two square plots of land which he utilizes for two different purposes – one for providing free education to the children below the age of 14 years and the other to provide free medical services for the needy villagers. The sum of the areas of two square plots is 15425 m\(^2\). If the difference of their perimeters is 60 m, find the sides of the two squares.
Answer: Let the sides of the squares be \( x \) m and \( y \) m, with \( x > y \).
\( x^2 + y^2 = 15425 \)
Difference of perimeters:
\( 4x - 4y = 60 \implies x - y = 15 \implies x = y + 15 \)
Substitute into the area equation:
\( (y + 15)^2 + y^2 = 15425 \implies 2y^2 + 30y - 15200 = 0 \)
\( \implies y^2 + 15y - 7600 = 0 \implies (y + 95)(y - 80) = 0 \)
Since length must be positive, \( y = 80 \) m.
Then \( x = 80 + 15 = 95 \) m.
The sides are 95 m and 80 m.

a) Which qualities of Mr. Ahuja are being depicted in the question?
He displays kindness, social service, compassion, and generosity towards the needy.
In simple words: Solving the equations for area and perimeter differences gives the sides of the two square plots as 95 m and 80 m.

Exam Tip: Clearly write down the units (meters) with your final side lengths to ensure a complete answer.

Question 6. In a class of 48 students, the number of regular students is more than the number of irregular students. Had two irregular students been regular, the product of the number of two types of students would be 380. Find the number of each type of students.
Answer: Let the regular students be \( x \) (where \( x > 24 \) since they are the majority).
Irregular students \( = 48 - x \).
If 2 irregular students were regular:
Regulars \( = x + 2 \), and Irregulars \( = (48 - x) - 2 = 46 - x \).
Their product is:
\( (x + 2)(46 - x) = 380 \implies x^2 - 44x + 288 = 0 \implies (x - 36)(x - 8) = 0 \)
Since \( x > 24 \), we choose \( x = 36 \).
Regular students \( = 36 \) and irregular students \( = 12 \).

a) Why is regularity essential in life?
Consistency helps build self-discipline, which is key to achieving targets in academics and work.
b) Write other values other than regularity that a student must possess.
Students should also focus on integrity, hard work, punctuality, and respect.
In simple words: Setting up the equation gives 36 regular students and 12 irregular students in the class.

Exam Tip: Use the "regular is more than irregular" condition to discard the incorrect mathematical root of 8.

Question 7. A school, apart from rewarding academically excellent student, rewards total 25 students for full attendance and creativity. If we subtract six times the number of creative students from the square of number of regular students we get 10.find the numberof each type of each type of students to be rewarded. Should there be more values for which the students can be rewarded? Justify your answer
Answer: Let the regular students be \( x \) and creative students be \( y \).
\( x + y = 25 \implies y = 25 - x \)
\( x^2 - 6y = 10 \implies x^2 - 6(25 - x) = 10 \implies x^2 + 6x - 160 = 0 \)
\( \implies (x + 16)(x - 10) = 0 \).
Since the count must be positive, \( x = 10 \).
Regular students \( = 10 \) and creative students \( = 25 - 10 = 15 \).

Yes, rewarding other values like sportsmanship, helpfulness, and civic action promotes a more holistic development of all students.
In simple words: Substituting \( y = 25 - x \) into the main equation gives 10 regular students and 15 creative students.

Exam Tip: Define your variables clearly at the beginning so that substitution is easy and error-free.

Question 8. In a class, 32 students are either hardworking or confident. If two more students were hardworking instead of confident, the product of the number of two types of students would be 240.Find the number of two types of students if the class has more hardworking students than confident ones. Which of the given two values is more helpful in the development of an individual? Justify your answer.
Answer: Let the hardworking students be \( x \) (where \( x > 16 \) since they are in the majority).
Confident students \( = 32 - x \).
If 2 more were hardworking:
Hardworking \( = x + 2 \), and Confident \( = 30 - x \).
\( (x + 2)(30 - x) = 240 \implies x^2 - 28x + 180 = 0 \implies (x - 18)(x - 10) = 0 \)
Since \( x > 16 \), we choose \( x = 18 \).
Hardworking students \( = 18 \), and Confident students \( = 14 \).

Both traits are important, but hard work is more fundamental since consistent effort builds competence, which naturally leads to true confidence.
In simple words: Setting up the product equation gives 18 hardworking students and 14 confident students in the class.

Exam Tip: Read the word problem carefully to identify all constraints on the final variables before choosing a root.

Question 9. A shopkeeper stopped using plastic bags and asked his customers to bring their own carry bags due to which his sales reduced by 20 units. If the square of new sale reduced by 400 gives 60 times the original sale, find his original sale.
Answer: Let the original sale be \( x \). New sale \( = x - 20 \).
\( (x - 20)^2 - 400 = 60x \)
\( \implies x^2 - 40x + 400 - 400 = 60x \implies x^2 - 100x = 0 \implies x(x - 100) = 0 \)
Since original sale cannot be zero, \( x = 100 \) units.

i) Why should we ban plastic bags?
They are non-biodegradable, damage the environment, and are toxic to animals.
ii) What steps can be taken in order to ban plastic bags?
We can encourage reusable canvas bags, levy charges on single-use plastics, and set strict policies.
iii) Which trait of shopkeeper’s personality is depicted here?
He displays strong social responsibility and eco-consciousness.
In simple words: The mathematical setup \( (x-20)^2 - 400 = 60x \) simplifies to \( x^2 - 100x = 0 \). Solving it gives the original sale as 100 units.

Exam Tip: Reject the zero root of a factored quadratic if it doesn't make physical sense in the context of the problem.