CBSE Class 10 Maths HOTs Quadratic Equations Set 05

Multiple Choice Questions

Question. Is \( -8 \) a solution of the equation \( 3x^2 + 8x + 2 = 0 \)? 
(a) Yes
(b) No
(c) Cannot be determined
(d) None of the above
Answer: (b)

Question. Solve the quadratic equation \( x^2 - 14x + 24 = 0 \) 
(a) 2, 12
(b) 3, 8
(c) 8, 3
(d) None of these
Answer: (a)

Question. The roots of a equation \( 2x^2 + 5\sqrt{2}x + 5 = 0 \) are
(a) \( \frac{-5\sqrt{2} \pm \sqrt{5}}{2} \)
(b) \( \frac{-5\sqrt{2} \pm \sqrt{10}}{4} \)
(c) \( \frac{-5\sqrt{2} \pm \sqrt{10}}{5} \)
(d) None of these
Answer: (b)

Question. The quadratic equation \( 7y^2 - 4y + 5 = 0 \) has
(a) Real and distinct
(b) Real and equal
(c) Imaginary
(d) More than 2 real roots
Answer: (c)

Question. If a number is added to twice its square, then the resultant is 21. The quadratic representation of this situation is 
(a) \( 2x^2 + x - 21 = 0 \)
(b) \( 2x^2 + x + 21 = 0 \)
(c) \( 2x^2 - x + 21 = 0 \)
(d) \( 2x^2 - x - 21 = 0 \)
Answer: (a)

Case Study MCQs

Sohan is preparing for UPSC exam. For this, he has to practice the chapter of quadratic equations. So, he started with factorisation method. Let two roots of \( ax^2 + bx + c \) be \( p \) and \( q \). \( \therefore ac = p \times q \) and \( p + q = b \). Now, factorize each of the following quadratic equations and find the roots.

Question. \( x^2 - 10x + 21 = 0 \)
(a) 9, 3
(b) 21, 1
(c) 3, 7
(d) 3, 9
Answer: (c)

Question. \( 15y^2 - 41y - 14 = 0 \)
(a) \( \frac{3}{7}, \frac{2}{3} \)
(b) \( \frac{3}{5}, \frac{7}{2} \)
(c) \( \frac{2}{3}, \frac{7}{5} \)
(d) \( \frac{2}{5}, \frac{7}{3} \)
Answer: (d)

Question. \( 21x^2 - 2x + \frac{1}{21} = 0 \)
(a) 21, 3
(b) \( \frac{1}{21}, \frac{1}{21} \)
(c) \( - 21, \frac{1}{21} \)
(d) \( \frac{1}{21}, \frac{3}{21} \)
Answer: (b)

Question. \( 6x^2 - 31x + 40 = 0 \)
(a) \( \frac{5}{2}, \frac{8}{3} \)
(b) \( -\frac{5}{2}, -\frac{8}{3} \)
(c) \( -\frac{5}{2}, \frac{3}{8} \)
(d) \( \frac{2}{3}, \frac{5}{8} \)
Answer: (a)

Question. \( 3x^2 + 2\sqrt{5}x - 5 = 0 \)
(a) \( 2\sqrt{5}, - 5 \)
(b) \( \sqrt{5}, -\frac{\sqrt{5}}{3} \)
(c) \( -\sqrt{5}, 3\sqrt{5} \)
(d) \( -\sqrt{5}, \frac{\sqrt{5}}{3} \)
Answer: (d)

Short Answer Type Questions

Question. Find the roots of the equation \( x^2 + 182 = 27x \)
Answer: 13, 14

Question. Find the roots of the quadratic equation \( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \) 
Answer: \( x = \frac{1}{b^2}, -\frac{1}{a^2} \)

Question. If the roots of the equation \( x^2 + 2cx + ab = 0 \) are real and unequal, then prove that the equation \( x^2 - 2(a + b)x + a^2 + b^2 + 2c^2 = 0 \) has no real roots.
Answer: For the first equation \( x^2 + 2cx + ab = 0 \), roots are real and unequal, so discriminant \( D_1 > 0 \).
\( D_1 = (2c)^2 - 4(1)(ab) = 4c^2 - 4ab > 0 \Rightarrow c^2 - ab > 0 \Rightarrow c^2 > ab \).
For the second equation \( x^2 - 2(a + b)x + a^2 + b^2 + 2c^2 = 0 \), discriminant \( D_2 = [-2(a + b)]^2 - 4(1)(a^2 + b^2 + 2c^2) \)
\( D_2 = 4(a^2 + b^2 + 2ab) - 4a^2 - 4b^2 - 8c^2 \)
\( D_2 = 4a^2 + 4b^2 + 8ab - 4a^2 - 4b^2 - 8c^2 \)
\( D_2 = 8ab - 8c^2 = 8(ab - c^2) \).
Since \( c^2 > ab \), \( ab - c^2 < 0 \). Therefore, \( D_2 < 0 \).
Hence, the equation has no real roots.

Long Answer Type Questions

Question. If \( x = - 5 \) is a root of the quadratic equation \( 2x^2 + px - 15 = 0 \) and the quadratic equation \( p(x^2 + x) + k = 0 \) has equal roots, then find the value of \( k \). 
Answer: 7/4

Question. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 sq m more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find its length and breadth of the rectangular park.
Answer: Length = 7 m and Breadth = 4 m

Short Answer Type Questions

Question. Check whether the following are quadratic equations or not.
(i) \( (x - 1)(x + 2) = (x - 3)(x + 1) \)
(ii) \( (x + 2)^2 = 4(x + 3) \)

Answer: (i) LHS = \( x^2 + x - 2 \), RHS = \( x^2 - 2x - 3 \). Equating gives \( 3x + 1 = 0 \). It is not of the form \( ax^2 + bx + c = 0 \). Hence, it is not a quadratic equation.
(ii) \( x^2 + 4x + 4 = 4x + 12 \Rightarrow x^2 - 8 = 0 \). It is of the form \( ax^2 + bx + c = 0 \). Hence, it is a quadratic equation.

Question. If \( x = \frac{1}{\sqrt{3}} \) is root of the equation \( Px^2 + (\sqrt{3} - \sqrt{2})x - 1 = 0 \), then find the value of \( P^2 + 1 \).

Answer: Substituting \( x = \frac{1}{\sqrt{3}} \) in \( Px^2 + (\sqrt{3} - \sqrt{2})x - 1 = 0 \):
\( P(\frac{1}{3}) + (\sqrt{3} - \sqrt{2})(\frac{1}{\sqrt{3}}) - 1 = 0 \)
\( \frac{P}{3} + 1 - \frac{\sqrt{2}}{\sqrt{3}} - 1 = 0 \Rightarrow \frac{P}{3} = \frac{\sqrt{2}}{\sqrt{3}} \Rightarrow P = \sqrt{6} \).
Value of \( P^2 + 1 = (\sqrt{6})^2 + 1 = 6 + 1 = 7 \).

Question. In each of the following equations, determine the value of \( k \) for which the given value is a solution of the equation.
(i) \( kx^2 + 2x - 3 = 0, x = 2 \)
(ii) \( x^2 + 2ax - k = 0, x = -a \)

Answer: (i) \( k(2)^2 + 2(2) - 3 = 0 \Rightarrow 4k + 1 = 0 \Rightarrow k = -\frac{1}{4} \).
(ii) \( (-a)^2 + 2a(-a) - k = 0 \Rightarrow a^2 - 2a^2 - k = 0 \Rightarrow -a^2 - k = 0 \Rightarrow k = -a^2 \).

Question. Find the value of \( k \) in the following equations
(i) \( x^2 - 2kx - 6 = 0 \), when \( x = 3 \)
(ii) \( x^2 - kx - \frac{5}{4} = 0 \), when \( x = \frac{1}{2} \)

Answer: (i) \( (3)^2 - 2k(3) - 6 = 0 \Rightarrow 9 - 6k - 6 = 0 \Rightarrow 3 = 6k \Rightarrow k = \frac{1}{2} \).
(ii) \( (\frac{1}{2})^2 - k(\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1}{4} - \frac{k}{2} - \frac{5}{4} = 0 \Rightarrow -1 = \frac{k}{2} \Rightarrow k = -2 \).

Question. Determine whether \( x = -\frac{1}{2}, x = \frac{1}{3} \) are the solutions of the given equation \( 6x^2 - x - 2 = 0 \), or not.

Answer: For \( x = -\frac{1}{2} \): \( 6(-\frac{1}{2})^2 - (-\frac{1}{2}) - 2 = 6(\frac{1}{4}) + \frac{1}{2} - 2 = \frac{3}{2} + \frac{1}{2} - 2 = 0 \). (Yes)
For \( x = \frac{1}{3} \): \( 6(\frac{1}{3})^2 - (\frac{1}{3}) - 2 = 6(\frac{1}{9}) - \frac{1}{3} - 2 = \frac{2}{3} - \frac{1}{3} - 2 = -\frac{5}{3} \neq 0 \). (No)

Question. Solve the quadratic equation by factorisation method. \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)

Answer: \( 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0 \)
\( 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0 \)
\( (4x - \sqrt{3})(\sqrt{3}x + 2) = 0 \)
\( x = \frac{\sqrt{3}}{4} \) or \( x = -\frac{2}{\sqrt{3}} \).

Question. Solve for \( x \): \( \frac{16}{x} - 1 = \frac{15}{x+1}; x \neq 0, -1 \).

Answer: \( \frac{16 - x}{x} = \frac{15}{x+1} \Rightarrow (16-x)(x+1) = 15x \)
\( 16x + 16 - x^2 - x = 15x \Rightarrow -x^2 + 15x + 16 = 15x \Rightarrow x^2 = 16 \)
\( x = \pm 4 \).

Question. Find the roots of the equation \( ax^2 + a = a^2x + x \). 

Answer: \( ax^2 - a^2x - x + a = 0 \Rightarrow ax(x - a) - 1(x - a) = 0 \)
\( (ax - 1)(x - a) = 0 \Rightarrow x = a, \frac{1}{a} \).

Question. Solve for \( x \), \( \sqrt{6x + 7} - (2x - 7) = 0 \) 

Answer: \( \sqrt{6x + 7} = 2x - 7 \). Squaring both sides: \( 6x + 7 = (2x - 7)^2 \)
\( 6x + 7 = 4x^2 - 28x + 49 \Rightarrow 4x^2 - 34x + 42 = 0 \Rightarrow 2x^2 - 17x + 21 = 0 \)
\( 2x^2 - 14x - 3x + 21 = 0 \Rightarrow 2x(x - 7) - 3(x - 7) = 0 \)
\( (2x - 3)(x - 7) = 0 \Rightarrow x = 7, 3/2 \).
Checking \( x = 3/2 \) in original: \( \sqrt{9+7} - (3-7) = 4 - (-4) = 8 \neq 0 \).
Checking \( x = 7 \) in original: \( \sqrt{42+7} - (14-7) = 7 - 7 = 0 \).
Hence, \( x = 7 \).

Question. Find the numerical difference of the roots of equation \( x^2 - 7x - 18 = 0 \).

Answer: \( x^2 - 9x + 2x - 18 = 0 \Rightarrow (x - 9)(x + 2) = 0 \). Roots are \( 9, -2 \).
Difference = \( |9 - (-2)| = 11 \).

Question. Using the quadratic formula, solve the quadratic equation. \( \sqrt{3}x^2 + 11x + 6\sqrt{3} = 0 \)

Answer: \( a = \sqrt{3}, b = 11, c = 6\sqrt{3} \).
\( D = b^2 - 4ac = 121 - 4(\sqrt{3})(6\sqrt{3}) = 121 - 72 = 49 \).
\( x = \frac{-11 \pm \sqrt{49}}{2\sqrt{3}} = \frac{-11 \pm 7}{2\sqrt{3}} \).
\( x = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}} \) and \( x = \frac{-18}{2\sqrt{3}} = -\frac{9}{\sqrt{3}} = -3\sqrt{3} \).

Question. If the discriminant of the equation \( 5x^2 - sx + 4 = 0 \) is \( 1 \), then find the value of \( s \).

Answer: \( D = b^2 - 4ac \Rightarrow 1 = (-s)^2 - 4(5)(4) \Rightarrow 1 = s^2 - 80 \)
\( s^2 = 81 \Rightarrow s = \pm 9 \).

Question. Show that \( (x^2 + 1)^2 - x^2 = 0 \) has no real roots. 

Answer: \( (x^2 + 1)^2 - x^2 = 0 \Rightarrow x^4 + 2x^2 + 1 - x^2 = 0 \Rightarrow x^4 + x^2 + 1 = 0 \).
Let \( y = x^2 \), then \( y^2 + y + 1 = 0 \).
\( D = (1)^2 - 4(1)(1) = -3 < 0 \).
Since discriminant is negative, there are no real roots for \( y \), and thus no real roots for \( x \).

Question. Find the value of \( k \) for which the quadratic equation \( 2x^2 - kx + k = 0 \) has equal roots.

Answer: For equal roots, \( D = 0 \). \( (-k)^2 - 4(2)(k) = 0 \Rightarrow k^2 - 8k = 0 \)
\( k(k - 8) = 0 \Rightarrow k = 0, 8 \).

Question. Find the values of \( k \) for which the equation \( 9x^2 + 3kx + 4 = 0 \) has real roots.

Answer: For real roots, \( D \geq 0 \). \( (3k)^2 - 4(9)(4) \geq 0 \Rightarrow 9k^2 - 144 \geq 0 \)
\( k^2 \geq 16 \Rightarrow k \geq 4 \) or \( k \leq -4 \).

Question. If the equation \( (1 + m^2)x^2 + (2mc)x + (c^2 - a^2) = 0 \) has equal roots, then prove that \( c^2 = a^2(1 + m^2) \).

Answer: For equal roots, \( D = 0 \). \( (2mc)^2 - 4(1+m^2)(c^2-a^2) = 0 \)
\( 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
\( m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \)
\( -c^2 + a^2(1 + m^2) = 0 \Rightarrow c^2 = a^2(1 + m^2) \). Hence proved.

Question. The sum of two numbers is \( 11 \) and the sum of their reciprocals is \( \frac{11}{28} \). Find the numbers. 

Answer: Let numbers be \( x \) and \( 11 - x \).
\( \frac{1}{x} + \frac{1}{11-x} = \frac{11}{28} \Rightarrow \frac{11-x+x}{x(11-x)} = \frac{11}{28} \Rightarrow \frac{11}{11x-x^2} = \frac{11}{28} \)
\( 11x - x^2 = 28 \Rightarrow x^2 - 11x + 28 = 0 \Rightarrow (x-7)(x-4) = 0 \).
Numbers are \( 4 \) and \( 7 \).

Question. In a cricket match. Harbhajan took three wickets less than twice the number of wickets taken by Zaheer. The product of the numbers of wickets taken by these two is \( 20 \). Represent the above situation in the form of a quadratic equation.

Answer: Let Zaheer took \( x \) wickets. Harbhajan took \( 2x - 3 \) wickets.
Product: \( x(2x - 3) = 20 \Rightarrow 2x^2 - 3x - 20 = 0 \).

Long Answer Type Questions

Question. If \( x = 2 \) and \( x = 3 \) are roots of the equation \( 3x^2 - 2ax + 2b = 0 \), then find the values of \( a \) and \( b \).

Answer: For \( x=2 \): \( 3(4) - 2a(2) + 2b = 0 \Rightarrow 12 - 4a + 2b = 0 \Rightarrow 2a - b = 6 \). (i)
For \( x=3 \): \( 3(9) - 2a(3) + 2b = 0 \Rightarrow 27 - 6a + 2b = 0 \Rightarrow 6a - 2b = 27 \). (ii)
Multiply (i) by 2: \( 4a - 2b = 12 \).
Subtracting from (ii): \( 2a = 15 \Rightarrow a = \frac{15}{2} \).
Substitute \( a \) in (i): \( 15 - b = 6 \Rightarrow b = 9 \).

Question. Find the nature of roots of the following quadratic equations. If the real roots exist, then also find the roots.
(i) \( 4x^2 + 12x + 9 = 0 \)
(ii) \( 3x^2 + 5x - 7 = 0 \)

Answer: (i) \( D = 12^2 - 4(4)(9) = 144 - 144 = 0 \). Real and equal roots.
Roots are \( x = -\frac{b}{2a} = -\frac{12}{8} = -\frac{3}{2} \).
(ii) \( D = 5^2 - 4(3)(-7) = 25 + 84 = 109 > 0 \). Real and distinct roots.
Roots are \( x = \frac{-5 \pm \sqrt{109}}{6} \).

Question. Find the value of \( k \) for which the given equation has equal roots. \( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \)

Answer: For equal roots, \( D = 0 \). \( [2(k - 12)]^2 - 4(k - 12)(2) = 0 \)
\( 4(k-12)^2 - 8(k-12) = 0 \Rightarrow 4(k-12)[(k-12) - 2] = 0 \)
\( 4(k-12)(k-14) = 0 \Rightarrow k = 12 \) or \( k = 14 \).
If \( k=12 \), equation becomes \( 2=0 \), which is impossible. So, \( k = 14 \).

Question. If \( x = -2 \) is a root of the equation \( 3x^2 + 7x + p = 0 \). Find the values of \( k \), so that the roots of the equation \( x^2 + k(4x + k - 1) + p = 0 \) are equal. 

Answer: \( x=-2 \) in \( 3x^2 + 7x + p = 0 \Rightarrow 3(4) + 7(-2) + p = 0 \Rightarrow 12 - 14 + p = 0 \Rightarrow p = 2 \).
Now, \( x^2 + 4kx + k^2 - k + 2 = 0 \). For equal roots, \( D = 0 \).
\( (4k)^2 - 4(1)(k^2 - k + 2) = 0 \Rightarrow 16k^2 - 4k^2 + 4k - 8 = 0 \)
\( 12k^2 + 4k - 8 = 0 \Rightarrow 3k^2 + k - 2 = 0 \Rightarrow (3k - 2)(k + 1) = 0 \)
\( k = \frac{2}{3}, -1 \).

Question. Find two consecutive odd natural numbers, sum of whose squares is \( 130 \). 

Answer: Let numbers be \( x \) and \( x+2 \).
\( x^2 + (x+2)^2 = 130 \Rightarrow x^2 + x^2 + 4x + 4 = 130 \Rightarrow 2x^2 + 4x - 126 = 0 \)
\( x^2 + 2x - 63 = 0 \Rightarrow (x+9)(x-7) = 0 \).
As numbers are natural, \( x = 7 \). Numbers are \( 7, 9 \).

Question. A piece of cloth costs ₹ \( 200 \). If the piece was \( 5 \) m longer and each metre of cloth costs ₹ \( 2 \) less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre? 

Answer: Let length be \( x \) and rate be \( \frac{200}{x} \).
New length \( = x+5 \), New rate \( = \frac{200}{x} - 2 \).
\( (x+5)(\frac{200}{x} - 2) = 200 \Rightarrow 200 - 2x + \frac{1000}{x} - 10 = 200 \)
\( -2x + \frac{1000}{x} - 10 = 0 \Rightarrow -2x^2 + 1000 - 10x = 0 \Rightarrow x^2 + 5x - 500 = 0 \)
\( (x+25)(x-20) = 0 \Rightarrow x = 20 \) m.
Original rate \( = \frac{200}{20} = Rs 10 \) per m.

Question. The difference of two numbers is \( 4 \). If the difference of their reciprocals is \( \frac{4}{21} \), find the two numbers. 

Answer: Let numbers be \( x \) and \( x+4 \).
\( \frac{1}{x} - \frac{1}{x+4} = \frac{4}{21} \Rightarrow \frac{x+4-x}{x(x+4)} = \frac{4}{21} \Rightarrow \frac{4}{x^2+4x} = \frac{4}{21} \)
\( x^2 + 4x = 21 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0 \).
Numbers are \( 3, 7 \) or \( -7, -3 \).

Question. The perimeter of a right angled triangle is \( 70 \) units and its hypotenuse is \( 29 \) units we would like to find the length of the other sides.

Answer: Let other sides be \( x \) and \( 70 - 29 - x = 41 - x \).
By Pythagoras: \( x^2 + (41 - x)^2 = 29^2 \Rightarrow x^2 + 1681 + x^2 - 82x = 841 \)
\( 2x^2 - 82x + 840 = 0 \Rightarrow x^2 - 41x + 420 = 0 \)
\( (x - 20)(x - 21) = 0 \Rightarrow x = 20, 21 \). Sides are \( 20 \) and \( 21 \).

Question. The sum of the reciprocals of Anjali’s age \( 3 \) yr ago and \( 5 \) yr from now is \( \frac{1}{3} \). Find the present age of Anjali.

Answer: Let present age be \( x \).
\( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \Rightarrow \frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3} \Rightarrow \frac{2x+2}{x^2+2x-15} = \frac{1}{3} \)
\( 6x + 6 = x^2 + 2x - 15 \Rightarrow x^2 - 4x - 21 = 0 \Rightarrow (x-7)(x+3) = 0 \).
Age is \( 7 \) yr.

Question. A two-digit number is such that the product of the digits is \( 12 \). When \( 36 \) is added to the number the digits interchange their places. Find the two-digit number.

Answer: Let digits be \( x \) and \( \frac{12}{x} \). Number \( = 10x + \frac{12}{x} \).
\( 10x + \frac{12}{x} + 36 = 10(\frac{12}{x}) + x \Rightarrow 9x - \frac{108}{x} + 36 = 0 \Rightarrow 9x^2 + 36x - 108 = 0 \)
\( x^2 + 4x - 12 = 0 \Rightarrow (x+6)(x-2) = 0 \Rightarrow x = 2 \).
Digits are \( 2 \) and \( 6 \). Number is \( 26 \).

Question. John and Janvi together have \( 45 \) marbles. Both of them lost \( 5 \) marbles each and the product of the number of marbles they now have, is \( 124 \). Find out how many marbles they had to start with?

Answer: Let John had \( x \), Janvi had \( 45-x \).
\( (x-5)(45-x-5) = 124 \Rightarrow (x-5)(40-x) = 124 \Rightarrow 40x - x^2 - 200 + 5x = 124 \)
\( x^2 - 45x + 324 = 0 \Rightarrow (x-36)(x-9) = 0 \).
They had \( 36 \) and \( 9 \) marbles.

Question. The hypotenuse of right angled triangle is \( 6 \) m more than twice the shortest side. If the third side is \( 2 \) m less than the hypotenuse, then find all sides of the triangle.

Answer: Let shortest side \( = x \). Hypotenuse \( = 2x + 6 \). Third side \( = 2x + 4 \).
\( x^2 + (2x + 4)^2 = (2x + 6)^2 \Rightarrow x^2 + 4x^2 + 16x + 16 = 4x^2 + 24x + 36 \)
\( x^2 - 8x - 20 = 0 \Rightarrow (x-10)(x+2) = 0 \Rightarrow x = 10 \).
Sides are \( 10 \) m, \( 24 \) m, \( 26 \) m.

Question. 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 of Nisha. Find the present ages of both Asha and Nisha. 

Answer: Let Nisha's age \( = x \), Asha's age \( = x^2 + 2 \).
Nisha grows to \( x^2 + 2 \) in \( (x^2 + 2 - x) \) years.
Asha's age then \( = (x^2 + 2) + (x^2 + 2 - x) = 2x^2 - x + 4 \).
According to problem: \( 2x^2 - x + 4 = 10x - 1 \Rightarrow 2x^2 - 11x + 5 = 0 \)
\( (2x - 1)(x - 5) = 0 \Rightarrow x = 5 \) (since age cannot be 0.5).
Nisha is \( 5 \) yr, Asha is \( 27 \) yr.

Question. The speed of a boat in still water is \( 15 \) km/h. It can go \( 30 \) km upstream and return downstream to the original point in \( 4 \) h and \( 30 \) min. Find the speed of stream.

Answer: Let speed of stream \( = x \).
\( \frac{30}{15-x} + \frac{30}{15+x} = 4.5 \Rightarrow \frac{450+30x+450-30x}{225-x^2} = 4.5 \Rightarrow \frac{900}{225-x^2} = 4.5 \)
\( 200 = 225 - x^2 \Rightarrow x^2 = 25 \Rightarrow x = 5 \) km/h.

Question. Two water taps together can fill a tank in \( 1 \frac{7}{8} \) h. The tap with longer diameter takes \( 2 \) h less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Answer: Let smaller tap take \( x \) h, larger tap take \( x-2 \) h.
\( \frac{1}{x} + \frac{1}{x-2} = \frac{8}{15} \Rightarrow \frac{2x-2}{x^2-2x} = \frac{8}{15} \Rightarrow 30x - 30 = 8x^2 - 16x \)
\( 8x^2 - 46x + 30 = 0 \Rightarrow 4x^2 - 23x + 15 = 0 \Rightarrow (4x-3)(x-5) = 0 \).
\( x = 5 \) h (as \( x=3/4 \) makes \( x-2 \) negative). Taps take \( 5 \) h and \( 3 \) h.