CBSE Class 10 Maths HOTs Quadratic Equations Set 14

Question. If \( x = 2 \) is a solution of the equation \( x^2 - 5x + 6k = 0 \), the value of \( k \) is
(a) 2
(b) 1
(c) 0
(d) 3
Answer: (b) 1

Question. The product of three consecutive integers is equal to 6 times the sum of the three integers. If the smallest integer is \( x \), which of the following equation represent the above situation?
(a) \( 2x^2 + x - 9 = 0 \)
(b) \( x^2 + 2x + 18 = 0 \)
(c) \( x^2 + 2x - 18 = 0 \)
(d) \( 2x^2 - x + 9 = 0 \)
Answer: (c) \( x^2 + 2x - 18 = 0 \)

Question. If \( x = -2 \) is a root of the equation \( x^2 - 2x + k = 0 \)? then \( k = ? \)
(a) 4
(b) 8
(c) -8
(d) -4
Answer: (c) -8

Question. If \( x = 3 \) is one root of the quadratic equation \( x^2 - 2kx - 6 = 0 \), then the value of \( k \) is
(a) \( 1/3 \)
(b) \( 2/3 \)
(c) 1
(d) \( 1/2 \)
Answer: (d) \( 1/2 \)

Question. If \( x^2 + x + 1 = 0 \) then
(a) \( x = 0 \) is a solution of this quadratic equation
(b) \( x = 2 \) is a solution of this quadratic equation
(c) \( x = 0 \) or \( x = 2 \) is a solution of this quadratic equation
(d) \( x = 0 \) and \( x = 2 \) are not solution of this quadratic equation
Answer: (d) \( x = 0 \) and \( x = 2 \) are not solution of this quadratic equation

Question. If \( (x - a) \) is one of the factors of the polynomial \( ax^2 + bx + c \), then one of the roots of \( ax^2 + bx + c = 0 \) is
(a) 1
(b) c
(c) a
(d) None of the options
Answer: (c) a

Question. Which of the following is a solution of the equation \( x^2 - 6x + 5 = 0 \)?
(a) 2
(b) 5
(c) 9
(d) 15
Answer: (b) 5

Question. Which of the following are quadratic equations:
(i) \( x^3 - x = x^2 + 2 \)
(ii) \( (x - 1) (x + 2) = x + 3 \)
(iii) \( x^2 + \frac{1}{x^2} + 1 = 0 \)
(iv) \( (x + 1) (x + 3) = (x - 1) (x - 4) \)
(v) \( \sqrt{x^2 + 4} = (x^2 + 1)^2 \)
(vi) \( \frac{x+1}{x-1} + \frac{x+2}{x+3} = 1 \) ( \( x \neq 1, x \neq -3 \) )
Answer: (ii) and (vi) are quadratic equations.

Question. If 2 is a root of the equation \( x^2 + bx + 12 = 0 \), find the value of \( b \).
Answer: Substituting \( x = 2 \) in \( x^2 + bx + 12 = 0 \)
\( (2)^2 + b(2) + 12 = 0 \)

\( \implies 4 + 2b + 12 = 0 \)

\( \implies 2b = -16 \)

\( \implies b = -8 \)

Question. A train travels 360 km at a uniform speed. If the speed had been 5 km/h. more it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.
Answer: Let the speed of the train be \( x \) km/h.
Time taken \( = \frac{360}{x} \)
According to the question,
\( \frac{360}{x} - \frac{360}{x+5} = 1 \)

\( \implies 360 \left( \frac{x+5-x}{x(x+5)} \right) = 1 \)

\( \implies 1800 = x^2 + 5x \)

\( \implies x^2 + 5x - 1800 = 0 \)

Question. The product of two consecutive even integers is 528. Represent the situation in the form of a quadratic equation.
Answer: Let the two consecutive even integers be \( x \) and \( x+2 \).
According to the question,
\( x(x+2) = 528 \)

\( \implies x^2 + 2x - 528 = 0 \)

Solution of a Quadratic Equation by Factorisation

To find the solution of a quadratic equation by factorisation, we first write the given quadratic equation as product of two linear factors by splitting the middle term. By equating each factor to zero we get possible solutions/roots of the given quadratic equation.

For Example: Find the roots of the equation \( x^2 - 5x + 6 = 0 \) by factorisation.
Answer: Let us 1st split the middle term
\( x^2 - 3x - 2x + 6 = 0 \)
\( x(x - 3) - 2(x - 3) = 0 \)

\( \implies (x - 3)(x - 2) = 0 \)
Put \( x - 3 = 0 \implies x = 3 \) and \( x - 2 = 0 \implies x = 2 \)
So, \( x = 2 \) and \( x = 3 \) are solutions of equation.

Solution of Quadratic Equation using Quadratic Formula or Discriminant Method

Let quadratic equation be \( ax^2 + bx + c = 0 \)
Step 1. Find \( D = b^2 - 4ac \).
Step 2. (i) If \( D > 0 \), Solution/roots of the quadratic equation are given by \( x = \frac{-b + \sqrt{D}}{2a}, \frac{-b - \sqrt{D}}{2a} \)
(ii) If \( D = 0 \), Solution/roots of the quadratic equation are given by \( x = \frac{-b}{2a} \).
(iii) If \( D < 0 \), equation has no real roots.

Question. Which of the following are the roots of the quadratic equation, \( x^2 - 9x + 20 = 0 \)?
(a) 3, 4
(b) 4, 5
(c) 5, 6
(d) 6, 7
Answer: (b) 4, 5

Question. Ruchir was asked his age by his friend. Ruchir said “The number you get when you subtract 25 times my age from twice the square of my age will be thrice your age. If the friend’s age is 14 years, then the age of Ruchir is
(a) 10 years
(b) 28 years
(c) 7 years
(d) 14 years
Answer: (d) 14 years

Question. Roots of the following quadratic equation \( \sqrt{3}x^2 + 10x + 7\sqrt{3} = 0 \) are
(a) \( \frac{7}{\sqrt{3}}, \sqrt{3} \)
(b) \( -\frac{7}{\sqrt{3}}, -\sqrt{3} \)
(c) \( \frac{7}{\sqrt{3}}, -\sqrt{3} \)
(d) \( -\frac{7}{\sqrt{3}}, \sqrt{3} \)
Answer: (b) \( -\frac{7}{\sqrt{3}}, -\sqrt{3} \)

Question. Divide 27 into two parts such that the sum of their reciprocals is \( \frac{3}{20} \).
Answer: Let one part be \( x \) and another part be \( 27 - x \)
So, \( \frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20} \)

\( \implies \frac{27-x+x}{x(27-x)} = \frac{3}{20} \)

\( \implies \frac{27}{x(27-x)} = \frac{3}{20} \)

\( \implies 180 = x(27 - x) \)

\( \implies x^2 - 27x + 180 = 0 \)

\( \implies (x - 15)(x - 12) = 0 \)

\( \implies x = 15, x = 12 \)
So, two parts of 27 are 15 and 12

Question. Solve the following quadratic equation by factorisation: \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \)
Answer: \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \)

\( \implies 12abx^2 - 9a^2x + 8b^2x - 6ab = 0 \)

\( \implies 3ax(4bx - 3a) + 2b(4bx - 3a) = 0 \)

\( \implies (4bx - 3a) (3ax + 2b) = 0 \)

\( \implies 4bx - 3a = 0 \) or \( 3ax + 2b = 0 \)

\( \implies x = \frac{3a}{4b} \) or \( x = \frac{-2b}{3a} \)

Question. Solve for \( x \): \( \frac{x+3}{x+2} = \frac{3x-7}{2x-3}, x \neq -2, \frac{3}{2} \)
Answer: \( (x + 3)(2x - 3) = (3x - 7)(x + 2) \)

\( \implies 2x^2 - 3x + 6x - 9 = 3x^2 + 6x - 7x - 14 \)

\( \implies x^2 - 4x - 5 = 0 \)

\( \implies x^2 - 5x + x - 5 = 0 \)

\( \implies x(x - 5) + 1(x - 5) = 0 \)

\( \implies (x - 5)(x + 1) = 0 \)
\( x = 5 \) or \( x = -1 \)

Question. Two numbers differ by 3 and their product is 504. Find the numbers.
Answer: Let one number be \( x \)
\( \therefore \) other number be \( (x + 3) \)
A.T.Q \( x(x + 3) = 504 \)

\( \implies x^2 + 3x - 504 = 0 \)

\( \implies x^2 + 24x - 21x - 504 = 0 \)

\( \implies (x + 24) (x - 21) = 0 \)

\( \implies x + 24 = 0 \) or \( x - 21 = 0 \)

\( \implies x = -24 \) or \( x = 21 \)
When \( x = -24 \), numbers are -24 and -24 + 3 = -21
When \( x = 21 \), numbers are 21 and 21 + 3 = 24.

Question. Solve the following equation using by quadratic formula: \( x^2 + 5x + 5 = 0 \).
Answer: Given equation is \( x^2 + 5x + 5 = 0 \)
Here \( a = 1, b = 5, c = 5 \).
\( D = b^2 - 4ac \)

\( \implies D = (5)^2 - 4 \times 1 \times 5 \)

\( \implies D = 5 > 0 \)
\( \therefore \) solution is given by \( x = \frac{-b \pm \sqrt{D}}{2a} \)

\( \implies x = \frac{-5 \pm \sqrt{5}}{2 \times 1} \)

\( \implies x = \frac{-5 + \sqrt{5}}{2}, \frac{-5 - \sqrt{5}}{2} \) are solutions.

Question. Solve for \( x \): \( 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \).
Answer: Here, \( A = 9, B = -9(a + b), C = 2a^2 + 5ab + 2b^2 \)
\( D = B^2 - 4AC \)
\( = [-9(a + b)]^2 - 4 \times 9(2a^2 + 5ab + 2b^2) \)
\( = 81(a + b)^2 - 36(2a^2 + 5ab + 2b^2) \)
\( = 81(a^2 + b^2 + 2ab) - 72a^2 - 180ab - 72b^2 \)
\( = 9a^2 + 9b^2 - 18ab \)
\( = 9(a^2 + b^2 - 2ab) = 9(a - b)^2 \)
\( x = \frac{-B \pm \sqrt{D}}{2A} \)
\( = \frac{9(a + b) \pm 3(a - b)}{18} \)
\( = \frac{2a+b}{3}, \frac{a+2b}{3} \)

Question. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Answer: Given distance is 1500 km. Let usual speed = \( s \).
We know, \( \text{time} = \frac{\text{distance}}{\text{speed}} \)
From question, \( \frac{1500}{s} - \frac{1500}{s+100} = \frac{1}{2} \) (half an hour late).

\( \implies 1500 \left[ \frac{s+100-s}{s(s+100)} \right] = \frac{1}{2} \)

\( \implies \frac{150000}{s^2+100s} = \frac{1}{2} \)

\( \implies s^2 + 100s - 300000 = 0 \)

\( \implies (s + 600)(s - 500) = 0 \)

\( \implies s = 500 \) or \( s = -600 \).
But speed cannot be negative.
The usual speed of the plane is 500 km/hr.

Question. Using quadratic formula, solve the following quadratic equation for \( x \): \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \).
Answer: \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \)
Here \( a = p^2, b = (p^2 - q^2), c = -q^2 \)
\( D = b^2 - 4ac = (p^2 - q^2)^2 - 4 \times p^2 \times (-q^2) \)
\( = (p^2 + q^2)^2 \)
Now \( x = \frac{-b \pm \sqrt{D}}{2a} \)

\( \implies x = \frac{-(p^2 - q^2) \pm \sqrt{(p^2 + q^2)^2}}{2 \times p^2} \)

\( \implies x = \frac{-p^2 + q^2 \pm (p^2 + q^2)}{2p^2} \)

\( \implies x = \frac{q^2}{p^2}, -1 \)

Question. Solve for \( x \): \( \frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3} (x \neq 2, 4) \)
Answer: \( \frac{x-1}{x-2} + \frac{x-3}{x-4} = \frac{10}{3} \)

\( \implies \frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{3} \)

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

\( \implies 3(2x^2 - 10x + 10) = 10(x^2 - 6x + 8) \)

\( \implies 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \)

\( \implies 4x^2 - 30x + 50 = 0 \)

\( \implies 2x^2 - 15x + 25 = 0 \)

\( \implies 2x^2 - 10x - 5x + 25 = 0 \)

\( \implies 2x(x - 5) - 5(x - 5) = 0 \)

\( \implies (x - 5) (2x - 5) = 0 \)
\( x = 5 \) or \( x = \frac{5}{2} \)

Question. A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the original speed of the train.
Answer: Let original speed of train be \( x \) km/h.
new speed be \( = (x - 8) \) km/h.

\( \implies \frac{480}{x - 8} - \frac{480}{x} = 3 \)

\( \implies 480 \left[ \frac{x - (x - 8)}{x(x - 8)} \right] = 3 \)

\( \implies \frac{480 \times 8}{x^2 - 8x} = 3 \)

\( \implies 3x(x - 8) = 3840 \)

\( \implies x^2 - 8x - 1280 = 0 \)

\( \implies x^2 - 40x + 32x - 1280 = 0 \)

\( \implies (x + 32) (x - 40) = 0 \)

\( \implies x = -32 \) (rejected) or \( x = 40 \)
Original speed of train \( = 40 \) km/h

Question. A speed of a boat in still water is 11 km/hour. It can go 12 km upstream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream.
Answer: Let speed of the stream be \( x \) km/h.
Upstream speed \( = (11 - x) \) km/h and downstream speed \( = (11 + x) \) km/h.
Distance \( = 12 \) km.
Time taken for downstream \( = \frac{12}{11+x} \) hours.
Time taken for upstream \( = \frac{12}{11-x} \) hours.
According to the question, \( \frac{12}{11+x} + \frac{12}{11-x} = 2\frac{3}{4} = \frac{11}{4} \)

\( \implies 12 \left( \frac{11-x+11+x}{(11+x)(11-x)} \right) = \frac{11}{4} \)

\( \implies \frac{12 \times 22}{121 - x^2} = \frac{11}{4} \)

\( \implies 4 \times 264 = 11 (121 - x^2) \)

\( \implies \frac{4 \times 264}{11} = 121 - x^2 \)

\( \implies 96 = 121 - x^2 \)

\( \implies x^2 = 25 \)

\( \implies x = \pm 5 \)
Hence, speed of the stream is \( x = 5 \) km/h

Question. A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km/h less than that of the fast train, find the speeds of the two trains.
Answer: Let speed of fast train be \( x \) km/h.

\( \implies \) speed of slow train be \( (x - 10) \) km/h.
According to question, \( \frac{600}{x - 10} - \frac{600}{x} = 3 \)

\( \implies \frac{600x - 600(x - 10)}{x(x - 10)} = 3 \)

\( \implies \frac{6000}{x^2 - 10x} = 3 \)

\( \implies 2000 = x^2 - 10x \)

\( \implies x^2 - 10x - 2000 = 0 \)

\( \implies x^2 - 50x + 40x - 2000 = 0 \)

\( \implies (x - 50) (x + 40) = 0 \)

\( \implies x = 50, -40 \) (rejected).
Speed of fast train \( = 50 \) km/h and speed of slow train \( = 50 - 10 = 40 \) km/h.

Long Answer Type Questions

Question. A train travels 360 km at a uniform speed. In the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train. 
Answer: Let, the actual speed of the train be ‘x’ km/hr. Time taken by the train at actual speed, \( t_1 = \frac{360}{x} \) hr Increased speed of the train = (x + 5) km / hr Time taken by the train at the increased speed \( t_2 = \frac{360}{x + 5} \) hr According to the given condition: \( t_1 - t_2 = 1 \)
\( \implies \frac{360}{x} - \frac{360}{x + 5} = 1 \)
\( \implies \frac{360(x + 5 - x)}{x(x + 5)} = 1 \)
\( \implies 360 \times 5 = x^2 + 5x \)
\( \implies x^2 + 5x - 1800 = 0 \)
\( \implies x^2 + 5x - 1800 = 0 \)
\( \implies x^2 + 45x - 40x - 1800 = 0 \)
\( \implies x(x + 45) - 40(x + 45) = 0 \)
\( \implies (x - 40) (x + 45) = 0 \)
\( \implies x = 40 \) [\( \because x = - 45 \), is not possible as speed cannot be negative] Hence, the actual speed of the train is 40 km/hr.

Question. Solve for x : \( \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \); \( a \neq b \neq 0, x \neq 0, x \neq -(a + b) \) 
Answer: \( \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \)
\( \implies \frac{1}{a + b + x} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \)
\( \implies \frac{x - a - b - x}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \implies \frac{-(a + b)}{x(a + b + x)} = \frac{a + b}{ab} \)
\( \implies - ab = ax + bx + x^2 \)
\( \implies x^2 + ax + bx + ab = 0 \)
\( \implies x(x + a) + b(x + a) = 0 \)
\( \implies (x + b) (x + a) = 0 \)
\( \implies x = - a, - b \) Hence, the values of x are \( - a \) and \( - b \).

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. 
Answer: Let n be the required natural number. According to the question: Square of natural number diminished by 84 gives \( n^2 - 84 \). Thrice of 8 more than given number = \( 3(8 + n) \). According to the question,
\( n^2 - 84 = 3(8 + n) \)
\( \implies n^2 - 84 = 24 + 3n \)
\( \implies n^2 - 3n - 108 = 0 \) Splitting the middle term, we have
\( \implies n^2 - 12n + 9n - 108 = 0 \)
\( \implies n(n - 12) + 9(n - 12) = 0 \)
\( \implies (n - 12)(n + 9) = 0 \)
\( n = 12 \) or \( n = -9 \) But \( n \neq -9 \) as n is a natural number. Hence, the required natural number is 12.

Question. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number. 
Answer: Let n be the required natural number. According to the question, number when increased by 12 is \( n + 12 \). 160 times number’s reciprocal = \( 160(\frac{1}{n}) = \frac{160}{n} \) Now, by the given condition
\( n + 12 = \frac{160}{n} \)
\( \implies n(n + 12) = 160 \)
\( \implies n^2 + 12n - 160 = 0 \) Splitting the middle term, we have
\( n^2 + 20n - 8n - 160 = 0 \)
\( n(n + 20) - 8(n + 20) = 0 \)
\( (n + 20)(n - 8) = 0 \)
\( n = -20 \) or \( 8 \) But \( n \neq -20 \) as n is a natural number. Hence, the required number is 8.

Question. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane. 
Answer: Let ‘x’ km / hr be the speed of the plane. Increased speed = (x + 250) km / hr Time taken at the usual speed = \( \left( \frac{1500}{x} \right) \) hr Time taken at increased speed = \( \left( \frac{1500}{x + 250} \right) \) hr Difference between the two times taken = 30 minutes = \( \frac{1}{2} \) hr
\( \therefore \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \)
\( \implies \frac{1}{x} - \frac{1}{x + 250} = \frac{1}{3000} \)
\( \implies \frac{x + 250 - x}{x(x + 250)} = \frac{1}{3000} \)
\( \implies x^2 + 250x = 250 \times 3000 \)
\( \implies x^2 + 250x - 750000 = 0 \)
\( \implies x^2 + 1000x - 750x - 750000 = 0 \)
\( \implies x (x + 1000) - 750 (x + 1000) = 0 \)
\( \implies (x - 750) (x + 1000) = 0 \)
\( \implies x = 750 \) or \( - 1000 \)
\( \implies x = 750 \) [\( \because \) speed cannot be negative] Hence, the usual speed of the plane is 750 km/hr.

Question. Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m\(^2 \). 
Answer: Let ‘l‘ be the length and ‘b’ be the breadth of the rectangular park. Perimeter of the park, \( p = 2(l + b) \) Area of the park, \( A = l \times b \) According to the given conditions:
\( 2(l + b) = 60 \)
\( \implies l + b = 30 \)
\( \implies l = 30 - b \) ....(i) and \( lb = 200 \)
\( \implies (30 - b)b = 200 \) [from (i)]
\( \implies 30b - b^2 = 200 \)
\( \implies b^2 - 30b + 200 = 0 \) on splitting the middle term, we get:
\( \implies b^2 - 20b - 10b + 200 = 0 \)
\( \implies (b - 20) (b - 10) = 0 \)
\( \implies b = 20 \) or \( 10 \) when \( b = 20m, l = 10 m \) when \( b = 10m, l = 20 m \) Hence, the length and breadth of the rectangle are 10 m and 20 m or 20 m and 10 m respectively.

Question. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now? 
Answer: Let actual age of Zeba be x years. Her age when she was 5 years younger = (x - 5) years. According to the condition given in question: Square of her age = 11 more than 5 times her actual age
\( (x - 5)^2 = 11 + 5(x) \)
\( \implies x^2 + 25 - 10x = 11 + 5x \) [ \( \because (a - b)^2 = a^2 + b^2 - 2ab \)]
\( \implies x^2 - 10x - 5x + 25 - 11 = 0 \)
\( \implies x^2 - 15x + 14 = 0 \) Splitting the middle term, we have
\( \implies x^2 - 14x - x + 14 = 0 \)
\( \implies x(x - 14) - 1(x - 14) = 0 \)
\( \implies (x - 14)(x - 1) = 0 \)
\( \implies x = 14 \) or \( x = 1 \) But \( x \neq 1 \) as in that case (x - 5) will not be possible
\( \implies x = 14 \) Hence, Zeba’s age now is 14 years.

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 present age be x years. Then, Asha’s present age = \( (2 + x^2) \) [By the given condition] Now, when Nisha grows to her mother’s present age i.e. after \( \{ (x^2 + 2) - x \} \) years. Then, Asha’s age will become \( \{ (x^2 + 2) + (x^2 + 2 - x) \} \) years. Now by the given condition, Asha’s age = 1 year less than 10 times present age of Nisha.
\( (2 + x^2) + \{ (x^2 + 2) - x \} = 10x - 1 \)
\( \implies 2 + x^2 + x^2 + 2 - x = 10x - 1 \)
\( \implies 2x^2 - 11x + 5 = 0 \) Splitting the middle term, we have
\( \implies 2x^2 - 10x - x + 5 = 0 \)
\( \implies 2x(x - 5) - 1(x - 5) = 0 \)
\( \implies (x - 5)(2x - 1) = 0 \)
\( \implies x = 5 \) or \( x = \frac{1}{2} \) But \( x \neq \frac{1}{2} \) as then Nisha's age = \( \frac{1}{2} \). This means that her mother Asha's age = \( (x^2 + 2) = (\frac{1}{4} + 2) = 2\frac{1}{4} \) years which is not possible. Hence, the present age of Nisha = 5 years and the present age of Asha = \( x^2 + 2 = 5^2 + 2 = 25 + 2 = 27 \) years.

Question. In a class test, the sum of Arun's marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects. 
Answer: Let, the Arun’s marks in Hindi be x. Then, marks in English = 30 - x According to the given condition,
\( \implies (x + 2) (30 - x - 3) = 210 \)
\( \implies (x + 2) (27 - x) = 210 \)
\( \implies 27x + 54 - x^2 - 2x = 210 \)
\( \implies x^2 - 25x + 156 = 0 \)
\( \implies x^2 - 13x - 12x + 156 = 0 \)
\( \implies x (x - 13) - 12 (x - 13) = 0 \)
\( \implies (x - 13) (x - 12) = 0 \)
\( \implies x = 13 \) or \( 12 \) When : x = 13 Marks in Hindi = 13 Marks in English = 30 - 13 = 17 When : x = 12 Marks in Hindi = 12 Marks in English = 30 - 12 = 18 Hence, the marks obtained in the two subjects are (13, 17) or (12, 18).

Question. A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. 
Answer: Let, the speed of the stream be x km/hr. Speed of the boat in still water = 18 km/hr \( \therefore \) The speed of the boat in upstream = (18 - x) km/hr The speed of the boat in downstream = (18 + x) km/hr. Total distance to be covered = 24 km \( \therefore \) Time taken in upstream, \( t_1 = \frac{24}{(18 - x)} \) hr Time taken in downstream, \( t_2 = \frac{24}{(18 + x)} \) hr According to the question,
\( \frac{24}{18 - x} - \frac{24}{18 + x} = 1 \)
\( \implies \frac{24(18 + x) - 24(18 - x)}{(18 - x)(18 + x)} = 1 \)
\( \implies 24 \times 2x = 324 - x^2 \)
\( \implies x^2 + 48x - 324 = 0 \)
\( \implies x^2 + 54x - 6x - 324 = 0 \)
\( \implies x(x + 54) - 6(x + 54) = 0 \)
\( \implies (x + 54) (x - 6) = 0 \)
\( x \neq - 54 \) [\( \because \) speed can’t be negative]
\( \therefore x = 6 \) Hence, the speed of the stream is 6 km/hr.

Question. A train travels at a certain average speed for a distance of 63 km and then travels at 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 total journey, what is the original average speed?
Answer: Let the original speed of the train be ‘x’ km/hr. Increased speed = (x + 6) km/hr Now, time taken to cover 60 km at original speed, \( t_1 = \frac{63}{x} \) hr Time taken to cover 72 km at increased speed: \( t_2 = \frac{72}{x + 6} \) hr
\( \frac{63}{x} + \frac{72}{x + 6} = 3 \)
\( \implies \frac{63(x + 6) + 72x}{x(x + 6)} = 3 \)
\( \implies 63x + 378 + 72x = 3 (x^2 + 6x) \)
\( \implies 135x + 378 = 3x^2 + 18x \)
\( \implies 3x^2 + 18x - 135x - 378 = 0 \)
\( \implies 3x^2 - 117x - 378 = 0 \)
\( \implies x^2 - 39x - 126 = 0 \)
\( \implies x^2 - 42x + 3x - 126 = 0 \)
\( \implies x(x - 42) + 3(x - 42) = 0 \)
\( \implies (x - 42) (x + 3) = 0 \)
\( x = - 3 \) [\( \because \) speed can’t be negative]
\( \therefore x = 42 \) Hence, the original average speed of train is 42 km/hr.

Question. The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm, then find the other two sides.
Answer: Let, the base of the right angled triangle be ‘x’ cm. Then, the altitude of a right-angled triangle is (x - 7) cm. And, the hypotenuse of right angled triangle = 13 cm Then, by the pythagoras theorem
\( H^2 = P^2 + B^2 \)
\( \implies 13^2 = (x - 7)^2 + x^2 \)
\( \implies 169 = x^2 + 49 - 14x + x^2 \)
\( \implies 2x^2 - 14x - 120 = 0 \)
\( \implies x^2 - 7x - 60 = 0 \)
\( \implies x^2 - 12x + 5x - 60 = 0 \)
\( \implies x(x - 12) + 5(x - 12) = 0 \)
\( \implies (x + 5) (x - 12) = 0 \)
\( \implies x = 12 \) [\( \because x = - 5 \) is not possible] The base of the right angled triangle = 12 cm and altitude = 12 - 7 = 5 cm. Hence, the other two sides of triangle are 5 cm and 12 cm.

Question. Solve for x : \( \frac{x + 3}{x - 2} - \frac{1 - x}{x} = \frac{17}{4} ; x \neq 0, 2 \)
Answer: \( \frac{x + 3}{x - 2} - \frac{1 - x}{x} = \frac{17}{4} \)
\( \implies \frac{x(x + 3) - (x - 2)(1 - x)}{x(x - 2)} = \frac{17}{4} \)
\( \implies \frac{(x^2 + 3x) - (x - x^2 - 2 + 2x)}{(x^2 - 2x)} = \frac{17}{4} \)
\( \implies (x^2 + 3x - 3x + 2 + x^2) \times 4 = 17(x^2 - 2x) \)
\( \implies 4(2x^2 + 2) = 17x^2 - 34x \)
\( \implies 8x^2 + 8 = 17x^2 - 34x \)
\( \implies 17x^2 - 8x^2 - 34x - 8 = 0 \)
\( \implies 9x^2 - 34x - 8 = 0 \)
\( \implies 9x^2 - 36x + 2x - 8 = 0 \)
\( \implies 9x(x - 4) + 2(x - 4) = 0 \)
\( \implies (9x + 2) (x - 4) = 0 \)
\( \implies x = -\frac{2}{9}, 4 \) Hence, the values of x are \( -\frac{2}{9}, 4 \).

Question. Find two consecutive odd natural numbers, the sum of whose squares is 394. 
Answer: Let, the first number be x and the second number be (x + 2). According to the given condition,
\( x^2 + (x + 2)^2 = 394 \)
\( \implies x^2 + x^2 + 4 + 4x = 394 \)
\( \implies 2x^2 + 4x - 390 = 0 \)
\( \implies x^2 + 2x - 195 = 0 \)
\( \implies x^2 + 15x - 13x - 195 = 0 \)
\( \implies x(x + 15) - 13 (x + 15) = 0 \)
\( \implies (x - 13) (x + 15) = 0 \)
\( \implies x = 13, - 15 \)
\( x \neq -15 \) [\( \because \) natural numbers are not negative] Hence, the two consecutive odd natural numbers are 13 and 15.

Question. A and B working together can do a work in 6 days. If A takes 5 days less than B to finish the work, in how many days can B can do the work alone?
Answer: Let B take ‘x’ days to complete the work. Then, A takes (x - 5) days to complete the work done. According to the given condition:
\( \frac{1}{x} + \frac{1}{x - 5} = \frac{1}{6} \)
\( \implies \frac{x - 5 + x}{x(x - 5)} = \frac{1}{6} \)
\( \implies 6(2x - 5) = x^2 - 5x \)
\( \implies 12x - 30 = x^2 - 5x \)
\( \implies x^2 - 17x + 30 = 0 \)
\( \implies x^2 - 15x - 2x + 30 = 0 \)
\( \implies x(x - 15) - 2(x - 15) = 0 \)
\( \implies (x - 15)(x - 2) = 0 \)
\( \implies x = 2 \) or \( 15 \) But \( x = 2 \) is not possible as \( x < 5 \).
\( \therefore x = 15 \) Hence, B takes 15 days to complete the work alone.

Question. Find x in terms of a, b and c : \( \frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c} \), \( x \neq a, b, c \) 
Answer: \( \frac{a}{x - a} + \frac{b}{x - b} = \frac{2c}{x - c} \)
\( \implies \frac{a(x - b) + b(x - a)}{(x - a)(x - b)} = \frac{2c}{x - c} \)
\( \implies \frac{ax - ab + bx - ab}{x^2 - ax - bx + ab} = \frac{2c}{x - c} \)
\( \implies (ax - 2ab + bx) (x - c) = (x^2 - ax - bx + ab) (2c) \)
\( \implies ax^2 - 2abx + bx^2 - acx + 2abc - bcx = 2cx^2 - 2acx - 2bcx + 2abc \)
\( \implies ax^2 + bx^2 - 2cx^2 - acx + 2acx - bcx + 2bcx = 0 \)
\( \implies (a + b - 2c)x^2 + acx + bcx = 0 \)
\( \implies x[(a + b - 2c)x + (ac + bc)] = 0 \)
\( \implies (a + b - 2c)x + (ac + bc) = 0 \), or \( x = 0 \)
\( \implies (a + b - 2c) x + (ac + bc) = 0 \)
\( \implies (a + b - 2c)x = - (ac + bc) \)
\( \implies x = \frac{-(ac + bc)}{a + b - 2c} = \frac{-c(a + b)}{a + b - 2c} \) Hence, the ‘x’ in terms of a, b, c is \( \frac{-c(a + b)}{(a + b - 2c)} \).

Question. At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find t. 
Answer: It is given that at t minutes past 2 pm, the time needed by the minute hand to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) min.
\( \implies t + \left( \frac{t^2}{4} - 3 \right) = 60 \) [ \( \because \) time between 2 pm and 3 pm = 1 hour = 60 min.]
\( \implies 4t + t^2 - 12 = 240 \)
\( \implies t^2 + 4t - 12 - 240 = 0 \)
\( \implies t^2 + 4t - 252 = 0 \) Splitting the middle term, we have
\( t^2 + 18t - 14t - 252 = 0 \)
\( \implies t(t + 18) - 14(t + 18) = 0 \)
\( \implies (t + 18)(t - 14) = 0 \)
\( t = -18 \) or \( t = 14 \). But \( t \neq -18 \) as time cannot be negative
\( \implies t = 14 \) Hence, the required value of t is 14 minutes.

Question. Solve for x : \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \), \( x \neq 3, -3/2 \) 
Answer: \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \)
\( \implies \frac{2x(2x + 3) + 1(x - 3) + 3x + 9}{(x - 3)(2x + 3)} = 0 \)
\( \implies 4x^2 + 6x + x - 3 + 3x + 9 = 0 \)
\( \implies 4x^2 + 10x + 6 = 0 \)
\( \implies 2x^2 + 5x + 3 = 0 \)
\( \implies 2x^2 + 2x + 3x + 3 = 0 \) (on splitting the middle term)
\( \implies 2x(x + 1) + 3(x + 1) = 0 \)
\( \implies (2x + 3) (x + 1) = 0 \)
\( \implies x = -\frac{3}{2} \) or \( - 1 \) Hence, the values of x are \( -\frac{3}{2} \) and \( - 1 \).