CBSE Class 10 Mathematics Quadratic Equations VBQs Set L

Question. Write all the values of \( p \) for which the quadratic equation \( x^2 + px + 16 = 0 \) has equal roots. Find the roots of the equation so obtained.*
Answer: Given, equation is \( x^2 + px + 16 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \)
where, \( a = 1, b = p \) and \( c = 16 \)
\( \therefore D = b^2 - 4ac \)
\( = p^2 - 4 \times 1 \times 16 \)
\( = p^2 - 64 \)
for equal roots, we have
\( D = 0 \Rightarrow p^2 - 64 = 0 \)
\( p^2 = 64 \Rightarrow p = \pm 8 \)
Putting \( p = 8 \) in given equation we have, \( x^2 + 8x + 16 = 0 \)
\( (x + 4)^2 = 0 \Rightarrow x = - 4 \)
Now, putting \( p = - 8 \) in the given equation, we get
\( x^2 - 8x + 16 = 0 \)
\( (x - 4)^2 = 0 \Rightarrow x = 4 \)
\( \therefore \) Required roots are – 4 and – 4 or 4 and 4.

Question. Two water taps together take 6 hours to fill a tank. If the tap with the larger diameter takes 9 hours lesser than the tap with the smaller diameter, then find the time in which each tap can separately fill the tap.
Answer: Let the smaller tap fill the tank in \( x \) hr and the total capacity of the tank be 1 unit.
Then in 1 hr, it can fill \( \frac{1}{x} \) unit.
So the larger tank takes \( (x - 9) \) hours to fill the tank.
Then in 1 hr, it can fill \( \frac{1}{x - 9} \) unit.
Together they can fill the tank in 6 hours.
Then in 1 hour, they can fill \( \frac{1}{6} \) unit of the tank.
Thus, \( \frac{1}{x} + \frac{1}{x - 9} = \frac{1}{6} \)
\( \Rightarrow \frac{x - 9 + x}{x(x - 9)} = \frac{1}{6} \)
\( \Rightarrow \frac{2x - 9}{x(x - 9)} = \frac{1}{6} \)
\( \Rightarrow 6(2x - 9) = x(x - 9) \)
\( \Rightarrow 12x - 54 = x^2 - 9x \)
\( \Rightarrow x^2 - 9x - 12x + 54 = 0 \)
\( \Rightarrow x^2 - 21x + 54 = 0 \)
\( \Rightarrow x^2 - 18x - 3x + 54 = 0 \)
\( \Rightarrow x(x - 18) - 3(x - 18) = 0 \)
\( \Rightarrow (x - 18)(x - 3) = 0 \)
\( \Rightarrow x = 3, 18 \)
As 9 hours less than 3 hours is – 6 which is not possible, so the smaller tap takes 18 hours to fill the tank and the larger takes 9 hours to do so.

Question. Solve for \( x \):* \( \frac{1}{x + 1} + \frac{3}{5x + 1} = \frac{5}{x + 4} \), \( x \neq - 1, -\frac{1}{5}, - 4 \)
Answer: Given,
\( \frac{1}{x + 1} + \frac{3}{5x + 1} = \frac{5}{x + 4} \)
\( \Rightarrow \frac{1}{x + 1} - \frac{5}{x + 4} = -\frac{3}{5x + 1} \)
\( \Rightarrow \frac{(x + 4) - 5(x + 1)}{(x + 1)(x + 4)} = -\frac{3}{5x + 1} \)
\( \Rightarrow \frac{x + 4 - 5x - 5}{x^2 + 5x + 4} = -\frac{3}{5x + 1} \)
\( \Rightarrow \frac{- (4x + 1)}{x^2 + 5x + 4} = -\frac{3}{5x + 1} \)
\( \Rightarrow (4x + 1) (5x + 1) = 3(x^2 + 5x + 4) \)
\( \Rightarrow 20x^2 + 4x + 5x + 1 = 3x^2 + 15x + 12 \)
\( \Rightarrow 17x^2 - 6x - 11 = 0 \)
\( \Rightarrow 17x^2 - 17x + 11x - 11 = 0 \)
\( \Rightarrow 17x(x - 1) + 11 (x - 1) = 0 \)
\( \Rightarrow (x - 1) (17x + 11) = 0 \)
Either \( x = 1 \) or \( x = -\frac{11}{17} \)
Given \( x \neq 1 \), so \( x = -\frac{11}{17} \). Ans.

Question. Solve \( \frac{x - 1}{x - 2} + \frac{x - 3}{x - 4} = \frac{10}{3} \), where \( x \neq 2, 4 \).*
Answer: Given,
\( \frac{(x - 1)(x - 4) + (x - 3)(x - 2)}{(x - 2)(x - 4)} = \frac{10}{3} \)
\( \Rightarrow 3[(x - 1) (x - 4) + (x - 3) (x - 2)] = 10[(x - 2) (x - 4)] \)
\( \Rightarrow 3[x^2 - x - 4x + 4 + x^2 - 3x - 2x + 6] = 10[x^2 - 2x - 4x + 8] \)
\( \Rightarrow 3[2x^2 - 10x + 10] = 10[x^2 - 6x + 8] \)
\( \Rightarrow 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \)
\( \Rightarrow 4x^2 - 30x + 50 = 0 \)
\( \Rightarrow 2x^2 - 15x + 25 = 0 \)
\( \Rightarrow 2x^2 - 10x - 5x + 25 = 0 \)
\( \Rightarrow 2x(x - 5) - 5(x - 5) = 0 \)
\( \Rightarrow (x - 5)(2x - 5) = 0 \)
\( \Rightarrow x = 5, \frac{5}{2} \). Ans.

Question. Solve for \( x \): \( \left(\frac{4x - 3}{2x + 1}\right) - 10\left(\frac{2x + 1}{4x - 3}\right) = 3 \), where \( x \neq -\frac{1}{2}, \frac{3}{4} \).
Answer: Given,
\( \left(\frac{4x - 3}{2x + 1}\right) - 10\left(\frac{2x + 1}{4x - 3}\right) = 3 \)
\( \Rightarrow \frac{(4x - 3)^2 - 10(2x + 1)^2}{(2x + 1)(4x - 3)} = 3 \)
\( \Rightarrow (4x - 3)^2 - 10(2x + 1)^2 = 3(2x + 1) (4x - 3) \)
\( \Rightarrow [16x^2 + 9 - 24x] - 10[4x^2 + 1 + 4x] = 3[8x^2 + 4x - 6x - 3] \)
\( \Rightarrow 16x^2 + 9 - 24x - 40x^2 - 10 - 40x = 24x^2 - 6x - 9 \)
\( \Rightarrow 48x^2 + 58x - 8 = 0 \)
\( \Rightarrow 24x^2 + 29x - 4 = 0 \)
\( \Rightarrow 24x^2 + 32x - 3x - 4 = 0 \)
\( \Rightarrow 8x(3x + 4) - 1(3x + 4) = 0 \)
\( \Rightarrow (8x - 1)(3x + 4) = 0 \)
\( \Rightarrow x = \frac{1}{8}, -\frac{4}{3} \). Ans.

Question. Solve : \( 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \).**
Answer: Given, \( 9x^2 - 9(a + b)x + (2a^2 + 5ab + 2b^2) = 0 \)
Comparing with \( Ax^2 + Bx + C = 0 \), we get
\( A = 9, B = - 9(a + b) \) and \( C = (2a^2 + 5ab + 2b^2) \)
\( \therefore D = B^2 - 4AC \)
\( \Rightarrow D = [- 9(a + b)]^2 - 4(9) (2a^2 + 5ab + 2b^2) \)
\( \Rightarrow D = 81(a + b)^2 - 36(2a^2 + 5ab + 2b^2) \)
\( \Rightarrow D = 81[a^2 + b^2 + 2ab] - 72a^2 - 180ab - 72b^2 \)
\( \Rightarrow D = 81a^2 + 81b^2 + 162ab - 72a^2 - 180ab - 72b^2 \)
\( \Rightarrow D = 9a^2 + 9b^2 - 18ab \)
\( \Rightarrow D = 9(a^2 + b^2 - 2ab) \)
\( \Rightarrow D = 9(a - b)^2 \geq 0 \)
Thus \( x = \frac{-B \pm \sqrt{D}}{2A} \)
\( x = \frac{9(a + b) + 3(a - b)}{18} \) and \( x = \frac{9(a + b) - 3(a - b)}{18} \)
\( \Rightarrow x = \frac{6(2a + b)}{18} \) and \( \frac{6(a + 2b)}{18} \)
\( \Rightarrow x = \frac{(2a + b)}{3} \) and \( \frac{(a + 2b)}{3} \). Ans.

Question. Solve for \( x \) : \( x^2 + 5x - (a^2 + a - 6) = 0 \).*
Answer: Taking \( (a^2 + a - 6) = a^2 + 3a - 2a - 6 \)
\( = a(a + 3) - 2(a + 3) \)
\( = (a + 3) (a - 2) \)
\( \therefore x^2 + 5x - (a + 3) (a - 2) = 0 \)
\( \Rightarrow x^2 + (a + 3)x - (a - 2)x - (a + 3)(a - 2) = 0 \)
\( \Rightarrow x[x + (a + 3)] - (a - 2)[x + (a + 3)] = 0 \)
\( \Rightarrow [x + (a + 3)][x - (a - 2)] = 0 \)
\( \Rightarrow x = -(a + 3), (a - 2) \). Ans.

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: Sol. Let Arun marks in hindi be \( x \) and marks in english be \( y \).
Then, according to question, we have \( x + y = 30 \) ...(i)
\( (x + 2) (y - 3) = 210 \) ...(ii)
From equation (i) put \( x = 30 - y \) in equation (ii),
\( (30 - y + 2) (y - 3) = 210 \)
\( (32 - y) (y - 3) = 210 \)
\( 32y - 96 - y^2 + 3y = 210 \)
\( y^2 - 35y + 306 = 0 \)
\( y^2 - 18y - 17y + 306 = 0 \)
\( y(y - 18) - 17(y - 18) = 0 \)
\( (y - 18)(y - 17) = 0 \)
\( y = 18, 17 \)
Put \( y = 18 \) and \( 17 \) in equation (i), we get
\( x = 12, 13 \)
Hence, his marks in hindi can be 12 and 13 and in english his marks can be 18 and 17. Ans.

Question. 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: Sol. Let the ten's digit be \( x \) and the unit's digit be \( y \). Thus, the number is \( 10x + y \).
Now \( xy = 18 \)
\( \Rightarrow x = \frac{18}{y} \) ...(i)
Also \( (10x + y) - 63 = (10y + x) \)
\( \Rightarrow 10x + y - 63 = 10y + x \)
\( \Rightarrow 9x - 9y = 63 \)
\( \Rightarrow x - y = 7 \)
\( \Rightarrow \frac{18}{y} - y = 7 \) [From (i)]
\( \Rightarrow 18 - y^2 = 7y \)
\( \Rightarrow y^2 + 7y - 18 = 0 \)
\( \Rightarrow y^2 + 9y - 2y - 18 = 0 \)
\( \Rightarrow y( y + 9) - 2(y + 9) = 0 \)
\( ( y - 2)( y + 9) = 0 \)
\( \Rightarrow y = 2 \) or \( - 9 \)
Thus, \( x = 9 \) or \( - 2 \)
As digit cannot be negative,
\( \therefore x = 9, y = 2 \)
So, the required number is
\( 10x + y = 10 \cdot 9 + 2 = 92 \) Ans.

Question. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is \( 2 \frac{16}{21} \), find the fraction.
Answer: Sol. Let the numerator be \( x \). So the denominator is \( 2x + 1 \).
Now \( \frac{x}{2x + 1} + \frac{2x + 1}{x} = 2 \frac{16}{21} \)
\( \Rightarrow \frac{x}{2x + 1} + \frac{2x + 1}{x} = \frac{58}{21} \)
\( \Rightarrow \frac{x^2 + (2x + 1)^2}{x(2x + 1)} = \frac{58}{21} \)
\( \Rightarrow \frac{x^2 + 4x^2 + 4x + 1}{x(2x + 1)} = \frac{58}{21} \)
\( \Rightarrow \frac{5x^2 + 4x + 1}{x(2x + 1)} = \frac{58}{21} \)
\( \Rightarrow 21(5x^2 + 4x + 1) = 58x(2x + 1) \)
\( \Rightarrow 105x^2 + 84x + 21 = 116x^2 + 58x \)
\( \Rightarrow 11x^2 - 26x - 21 = 0 \)
\( \Rightarrow 11x^2 - 33x + 7x - 21 = 0 \)
\( \Rightarrow 11x(x - 3) + 7(x - 3) = 0 \)
\( \Rightarrow (x - 3)(11x + 7) = 0 \)
\( \Rightarrow x = 3 \) and \( -\frac{7}{11} \)
So the numerator is 3. [As a fraction cannot be negative]
Hence, the fraction is \( \frac{3}{7} \). Ans.

Question. 7 years ago, Varun's age was five times the square of Swati's age. 3 years hence, Swati's age will be two-fifths of Varun's age. Find their present ages.
Answer: Sol. Let Swati's present age be \( x \) years and Varun's be \( y \) years.
7 years ago, Swati's age was \( (x - 7) \) years and Varun's was \( ( y - 7) \).
According to the question,
\( \Rightarrow y - 7 = 5(x - 7)^2 \) ...(i)
3 years later, Swati's age will be \( (x + 3) \) and Varun's will be \( ( y + 3) \).
According to the question,
\( x + 3 = \frac{2}{5} ( y + 3) \)
\( \Rightarrow 5x + 15 = 2y + 6 \)
\( \Rightarrow 2y = 5x + 9 \)
\( \Rightarrow y = \frac{5x + 9}{2} \) ...(ii)
Substituting the value of \( y \) in equation (i),
Hence, \( 5(x - 7)^2 = \frac{5x + 9}{2} - 7 \)
\( \Rightarrow 10(x - 7)^2 = 5x + 9 - 14 \)
\( \Rightarrow 10[x^2 - 14x + 49] = 5x - 5 \)
\( \Rightarrow 10x^2 - 140x + 490 = 5x - 5 \)
\( \Rightarrow 10x^2 - 145x + 495 = 0 \)
\( \Rightarrow 2x^2 - 29x + 99 = 0 \)
\( \Rightarrow 2x^2 - 18x - 11x + 99 = 0 \)
\( \Rightarrow 2x(x - 9) - 11(x - 9) = 0 \)
\( \Rightarrow (x - 9)(2x - 11) = 0 \)
\( \Rightarrow x = 9 \) and \( \frac{11}{2} \)
If Swati's present age is \( 5 \frac{1}{2} \) years, then 7 years ago it would be \( - 1 \frac{1}{2} \) years, which is not possible. Thus, her present age is 9 years.
Thus, Varun's present age
\( = \frac{5(9) + 9}{2} = \frac{45 + 9}{2} \)
\( = \frac{54}{2} = 27 \) years. Ans.

Question. A train travels 288 km at a uniform speed. If the speed has been 4 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Answer: Sol. Let the speed of the train be \( x \) km/hr.
The distance travelled by the train is 288 km.
Thus, time \( = \frac{288}{x} \) hr.
and when speed is \( (x + 4) \) km/hr, time taken \( = \frac{288}{x + 4} \)
According to the question,
\( \frac{288}{x} = \frac{288}{x + 4} + 1 \)
\( \Rightarrow 288 \left( \frac{1}{x} - \frac{1}{x + 4} \right) = 1 \)
\( \Rightarrow 288 \left( \frac{x + 4 - x}{x(x + 4)} \right) = 1 \)
\( \Rightarrow 288(4) = x^2 + 4x \)
\( \Rightarrow x^2 + 4x - 1152 = 0 \)
\( \Rightarrow x^2 + 36x - 32x - 1152 = 0 \)
\( \Rightarrow x(x + 36) - 32(x + 36) = 0 \)
\( \Rightarrow (x + 36)(x - 32) = 0 \)
\( \Rightarrow x = 32 \) and \( - 36 \)
As speed cannot be negative, so the speed of the train is 32 km/hr. Ans.

Question. An aeroplane left 30 minutes later than its scheduled time and in order to reach its destination 1500 km away on time, it had to increase its speed by 250 km/hr. Determine its usual speed.
Answer: Sol. Let the speed of the flight be \( x \) km/hr.
Distance to be covered is 1500 km.
Thus, time taken \( = \frac{1500}{x} \) hr.
If flight is delayed by 30 minutes, i.e., \( \frac{1}{2} \) hour,
Then, speed of flight \( = x + 250 \)
... Time taken \( = \frac{1500}{x + 250} \)
Now, according to the question,
\( \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{1}{x} - \frac{1}{x + 250} \right) = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{x + 250 - x}{x(x + 250)} \right) = \frac{1}{2} \)
\( \Rightarrow 1500 \times 250 \times 2 = x (x + 250) \)
\( \Rightarrow 750000 = x^2 + 250x \)
\( \Rightarrow x^2 + 250x - 750000 = 0 \)
\( \Rightarrow x^2 + 1000x - 750x - 750000 = 0 \)
\( \Rightarrow x(x + 1000) - 750(x + 1000) = 0 \)
\( \Rightarrow (x + 1000)(x - 750) = 0 \)
\( \Rightarrow x = 750, - 1000 \)
Since, speed cannot be negative.
So, the speed of the aeroplane is 750 km/hr. Ans.

Question. The sum of the areas of two squares is 260 \( m^2 \). If the difference of their perimeters is 24 m, then find the sides of the two squares.
Answer: Sol. Let the sides of the two squares be \( x \) m and \( y \) m respectively.
Thus area of first square \( = x^2 \text{ m}^2 \)
and perimeter of first square \( = 4x \) m
Also, area of second square \( = y^2 \text{ m}^2 \)
and perimeter of second square \( = 4y \) m
Now, \( x^2 + y^2 = 260 \) ...(i)
and \( 4(x - y) = 24 \)
\( \Rightarrow y = x - 6 \) ...(ii)
Now, \( x^2 + y^2 = 260 \) [From (i)]
\( \Rightarrow x^2 + (x - 6)^2 = 260 \) [Using (ii)]
\( \Rightarrow x^2 + x^2 - 12x + 36 = 260 \)
\( \Rightarrow 2x^2 - 12x - 224 = 0 \)
\( \Rightarrow x^2 - 6x - 112 = 0 \)
\( \Rightarrow x^2 - 14x + 8x - 112 = 0 \)
\( \Rightarrow x(x - 14) + 8(x - 14) = 0 \)
\( \Rightarrow (x + 8)(x - 14) = 0 \)
\( \Rightarrow x = - 8 \) and 14
As length cannot be negative, so we take \( x = 14 \) m
and \( y = 14 - 6 = 8 \) m
Thus, the sides of the two squares are 14 m and 8 m long. Ans.

Question. 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: Sol. As the field is rectangular, so all the angles are of 90°.
Hence the triangle formed by the diagonal and the two sides is a right-angled triangle.
Let the length of the shorter side be \( x \) m.
Thus the length of the longer side is \( (x + 30) \) m and the diagonal is \( (x + 60) \) m.
Applying Pythagoras' theorem,
\( (x + 60)^2 = (x + 30)^2 + x^2 \)
\( \Rightarrow x^2 = (x + 60)^2 - (x + 30)^2 \)
\( \Rightarrow x^2 = (x + 60 - x - 30)(x + 60 + x + 30) \)
\( \Rightarrow x^2 = 30(2x + 90) \)
\( \Rightarrow x^2 = 60x + 2700 \)
\( \Rightarrow x^2 - 60x - 2700 = 0 \)
\( \Rightarrow x^2 - 90x + 30x - 2700 = 0 \)
\( \Rightarrow x(x - 90) + 30(x - 90) = 0 \)
\( \Rightarrow (x - 90)(x + 30) = 0 \)
\( \Rightarrow x = 90 \) and \( - 30 \)
As length cannot be negative, so the length of the smaller side is 90 m and the length of the larger side is 120 m. Ans.

Question. The hypotenuse of a right-angled triangle is 6 cm more than twice the shortest side. If the third side is 2 cm less than the hypotenuse, find the sides of the triangle.
Answer: Sol. Let the shortest side be \( x \) cm.
So the length of the hypotenuse is \( (2x + 6) \) cm and the third side is \( (2x + 4) \) cm.
Now,
\( x^2 + (2x + 4)^2 = (2x + 6)^2 \)
\( \Rightarrow x^2 = (2x + 6)^2 - (2x + 4)^2 \)
\( \Rightarrow x^2 = (2x + 6 - 2x - 4)(2x + 6 + 2x + 4) \)
\( \Rightarrow x^2 = 2(4x + 10) \)
\( \Rightarrow x^2 - 8x - 20 = 0 \)
\( \Rightarrow x^2 - 10x + 2x - 20 = 0 \)
\( \Rightarrow x(x - 10) + 2(x - 10) = 0 \)
\( \Rightarrow (x - 10)(x + 2) = 0 \)
\( \Rightarrow x = 10 \) and \( - 2 \)
As length cannot be negative, so we take \( x = 10 \) cm.
Hence length of shortest side is 10 cm, length of hypotenuse is 26 cm and length of the third side is 24 cm. Ans.

Question. A person on tour has Rs 4,200 for his expenses. If he extends his tour by 3 days, he has to cut down his daily expenses by Rs 70. Find the original duration of the tour.
Answer: Sol. Let the original duration of the tour be \( x \) days.
Thus, daily expenses \( = Rs \frac{4200}{x} \)
Now, extended duration \( = x + 3 \) days
Thus, new daily expenses \( = Rs \frac{4200}{x + 3} \)
According to the question,
\( \frac{4200}{x + 3} = \frac{4200}{x} - 70 \)
\( \Rightarrow \frac{4200}{x} - \frac{4200}{x + 3} = 70 \)
\( \Rightarrow 4200 \left( \frac{1}{x} - \frac{1}{x + 3} \right) = 70 \)
\( \Rightarrow 60 \left( \frac{x + 3 - x}{x(x + 3)} \right) = 1 \)
\( \Rightarrow 180 = x^2 + 3x \)
\( \Rightarrow x^2 + 3x - 180 = 0 \)
\( \Rightarrow x^2 + 15x - 12x - 180 = 0 \)
\( \Rightarrow x(x + 15) - 12(x + 15) = 0 \)
\( \Rightarrow (x + 15)(x - 12) = 0 \)
\( \Rightarrow x = 12 \) and \( - 15 \)
As the number of days cannot be negative, so the original duration of the tour is 12 days. Ans.

Question. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres 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 the length and breadth of the rectangular park.*
Answer: Sol. Let the length of the rectangular park be \( x \) m. Then, breadth \( = (x - 3) \) m
Area of rectangular park \( = x (x - 3) \text{ m}^2 \)
and area of isosceles triangular park
\( = \frac{1}{2} (x - 3) \times 12 \text{ m}^2 \)
\( = 6 (x - 3) \text{ m}^2 \)
According to the question,
\( x(x - 3) - 6(x - 3) = 4 \)
\( x^2 - 3x - 6x + 18 = 4 \)
\( x^2 - 9x + 14 = 0 \)
\( x^2 - 7x - 2x + 14 = 0 \)
\( x(x - 7) - 2(x - 7) = 0 \)
\( (x - 2) (x - 7) = 0 \)
\( x = 2 \) or \( 7 \)
\( \therefore x = 7 \) m (as breadth can't be negative)
and \( x - 3 = (7 - 3) \text{ m} = 4 \) m
Hence, length and breadth of the rectangular park is 7 m and 4 m respectively. Ans.

Question. 300 apples were distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.
Answer: Sol. Let the number of students be \( x \).
So, the number of apples per student is \( \frac{300}{x} \).
If the number of students were \( x + 10 \) then the number of apples per student would have been \( \frac{300}{x + 10} \).
According to the question,
\( \frac{300}{x + 10} + 1 = \frac{300}{x} \)
\( \Rightarrow \frac{300}{x} - \frac{300}{x + 10} = 1 \)
\( \Rightarrow 300 \left( \frac{x + 10 - x}{x(x + 10)} \right) = 1 \)
\( \Rightarrow 300 \left( \frac{10}{x(x + 10)} \right) = 1 \)
\( \Rightarrow x(x + 10) = 3000 \)
\( \Rightarrow x^2 + 10x = 3000 \)
\( \Rightarrow x^2 + 10x - 3000 = 0 \)
\( \Rightarrow x^2 + 60x - 50x - 3000 = 0 \)
\( \Rightarrow x(x + 60) - 50(x + 60) = 0 \)
\( \Rightarrow (x + 60)(x - 50) = 0 \)
\( \Rightarrow x = 50 \) and \( - 60 \)
Since number of students cannot be negative, so the correct answer is 50. Ans.

Question. The total cost of a certain length of a piece of cloth is ₹ 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 its original rate per metre?*
Answer: Sol. Let the original length of piece of cloth is \( x \) m and rate of cloth is ₹ \( y \) per metre.
Then according to question, we have \( x \times y = 200 \) ...(i)
and if length be 5 m longer and each meter of cloth be ₹ 2 less then
\( (x + 5) (y - 2) = 200 \)
\( xy - 2x + 5y - 10 = 200 \) ...(ii)
On equating equation (i) and (ii), we have
\( xy = xy - 2x + 5y - 10 \)
\( \Rightarrow 2x - 5y = - 10 \) ...(iii)
\( y = \frac{200}{x} \) from equation (i)
\( \Rightarrow 2x - 5 \times \frac{200}{x} = - 10 \)
\( \Rightarrow 2x - \frac{1000}{x} = - 10 \)
\( \Rightarrow 2x^2 - 1000 = - 10x \)
\( \Rightarrow 2x^2 + 10x - 1000 = 0 \)
\( \Rightarrow x^2 + 5x - 500 = 0 \)
\( \Rightarrow x^2 + 25x - 20x - 500 = 0 \)
\( \Rightarrow x(x + 25) - 20 (x + 25) = 0 \)
\( (x + 25) (x - 20) = 0 \Rightarrow x = 20 \)
(\( x \neq - 25 \) length of cloth can never be negative)
\( \therefore x \times y = 200 \Rightarrow 20 \times y = 200 \)
\( y = 10 \)
Thus, length of the piece of cloth is 20 m and original price per metre is Rs 10.

Question. Two taps running together can fill a tank in \( 3 \frac{1}{13} \) hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank?*
Answer: Sol. Let one tap fill the tank in \( x \) hrs.
So, other tap will take \( (x + 3) \) hrs
Together they fill the tank in \( 3 \frac{1}{13} = \frac{40}{13} \) hours
According to the question,
\( \frac{1}{x} + \frac{1}{x + 3} = \frac{13}{40} \)
\( \Rightarrow \frac{x + 3 + x}{(x)(x + 3)} = \frac{13}{40} \)
\( \Rightarrow \frac{2x + 3}{x^2 + 3x} = \frac{13}{40} \)
\( 13x^2 + 39x = 80x + 120 \)
\( 13x^2 - 41x - 120 = 0 \)
\( 13x^2 - 65x + 24x - 120 = 0 \)
\( 13x(x - 5) + 24(x - 5) = 0 \)
\( (x - 5) (13x + 24) = 0 \)
Either \( x - 5 = 0 \) or \( 13x + 24 = 0 \)
\( x = 5, x = - 24/13 \) (Rejected)
Thus, one tap fill the tank in 5 hrs.
So, other tap fill the tank in \( 5 + 3 = 8 \) hrs.

Question. The sum of the areas of 2 squares is 640 m\(^2\). If the difference of their perimeters is 64 m, then find the sides of the two squares. [NCERT]
Answer: Sol. Let the sides of the two squares be \( x \) m and \( y \) m respectively.
Thus, area of first square \( = x^2 \text{ m}^2 \)
and perimeter of first square \( = 4x \) m
Also area of second square \( = y^2 \text{ m}^2 \)
and perimeter of second square \( = 4y \) m
Now, \( x^2 + y^2 = 640 \) ...(i)
and \( 4(x - y) = 64 \)
\( \Rightarrow y = x - 16 \) ...(ii)
From equation (i),
\( x^2 + y^2 = 640 \)
\( \Rightarrow x^2 + (x - 16)^2 = 640 \)
\( \Rightarrow x^2 + x^2 - 32x + 256 = 640 \) [Using (i)]
\( \Rightarrow 2x^2 - 32x - 384 = 0 \)
\( \Rightarrow x^2 - 16x - 192 = 0 \)
\( \Rightarrow x^2 - 24x + 8x - 192 = 0 \)
\( \Rightarrow x(x - 24) + 8(x - 24) = 0 \)
\( \Rightarrow (x + 8)(x - 24) = 0 \)
\( \Rightarrow x = - 8 \) and 24
As length cannot be negative so we take \( x = 24 \) m
and \( y = 24 - 16 = 8 \) m
Thus, the sides of the two squares are 24 m and 8 m long.

Question. A passenger while boarding a plane hurt herself and the captain called for immediate medical attention. Thus the plane left 30 minutes behind schedule. In order to reach its destination 1500 km away on time, the speed was increased by 100 km/hr from its usual speed. Find the usual speed.*
Answer: Sol. Let the usual speed be \( x \) km/hr.
We know, \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
So, Usual time taken \( = \frac{1500}{x} \) hr.
Increased speed \( = (x + 100) \text{ km/hr} \)
Thus, new time taken \( = \frac{1500}{x + 100} \)
Now, according to the question
\( \frac{1500}{x + 100} + \frac{1}{2} = \frac{1500}{x} \)
\( \Rightarrow \frac{1500}{x} - \frac{1500}{x + 100} = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{1}{x} - \frac{1}{x + 100} \right) = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{x + 100 - x}{x(x + 100)} \right) = \frac{1}{2} \)
\( \Rightarrow 1500 \left( \frac{100}{x(x + 100)} \right) = \frac{1}{2} \)
\( \Rightarrow x(x + 100) = 300000 \)
\( \Rightarrow x^2 + 100x - 300000 = 0 \)
\( \Rightarrow x^2 + 600x - 500x - 300000 = 0 \)
\( \Rightarrow x(x + 600) - 500(x + 600) = 0 \)
\( \Rightarrow (x + 600) (x - 500) = 0 \)
\( \Rightarrow x = 500 \) and \( - 600 \)
As speed cannot be negative, so the original speed of the plane is 500 km/hr.

Question. A bus travels at a certain average speed for a distance of 75 km and then travels a distance of 90 km at an average speed of 10 km/h more than the first speed. If it takes 3 hours to complete the total journey, find its first speed.*
Answer: Sol. Let \( x \) be the initial speed of the bus.
We know that \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \) or \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
So, according to the question,
\( 3 = \frac{75}{x} + \frac{90}{x + 10} \)
\( \Rightarrow 3 = \frac{75(x + 10) + 90x}{x(x + 10)} \)
\( \Rightarrow 3 (x) (x + 10) = 75x + 750 + 90x \)
\( \Rightarrow 3x^2 + 30x = 165x + 750 \)
\( \Rightarrow 3x^2 - 135x - 750 = 0 \)
\( \Rightarrow x^2 - 45x - 250 = 0 \)
\( \Rightarrow x^2 - 50x + 5x - 250 = 0 \)
\( \Rightarrow x(x - 50) + 5(x - 50) = 0 \)
\( \Rightarrow (x + 5) (x - 50) = 0 \)
\( \Rightarrow x = - 5 \) or \( x = 50 \)
Since, speed cannot be negative
So, \( x = 50 \)
Hence, the initial speed of bus is 50 km/hr.

Question. Two pipes together can fill a tank in \( 11 \frac{1}{9} \) minutes. If one pipe takes 5 minutes more than the other to fill tank separately, then find the time in which each pipe can fill the tank separately.*
Answer: Sol. Let the smaller tap fill the tank in \( x \) minutes and the total capacity of the tank be 1 sq. unit.
Then in 1 minute, smaller tap can fill \( \frac{1}{x} \) sq. unit.
So the other pipe takes \( (x - 5) \) minutes to fill the tank.
Then in 1 minute, it can fill \( \frac{1}{x - 5} \) sq. unit.
Together they can fill the tank is \( 11 \frac{1}{9} = \frac{100}{9} \) minutes
Then in 1 minute, they can fill \( \frac{9}{100} \) of the tank.
Thus, \( \frac{1}{x} + \frac{1}{x - 5} = \frac{9}{100} \)
\( \Rightarrow \frac{x - 5 + x}{x(x - 5)} = \frac{9}{100} \)
\( \frac{2x - 5}{x(x - 5)} = \frac{9}{100} \)
\( 200x - 500 = 9x^2 - 45x \)
\( \Rightarrow 9x^2 - 245x + 500 = 0 \)
\( \Rightarrow 9x^2 - 225x - 20x + 500 = 0 \)
\( \Rightarrow 9x(x - 25) - 20(x - 25) = 0 \)
\( \Rightarrow (x - 25)(9x - 20) = 0 \)
\( \Rightarrow x = 25 \) or \( \frac{20}{9} \).
As 5 minutes less than \( \frac{20}{9} \) minutes is less than 0 which is not possible, so the first pipe takes 25 minutes to fill the tank and the second takes 20 minutes to so do.

Question. The difference between two natural numbers is 5 and the difference between their reciprocals is \( \frac{5}{14} \). Find the numbers.*
Answer: Sol. Let the natural numbers be \( x \) and \( y \) respectively.
Now, \( x - y = 5 \dots \text{(i)} \)
Also, \( \frac{1}{y} - \frac{1}{x} = \frac{5}{14} \)
\( \Rightarrow \frac{x - y}{xy} = \frac{5}{14} \)
\( \Rightarrow 14x - 14y = 5xy \)
\( \Rightarrow 14(x - y) = 5xy \)
\( \Rightarrow 14 \times 5 = 5xy \) [From (i)]
\( \Rightarrow xy = 14 \)
\( \Rightarrow x = \frac{14}{y} \)
Substituting \( x = \frac{14}{y} \) in equation (i), we get
\( \frac{14}{y} - y = 5 \)
\( \Rightarrow y - \frac{14}{y} + 5 = 0 \)
\( \Rightarrow y^2 + 5y - 14 = 0 \)
\( \Rightarrow y^2 + 7y - 2y - 14 = 0 \)
\( \Rightarrow y( y + 7) - 2( y + 7) = 0 \)
\( \Rightarrow (y + 7) ( y - 2) = 0 \)
\( \Rightarrow y = 2, - 7 \)
Since natural numbers cannot be \( -ve \), so \( y = 2 \).
Thus, from equation (i),
\( x - y = 5 \)
\( \Rightarrow x - 2 = 5 \)
\( \Rightarrow x = 7 \)
So, the two numbers are 7 and 2.

Question. A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digit interchange their places. Find the number.*
Answer: Sol. Let the ten's digit be \( x \) and the unit's digit be \( y \).
Thus, the given number is \( (10x + y) \) and its reverse is \( (10y + x) \).
So, \( xy = 20 \)
\( \Rightarrow x = \frac{20}{y} \dots \text{(i)} \)
and \( 10x + y + 9 = 10y + x \)
\( \Rightarrow 9x - 9y = - 9 \)
\( \Rightarrow x - y = - 1 \)
\( \Rightarrow \frac{20}{y} - y = - 1 \) [From (i)]
\( \Rightarrow 20 - y^2 = - y \)
\( \Rightarrow y^2 - y - 20 = 0 \)
\( \Rightarrow y^2 - 5y + 4y - 20 = 0 \)
\( \Rightarrow y(y - 5) + 4(y - 5) = 0 \)
\( \Rightarrow (y - 5)(y + 4) = 0 \)
\( \Rightarrow y = 5, - 4 \)
So, \( x = \frac{20}{5} = 4 \) or \( x = \frac{20}{-4} = - 5 \)
Thus, the number is 45 or - 54.

Evaluation and Analysis Based Questions

Question. An equation has been given as \( \frac{c}{x^2} + \frac{k}{x} = 1 \).
Find the relation between \( c \) and \( k \), if
(i) \( x \) has real values
(ii) \( x \) has no real values.

Answer: Sol. We have,
\( \frac{c}{x^2} + \frac{k}{x} = 1 \)
Multiplying both the sides by \( x^2 \), we get
\( c + kx = x^2 \)
\( \Rightarrow x^2 - kx - c = 0 \)
(i) For real values:
\( (- k)^2 - 4.1(- c) \geq 0 \Rightarrow k^2 + 4c \geq 0 \)
\( \Rightarrow k^2 \geq - 4c \Rightarrow - k^2 \leq 4c \Rightarrow c \geq \frac{-k^2}{4} \)
(ii) For non-real values:
\( (- k)^2 - 4.1(- c) \leq 0 \)
\( \Rightarrow k^2 + 4c \leq 0 \Rightarrow k^2 \leq - 4c \)
\( \Rightarrow - k^2 \geq 4c \Rightarrow 4c \leq -k^2 \Rightarrow c \leq \frac{-k^2}{4} \).

Question. If \( \alpha \) and \( \beta \) are roots of a quadratic equation such that \( \alpha + \beta = 2 \) and \( \alpha^4 + \beta^4 = 272 \), then find the equation.
Answer: Sol. \( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 \)
\( = \{(\alpha + \beta)^2 - 2\alpha\beta\}^2 - 2\alpha^2\beta^2 \)
\( = (2^2 - 2\alpha\beta)^2 - 2\alpha^2\beta^2 \)
Let \( \alpha\beta = t \),
Then \( (4 - 2t)^2 - 2t^2 = 272 \)
\( \Rightarrow 16 - 16t + 4t^2 - 2t^2 = 272 \)
\( \Rightarrow 2t^2 - 16t - 256 = 0 \)
\( \Rightarrow t^2 - 8t - 128 = 0 \)
\( \Rightarrow t^2 - 16t + 8t - 128 = 0 \)
\( \Rightarrow t(t -16) + 8(t -16) = 0 \)
\( \Rightarrow (t -16)(t + 8) = 0 \)
\( \Rightarrow t = 16 \) or - 8
For \( t = 16 \)
\( \alpha + \beta = 2 \) and \( \alpha\beta = 16 \Rightarrow x^2 - 2x + 16 = 0 \)
For \( t = - 8 \)
\( \alpha + \beta = 2 \) and \( \alpha\beta = - 8 \Rightarrow x^2 - 2x - 8 = 0 \)

Question. If \( p, q, r \) and \( s \) are real numbers, such that \( pr = 2 (q + s) \), then show that at least one of the two equations \( x^2 + px + q = 0 \) and \( x^2 + rx + s = 0 \) has real roots.
Answer: Sol. We have,
\( x^2 + px + q = 0 \dots \text{(i)} \)
and \( x^2 + rx + s = 0 \dots \text{(ii)} \)
Let \( D_1 \) and \( D_2 \) be the discriminants of equation (i) and (ii) respectively. Then,
\( D_1 = p^2 - 4q \) and \( D_2 = r^2 - 4s \)
\( \Rightarrow D_1 + D_2 = p^2 - 4q + r^2 - 4s \)
\( = ( p^2 + r^2) - 4(q + s) \)
\( \Rightarrow D_1 + D_2 = p^2 + r^2 - 4 \left( \frac{pr}{2} \right) \)
[\( \because pr = 2(q + s) \therefore q + s = \frac{pr}{2} \)]
\( \Rightarrow D_1 + D_2 = p^2 + r^2 - 2pr = (p - r)^2 \geq 0 \)
[\( \because (p - r)^2 \geq 0 \) for all real \( p, r \)]
\( \Rightarrow \) At least one of \( D_1 \) and \( D_2 \) is greater than or equal to zero.
\( \Rightarrow \) At least one of the two equations has real roots. Hence Proved.

Question. Find the values of \( k \) for which the equation \( x^2 + 5kx + 16 = 0 \) has no real roots.
Answer: Sol. The given equation is \( x^2 + 5kx + 16 = 0 \).
Here, \( a = 1, b = 5k \) and \( c = 16 \)
\( \therefore D = b^2 - 4ac = (5k)^2 - 4 \times 1 \times 16 \)
\( = 25k^2 - 64 \)
The given equation will have no real roots, if
\( D < 0 \)
\( \Rightarrow 25k^2 - 64 < 0 \)
\( \Rightarrow 25 \left( k^2 - \frac{64}{25} \right) < 0 \)
\( \Rightarrow k^2 - \frac{64}{25} < 0 \)
\( \Rightarrow K^2 < \frac{64}{25} \)
\( \Rightarrow K^2 < \left( \frac{8}{5} \right)^2 \)
\( \Rightarrow - \frac{8}{5} < k < \frac{8}{5} \)
Thus, \( - \frac{8}{5} < k < \frac{8}{5} \).

Question. If the roots of the equation \( x^2 + 2cx + ab = 0 \) are real and unequal, prove that the equation \( x^2 - 2(a + b) x + a^2 + b^2 + 2c^2 = 0 \) has no real roots.
Answer: Sol. The two equations are
\( x^2 + 2cx + ab = 0 \dots \text{(i)} \)
and \( x^2 - 2(a + b) x + a^2 + b^2 + 2c^2 = 0 \dots \text{(ii)} \)
Let \( D_1 \) and \( D_2 \) be the discriminants of equations (i) and (ii) respectively. Then,
\( D_1 = (2c)^2 - 4 \times 1 \times ab \)
\( = 4c^2 - 4ab = 4(c^2 - ab) \)
\( D_2 = \{-2(a + b)\}^2 - 4 \times 1 \times (a^2 + b^2 + 2c^2) \)
\( D_2 = 4(a + b)^2 - 4(a^2 + b^2 + 2c^2) \)
\( D_2 = 4\{a^2 + b^2 + 2ab - a^2 - b^2 - 2c^2\} \)
\( D_2 = 4(2ab - 2c^2) \)
\( = - 8(c^2 - ab) \)
Since, the roots of equation (i) are real and unequal Therefore,
\( D_1 > 0 \)
\( \Rightarrow 4(c^2 - ab) > 0 \)
\( \Rightarrow c^2 - ab > 0 \)
\( \Rightarrow - 8(c^2 - ab) < 0 \)
\( \Rightarrow D_2 < 0 \)
i.e., roots of equation (ii) are not real. Hence Proved.

Question. Prove that the equation \( x^2(a^2 + b^2) + 2x (ac + bd) + (c^2 + d^2) = 0 \) has no real root, if \( ad \neq bc \).
Answer: Sol. The discriminant of the given equation is given by
\( D = \{2 (ac + bd)\}^2 - 4(a^2 + b^2) (c^2 + d^2) \)
\( D = 4(ac + bd )^2 - 4(a^2 + b^2)(c^2 + d^2) \)
\( D = 4[(ac + bd )^2 - (a^2 + b^2) (c^2 + d^2)] \)
\( D = 4[a^2 c^2 + b^2d^2 + 2ac \cdot bd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2] \)
\( D = 4[2ac \cdot bd - a^2d^2 - b^2c^2] \)
\( D = - 4[a^2d^2 + b^2c^2 - 2ad \cdot bc] = - 4(ad - bc)^2 \)
We have, \( ad \neq bc \)
\( ad - bc \neq 0 \Rightarrow (ad - bc)^2 > 0 \)
\( \Rightarrow - 4(ad - bc)^2 < 0 \Rightarrow D < 0 \)
Hence, the given equation has no real roots. Hence Proved.

Question. Find the value of \( p \) if \( \alpha^2 + \beta^2 - \alpha\beta = \frac{13}{4} \) where \( \alpha \) and \( \beta \) are roots of quadratic equation \( x^2 + px + 1 = 0 \).
Answer: Sol. Given, \( \alpha^2 + \beta^2 - \alpha\beta = \frac{13}{4} \dots \text{(i)} \)
If \( \alpha \) and \( \beta \) are roots of the equation \( x^2 + px + 1 = 0 \)
Then,
Sum of roots, \( \alpha + \beta = \frac{- b}{a} = - p \)
Product of roots, \( \alpha \beta = \frac{c}{a} = 1 \)
Substituting \( \alpha \beta = 1 \) in equation (i), we get
\( \alpha^2 + \beta^2 - 1 = \frac{13}{4} \)
\( \Rightarrow \alpha^2 + \beta^2 = \frac{13}{4} + 1 = \frac{17}{4} \)
\( \Rightarrow (\alpha + \beta)^2 - 2\alpha\beta = \frac{17}{4} \)
\( \Rightarrow (\alpha + \beta)^2 - 2 = \frac{17}{4} \) [\dots \( \alpha\beta = 1 \)]
\( \Rightarrow (\alpha + \beta)^2 = \frac{17}{4} + 2 = \frac{25}{4} \)
\( \therefore \alpha + \beta = \sqrt{\frac{25}{4}} = \frac{5}{2} = - p \)
Thus, \( p = - \frac{5}{2} \).

Question. Out of a certain number of Saras birds, one-fourth the number are moving about in lotus plants, \( \frac{1}{9} \text{th} \) coupled (along) with \( \frac{1}{4} \text{th} \) as well as 7 times the square root of the number move on a hill, 56 birds remain in Vacula tree. What is the total number of birds?
Answer: Sol. Let the total number of birds be \( x \).
\( \therefore \) Number of birds moving in lotus plants \( = \frac{x}{4} \)
Number of birds moving on a hill
\( = \frac{x}{9} + \frac{x}{4} + 7\sqrt{x} \)
Number of birds in Vakula tree \( = 56 \).
Using the given informations, we have
\( \frac{x}{4} + \left( \frac{x}{9} + \frac{x}{4} + 7\sqrt{x} \right) + 56 = x \)
\( \Rightarrow x - \frac{x}{4} - \frac{x}{9} - \frac{x}{4} - 7\sqrt{x} - 56 = 0 \)
\( \Rightarrow \frac{36x - 9x - 4x - 9x}{36} - 7\sqrt{x} - 56 = 0 \)
\( \Rightarrow \frac{14x}{36} - 7\sqrt{x} - 56 = 0 \)
\( \Rightarrow \frac{7x}{18} - 7\sqrt{x} - 56 = 0 \)
\( \Rightarrow \frac{x}{18} - \sqrt{x} - 8 = 0 \)
\( \Rightarrow x - 18\sqrt{x} - 144 = 0 \)
Putting \( \sqrt{x} = y \), we have
\( y^2 - 18y - 144 = 0 \)
\( \Rightarrow y^2 - 24y + 6y - 144 = 0 \)
\( \Rightarrow y(y - 24) + 6(y - 24) = 0 \)
\( \Rightarrow (y - 24) (y + 6) = 0 \)
\( \Rightarrow y = 24, - 6 \)
But \( y \neq - 6 \), since \( \sqrt{x} = y \) is positive.
\( \therefore y = 24 \Rightarrow \sqrt{x} = 24 \Rightarrow x = 576 \)
Hence, the total number of birds are 576.

Question. Out of a group of swans, \( \frac{7}{2} \) times the square root of the number are playing on the shore of a tank. The two remaining ones are playing, with amorous fight, in the water. What is the total number of swans?
Answer: Solution : Let the total number of swans be \( x \). Then,
Number of swans playing on the shore of the tank \( = \frac{7}{2}\sqrt{x} \)
It is given that there are two remaining swans.
\( \therefore x = \frac{7}{2}\sqrt{x} + 2 \)
\( x - \frac{7}{2}\sqrt{x} - 2 = 0 \)
Putting \( \sqrt{x} = y \), we have
\( y^2 - \frac{7}{2}y - 2 = 0 \)
\( 2y^2 - 7y - 4 = 0 \)
\( 2y^2 - 8y + y - 4 = 0 \)
\( 2y (y - 4) + (y - 4) = 0 \)
\( (y - 4) (2y + 1) = 0 \)
\( \Rightarrow y = 4 \) or \( y = - \frac{1}{2} \)
\( \Rightarrow y = 4 \) [\dots \( y = - \frac{1}{2} \) is not possible]
\( \Rightarrow x = y^2 = 4^2 = 16 \)
Hence, the total number of swans is 16.

Question. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Answer: Sol. Let \( x \) be the total number of camels.
Then, number of camels in the forest \( = \frac{x}{4} \)
Number of camels on mountains \( = 2\sqrt{x} \)
and number of camels on the bank of river \( = 15 \)
Thus, total number of camels \( = \frac{x}{4} + 2\sqrt{x} + 15 \)
Now, by hypothesis, we have
\( \frac{x}{4} + 2\sqrt{x} + 15 = x \)
\( \Rightarrow x + 8\sqrt{x} + 60 = 4x \)
\( \Rightarrow 3x - 8\sqrt{x} - 60 = 0 \)
Let \( \sqrt{x} = y \), then \( x = y^2 \)
\( \Rightarrow 3y^2 - 8y - 60 = 0 \)
\( \Rightarrow 3y^2 - 18y + 10y - 60 = 0 \)
\( 3y(y - 6) + 10(y - 6) = 0 \)
\( \Rightarrow (3y + 10) ( y - 6) = 0 \)
\( \Rightarrow y = 6 \) or \( y = - \frac{10}{3} \)
If, \( y = - \frac{10}{3} \)
Then, \( x = \left( - \frac{10}{3} \right)^2 = \frac{100}{9} \) (\dots \( x = y^2 \))
But, the number of camels cannot be a fraction.
\( \therefore y = 6 \Rightarrow x = 6^2 = 36 \)
Hence, the number of camels = 36.

Question. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is \( 2 \frac{16}{21} \), find the fraction.
Answer: Sol. Let the numerator of the fraction be \( x \).
Then, Denominator \( = 2x + 1 \) [Given]
\( \therefore \) Fraction \( = \frac{x}{2x + 1} \)
\( \Rightarrow \) Required of the fraction \( = \frac{2x + 1}{x} \)
It is given that the sum of the fraction and its reciprocal is \( 2 \frac{16}{21} \).
\( \Rightarrow \frac{x}{2x + 1} + \frac{2x + 1}{x} = 2 \frac{16}{21} \)
\( \Rightarrow \frac{x^2 + (2x + 1)^2}{x(2x + 1)} = \frac{58}{21} \)
\( \Rightarrow \frac{5x^2 + 4x + 1}{2x^2 + x} = \frac{58}{21} \)
\( \Rightarrow 21(5x^2 + 4x + 1) = 58(2x^2 + x) \)
\( \Rightarrow 105x^2 + 84x + 21 = 116x^2 + 58x \)
\( \Rightarrow 11x^2 - 26x - 21 = 0 \).
\( \Rightarrow 11x^2 - (33 - 7)x - 21 = 0 \)
\( \Rightarrow 11x^2 - 33x + 7x - 21 = 0 \)
\( \Rightarrow 11x(x - 3) + 7(x - 3) = 0 \)
\( \Rightarrow (11x + 7) (x - 3) = 0 \)
\( \Rightarrow x = \frac{-7}{11} \) or \( x = 3 \)
Since numerator of a fraction cannot be negative
\( \therefore x = \frac{-7}{11} \) (Rejected)
\( \therefore x = 3 \)
Denominator \( = 2x + 1 \)
\( = 2(3) + 1 \)
\( = 6 + 1 = 7 \)
\( \therefore \) Required fraction \( = \frac{3}{7} \). Ans.