CBSE Class 10 Mathematics Quadratic Equations VBQs Set K

Question. Write the sum of the real roots of the equation \( x^2 + |x| - 6 = 0 \).
Answer: \( x^2 + |x| - 6 = 0 \)
\( \Rightarrow x^2 + x - 6 = 0 \)(... \( |x| = x \))
Here, \( a = 1, b = 1, c = - 6 \)
\( \therefore \) Sum of roots \( = -\frac{b}{a} = \frac{- 1}{1} = - 1 \)

Question. In there any real value of ‘a’ for which the equation \( x^2 + 2x + (a^2 + 1) = 0 \) has real roots?
Answer: \( x^2 + 2x + (a^2 + 1) = 0 \)
\( D = (b)^2 - 4ac \)
\( = (2)^2 - 4 \times 1(a^2 + 1) \)
\( = 4 - 4a^2 - 4 = - 4a^2 \)
For real value of \( x \), \( D \geq 0 \)
But \( - 4a^2 \leq 0 \)
So, it is not possible.
There is no real value of \( a \).

Question. Write the condition to be satisfied for which equation \( ax^2 + 2bx + c = 0 \) and \( bx^2 - 2\sqrt{ac}x + b = 0 \) have equal roots.
Answer: In \( ax^2 + 2bx + c = 0 \)
\( D_1 = (2b)^2 - 4 \times a \times c \)
\( = 4b^2 - 4ac \)
\( \therefore \) Roots are equal, \( D_1 = 0 \)
\( \therefore 4b^2 - 4ac = 0 \)
\( \Rightarrow 4b^2 = 4ac \)
\( \Rightarrow b^2 = ac \)
and in \( bx^2 - 2\sqrt{ac}x + b = 0 \)
\( D_2 = (- 2\sqrt{ac})^2 - 4 \times b \times b \)
\( = 4ac - 4b^2 \)
\( \therefore \) Roots are equal, \( D_2 = 0 \)
\( \therefore 4ac = 4b^2 \)
\( \Rightarrow b^2 = ac \)
\( \therefore \) The required condition is \( b^2 = ac \).

Question. Show that \( x = - 3 \) is a solution of \( x^2 + 6x + 9 = 0 \).
Answer: The given equation is \( x^2 + 6x + 9 = 0 \)
If \( x = - 3 \) is its solution then it will satisfy it
L.H.S. \( = (- 3)^2 + 6(- 3) + 9 \)
\( = 9 - 18 + 9 \)
\( = 18 - 18 \)
\( = 0 = \) R.H.S.
Hence, \( x = - 3 \) is its one root.

Question. In a rectangular park of dimensions 50 m × 40 m, a rectangular pond is constructed so that the area of grass strip of uniform width surrounding the pond would be \( 1184 \text{ m}^2 \). Find the length and breadth of the pond.
Answer: Let ABCD be rectangular lawn and EFGH be rectangular pond. Let \( x \) m be the width of grass area, which is same around the pond.
Given, Length of lawn \( = 50 \) m, Width of lawn \( = 40 \) m
\( \therefore \) Length of pond \( = (50 - 2x) \) m and Breadth of pond \( = (40 - 2x) \) m
Also given,
Area of grass surrounding the pond \( = 1184 \text{ m}^2 \)
\( \Rightarrow \) Area of rectangular lawn – Area of pond \( = 1184 \text{ m}^2 \)
\( \Rightarrow 50 \times 40 - \{(50 - 2x) \times (40 - 2x)\} = 1184 \)
\( \Rightarrow 2000 - (2000 - 80x - 100x + 4x^2) = 1184 \)
\( \Rightarrow 2000 - 2000 + 180x - 4x^2 = 1184 \)
\( \Rightarrow 4x^2 - 180x + 1184 = 0 \)
\( \Rightarrow x^2 - 45x + 296 = 0 \)
\( \Rightarrow x^2 - 37x - 8x + 296 = 0 \)
\( \Rightarrow x(x - 37) - 8(x - 37) = 0 \)
\( \Rightarrow (x - 37) (x - 8) = 0 \)
\( \Rightarrow x - 37 = 0 \) or \( x - 8 = 0 \)
\( \Rightarrow x = 37 \) or \( x = 8 \)
\( x = 37 \) is not possible as in this case length of pond becomes \( 50 - 2 \times 37 = - 24 \) (not possible)
Hence, \( x = 8 \) is acceptable
\( \therefore \) Length of pond \( = 50 - 2 \times 8 = 34 \) m
Breadth of pond \( = 40 - 2 \times 8 = 24 \) m

Short Answer Type Questions

Question. Solve : \( \frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1} \), where \( x \neq - 1, \frac{1}{3} \).*
Answer: Given, \( \frac{3}{x + 1} - \frac{1}{2} = \frac{2}{3x - 1} \)
\( \Rightarrow \frac{6 - (x + 1)}{2(x + 1)} = \frac{2}{3x - 1} \)
\( \Rightarrow [6 - (x + 1)] (3x - 1) = 4(x + 1) \)
\( \Rightarrow 6(3x - 1) - (3x - 1) (x + 1) = 4(x + 1) \)
\( \Rightarrow 18x - 6 - (3x^2 - x + 3x - 1) = 4x + 4 \)
\( \Rightarrow 14x - 10 - 3x^2 - 2x + 1 = 0 \)
\( \Rightarrow - 3x^2 + 12x - 9 = 0 \)
\( \Rightarrow 3x^2 - 12x + 9 = 0 \)
\( \Rightarrow x^2 - 4x + 3 = 0 \)
\( \Rightarrow x^2 - 3x - x + 3 = 0 \)
\( \Rightarrow x(x - 3) - 1(x - 3) = 0 \)
\( \Rightarrow (x - 3) (x - 1) = 0 \)
\( \Rightarrow x = 1, 3 \).

Question. Solve : \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \), where \( x \neq 0, -\frac{3}{2} \).**
Answer: Given, \( \frac{4}{x} - 3 = \frac{5}{2x + 3} \)
\( \Rightarrow \frac{4 - 3x}{x} = \frac{5}{2x + 3} \)
\( \Rightarrow 4(2x + 3) - 3x(2x + 3) = 5x \)
\( \Rightarrow 8x + 12 - 6x^2 - 9x = 5x \)
\( \Rightarrow - 6x^2 - 6x + 12 = 0 \)
\( \Rightarrow 6x^2 + 6x - 12 = 0 \)
\( \Rightarrow x^2 + x - 2 = 0 \)
\( \Rightarrow x^2 + 2x - x - 2 = 0 \)
\( \Rightarrow x(x + 2) - 1(x + 2) = 0 \)
\( \Rightarrow (x + 2) (x - 1) = 0 \)
\( \Rightarrow x = 1, - 2 \).

Question. Solve for \( x \) :* \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \), \( x \neq 3, -\frac{3}{2} \)
Answer: We have, \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \)
\( \Rightarrow 2x(2x + 3) + (x - 3) + (3x + 9) = 0 \)
\( \Rightarrow 4x^2 + 6x + x - 3 + 3x + 9 = 0 \)
\( \Rightarrow 4x^2 + 10x + 6 = 0 \)
\( \Rightarrow 2x^2 + 5x + 3 = 0 \)
\( \Rightarrow 2x^2 + 2x + 3x + 3 = 0 \)
\( \Rightarrow 2x(x + 1) + 3 (x + 1) = 0 \)
\( \Rightarrow (2x + 3) (x + 1) = 0 \)
\( \Rightarrow x = - 1, -\frac{3}{2} \)
\( \Rightarrow x = - 1 \) [\( \because \) Given \( x \neq -3/2 \)]

Question. Solve : \( \frac{14}{x + 3} - 1 = \frac{5}{x + 1} \), where \( x \neq - 3, - 1 \).
Answer: Given, \( \frac{14}{x + 3} - 1 = \frac{5}{x + 1} \)
\( \Rightarrow \frac{14 - (x + 3)}{x + 3} = \frac{5}{x + 1} \)
\( \Rightarrow 14(x + 1) - (x + 3) (x + 1) = 5(x + 3) \)
\( \Rightarrow 14x + 14 - (x^2 + x + 3x + 3) = 5x + 15 \)
\( \Rightarrow 9x - 1 - x^2 - 4x - 3 = 0 \)
\( \Rightarrow - x^2 + 5x - 4 = 0 \)
\( \Rightarrow x^2 - 5x + 4 = 0 \)
\( \Rightarrow x^2 - 4x - x + 4 = 0 \)
\( \Rightarrow x(x - 4) - 1(x - 4) = 0 \)
\( \Rightarrow (x - 4) (x - 1) = 0 \)
\( \Rightarrow x = 4, 1 \).

Question. Solve : \( x^2 - 4ax - b^2 + 4a^2 = 0 \).*
Answer: Given, \( x^2 - 4ax - b^2 + 4a^2 = 0 \)
\( \Rightarrow x^2 - 4ax - (b^2 - 4a^2) = 0 \)
\( \Rightarrow x^2 - (b + 2a)x + (b - 2a)x - (b - 2a) (b + 2a) = 0 \)
\( \Rightarrow x[x - (b + 2a)] + (b - 2a) [x - (b + 2a)] = 0 \)
\( \Rightarrow x[x - b - 2a] + (b - 2a) [x - b - 2a] = 0 \)
\( \Rightarrow [x - b - 2a] [x + b - 2a] = 0 \)
\( \Rightarrow x = 2a + b \) and \( 2a - b \). Ans.

Question. The sum of two natural numbers is 8 and their product is 15. Find the numbers.*
Answer: Let one of the numbers be \( x \).
So, the other number is \( (8 - x) \).
Now, \( x(8 - x) = 15 \)
\( \Rightarrow 8x - x^2 = 15 \)
\( \Rightarrow x^2 - 8x + 15 = 0 \)
\( \Rightarrow x^2 - 5x - 3x + 15 = 0 \)
\( \Rightarrow x(x - 5) - 3(x - 5) = 0 \)
\( \Rightarrow (x - 5) (x - 3) = 0 \)
\( \Rightarrow x = 3 \) or \( 5 \)
\( \therefore 8 - x = 5 \) or \( 3 \)
Thus, the two numbers are 3 and 5. Ans.

Question. Solve : \( 4x^2 - 4ax + (a^2 - b^2) = 0 \).*
Answer: Given, \( 4x^2 - 4ax + (a^2 - b^2) = 0 \)
\( \Rightarrow 4x^2 - 4ax + (a - b) (a + b) = 0 \)
\( \Rightarrow 4x^2 - 2(a - b)x - 2(a + b)x + (a - b) (a + b) = 0 \)
\( \Rightarrow 2x[2x - (a - b)] - (a + b) [2x - (a - b)] = 0 \)
\( \Rightarrow [2x - (a - b)] [2x - (a + b)] = 0 \)
\( \Rightarrow 2x = a - b \) and \( a + b \)
\( \Rightarrow x = \frac{a - b}{2} \) and \( \frac{a + b}{2} \). Ans.

Question. Solve : \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)** 
Answer: Given, \( 3x^2 - 2\sqrt{6}x + 2 = 0 \)
\( \Rightarrow 3x^2 - \sqrt{6}x - \sqrt{6}x + 2 = 0 \)
\( \Rightarrow \sqrt{3}x(\sqrt{3}x - \sqrt{2}) - \sqrt{2}(\sqrt{3}x - \sqrt{2}) = 0 \)
\( \Rightarrow (\sqrt{3}x - \sqrt{2})^2 = 0 \)
\( \Rightarrow \sqrt{3}x - \sqrt{2} = 0 \)
\( \Rightarrow \sqrt{3}x = \sqrt{2} \)
\( \Rightarrow x = \sqrt{\frac{2}{3}} \)
\( \therefore x = \sqrt{\frac{2}{3}} \) and \( \sqrt{\frac{2}{3}} \) Ans.

Question. Solve : \( \sqrt{7}y^2 - 6y - 13\sqrt{7} = 0 \).
Answer: Given, \( \sqrt{7}y^2 - 6y - 13\sqrt{7} = 0 \)
\( \Rightarrow \sqrt{7}y^2 - 13y + 7y - 13\sqrt{7} = 0 \)
\( \Rightarrow y(\sqrt{7}y - 13) + \sqrt{7}(\sqrt{7}y - 13) = 0 \)
\( \Rightarrow (\sqrt{7}y - 13) (y + \sqrt{7}) = 0 \)
\( \Rightarrow y = \frac{13}{\sqrt{7}} \) and \( -\sqrt{7} \). Ans.

Question. Divide 12 into two parts such that the sum of their squares is 74.
Answer: Let the two parts be \( x \) and \( (12 - x) \).
So, \( x^2 + (12 - x)^2 = 74 \)
\( \Rightarrow x^2 + 144 + x^2 - 24x - 74 = 0 \)
\( \Rightarrow 2x^2 - 24x + 70 = 0 \)
\( \Rightarrow x^2 - 12x + 35 = 0 \)
\( \Rightarrow x^2 - 7x - 5x + 35 = 0 \)
\( \Rightarrow x(x - 7) - 5(x - 7) = 0 \)
\( \Rightarrow (x - 7) (x - 5) = 0 \)
\( \Rightarrow x = 5 \) and \( 7 \). Ans.

Question. The sum of two numbers is 15 and their reciprocals is \( \frac{3}{10} \). Find the numbers.
Answer: Let the numbers be \( x \) and \( (15 - x) \). So, their reciprocals are \( \frac{1}{x} \) and \( \frac{1}{15 - x} \) respectively.
Now, \( \frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \)
\( \Rightarrow \frac{15 - x + x}{x(15 - x)} = \frac{3}{10} \)
\( \Rightarrow \frac{15}{x(15 - x)} = \frac{3}{10} \)
\( \Rightarrow \frac{5}{x(15 - x)} = \frac{1}{10} \)
\( \Rightarrow x(15 - x) = 50 \)
\( \Rightarrow 15x - x^2 = 50 \)
\( \Rightarrow x^2 - 15x + 50 = 0 \)
\( \Rightarrow x^2 - 10x - 5x + 50 = 0 \)
\( \Rightarrow x(x - 10) - 5(x - 10) = 0 \)
\( \Rightarrow (x - 5) (x - 10) = 0 \)
\( \Rightarrow x = 5 \) and \( 10 \). Ans.

Question. The sum of the squares of two consecutive natural numbers is 421. Find the numbers.
Answer: Let the two consecutive natural numbers be \( x \) and \( (x + 1) \).
Thus, \( x^2 + (x + 1)^2 = 421 \)
\( \Rightarrow x^2 + x^2 + 2x + 1 = 421 \)
\( \Rightarrow 2x^2 + 2x - 420 = 0 \)
\( \Rightarrow x^2 + x - 210 = 0 \)
\( \Rightarrow x^2 + 15x - 14x - 210 = 0 \)
\( \Rightarrow x(x + 15) - 14(x + 15) = 0 \)
\( \Rightarrow (x + 15) (x - 14) = 0 \)
\( \Rightarrow x = 14 \) and \( - 15 \)
As \( - 15 \) is not a natural number so, the two numbers will be 14 and 15. Ans.

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: Let the two consecutive odd numbers be \( x \) and \( (x + 2) \).
Thus \( x^2 + (x + 2)^2 = 394 \)
\( \Rightarrow x^2 + x^2 + 4x + 4 = 394 \)
\( \Rightarrow 2x^2 + 4x - 390 = 0 \)
\( \Rightarrow x^2 + 2x - 195 = 0 \)
\( \Rightarrow x^2 + 15x - 13x - 195 = 0 \)
\( \Rightarrow x(x + 15) - 13(x + 15) = 0 \)
\( \Rightarrow (x + 15) (x - 13) = 0 \)
\( \Rightarrow x = 13 \) and \( - 15 \)
So, the numbers are either 13 and 15 or \( - 15 \) and \( - 13 \). Ans.

Question. In a class test, the sum of Kamal’s marks in Mathematics and English is 40. Had he secured 3 more marks in Mathematics and 4 less in English then, the product of the marks in both the tests would have been 360. Find the marks obtained by him in the two subjects separately.
Answer: Let the marks obtained in Mathematics be \( x \).
So the marks obtained in English is \( 40 - x \).
Now, \( (x + 3) (40 - x - 4) = 360 \)
\( \Rightarrow (x + 3) (36 - x) = 360 \)
\( \Rightarrow 36x + 108 - x^2 - 3x = 360 \)
\( \Rightarrow x^2 - 33x + 252 = 0 \)
\( \Rightarrow x^2 - 21x - 12x + 252 = 0 \)
\( \Rightarrow x(x - 21) - 12(x - 21) = 0 \)
\( \Rightarrow (x - 12) (x - 21) = 0 \)
\( \Rightarrow x = 12 \) and \( 21 \)
Thus, the marks obtained in Mathematics are either 12 or 21 and in Science are 28 or 19. Ans.

Question. A dealer sells a toy for ₹ 24 and gains as much percent as the cost price of the toy. Find the cost price of the toy.
Answer: Suppose the cost price of the toy be ₹ \( x \) then gain is \( x\% \)
We know Gain \( = \left( x \times \frac{x}{100} \right) \)
\( = \left( \frac{x^2}{100} \right) \)
Now, S.P. = C.P. + Gain
\( = x + \frac{x^2}{100} \)
As, S.P. = ₹ 24 (Given)
\( \therefore x + \frac{x^2}{100} = 24 \)
\( \Rightarrow x^2 + 100x - 2400 = 0 \)
\( x^2 + 120x - 20x - 2400 = 0 \)
\( x(x + 120) - 20(x + 120) = 0 \)
\( (x - 20)(x + 120) = 0 \)
\( x = 20 \) (\( \therefore x > 0 \))
Hence, the cost price of the toy will be ₹ 20. Ans.

Question. The difference of the squares of two natural numbers is 45. The square of the smaller number is 4 times the larger number. Find the numbers.
Answer: Let the smaller number be \( x \) and the larger be \( y \).
Now, \( x^2 = 4y \)
Also, \( y^2 - x^2 = 45 \)
\( \Rightarrow y^2 - 4y = 45 \)
\( \Rightarrow y^2 - 4y - 45 = 0 \)
\( \Rightarrow y^2 - 9y + 5y - 45 = 0 \)
\( \Rightarrow y(y - 9) + 5(y - 9) = 0 \)
\( \Rightarrow (y - 9) (y + 5) = 0 \)
\( \Rightarrow y = 9 \) or \( - 5 \)
As natural numbers are not negative, so the larger number is 9.
Thus, \( x^2 = 4(9) = 36 \)
\( \Rightarrow x = 6 \) or \( - 6 \)
Again as natural numbers are not negative, so the smaller number is 6.
So, the two natural numbers are 9 and 6. Ans.

Question. Three-eighth of the students of a class opted for visiting an old age home. Sixteen students opted for having a nature walk. Square root of total number of students in the class opted for tree plantation in the school. The number of students who visited the old age home is same as the number of students who went for a nature walk and did tree plantation. Find the total number of students.
Answer: Suppose the total number of students is \( x \)
Number of students who opted for visiting old age home \( = \frac{3}{8}x \)
Number of students who opted for having a nature walk = 16
Number of students who opted for tree plantation \( = \sqrt{x} \)
So, \( \frac{3}{8}x = 16 + \sqrt{x} \)
\( \therefore 3x - 8\sqrt{x} - 128 = 0 \)
Suppose, \( \sqrt{x} = y \)
\( \therefore 3y^2 - 8y - 128 = 0 \)
\( \therefore ( y - 8)(3y + 16) = 0 \)
\( y = 8 \) [\( \therefore y > 0 \)]
Hence, \( x = 64 \)
Number of students = 64. Ans.

Question. A charity trust decides to build the player hall having a carpet area of 300 sq m with its length 1 m more than twice its breadth. Find the length and breadth of the hall.
Answer: Let breadth of the hall be \( x \) m
\( \therefore \) Length of the hall \( = (2x + 1) \) m
Since, Area = \( l \times b \)
\( \Rightarrow (2x + 1) \times x = 300 \)
\( \Rightarrow 2x^2 + x - 300 = 0 \)
\( \Rightarrow 2x^2 + 25x - 24x - 300 = 0 \)
\( \Rightarrow x(2x + 25) - 12(2x + 25) = 0 \)
\( \Rightarrow (x - 12) (2x + 25) = 0 \)
\( \Rightarrow x = 12 \), \( x = -\frac{25}{2} \) (Rejected)
\( \therefore x = 12 \) m \( \Rightarrow \) breadth = 12 m,
Length \( = (2 \times 12 + 1) = 25 \) m. Ans.

Question. Students of class X collected ₹ 18000. They wanted to divide it equally among a certain number of students residing in slums area. When they started distributing the amount, 20 more students from nearby slums also joined. Now each student got ₹ 240 less. Find the number of students living in the slum.
Answer: Let number of students living in slums area be \( x \)
\( \therefore \) Share per student = ₹ \( \frac{18000}{x} \)
When number of students \( = x + 20 \)
Then, share per student = ₹ \( \frac{18000}{x + 20} \)
According to question,
\( \frac{18000}{x} - \frac{18000}{x + 20} = 240 \)
\( \Rightarrow \frac{18000x + 360000 - 18000x}{(x + 20)(x)} = 240 \)
\( \Rightarrow 360000 = 240(x^2 + 20x) \)
\( \Rightarrow 1500 = x^2 + 20x \)
\( \Rightarrow x^2 + 20x - 1500 = 0 \)
\( \Rightarrow x^2 + 50x - 30x - 1500 = 0 \)
\( \Rightarrow x(x + 50) - 30(x + 50) = 0 \)
\( \Rightarrow (x + 50) (x - 30) = 0 \)
\( \Rightarrow x = 30 \) or \( x = - 50 \) (Rejected)
\( \Rightarrow x = 30 \)
\( \therefore \) Number of students living in the slum area = 30. Ans.

Question. A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.
Answer: Let the usual speed be \( x \) km/hr.
Thus Time taken \( = \frac{300}{x} \) hr.
Increased speed is \( (x + 5) \) km/hr.
Thus New time \( = \frac{300}{x + 5} \) hr.
According to question,
\( \frac{300}{x} - \frac{300}{x + 5} = 2 \)
\( \Rightarrow 300 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 2 \)
\( \Rightarrow 1500 = 2x(x + 5) \)
\( \Rightarrow x(x + 5) = 750 \)
\( \Rightarrow x^2 + 5x - 750 = 0 \)
\( \Rightarrow x^2 + 30x - 25x - 750 = 0 \)
\( \Rightarrow x(x + 30) - 25(x + 30) = 0 \)
\( \Rightarrow (x + 30) (x - 25) = 0 \)
\( \Rightarrow x = 25 \) or \( - 30 \)
As speed cannot be negative, so the correct answer is 25 km/hr. Ans.

Question. A train travels a distance of 360 km at uniform speed. If its speed is increased by 5 km/hr, the journey will take 1 hour less. Find the original speed of the train.[NCERT]
Answer: Let the original speed be \( x \) km/hr.
Thus Time taken \( = \frac{360}{x} \) hr.
Increased speed is \( (x + 5) \) km/hr.
Thus New time \( = \frac{360}{x + 5} \) hr.
According to question,
\( \Rightarrow \frac{360}{x} - \frac{360}{x + 5} = 1 \)
\( \Rightarrow 360 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 1 \)
\( \Rightarrow 1800 = x(x + 5) \)
\( \Rightarrow x(x + 5) = 1800 \)
\( \Rightarrow x^2 + 5x - 1800 = 0 \)
\( \Rightarrow x^2 + 45x - 40x - 1800 = 0 \)
\( \Rightarrow x(x + 45) - 40(x + 45) = 0 \)
\( \Rightarrow (x + 45) (x - 40) = 0 \)
\( \Rightarrow x = 40 \) or \( - 45 \)
As speed cannot be negative, so the correct answer is 40 km/hr. Ans.

Question. The distance between Mumbai and Pune is 192 km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speeds of the two trains differ by 20 km/hr.*
Answer: Let the speed of the other train be \( x \) km/hr.
Thus, the speed of the Deccan Queen is \( (x + 20) \) km/hr.
Time taken by the other train \( = \frac{192}{x} \) hr.
Time taken by the Deccan Queen \( = \frac{192}{x + 20} \) hr.
According to question,
\( \Rightarrow \frac{192}{x} - \frac{192}{x + 20} = \frac{48}{60} \)
\( \Rightarrow 192 \left( \frac{x + 20 - x}{x(x + 20)} \right) = \frac{48}{60} \)
\( \Rightarrow 3840 \left( \frac{60}{48} \right) = x(x + 20) \)
\( \Rightarrow 4800 = x^2 + 20x \)
\( \Rightarrow x^2 + 20x - 4800 = 0 \)
\( \Rightarrow x^2 + 80x - 60x - 4800 = 0 \)
\( \Rightarrow x(x + 80) - 60(x + 80) = 0 \)
\( \Rightarrow (x + 80) (x - 60) = 0 \)
\( \Rightarrow x = 60 \) or \( - 80 \)
As speed cannot be negative, so the correct answer is 60 km/hr. Ans.

Question. The difference of the squares of two numbers is 88. If the larger number is 5 more than twice the smaller number then find the two numbers.*
Answer: Let the larger number be \( x \) and the smaller number be \( y \).
Given, \( x = 2y + 5 \)
and \( x^2 - y^2 = 88 \)
\( \Rightarrow (2y + 5)^2 - y^2 = 88 \)
\( \Rightarrow 4y^2 + 25 + 20y - y^2 = 88 \)
\( \Rightarrow 3y^2 + 20y - 63 = 0 \)
\( \Rightarrow 3y^2 + 27y - 7y - 63 = 0 \)
\( \Rightarrow 3y(y + 9) - 7(y + 9) = 0 \)
\( \Rightarrow (y + 9) (3y - 7) = 0 \)
\( \Rightarrow y = - 9 \) or \( \frac{7}{3} \)
Thus, \( x = - 18 + 5 = - 13 \)
or \( x = 2\left(\frac{7}{3}\right) + 5 = \frac{29}{3} \)
Hence, the two numbers are \( - 13 \) and \( - 9 \) or \( \frac{7}{3} \) and \( \frac{29}{3} \). Ans.

Question. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.*
Answer: Let the three consecutive natural numbers be \( x, x + 1 \) and \( x + 2 \).
According to the question,
\( (x + 1)^2 - [(x + 2)^2 - x^2] = 60 \)
\( x^2 + 1 + 2x - [(x + 2 - x) (x + 2 + x)] = 60 \)
\( x^2 + 2x + 1 - [2(2 + 2x)] = 60 \)
\( x^2 + 2x + 1 - 4 - 4x = 60 \)
\( x^2 - 2x - 63 = 0 \)
\( x^2 - 9x + 7x - 63 = 0 \)
\( x (x - 9) + 7 (x - 9) = 0 \)
\( (x + 7) (x - 9) = 0 \)
\( \therefore x = 9 \) or \( -7 \)
\( \Rightarrow x = 9 \) (\( \therefore \) numbers are natural)
\( \therefore \) Numbers are 9, 10, 11 Ans.

Question. The sum of two numbers is 16 and sum of their reciprocals is \( \frac{1}{3} \). Find the numbers.
Answer: Let the numbers be \( x \) and \( (16 - x) \). So, their reciprocals are \( \frac{1}{x} \) and \( \frac{1}{16 - x} \) respectively.
Now, \( \frac{1}{x} + \frac{1}{16 - x} = \frac{1}{3} \)
\( \Rightarrow \frac{16 - x + x}{x(16 - x)} = \frac{1}{3} \)
\( \Rightarrow \frac{16}{x(16 - x)} = \frac{1}{3} \)
\( \Rightarrow x(16 - x) = 48 \)
\( \Rightarrow 16x - x^2 = 48 \)
\( \Rightarrow x^2 - 16x + 48 = 0 \)
\( \Rightarrow x^2 - 12x - 4x + 48 = 0 \)
\( \Rightarrow x(x - 12) - 4(x - 12) = 0 \)
\( \Rightarrow (x - 4) (x - 12) = 0 \)
\( \Rightarrow x = 4 \) and 12. Ans.

Question. If the roots of the quadratic equation \( (a - b) x^2 + (b - c) x + (c - a) = 0 \) are equal, prove that \( 2a = b + c \).*
Answer: Given equation is, \( (a - b) x^2 + (b - c) x + (c - a) = 0 \)
By comparing the given equation with \( Ax^2 + Bx + C = 0 \), we get
\( A = a - b, B = b - c, C = c - a \)
Since the roots of the given quadratic equation are equal,
\( \therefore (b - c)^2 - 4(c - a) (a - b) = 0 \) [\( \because B^2 - 4AC = 0 \)]
\( b^2 + c^2 - 2bc - 4(ac - a^2 - bc + ab) = 0 \)
\( b^2 + c^2 - 2bc - 4ac + 4a^2 + 4bc - 4ab = 0 \)
\( (b^2 + c^2 + 2bc) - 4a(b + c) + 4a^2 = 0 \)
\( (b + c)^2 - 4a(b + c) + (2a)^2 = 0 \)
\( [(b + c) - 2a]^2 = 0 \)
\( b + c - 2a = 0 \)
i.e. \( 2a = b + c \)
Hence Proved.

Question. A girl is twice as old as her sister. Four years hence the product of their ages (in years) will be 160. Find their present ages.*
Answer: Let the sister’s age be \( x \). Thus, the girl’s age is \( 2x \).
Four years hence,
Sister’s age \( = (x + 4) \)
and Girl’s age \( = (2x + 4) \)
Now \( (x + 4) (2x + 4) = 160 \)
\( \Rightarrow 2x^2 + 8x + 4x + 16 = 160 \)
\( \Rightarrow 2x^2 + 12x + 16 = 160 \)
\( \Rightarrow 2x^2 + 12x - 144 = 0 \)
\( \Rightarrow x^2 + 6x - 72 = 0 \)
\( \Rightarrow x^2 + 12x - 6x - 72 = 0 \)
\( \Rightarrow x(x + 12) - 6(x + 12) = 0 \)
\( \Rightarrow (x + 12) (x - 6) = 0 \)
\( \Rightarrow x = 6 \) or \( - 12 \)
\( \Rightarrow x = 6 \) (as age is always positive)
Thus, the sister’s age is 6 years and the girl’s age is \( 2 \times 6 = 12 \) year. Ans.

Question. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present?*
Answer: Let age of Sumit be \( x \) years and age of his son be \( y \) years.
Then, according to question we have, \( x = 3y \)...(i)
Five years later,
\( x + 5 = 2\frac{1}{2} (y + 5) \)...(ii)
On putting \( x = 3y \) in equation (ii)
\( 3y + 5 = \frac{5}{2} (y + 5) \)
\( 3y + 5 = \frac{5y}{2} + \frac{25}{2} \)
\( 3y - \frac{5y}{2} = \frac{25}{2} - 5 \)
\( \frac{6y - 5y}{2} = \frac{25 - 10}{2} = \frac{15}{2} \)
\( \frac{y}{2} = \frac{15}{2} \)
\( y = 15 \)
So Sumit's age is \( 3 \times 15 = 45 \) years. Ans.

Question. If the price of a toy is reduced by ₹ 2, a person can buy two more toys for ₹360. Find the original price of the toy.
Answer: Let the original price of the toy be ₹ \( x \).
Thus, number of toys that can be bought for ₹360 \( = \frac{360}{x} \)
Now, reduced price of the toy \( = ₹ (x - 2) \)
Hence, number of toys \( = \frac{360}{x - 2} \)
Thus \( \frac{360}{x} + 2 = \frac{360}{x - 2} \)
\( \Rightarrow \frac{360}{x - 2} - \frac{360}{x} = 2 \)
\( \Rightarrow 360 \left( \frac{x - x + 2}{x(x - 2)} \right) = 2 \)
\( \Rightarrow 360 \left( \frac{2}{x(x - 2)} \right) = 2 \)
\( \Rightarrow 360 \left( \frac{1}{x(x - 2)} \right) = 1 \)
\( \Rightarrow x(x - 2) = 360 \)
\( \Rightarrow x^2 - 2x - 360 = 0 \)
\( \Rightarrow x^2 - 20x + 18x - 360 = 0 \)
\( \Rightarrow x(x - 20) + 18(x - 20) = 0 \)
\( \Rightarrow (x - 20) (x + 18) = 0 \)
\( \Rightarrow x = 20 \) or \( - 18 \)
\( \Rightarrow x = 20 \) (as price is always positive)
Hence, the original price of the toy is ₹ 20.

Question. If \( ad \neq bc \), then prove that the equation \( (a^2 + b^2) x^2 + 2(ac + bd) x + (c^2 + d^2) = 0 \) has no real roots.*
Answer: Given, \( (a^2 + b^2) x^2 + 2(ac + bd) x + (c^2 + d^2) = 0, ad \neq bc \)
\( D = b^2 - 4ac \)
\( = [2(ac + bd)]^2 - 4(a^2 + b^2) (c^2 + d^2) \)
\( = 4 [a^2c^2 + b^2d^2 + 2abcd] - 4[a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2] \)
\( = 4[a^2c^2 + b^2d^2 + 2abcd - a^2c^2 - a^2d^2 - b^2c^2 - b^2d^2] \)
\( = 4[- a^2d^2 - b^2c^2 + 2abcd] \)
\( = - 4(a^2d^2 + b^2c^2 - 2abcd) \)
\( = - 4(ad - bc)^2 \)
\( \therefore D \) is negative
Hence, given equation has no real roots.
Hence Proved.

Long Answer Type Questions

Question. Solve for \( x \) :* \( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4}, x \neq - 1, - 2, - 4 \)
Answer: We have,
\( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4}, x \neq - 1, - 2, - 4 \)
\( (x + 2) (x + 4) + 2(x + 1) (x + 4) = 4 (x + 1) (x + 2) \)
\( x^2 + 2x + 4x + 8 + 2 (x^2 + x + 4x + 4) = 4 (x^2 + x + 2x + 2) \)
\( x^2 + 6x + 8 + 2x^2 + 10x + 8 = 4x^2 + 12x + 8 \)
\( 3x^2 + 16x + 16 = 4x^2 + 12x + 8 \)
\( x^2 - 4x - 8 = 0 \)
\( \therefore x = \frac{4 \pm \sqrt{16 + 32}}{2} \)
\( x = \frac{4 \pm \sqrt{48}}{2} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3} \)
\( \therefore x = 2 \pm 2\sqrt{3} \). Ans.

Question. Solve for \( x \) : \( 3\left(\frac{7x+1}{5x-3}\right) - 4\left(\frac{5x-3}{7x+1}\right) = 11 \), where \( x \neq \frac{3}{5}, -\frac{1}{7} \)*
Answer: Given, \( 3\left(\frac{7x+1}{5x-3}\right) - 4\left(\frac{5x-3}{7x+1}\right) = 11 \)
\( \Rightarrow \frac{3(7x+1)^2 - 4(5x-3)^2}{(7x+1)(5x-3)} = 11 \)
\( \Rightarrow 3(7x + 1)^2 - 4(5x - 3)^2 = 11(7x + 1) (5x - 3) \)
\( \Rightarrow 3[49x^2 + 1 + 14x] - 4[25x^2 + 9 - 30x] = 11[35x^2 - 21x + 5x - 3] \)
\( \Rightarrow 147x^2 + 3 + 42x - 100x^2 - 36 + 120x = 11[35x^2 - 16x - 3] \)
\( \Rightarrow 47x^2 + 162x - 33 = 385x^2 - 176x - 33 \)
\( \Rightarrow 338x^2 = 338x \)
\( \Rightarrow 338x (x - 1) = 0 \)
\( \Rightarrow x = 1, 0 \). Ans.

Question. Solve for \( x \) : \( 2\left(\frac{2x-1}{x+3}\right) - 3\left(\frac{x+3}{2x-1}\right) = 5 \), where \( x \neq \frac{1}{2}, - 3 \).**
Answer: Given,
\( 2\left(\frac{2x-1}{x+3}\right) - 3\left(\frac{x+3}{2x-1}\right) = 5 \)
\( \Rightarrow \frac{2(2x-1)^2 - 3(x+3)^2}{(x+3)(2x-1)} = 5 \)
\( \Rightarrow 2(2x - 1)^2 - 3(x + 3)^2 = 5(x + 3) (2x - 1) \)
\( \Rightarrow 2[4x^2 + 1 - 4x] - 3[x^2 + 9 + 6x] = 5[2x^2 + 6x - x - 3] \)
\( \Rightarrow 8x^2 + 2 - 8x - 3x^2 - 27 - 18x = 10x^2 + 25x - 15 \)
\( \Rightarrow 5x^2 - 26x - 25 = 10x^2 + 25x - 15 \)
\( \Rightarrow 5x^2 + 51x + 10 = 0 \)
\( \Rightarrow 5x^2 + 50x + x + 10 = 0 \)
\( \Rightarrow 5x(x + 10) + 1(x + 10) = 0 \)
\( \Rightarrow (5x + 1)(x + 10) = 0 \)
\( \Rightarrow x = - 10, -\frac{1}{5} \). Ans.

Question. Solve for \( x \) : \( \frac{1}{2a+b+2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \).
Answer: Given,
\( \frac{1}{2a+b+2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \)
\( \Rightarrow \frac{1}{2a + b + 2x} - \frac{1}{2x} = \frac{1}{2a} + \frac{1}{b} \)
\( \Rightarrow \frac{2x - (2a + b + 2x)}{2x(2a + b + 2x)} = \frac{b + 2a}{2ab} \)
\( \Rightarrow \frac{-(2a + b)}{2x(2a + b + 2x)} = \frac{2a + b}{2ab} \)
\( \Rightarrow - 2ab = 2x(2a + b + 2x) \)
\( \Rightarrow - 2ab = 4ax + 2bx + 4x^2 \)
\( \Rightarrow 4x^2 + 4ax + 2bx + 2ab = 0 \)
\( \Rightarrow (4x^2 + 2bx) + (4ax + 2ab) = 0 \)
\( \Rightarrow 2x(2x + b) + 2a(2x + b) = 0 \)
\( \Rightarrow (2x + 2a)(2x + b) = 0 \)
\( \Rightarrow x = - \frac{2a}{2}, - \frac{b}{2} \)
\( \Rightarrow x = - a, - \frac{b}{2} \). Ans.

Question. A shopkeeper buys some books for ₹ 80. If he had bought 4 more books for the same amount then the price of each book would have been reduced by ₹ 1. Find the number of books he bought.
Answer: Let the number of books be \( x \).
Thus, the price of 1 book \( = ₹ \frac{80}{x} \).
If the number of books were \( x + 4 \), then the price per book \( = ₹ \frac{80}{x + 4} \)
According to the question,
\( \frac{80}{x} - \frac{80}{x + 4} = 1 \)
\( \Rightarrow 80 \left[ \frac{1}{x} - \frac{1}{x + 4} \right] = 1 \)
\( \Rightarrow 80 \left[ \frac{x + 4 - x}{x(x + 4)} \right] = 1 \)
\( \Rightarrow x(x + 4) = 320 \)
\( \Rightarrow x^2 + 4x - 320 = 0 \)
\( \Rightarrow x^2 + 20x - 16x - 320 = 0 \)
\( \Rightarrow x(x + 20) - 16(x + 20) = 0 \)
\( \Rightarrow (x + 20)(x - 16) = 0 \)
\( \Rightarrow x = 16, - 20 \)
As the number of books bought cannot be negative, so the correct answer is 16.

Question. A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?
Answer: Let B can finish a work in \( x \) days
So, A can finish work in \( (x - 6) \) days.
Together they finish the work in 4 days
According to the question,
\( \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4} \)
\( \frac{x - 6 + x}{x(x - 6)} = \frac{1}{4} \)
\( 4(2x - 6) = x^2 - 6x \)
\( 8x - 24 = x^2 - 6x \)
\( x^2 - 14x + 24 = 0 \)
\( x^2 - 12x - 2x + 24 = 0 \)
\( x(x - 12) - 2(x - 12) = 0 \)
\( (x - 12) (x - 2) = 0 \)
Either \( x - 12 = 0 \) or \( x - 2 = 0 \)
\( x = 12 \) or \( x = 2 \); Rejected
Hence, B can finish work in 12 days.
A can finish work in 6 days.