CBSE Class 10 Quadratic Equations Sure Shot Questions Set H

Short Answer Type Questions 

Question. Find the value of \(p\), for which one root of the quadratic equation \(px^2 - 14x + 8 = 0\) is 6 times the other.
Answer: Given equation is, \(px^2 - 14x + 8 = 0\).
Let roots of equation be \(\alpha\) and \(\beta\) such that \(\beta = 6\alpha \Rightarrow 6\alpha - \beta = 0\)...(i)
Now, sum of roots \(= \alpha + \beta = -\left(\frac{-14}{p}\right) = \frac{14}{p}\)...(ii)
and product of roots \(= \alpha \beta = \frac{8}{p}\)...(iii)
Solving (i) and (ii), we get \(\alpha = \frac{2}{p}\) and \(\beta = \frac{12}{p}\)
Putting these values in (iii) we get
\(\left(\frac{2}{p}\right) \times \left(\frac{12}{p}\right) = \frac{8}{p} \Rightarrow 8p = 24\)
\(\Rightarrow p = 3\) (\(\because p \neq 0\))

Question. Solve for \(x\) (in terms of \(a\) and \(b\)): \(\frac{a}{x-b} + \frac{b}{x-a} = 2, x \neq a, b\)
Answer: We have, \(\frac{a}{x - b} + \frac{b}{x - a} = 2\)
\(\Rightarrow \frac{a(x - a) + b(x - b)}{(x - b)(x - a)} = 2\)
\(\Rightarrow ax - a^2 + bx - b^2 = 2(x^2 - bx - ax + ab)\)
\(\Rightarrow 2x^2 - 3bx - 3ax + 2ab + a^2 + b^2 = 0\)
\(\Rightarrow 2x^2 - 3(a + b)x + (a + b)^2 = 0\)
\(\Rightarrow 2x^2 - 2(a + b)x - (a + b)x + (a + b)^2 = 0\)
\(\Rightarrow 2x[x - (a + b)] - (a + b)[x - (a + b)] = 0\)
\(\Rightarrow [x - (a + b)] [2x - (a + b)] = 0\)
\(\Rightarrow x = a + b\) or \(x = \frac{a + b}{2}\).

Question. Find the value of \(m\) so that the quadratic equation \(mx(5x - 6) + 9 = 0\) has two equal roots.
Answer: Given, \(mx(5x - 6) + 9 = 0 \Rightarrow 5mx^2 - 6mx + 9 = 0\)
For equal roots, discriminant, \(D = 0\)
\(\therefore (-6m)^2 - 4(5m)(9) = 0\)
\(\Rightarrow 36m^2 - 180m = 0 \Rightarrow 36m(m - 5) = 0\)
\(\Rightarrow m = 5\) (\(\because m \neq 0\))

Question. If \(x = 2/3\) and \(x = -3\) are roots of the quadratic equation \(ax^2 + 7x + b = 0\), then find the values of \(a\) and \(b\).
Answer: Given, roots of quadratic equation \(ax^2 + 7x + b = 0\) are \(\frac{2}{3}\) and \(-3\)
Sum of roots = \(\frac{-7}{a} \Rightarrow \frac{2}{3} + (-3) = \frac{-7}{a} \Rightarrow \frac{-7}{3} = \frac{-7}{a} \Rightarrow a = 3\)
Product of roots = \(\frac{b}{a} \Rightarrow \left(\frac{2}{3}\right)(-3) = \frac{b}{3} \Rightarrow -2 = \frac{b}{3} \Rightarrow b = -6\)
\(\therefore a = 3, b = -6\)

Question. For what value of \(k\) does the quadratic equation \((k - 5)x^2 + 2(k - 5)x + 2 = 0\) have equal roots?
Answer: We have, \((k - 5)x^2 + 2(k - 5)x + 2 = 0\)
Given equation has equal roots. \(\therefore\) Discriminant, \(D = 0\)
\(\Rightarrow [2(k - 5)]^2 - 4(k - 5)(2) = 0\)
\(\Rightarrow k^2 + 25 - 10k - 2k + 10 = 0\)
\(\Rightarrow k^2 - 12k + 35 = 0 \Rightarrow k^2 - 7k - 5k + 35 = 0\)
\(\Rightarrow (k - 7)(k - 5) = 0 \Rightarrow k = 7\) [\(\because k \neq 5\)]

Question. Find the value of \(p\) for which the roots of the equation \(px(x - 2) + 6 = 0\), are equal.
Answer: Given, \(px(x - 2) + 6 = 0 \Rightarrow px^2 - 2px + 6 = 0\)
Since, the roots are equal \(\therefore D = 0\)
\(\Rightarrow (-2p)^2 - 4 \times p \times 6 = 0 \Rightarrow 4p^2 - 24p = 0\)
\(\Rightarrow 4p(p - 6) = 0 \Rightarrow 4p = 0\) or \(p - 6 = 0\)
\(\Rightarrow p = 6\) [\(\because p \neq 0\)]

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 three consecutive natural numbers be \(x, x + 1, x + 2\).
According to question, \((x + 1)^2 - [(x + 2)^2 - (x)^2] = 60\)
\(\Rightarrow (x + 1)^2 - [x^2 + 4 + 4x - x^2] = 60\)
\(\Rightarrow x^2 + 1 + 2x - 4 - 4x = 60\)
\(\Rightarrow x^2 - 2x - 63 = 0 \Rightarrow x^2 - 9x + 7x - 63 = 0\)
\(\Rightarrow x(x - 9) + 7(x - 9) = 0 \Rightarrow (x + 7)(x - 9) = 0\)
\(\Rightarrow x = 9\) (\(\because x \neq -7\) as \(x\) is a natural number)
\(\therefore\) The numbers are 9, 10, 11.

Question. If 1 is a root of the quadratic equation \(3x^2 + ax - 2 = 0\) and the quadratic equation \(a(x^2 + 6x) - b = 0\) has equal roots, find the value of \(b\).
Answer: Since, 1 is a root of \(3x^2 + ax - 2 = 0\)
\(\therefore 3(1)^2 + a(1) - 2 = 0\)
\(\Rightarrow a + 1 = 0 \Rightarrow a = -1\)
Putting \(a = -1\) in \(a(x^2 + 6x) - b = 0\), we get
\(x^2 + 6x + b = 0\) ...(i)
Since, (i) has equal roots. \(\therefore D = 0\)
\(\Rightarrow 6^2 - 4(b) = 0 \Rightarrow 36 = 4b \Rightarrow b = 9\)

Question. Solve the quadratic equation \(16x^2 - 24x - 1 = 0\) by using the quadratic formula.
Answer: Here, \(a = 16, b = -24, c = -1\)
\(\therefore D = b^2 - 4ac = (-24)^2 - 4(16)(-1) = 576 + 64 = 640 > 0\).
\(\therefore \sqrt{D} = \sqrt{640} = \sqrt{64 \times 10} = 8\sqrt{10}\)
By quadratic formula, we have
\(x = \frac{-b \pm \sqrt{D}}{2a} = \frac{24 \pm 8\sqrt{10}}{32} = \frac{3 \pm \sqrt{10}}{4}\)
Therefore, the roots are \(\frac{3 + \sqrt{10}}{4}\) and \(\frac{3 - \sqrt{10}}{4}\)

Question. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.
Answer: Let the average speed of express train be \(x\) km/hr.
\(\therefore\) The average speed of passenger train \(= (x - 11)\) km/hr
Time taken by express train to travel 132 km = \(\frac{132}{x}\) hr
Time taken by passenger train to travel 132 km = \(\frac{132}{x - 11}\) hr
According to question, \(\frac{132}{x - 11} - \frac{132}{x} = 1\)
\(\Rightarrow 132x - 132x + 132 \times 11 = x(x - 11)\)
\(\Rightarrow 1452 = x^2 - 11x\)
\(\Rightarrow x^2 - 11x - 1452 = 0\), which is the required equation.

Short Answer Type Questions 

Question. Solve for \(x\): \(\frac{63}{x} + \frac{72}{x+6} = 3\)
Answer: Given, \(\frac{63}{x} + \frac{72}{x+6} = 3\)
\(\Rightarrow \frac{7}{x} + \frac{8}{x+6} = \frac{3}{9} = \frac{1}{3} \Rightarrow \frac{7(x+6) + 8x}{x(x+6)} = \frac{1}{3}\)
\(\Rightarrow 3(7x + 42 + 8x) = x^2 + 6x\)
\(\Rightarrow 45x + 126 = x^2 + 6x\)
\(\Rightarrow x^2 - 39x - 126 = 0 \Rightarrow x^2 - 42x + 3x - 126 = 0\)
\(\Rightarrow x(x - 42) + 3(x - 42) = 0 \Rightarrow (x - 42)(x + 3) = 0\)
\(\Rightarrow\) Either \(x - 42 = 0\) or \(x + 3 = 0 \Rightarrow x = 42\) or \(x = -3\)

Question. If the sum of two natural numbers is 8 and their product is 15, then find the numbers.
Answer: Let the two natural numbers be \(a\) and \(b\).
So, \(a + b = 8\) (Given) \(\Rightarrow a = 8 - b\) ...(1)
and \(ab = 15\) (given)
\(\Rightarrow (8 - b)b = 15\) [Using (1)]
\(\Rightarrow 8b - b^2 = 15 \Rightarrow b^2 - 8b + 15 = 0\)
\(\Rightarrow b^2 - 5b - 3b + 15 = 0 \Rightarrow b(b - 5) - 3(b - 5) = 0\)
\(\Rightarrow (b - 3)(b - 5) = 0 \Rightarrow b = 3\) or \(b = 5\)
When \(b = 3\) then \(a = 8 - 3 = 5\)
When \(b = 5\) then \(a = 8 - 5 = 3\)

Question. A two digit number is such that the product of the digits is 24. When 18 is added to the number, the digits interchange their places. Formulate the quadratic equation whose roots are the digits of the number.
Answer: Let the digit at ten’s place be \(y\) and digit at unit’s place be \(x\).
\(\therefore\) Number \(= 10y + x\)
Reversed Number \(= 10x + y\)
According to the question, \(10x + y = 18 + 10y + x\)
\(\Rightarrow 10x + y - 10y - x = 18 \Rightarrow 9x - 9y = 18 \Rightarrow x - y = 2\)
Also, \(xy = 24\)
Now, \((x + y)^2 = (x - y)^2 + 4xy = 2^2 + 4(24) = 100\)
\(\Rightarrow x + y = 10\) (\(\because x, y > 0\))
\(\therefore\) Required equation is \(t^2 - 10t + 24 = 0\)

Question. If the sum of first \(n\) even natural numbers is 420, then find the value of \(n\).
Answer: We have, \(2 + 4 + 6 + 8 + ...\) to \(n\) terms \(= 420\)
\(\Rightarrow n/2[2 \times 2 + (n - 1) \times 2] = 420\)
\(\Rightarrow n(2 + n - 1) = 420 \Rightarrow n(n + 1) = 420\)
\(\Rightarrow n^2 + n - 420 = 0 \Rightarrow n^2 + 21n - 20n - 420 = 0\)
\(\Rightarrow n(n + 21) - 20(n + 21) = 0\)
\(\Rightarrow (n + 21)(n - 20) = 0 \Rightarrow n = 20, -21\)
\(n\) is a natural number \(\therefore n > 0\)
\(\therefore n = 20\)

Question. If the ratio of the roots of the equation \(lx^2 + nx + n = 0\) is \(p : q\), prove that \(\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} = 0\).
Answer: Roots of the equation \(lx^2 + nx + n = 0\) are \(\frac{-n + \sqrt{n^2 - 4nl}}{2l}\) and \(\frac{-n - \sqrt{n^2 - 4nl}}{2l}\)
Also, \(\frac{p}{q} = \frac{-n + \sqrt{n^2 - 4nl}}{-n - \sqrt{n^2 - 4nl}}\) [Given]
Now, \(\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}} = \frac{\sqrt{p}}{\sqrt{q}} + \frac{\sqrt{q}}{\sqrt{p}} + \sqrt{\frac{n}{l}} = \frac{p + q}{\sqrt{pq}} + \sqrt{\frac{n}{l}}\)
\(= \frac{-n + \sqrt{n^2 - 4nl} - n - \sqrt{n^2 - 4nl}}{\sqrt{(-n + \sqrt{n^2 - 4nl})(-n - \sqrt{n^2 - 4nl})}} + \sqrt{\frac{n}{l}} = \frac{-2n}{\sqrt{n^2 - (n^2 - 4nl)}} + \sqrt{\frac{n}{l}}\)
\(= \frac{-2n}{2\sqrt{nl}} + \sqrt{\frac{n}{l}} = -\sqrt{\frac{n}{l}} + \sqrt{\frac{n}{l}} = 0\)

Question. Is the following situation possible? If so, then determine the present ages of the mother and her daughter in the problem given below: The sum of the ages of a mother and her daughter is 25 years. Five years ago, the product of their ages was 58.
Answer: Let the present age of the daughter be \(x\) years.
Then, the present age of the mother \(= (25 - x)\) years.
Five years ago, the ages of the daughter and the mother was \((x - 5)\) years and \((25 - x - 5)\) years respectively.
According to the given condition, we have \((x - 5)(20 - x) = 58\)
\(\Rightarrow -x^2 + 20x + 5x - 100 = 58 \Rightarrow x^2 - 25x + 158 = 0\)
Here, \(a = 1, b = -25, c = 158\)
\(\therefore D = b^2 - 4ac = (-25)^2 - 4 \times 1 \times 158 = 625 - 632 = -7 < 0\)
Thus, no real value of \(x\) is possible in the above problem. Hence, the given situation is not possible.

Question. Check whether the following equations are quadratic or not
(i) \(x(2x + 3) = x + 2\)
(ii) \(y(2y + 15) = 2(y^2 + y + 8)\).
Answer: (i) Given equation \(x(2x + 3) = x + 2\) can be written as \(2x^2 + 3x = x + 2 \Rightarrow 2x^2 + 3x - x - 2 = 0 \Rightarrow 2x^2 + 2x - 2 = 0\), which is of the standard form \(ax^2 + bx + c = 0, a \neq 0\). \(\therefore\) It is a quadratic equation.
(ii) Given equation, \(y(2y + 15) = 2(y^2 + y + 8)\) can be written as \(2y^2 + 15y = 2y^2 + 2y + 16 \Rightarrow 2y^2 - 2y^2 + 15y - 2y - 16 = 0 \Rightarrow 13y - 16 = 0\), which is not of the standard form \(ax^2 + bx + c = 0, a \neq 0\). \(\therefore\) It is not a quadratic equation.

Question. A polygon of \(n\) sides has \(\frac{n(n-3)}{2}\) diagonals. How many sides a polygon has with 54 diagonals?
Answer: Given, when number of sides is \(n\), then the number of diagonals is \(\frac{n(n-3)}{2}\).
It is given that the number of diagonals \(= 54\)
\(\Rightarrow \frac{n(n-3)}{2} = 54 \Rightarrow n^2 - 3n = 108 \Rightarrow n^2 - 3n - 108 = 0\)
\(\Rightarrow n^2 - 12n + 9n - 108 = 0 \Rightarrow n(n - 12) + 9(n - 12) = 0 \Rightarrow (n - 12)(n + 9) = 0\)
\(\Rightarrow n = 12\) or \(n = -9 \Rightarrow n = 12\) (\(\because n \neq -9\), as number of sides cannot be negative)
\(\therefore\) The number of sides of the polygon is 12.

Question. If we buy 2 tickets from station A to station B and 3 from station A to station C, we have to pay ₹ 795. But 3 tickets from A to B and 5 tickets from A to C cost a total of ₹ 1300. What is the fare from A to B and from A to C?
Answer: Let fare from station A to B be ₹ \(x\) and fare from station A to C be ₹ \(y\).
Then, according to the question, \(2x + 3y = 795\) ...(i) and \(3x + 5y = 1300\) ...(ii)
Multiplying (i) by 3 and (ii) by 2, we get \(6x + 9y = 2385\) ...(iii) and \(6x + 10y = 2600\) ...(iv)
Subtracting (iii) from (iv), we get \(y = 215\).
Putting \(y = 215\) in (i), we get \(2x + 3(215) = 795 \Rightarrow 2x = 150 \Rightarrow x = 75\)
Hence, fare from station A to B is ₹ 75 and fare from station A to C is ₹ 215.

Question. If roots of quadratic equation \(x^2 + 2px + mn = 0\) are real and equal, then show that the roots of the quadratic equation \(x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0\) are also equal.
Answer: \(x^2 + 2px + mn = 0\) ...(i) and \(x^2 - 2(m + n)x + (m^2 + n^2 + 2p^2) = 0\) ...(ii)
Equation (i) has equal roots. \(\therefore D = 0 \Rightarrow (2p)^2 - 4mn = 0 \Rightarrow 4p^2 = 4mn \Rightarrow p^2 = mn\) ...(iii)
Now, for equation (ii), discriminant, \(D = (-2(m + n))^2 - 4(m^2 + n^2 + 2p^2)\)
\(= 4(m^2 + n^2 + 2mn) - 4(m^2 + n^2 + 2mn)\) [Using (iii)]
\(= 0\). Thus, equation (ii) also has equal roots.

Question. The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.
Answer: Let two consecutive multiples of 7 be \(k, k + 7\).
According to question, \((k)^2 + (k + 7)^2 = 637 \Rightarrow k^2 + k^2 + 49 + 14k = 637\)
\(\Rightarrow 2k^2 + 14k - 588 = 0 \Rightarrow k^2 + 7k - 294 = 0 \Rightarrow k^2 + 21k - 14k - 294 = 0\)
\(\Rightarrow k(k + 21) - 14(k + 21) = 0 \Rightarrow (k + 21)(k - 14) = 0 \Rightarrow k = 14\) or \(k = -21\)
Hence, required multiples are 14, 21 or -14, -21.

Question. A farmer wishes to fence a 100 \(m^2\) rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden, letting compound wall of adjoining house act as the fourth side fence. Find the dimensions of his garden.
Answer: Let the length of one side of the garden be \(x\) m.
Then, \(2x +\) length of other side \(= 30\) m \(\Rightarrow\) Length of other side \(= (30 - 2x)\) m
\(\therefore x(30 - 2x) = 100\) [\(\because\) Area \(= 100 m^2\)]
\(\Rightarrow 2x^2 - 30x + 100 = 0 \Rightarrow x^2 - 15x + 50 = 0 \Rightarrow x^2 - 10x - 5x + 50 = 0\)
\(\Rightarrow x(x - 10) - 5(x - 10) = 0 \Rightarrow (x - 10)(x - 5) = 0 \Rightarrow x = 10\) or \(x = 5\)
Length of other side \(= (30 - 20)\) m or \((30 - 10)\) m \(\Rightarrow 10\) m or \(20\) m
Hence, dimensions of the rectangular garden are 20 m and 5 m.

Question. The ratio of the roots of the equation \(ax^2 + bx + c = 0\) is same as the ratio of the roots of the equation \(px^2 + qx + r = 0\). If \(D_1\) and \(D_2\) are the discriminants of \(ax^2 + bx + c = 0\) and \(px^2 + qx + r = 0\) respectively, then \(D_1 : D_2 = \)
Answer: Let \(\alpha_1, \beta_1\) be roots of \(ax^2 + bx + c = 0\) and \(\alpha_2, \beta_2\) be roots of \(px^2 + qx + r = 0\).
Given \(\frac{\alpha_1}{\beta_1} = \frac{\alpha_2}{\beta_2} \Rightarrow \frac{\alpha_1 + \beta_1}{\alpha_1 - \beta_1} = \frac{\alpha_2 + \beta_2}{\alpha_2 - \beta_2}\)
\(\Rightarrow \frac{\alpha_1 + \beta_1}{\sqrt{(\alpha_1 + \beta_1)^2 - 4\alpha_1 \beta_1}} = \frac{\alpha_2 + \beta_2}{\sqrt{(\alpha_2 + \beta_2)^2 - 4\alpha_2 \beta_2}} \Rightarrow \frac{-b/a}{\sqrt{D_1}/a} = \frac{-q/p}{\sqrt{D_2}/p}\)
\(\Rightarrow \frac{b^2}{D_1} = \frac{q^2}{D_2} \Rightarrow \frac{D_1}{D_2} = \frac{b^2}{q^2}\)

Question. If 8 and 2 are roots of \(x^2 + ax + \beta = 0\), whereas 3 and -3 are roots of \(x^2 + ax - b = 0\), then find the roots of \(x^2 + ax + b = 0\).
Answer: Since 8 and 2 are roots of \(x^2 + ax + \beta = 0\), sum of roots \(= 8+2 = -a \Rightarrow a = -10\).
Also, 3 and -3 are roots of \(x^2 + ax - b = 0\), product of roots \(= 3(-3) = -b \Rightarrow b = 9\).
Now, consider \(x^2 + ax + b = 0 \Rightarrow x^2 - 10x + 9 = 0 \Rightarrow x^2 - 9x - x + 9 = 0\)
\(\Rightarrow (x - 9)(x - 1) = 0 \Rightarrow x = 9\) or \(x = 1\)

Question. In a class test, the sum of Kamal’s marks in Maths and English is 40. Had he got 3 marks more in Maths and 4 marks less in English, the product of their marks would have been 360. Find his marks in two subjects.
Answer: Let marks in Maths be \(x\), then marks in English \(= 40 - x\).
By given condition, \((x + 3)(40 - x - 4) = 360 \Rightarrow (x + 3)(36 - x) = 360\)
\(\Rightarrow -x^2 + 33x + 108 = 360 \Rightarrow x^2 - 33x + 252 = 0\)
\(\Rightarrow x(x - 21) - 12(x - 21) = 0 \Rightarrow (x - 21)(x - 12) = 0 \Rightarrow x = 21\) or \(x = 12\).
If \(x = 21\), Maths \(= 21\), English \(= 19\). If \(x = 12\), Maths \(= 12\), English \(= 28\).

Long Answer Type Questions 

Question. A motor boat whose speed is 20 km/h in still water, takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.
Answer: Let speed of stream be \(x\) km/h. Upstream speed \(= (20 - x)\) km/h, Downstream speed \(= (20 + x)\) km/h.
Time taken upstream \(= \frac{48}{20 - x}\), Time taken downstream \(= \frac{48}{20 + x}\).
According to question, \(\frac{48}{20 - x} - \frac{48}{20 + x} = 1 \Rightarrow 48(20 + x - 20 + x) = (20 - x)(20 + x)\)
\(\Rightarrow 96x = 400 - x^2 \Rightarrow x^2 + 96x - 400 = 0 \Rightarrow (x + 100)(x - 4) = 0\)
\(\Rightarrow x = 4\) (\(\because\) Speed cannot be negative). Speed of stream is 4 km/h.

Question. Find the roots of the equation \(\frac{1}{2x - 3} + \frac{1}{x - 5} = 1, x \neq \frac{3}{2}, 5\).
Answer: \(\frac{x - 5 + 2x - 3}{(2x - 3)(x - 5)} = 1 \Rightarrow 3x - 8 = 2x^2 - 13x + 15 \Rightarrow 2x^2 - 16x + 23 = 0\).
Using quadratic formula, \(x = \frac{16 \pm \sqrt{256 - 184}}{4} = \frac{16 \pm \sqrt{72}}{4} = \frac{16 \pm 6\sqrt{2}}{4} = \frac{8 \pm 3\sqrt{2}}{2}\).

Question. Solve for \(x\): \(\frac{1}{(x - 1)(x - 2)} + \frac{1}{(x - 2)(x - 3)} + \frac{1}{(x - 3)(x - 4)} = \frac{1}{6}\).
Answer: \(\frac{1}{x-2} - \frac{1}{x-1} + \frac{1}{x-3} - \frac{1}{x-2} + \frac{1}{x-4} - \frac{1}{x-3} = \frac{1}{6}\)
\(\Rightarrow \frac{1}{x - 4} - \frac{1}{x - 1} = \frac{1}{6} \Rightarrow \frac{x - 1 - (x - 4)}{(x - 4)(x - 1)} = \frac{1}{6} \Rightarrow \frac{3}{x^2 - 5x + 4} = \frac{1}{6}\)
\(\Rightarrow x^2 - 5x + 4 = 18 \Rightarrow x^2 - 5x - 14 = 0 \Rightarrow (x - 7)(x + 2) = 0 \Rightarrow x = 7, -2\).

Question. By increasing the list price of a book by ₹ 10 a person can buy 10 books less for ₹ 1200. Find the original list price of the book.
Answer: Let original price be ₹ \(x\). New price \(= x + 10\).
\(\frac{1200}{x} - \frac{1200}{x + 10} = 10 \Rightarrow 1200\left[\frac{x + 10 - x}{x(x + 10)}\right] = 10 \Rightarrow \frac{12000}{x^2 + 10x} = 10\)
\(\Rightarrow x^2 + 10x - 1200 = 0 \Rightarrow (x + 40)(x - 30) = 0 \Rightarrow x = 30\) (\(\because\) Price cannot be negative). Original price is ₹ 30.

Question. A piece of cloth costs ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, then the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
Answer: Let length be \(x\) m. Rate \(= 200/x\). New length \(= x + 5\), new rate \(= 200/(x + 5)\).
\(\frac{200}{x} - \frac{200}{x + 5} = 2 \Rightarrow 200\left[\frac{x + 5 - x}{x(x + 5)}\right] = 2 \Rightarrow \frac{1000}{x^2 + 5x} = 2 \Rightarrow x^2 + 5x - 500 = 0\)
\(\Rightarrow (x + 25)(x - 20) = 0 \Rightarrow x = 20\) m. Original rate \(= 200/20 = \) ₹ 10 per metre.