CBSE Class 10 Maths HOTs Quadratic Equations Set 15

Question. A 2-digit number is such that product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Answer: Let digit at unit’s place = \( x \) and digit at ten’s place = \( y \)
\( \therefore \) Number = \( 10y + x \)
According to the question, \( xy = 18 \)

\( \implies y = \frac{18}{x} \) ... (i)
and \( 10y + x - 63 = 10x + y \)

\( \implies 9y - 9x - 63 = 0 \)

\( \implies y - x - 7 = 0 \)

\( \implies \frac{18}{x} - x - 7 = 0 \) [Using eq. (i)]

\( \implies 18 - x^2 - 7x = 0 \)

\( \implies x^2 + 7x - 18 = 0 \)

\( \implies (x + 9) (x - 2) = 0 \)

\( \implies x = -9, x = 2 \)
When \( x = 2, y = \frac{18}{2} = 9 \)
\( \therefore \) Number = 92.

Question. If the price of a book is reduced by Rs. 5, a person can buy 5 more books for Rs. 300. Find the original list price of a book.
Answer: Let price of a book be Rs. \( x \); Total cost = Rs. 300
\( \therefore \) Number of books = \( \frac{300}{x} \)
If price of a book = Rs. \( (x - 5) \)
then number of books = \( \frac{300}{x - 5} \)
According to the question,
\( \frac{300}{x - 5} - \frac{300}{x} = 5 \)

\( \implies \frac{300x - 300(x - 5)}{x(x - 5)} = 5 \)

\( \implies 1500 = 5(x^2 - 5x) \)

\( \implies x^2 - 5x - 300 = 0 \)

\( \implies (x - 20)(x + 15) = 0 \)

\( \implies x = 20, -15 \) (rejected)
Hence, original list price of a book is Rs. 20.

Question. If \( \alpha \) is a positive integer and the roots of the equation \( 6x^2 - 11x + \alpha = 0 \) are rational number, find the smallest value of \( \alpha \).
Answer: Given quadratic equation is \( 6x^2 - 11x + \alpha = 0 \)
Here, \( a = 6, b = -11, c = \alpha \)
\( D = b^2 - 4ac = (-11)^2 - 4 \times 6 \times \alpha \)

\( \implies D = 121 - 24\alpha \)
For rational roots, \( D \) has to be a perfect square

\( \implies 121 - 24\alpha = 0, 1, 4, 9, \dots \)
When \( D = 0 \)

\( \implies 121 - 24\alpha = 0 \)

\( \implies \alpha = \frac{121}{24} \)
but \( \alpha \) is a positive integer
\( \therefore \alpha \neq \frac{121}{24} \)
When \( D = 1 \)

\( \implies 121 - 24\alpha = 1 \)

\( \implies 24\alpha = 120 \)

\( \implies \alpha = \frac{120}{24} \)

\( \implies \alpha = 5 \)
\( \therefore \) Smallest value of \( \alpha = 5 \)

Question. The power P in megawatt (MW), produced between mid-night and noon by a nuclear power plant is given by \( P = 3x^2 - 42x + 349 \), where \( x \) is the hours of the day. At what time is the power 250 MW?
Answer: According to the question,
\( 3x^2 - 42x + 349 = 250 \)

\( \implies 3x^2 - 42x + 99 = 0 \)

\( \implies 3(x^2 - 14x + 33) = 0 \)

\( \implies x^2 - 14x + 33 = 0 \)

\( \implies x^2 - 11x - 3x + 33 = 0 \)

\( \implies x(x - 11) - 3(x - 11) = 0 \)

\( \implies (x - 11) (x - 3) = 0 \)

\( \implies x = 11, x = 3 \)
The power is 250 MW at 3:00 am and 11:00 am.

Question. Solve for \( x \): \( x^{2/3} + x^{1/3} - 2 = 0 \)
Answer: Given equation is \( x^{2/3} + x^{1/3} - 2 = 0 \)
Let \( x^{1/3} = y \)

\( \implies y^2 + y - 2 = 0 \)

\( \implies y^2 + 2y - y - 2 = 0 \)

\( \implies y(y + 2) - 1(y + 2) = 0 \)

\( \implies (y + 2)(y - 1) = 0 \)

\( \implies y = -2, y = 1 \)
When \( y = -2 \)

\( \implies x^{1/3} = -2 \)

\( \implies x = (-2)^3 \)

\( \implies x = -8 \)
When \( y = 1 \)

\( \implies x^{1/3} = 1 \)

\( \implies x = (1)^3 = 1 \)
\( \therefore x = -8, x = 1 \)

Question. If \( x = \sqrt{7 + 4\sqrt{3}} \), then find the value of \( x^2 + \frac{1}{x^2} \).
Answer: Here \( x = \sqrt{7 + 4\sqrt{3}} \)

\( \implies x = \sqrt{(\sqrt{3})^2 + (2)^2 + 2 \times 2 \times \sqrt{3}} \)

\( \implies x = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3} \)

\( \implies \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \)

\( \implies \frac{1}{x} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = 2 - \sqrt{3} \)
Now \( x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \)
\( (x + \frac{1}{x})^2 = (4)^2 \)

\( \implies x^2 + \frac{1}{x^2} + 2 = 16 \)

\( \implies x^2 + \frac{1}{x^2} = 14 \)

Question. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream.
Answer: Speed of boat in still water = 15 km/h
Let speed of the stream be \( x \) km/h
Speed of the boat for downstream = \( (15 + x) \) km/h and the speed of the boat for upstream = \( (15 - x) \) km/h
Distance = 30 km
According to the question,
\( \frac{30}{15 - x} + \frac{30}{15 + x} = 4 \frac{30}{60} = \frac{9}{2} \)

\( \implies 30 \left[ \frac{15 + x + 15 - x}{(15 - x)(15 + x)} \right] = \frac{9}{2} \)

\( \implies 30 \left[ \frac{30}{225 - x^2} \right] = \frac{9}{2} \)

\( \implies \frac{900}{225 - x^2} = \frac{9}{2} \)

\( \implies \frac{100}{225 - x^2} = \frac{1}{2} \)

\( \implies 225 - x^2 = 200 \)

\( \implies x^2 = 25 \)

\( \implies x = 5 \) or \( -5 \)
\( x = -5 \) is rejected because speed of stream cannot be negative.
\( \therefore x = 5 \) km/h
Hence, speed of the stream is 5 km/h.

Question. Solve for \( x \): \( \frac{1}{x + 1} + \frac{3}{5x + 1} = \frac{5}{x + 4}, x \neq -1, -\frac{1}{5}, -4 \)
Answer: The given equation is:
\( \frac{1}{x + 1} + \frac{3}{5x + 1} = \frac{5}{x + 4} \)

\( \implies \frac{(5x + 1) + 3(x + 1)}{(x + 1)(5x + 1)} = \frac{5}{x + 4} \)

\( \implies \frac{5x + 1 + 3x + 3}{5x^2 + 6x + 1} = \frac{5}{x + 4} \)

\( \implies \frac{8x + 4}{5x^2 + 6x + 1} = \frac{5}{x + 4} \)

\( \implies (8x + 4)(x + 4) = 5(5x^2 + 6x + 1) \)

\( \implies 8x^2 + 36x + 16 = 25x^2 + 30x + 5 \)

\( \implies 17x^2 - 6x - 11 = 0 \)

\( \implies 17x^2 - 17x + 11x - 11 = 0 \)

\( \implies 17x(x - 1) + 11(x - 1) = 0 \)

\( \implies (17x + 11)(x - 1) = 0 \)

\( \implies x = -\frac{11}{17}, 1 \)

Question. Find two consecutive odd natural numbers, the sum of whose squares is 394.
Answer: Let first odd natural number be \( (2x - 1) \) and other consecutive odd natural number is \( (2x + 1) \)
so,
\( (2x - 1)^2 + (2x + 1)^2 = 394 \)

\( \implies 4x^2 + 1 - 4x + 4x^2 + 1 + 4x = 394 \)

\( \implies 8x^2 + 2 = 394 \)

\( \implies 8x^2 = 392 \)

\( \implies x^2 = 49 \)

\( \implies x = 7 \)
\( \therefore \) First odd natural number = \( (2 \times 7 - 1) = 13 \)
Next odd natural number = \( (2 \times 7 + 1) = 15 \)

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 speed of boat = 18 km/hr in still water. Let speed of stream = \( s \).
Time upstream = \( \frac{24}{18 - s} \); Time downstream = \( \frac{24}{18 + s} \)
According to the question, \( \frac{24}{18 - s} = 1 + \frac{24}{18 + s} \)

\( \implies \frac{24}{18 - s} - \frac{24}{18 + s} = 1 \)

\( \implies 24 \left( \frac{18 + s - (18 - s)}{(18 - s)(18 + s)} \right) = 1 \)

\( \implies 24(2s) = 18^2 - s^2 \)

\( \implies 48s = 324 - s^2 \)

\( \implies s^2 + 48s - 324 = 0 \)

\( \implies s^2 + 54s - 6s - 324 = 0 \)

\( \implies s(s + 54) - 6(s + 54) = 0 \)

\( \implies (s - 6)(s + 54) = 0 \)

\( \implies s = 6 \) or \( s = -54 \).
But speed cannot be negative.
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 original average speed of the train be \( x \) km/h
New average speed be \( (x + 6) \) km/h
Time taken for distance of 63 km = \( \frac{63}{x} \) hours
Time taken for distance of 72 km = \( \frac{72}{x + 6} \) hours
According to the question, \( \frac{63}{x} + \frac{72}{x + 6} = 3 \)

\( \implies 63(x + 6) + 72x = 3x(x + 6) \)

\( \implies 63x + 378 + 72x = 3x^2 + 18x \)

\( \implies 135x + 378 = 3x^2 + 18x \)

\( \implies 3x^2 - 117x - 378 = 0 \)

\( \implies x^2 - 39x - 126 = 0 \)

\( \implies x^2 - 42x + 3x - 126 = 0 \)

\( \implies (x - 42)(x + 3) = 0 \)

\( \implies x = 42 \) or \( x = -3 \) (rejected)
Therefore, original average speed of the train is 42 km/h.

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: Suppose the faster tap takes \( x \) hours to fill the tank.
Therefore, the slower tap will take \( (x + 3) \) hours to fill the tank
Portion filled by faster tap in 1 hour = \( \frac{1}{x} \)
Portion filled in \( \frac{40}{13} \) hours = \( \frac{40}{13x} \)
Similarly, portion filled by slower tap in \( \frac{40}{13} \) hours = \( \frac{40}{13(x + 3)} \)
According to the question, \( \frac{40}{13x} + \frac{40}{13(x + 3)} = 1 \)

\( \implies \frac{1}{x} + \frac{1}{x + 3} = \frac{13}{40} \)

\( \implies \frac{x + 3 + x}{x(x + 3)} = \frac{13}{40} \)

\( \implies 40(2x + 3) = 13x(x + 3) \)

\( \implies 80x + 120 = 13x^2 + 39x \)

\( \implies 13x^2 - 41x - 120 = 0 \)

\( \implies 13x^2 - 65x + 24x - 120 = 0 \)

\( \implies (13x + 24)(x - 5) = 0 \)

\( \implies x = 5 \) or \( x = -\frac{24}{13} \) (rejected)
Hence, the faster tap fills the cistern in 5 hours and the slower tap fills in 8 hours.

Question. Solve the following for \( x \): \( \frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \)
Answer: \( \frac{1}{2a + b + 2x} - \frac{1}{2x} = \frac{1}{2a} + \frac{1}{b} \)

\( \implies \frac{2x - (2a + b + 2x)}{2x(2a + b + 2x)} = \frac{a + 2b}{2ab} \)

\( \implies \frac{-(2a + b)}{2x(2a + b + 2x)} = \frac{2a + b}{2ab} \)

\( \implies \frac{-1}{2ax + bx + 2x^2} = \frac{1}{ab} \)

\( \implies 2x^2 + 2ax + bx + ab = 0 \)

\( \implies 2x(x + a) + b(x + a) = 0 \)

\( \implies (x + a)(2x + b) = 0 \)

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

Question. In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in maths and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.
Answer: Let marks obtained in mathematics be \( x \) and marks obtained in science be \( y \).
According to the question \( x + y = 28 \)

\( \implies y = 28 - x \) ... (i)
Also \( (x + 3)(y - 4) = 180 \)

\( \implies (x + 3)(28 - x - 4) = 180 \) [Using eq. (i)]

\( \implies (x + 3)(24 - x) = 180 \)

\( \implies 24x - x^2 + 72 - 3x = 180 \)

\( \implies x^2 - 21x + 108 = 0 \)

\( \implies (x - 12)(x - 9) = 0 \)

\( \implies x = 12 \) or \( x = 9 \).
If marks in maths = 12, then marks in science = \( 28 - 12 = 16 \).
If marks in maths = 9, then marks in science = \( 28 - 9 = 19 \).

Question. A trader bought a number of articles for Rs. 900, five articles were found damaged. He sold each of the remaining articles at Rs. 2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought.
Answer: Let number of articles be \( x \), and cost of each article be \( y \).
1st condition: \( xy = 900 \implies y = \frac{900}{x} \)
2nd condition: \( (x - 5)(y + 2) = 900 + 80 \)

\( \implies (x - 5)\left(\frac{900}{x} + 2\right) = 980 \)

\( \implies (x - 5)\left(\frac{900 + 2x}{x}\right) = 980 \)

\( \implies 900x + 2x^2 - 4500 - 10x = 980x \)

\( \implies 2x^2 - 90x - 4500 = 0 \)

\( \implies x^2 - 45x - 2250 = 0 \)

\( \implies (x - 75)(x + 30) = 0 \)

\( \implies x = 75 \) or \( -30 \) (rejected).
Number of articles = 75.

Question. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/hr and time of flight increased by 30 minutes. Find the original duration of flight.
Answer: Let original speed of the aircraft be \( x \) km/hr.
According to the question, \( \frac{600}{x - 200} - \frac{600}{x} = \frac{30}{60} = \frac{1}{2} \)

\( \implies 600 \left[ \frac{x - (x - 200)}{x(x - 200)} \right] = \frac{1}{2} \)

\( \implies \frac{600 \times 200}{x^2 - 200x} = \frac{1}{2} \)

\( \implies x^2 - 200x - 240000 = 0 \)

\( \implies (x - 600)(x + 400) = 0 \)

\( \implies x = 600 \) or \( -400 \) (rejected).
Original speed = 600 km/hr.
Original duration of flight = \( \frac{600}{600} = 1 \) hour.

Question. A takes 6 days less than the time taken by B to finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.
Answer: Let number of days taken by B = \( x \). Then A takes \( (x - 6) \).
According to the question, \( \frac{1}{x - 6} + \frac{1}{x} = \frac{1}{4} \)

\( \implies \frac{x + x - 6}{x(x - 6)} = \frac{1}{4} \)

\( \implies 4(2x - 6) = x^2 - 6x \)

\( \implies 8x - 24 = x^2 - 6x \)

\( \implies x^2 - 14x + 24 = 0 \)

\( \implies (x - 12)(x - 2) = 0 \)

\( \implies x = 12 \) or \( 2 \).
If \( x = 2 \), \( x - 6 \) is negative, which is not possible. So \( x = 12 \).
Number of days taken by B = 12.

Question. Find three consecutive positive integers whose product is equal to sixteen times their sum.
Answer: Let integers be \( x, x + 1, x + 2 \).
According to the question, \( x(x + 1)(x + 2) = 16(3x + 3) = 48(x + 1) \).
Since \( x \) is a positive integer, \( x + 1 \neq 0 \).

\( \implies x(x + 2) = 48 \)

\( \implies x^2 + 2x - 48 = 0 \)

\( \implies (x + 8)(x - 6) = 0 \)

\( \implies x = 6 \) or \( -8 \).
Integers are 6, 7 and 8.

Question. Solve for \( x \): \( 3^{x + 1} + 3^{2x + 1} = 270 \)
Answer: \( 3 \cdot 3^x + 3 \cdot (3^x)^2 = 270 \). Let \( 3^x = y \).

\( \implies 3y^2 + 3y - 270 = 0 \)

\( \implies y^2 + y - 90 = 0 \)

\( \implies (y + 10)(y - 9) = 0 \)

\( \implies y = -10 \) (not possible) or \( y = 9 \).
\( 3^x = 9 \implies 3^x = 3^2 \implies x = 2 \).

PRACTICE QUESTIONS

Question. If \( (1 - p) \) is a root of the equation \( x^2 + px + 1 - p = 0 \), then roots are
(a) 0, 1
(b) – 1, 1
(c) 0, – 1
(d) – 1, 2
Answer: (c) 0, – 1

Question. If \( p, q \) and \( r \) are rational numbers and \( p \neq q \neq r \), then roots of the equation \( (p^2 - q^2)x^2 - (q^2 - r^2)x + (r^2 - p^2) = 0 \) are
(a) \( \frac{p}{q}, \frac{r}{p} \)
(b) \( \frac{p^2}{q^2}, \frac{r^2}{q^2} \)
(c) \( 1, \frac{r^2 - p^2}{p^2 - q^2} \)
(d) \( -1, \frac{p^2 - q^2}{p^2 - r^2} \)
Answer: (c) 1, \frac{r^2 - p^2}{p^2 - q^2}

Question. The number of integral values of \( x \) so that \( 2x^2 - 7x + 6 = 1 \) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2

Question. Find the discriminant of the quadratic equation: \( 3\sqrt{3}x^2 + 10x + \sqrt{3} = 0 \).
Answer: \( D = 10^2 - 4(3\sqrt{3})(\sqrt{3}) = 100 - 36 = 64 \).

Question. Solve for \( x \): \( 4x^2 - 2(a^2 + b^2)x + a^2b^2 = 0 \).
Answer: \( x = \frac{a^2}{2}, \frac{b^2}{2} \).

Question. Solve for x : \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \).
Answer: \( 4x^2 - 4a^2x + (a^4 - b^4) = 0 \)
Comparing with \( Ax^2 + Bx + C = 0 \), we have \( A = 4, B = -4a^2, C = (a^4 - b^4) \)
\( D = B^2 - 4AC \)

\( \implies D = (-4a^2)^2 - 4(4)(a^4 - b^4) \)

\( \implies D = 16a^4 - 16a^4 + 16b^4 \)

\( \implies D = 16b^4 \)
Using quadratic formula, \( x = \frac{-B \pm \sqrt{D}}{2A} \)

\( \implies x = \frac{4a^2 \pm \sqrt{16b^4}}{2 \times 4} \)

\( \implies x = \frac{4a^2 \pm 4b^2}{8} \)

\( \implies x = \frac{a^2 + b^2}{2}, \frac{a^2 - b^2}{2} \)

Question. Solve the following quadratic equations by the method of completing square:
(i) \( 3x^2 - 8x - 3 = 0 \)
(ii) \( x^2 + 10x + 25 = 0 \)
Answer:
(i) \( 3x^2 - 8x - 3 = 0 \)
Dividing by 3, \( x^2 - \frac{8}{3}x - 1 = 0 \)

\( \implies x^2 - \frac{8}{3}x = 1 \)
Adding \( (\frac{4}{3})^2 \) to both sides:

\( \implies x^2 - \frac{8}{3}x + (\frac{4}{3})^2 = 1 + \frac{16}{9} \)

\( \implies (x - \frac{4}{3})^2 = \frac{25}{9} \)

\( \implies x - \frac{4}{3} = \pm \frac{5}{3} \)

\( \implies x = \frac{4}{3} \pm \frac{5}{3} \)

\( \implies x = 3, -\frac{1}{3} \)

(ii) \( x^2 + 10x + 25 = 0 \)

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

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

\( \implies x + 5 = 0 \)

\( \implies x = -5, -5 \)

Question. Using quadratic formula solve the following quadratic equations:
(i) \( 2x^2 - 2\sqrt{2}x + 1 = 0 \)
(ii) \( 2x^2 - 11x + 9 = 0 \)
(iii) \( 5x^2 - 9x - 14 = 0 \)
Answer:
(i) \( 2x^2 - 2\sqrt{2}x + 1 = 0 \)
\( D = (-2\sqrt{2})^2 - 4(2)(1) = 8 - 8 = 0 \)

\( \implies x = \frac{2\sqrt{2} \pm 0}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \)
Roots are \( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \).

(ii) \( 2x^2 - 11x + 9 = 0 \)
\( D = (-11)^2 - 4(2)(9) = 121 - 72 = 49 \)

\( \implies x = \frac{11 \pm \sqrt{49}}{2 \times 2} = \frac{11 \pm 7}{4} \)

\( \implies x = \frac{18}{4}, \frac{4}{4} \)

\( \implies x = 4.5, 1 \)

(iii) \( 5x^2 - 9x - 14 = 0 \)
\( D = (-9)^2 - 4(5)(-14) = 81 + 280 = 361 \)

\( \implies x = \frac{9 \pm \sqrt{361}}{2 \times 5} = \frac{9 \pm 19}{10} \)

\( \implies x = \frac{28}{10}, \frac{-10}{10} \)

\( \implies x = 2.8, -1 \)

Question. Solve the following equation for x:
\( \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4} \), \( x \neq -1, -2, -4 \)
Answer:
\( \frac{(x + 2) + 2(x + 1)}{(x + 1)(x + 2)} = \frac{5}{x + 4} \)

\( \implies \frac{x + 2 + 2x + 2}{x^2 + 3x + 2} = \frac{5}{x + 4} \)

\( \implies \frac{3x + 4}{x^2 + 3x + 2} = \frac{5}{x + 4} \)

\( \implies (3x + 4)(x + 4) = 5(x^2 + 3x + 2) \)

\( \implies 3x^2 + 12x + 4x + 16 = 5x^2 + 15x + 10 \)

\( \implies 3x^2 + 16x + 16 = 5x^2 + 15x + 10 \)

\( \implies 2x^2 - x - 6 = 0 \)

\( \implies 2x^2 - 4x + 3x - 6 = 0 \)

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

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

\( \implies x = 2, -\frac{3}{2} \)

Question. Solve for x : \( 2\left(\frac{x - 1}{x + 3}\right) - 7\left(\frac{x + 3}{x - 1}\right) = 5 \); given that \( x \neq -3, 1 \)
Answer: Let \( \frac{x - 1}{x + 3} = y \). Then the equation becomes:
\( 2y - \frac{7}{y} = 5 \)

\( \implies 2y^2 - 7 = 5y \)

\( \implies 2y^2 - 5y - 7 = 0 \)

\( \implies 2y^2 - 7y + 2y - 7 = 0 \)

\( \implies y(2y - 7) + 1(2y - 7) = 0 \)

\( \implies (2y - 7)(y + 1) = 0 \)

\( \implies y = \frac{7}{2} \) or \( y = -1 \)
Case 1: \( \frac{x - 1}{x + 3} = \frac{7}{2} \)

\( \implies 2x - 2 = 7x + 21 \)

\( \implies 5x = -23 \)

\( \implies x = -4.6 \)
Case 2: \( \frac{x - 1}{x + 3} = -1 \)

\( \implies x - 1 = -x - 3 \)

\( \implies 2x = -2 \)

\( \implies x = -1 \)
The solutions are \( x = -1, -4.6 \).

Question. Solve for x : \( \frac{2x}{x - 3} + \frac{1}{2x + 3} + \frac{3x + 9}{(x - 3)(2x + 3)} = 0 \).
Answer: Multiplying the entire equation by \( (x - 3)(2x + 3) \):
\( 2x(2x + 3) + 1(x - 3) + (3x + 9) = 0 \)

\( \implies 4x^2 + 6x + x - 3 + 3x + 9 = 0 \)

\( \implies 4x^2 + 10x + 6 = 0 \)
Dividing by 2:

\( \implies 2x^2 + 5x + 3 = 0 \)

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

\( \implies 2x(x + 1) + 3(x + 1) = 0 \)

\( \implies (x + 1)(2x + 3) = 0 \)

\( \implies x = -1, -\frac{3}{2} \)
As per given constraints, check if denominators become zero. If not specified, \( x = -1, -1.5 \).

Question. A two-digit positive number is six times the sum of its digits and is also equal to 6 less than thrice the product of its digits. Find the number.
Answer: Let the tens digit be \( x \) and unit digit be \( y \).
Number \( = 10x + y \)
According to first condition: \( 10x + y = 6(x + y) \)

\( \implies 10x + y = 6x + 6y \)

\( \implies 4x = 5y \)

\( \implies x = \frac{5}{4}y \)...(1)
According to second condition: \( 10x + y = 3xy - 6 \)
Substituting \( x \):
\( 10(\frac{5}{4}y) + y = 3(\frac{5}{4}y)y - 6 \)

\( \implies \frac{25}{2}y + y = \frac{15}{4}y^2 - 6 \)

\( \implies \frac{27}{2}y = \frac{15y^2 - 24}{4} \)

\( \implies 54y = 15y^2 - 24 \)

\( \implies 15y^2 - 54y - 24 = 0 \)
Dividing by 3:

\( \implies 5y^2 - 18y - 8 = 0 \)

\( \implies 5y^2 - 20y + 2y - 8 = 0 \)

\( \implies 5y(y - 4) + 2(y - 4) = 0 \)

\( \implies y = 4 \) (Digit must be positive integer)
Then \( x = \frac{5}{4}(4) = 5 \).
The number is 54.

Question. The difference of two numbers is 5 and the difference of their reciprocals is \( \frac{1}{10} \). Find the numbers.
Answer: Let numbers be \( x \) and \( x - 5 \).
\( \frac{1}{x-5} - \frac{1}{x} = \frac{1}{10} \)

\( \implies \frac{x - (x - 5)}{x(x - 5)} = \frac{1}{10} \)

\( \implies \frac{5}{x^2 - 5x} = \frac{1}{10} \)

\( \implies x^2 - 5x = 50 \)

\( \implies x^2 - 5x - 50 = 0 \)

\( \implies (x - 10)(x + 5) = 0 \)

\( \implies x = 10, -5 \)
If \( x = 10 \), numbers are 10, 5.
If \( x = -5 \), numbers are -5, -10.

Question. By increasing the list price of a book by Rs. 10 a person can buy 10 less books for Rs. 1,200. Find the original list price of the book.
Answer: Let original price be Rs. \( x \).
Number of books \( = \frac{1200}{x} \)
New price \( = x + 10 \)
Number of books \( = \frac{1200}{x + 10} \)

\( \implies \frac{1200}{x} - \frac{1200}{x + 10} = 10 \)

\( \implies \frac{120}{x} - \frac{120}{x + 10} = 1 \)

\( \implies 120(x + 10) - 120x = x^2 + 10x \)

\( \implies 1200 = x^2 + 10x \)

\( \implies x^2 + 10x - 1200 = 0 \)

\( \implies (x + 40)(x - 30) = 0 \)

\( \implies x = 30 \) (Price cannot be negative)
Original list price is Rs. 30.

Question. The numerator of a fraction is one less than its denominator. If three is added to each of the numerator and denominator, the fraction is increased by \( \frac{3}{28} \). Find the fraction.
Answer: Let denominator be \( x \). Then numerator is \( x - 1 \).
Original fraction \( = \frac{x - 1}{x} \)
New fraction \( = \frac{x - 1 + 3}{x + 3} = \frac{x + 2}{x + 3} \)

\( \implies \frac{x + 2}{x + 3} - \frac{x - 1}{x} = \frac{3}{28} \)

\( \implies \frac{x(x + 2) - (x - 1)(x + 3)}{x(x + 3)} = \frac{3}{28} \)

\( \implies \frac{x^2 + 2x - (x^2 + 2x - 3)}{x^2 + 3x} = \frac{3}{28} \)

\( \implies \frac{3}{x^2 + 3x} = \frac{3}{28} \)

\( \implies x^2 + 3x = 28 \)

\( \implies x^2 + 3x - 28 = 0 \)

\( \implies (x + 7)(x - 4) = 0 \)

\( \implies x = 4, -7 \)
For a positive fraction, \( x = 4 \). Numerator \( = 3 \). Fraction is \( \frac{3}{4} \).

Question. A piece of cloth costs Rs. 200. If the piece were 5 m longer and each metre of cloth costed Rs. 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: Let length be \( x \) metres and rate be Rs. \( y \)/m.
\( xy = 200 \implies y = \frac{200}{x} \)
Also, \( (x + 5)(y - 2) = 200 \)

\( \implies (x + 5)(\frac{200}{x} - 2) = 200 \)

\( \implies 200 - 2x + \frac{1000}{x} - 10 = 200 \)

\( \implies -2x + \frac{1000}{x} - 10 = 0 \)

\( \implies -2x^2 + 1000 - 10x = 0 \)

\( \implies x^2 + 5x - 500 = 0 \)

\( \implies (x + 25)(x - 20) = 0 \)

\( \implies x = 20 \) (Length cannot be negative)
Rate \( y = \frac{200}{20} = Rs. 10 \).
Length = 20 m, Rate = Rs. 10 per metre.

Question. A person goes a distance of 30 km on his bicycle. The number of hours taken by him is one less than his average speed in km/h. Find the time taken by him to complete the journey.
Answer: Let speed be \( x \) km/h. Time taken \( = \frac{30}{x} \) hours.
According to question: \( \frac{30}{x} = x - 1 \)

\( \implies 30 = x^2 - x \)

\( \implies x^2 - x - 30 = 0 \)

\( \implies (x - 6)(x + 5) = 0 \)

\( \implies x = 6 \) (Speed cannot be negative)
Time taken \( = \frac{30}{6} = 5 \) hours.

Question. A party of tourists booked a room in a hotel for Rs. 1200. Three of the members failed to pay, as a result others had to pay Rs. 20 more (each). How many tourists were there in the party?
Answer: Let the number of tourists be \( x \). Original share \( = \frac{1200}{x} \).
Remaining tourists \( = x - 3 \). New share \( = \frac{1200}{x - 3} \).

\( \implies \frac{1200}{x - 3} - \frac{1200}{x} = 20 \)

\( \implies \frac{60}{x - 3} - \frac{60}{x} = 1 \)

\( \implies 60x - 60(x - 3) = x^2 - 3x \)

\( \implies 180 = x^2 - 3x \)

\( \implies x^2 - 3x - 180 = 0 \)

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

\( \implies x = 15 \).
There were 15 tourists.

Question. A man sells a table for Rs. 96 and gains as much per cent as the cost of table. Find the cost price of table.
Answer: Let CP be Rs. \( x \). Gain % \( = x\% \).
SP \( = CP + Gain \)
\( 96 = x + \frac{x}{100} \times x \)

\( \implies 9600 = 100x + x^2 \)

\( \implies x^2 + 100x - 9600 = 0 \)

\( \implies (x + 160)(x - 60) = 0 \)

\( \implies x = 60 \).
Cost price is Rs. 60.

Question. Out of a group of swans, 7/2 times the square root of the total number are playing on the shore of a pond. The two remaining ones are swimming in water. Find the total number of swans.
Answer: Let total number of swans be \( x \).
\( \frac{7}{2}\sqrt{x} + 2 = x \)

\( \implies 7\sqrt{x} = 2x - 4 \)
Squaring both sides:
\( 49x = (2x - 4)^2 \)

\( \implies 49x = 4x^2 - 16x + 16 \)

\( \implies 4x^2 - 65x + 16 = 0 \)

\( \implies 4x^2 - 64x - x + 16 = 0 \)

\( \implies 4x(x - 16) - 1(x - 16) = 0 \)

\( \implies x = 16 \) or \( 0.25 \) (Total must be an integer)
Total number of swans is 16.

Question. A farmer wishes to start a 100 sq m ‘rectangular’ vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the rectangular garden letting his house compound wall act as the fourth side fence. Find the dimensions of his garden.
Answer: Let dimensions be \( l \) (length parallel to wall) and \( b \) (width).
Area \( lb = 100 \implies l = \frac{100}{b} \).
Fencing for three sides \( = l + 2b = 30 \).

\( \implies \frac{100}{b} + 2b = 30 \)

\( \implies 100 + 2b^2 = 30b \)

\( \implies 2b^2 - 30b + 100 = 0 \)

\( \implies b^2 - 15b + 50 = 0 \)

\( \implies (b - 10)(b - 5) = 0 \)

\( \implies b = 10 \) or \( 5 \).
If \( b = 10 \), \( l = 10 \). Dimensions: 10 m by 10 m.
If \( b = 5 \), \( l = 20 \). Dimensions: 20 m by 5 m.

Question. The length of the hypotenuse of a right triangle is one unit more than twice the length of the shortest side and the other side is one unit less than twice the length of the shortest side. Find the lengths of the other two sides.
Answer: Let shortest side be \( x \).
Hypotenuse \( = 2x + 1 \). Other side \( = 2x - 1 \).
By Pythagoras theorem: \( x^2 + (2x - 1)^2 = (2x + 1)^2 \)

\( \implies x^2 + 4x^2 - 4x + 1 = 4x^2 + 4x + 1 \)

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

\( \implies x(x - 8) = 0 \)

\( \implies x = 8 \) (Length cannot be zero).
Other sides: \( 2(8) + 1 = 17 \) and \( 2(8) - 1 = 15 \).
Sides are 8, 15, 17.

Question. The sum of the areas of two squares is 640 m\( ^2 \). If the difference in their perimeters be 64 m find the sides of the two squares.
Answer: Let sides be \( a \) and \( b \).
\( a^2 + b^2 = 640 \)
\( 4a - 4b = 64 \implies a - b = 16 \implies a = b + 16 \)

\( \implies (b + 16)^2 + b^2 = 640 \)

\( \implies b^2 + 32b + 256 + b^2 = 640 \)

\( \implies 2b^2 + 32b - 384 = 0 \)

\( \implies b^2 + 16b - 192 = 0 \)

\( \implies (b + 24)(b - 8) = 0 \)

\( \implies b = 8 \) (Side cannot be negative).
\( a = 8 + 16 = 24 \).
Sides are 24 m and 8 m.

Question. Two water taps together can fill a tank is 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: Let smaller tap take \( x \) hours. Larger tap takes \( x - 9 \) hours.
\( \frac{1}{x} + \frac{1}{x - 9} = \frac{1}{6} \)

\( \implies \frac{x - 9 + x}{x^2 - 9x} = \frac{1}{6} \)

\( \implies 6(2x - 9) = x^2 - 9x \)

\( \implies 12x - 54 = x^2 - 9x \)

\( \implies x^2 - 21x + 54 = 0 \)

\( \implies (x - 18)(x - 3) = 0 \)

\( \implies x = 18 \) (If \( x = 3 \), \( x - 9 \) is negative).
Smaller tap: 18 hours. Larger tap: 9 hours.

Question. A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.
Answer: Let B take \( x \) days. A takes \( x - 10 \) days.
\( \frac{1}{x} + \frac{1}{x - 10} = \frac{1}{12} \)

\( \implies \frac{2x - 10}{x^2 - 10x} = \frac{1}{12} \)

\( \implies 24x - 120 = x^2 - 10x \)

\( \implies x^2 - 34x + 120 = 0 \)

\( \implies (x - 30)(x - 4) = 0 \)

\( \implies x = 30 \) (If \( x = 4 \), \( x - 10 \) is negative).
B takes 30 days.

Question. A man bought a certain number of toys for Rs. 180, he kept one for his own use and sold the rest for one rupee each more than he gave for them, besides getting his own toy for nothing he made a profit of Rs. 10. Find the number of toys.
Answer: Let number of toys be \( x \). CP per toy \( = \frac{180}{x} \).
Toys sold \( = x - 1 \). SP per toy \( = \frac{180}{x} + 1 \).
Total SP \( = (x - 1)(\frac{180}{x} + 1) \).
Profit \( = Total SP - Total CP = 10 \)
\( (x - 1)(\frac{180 + x}{x}) - 180 = 10 \)

\( \implies \frac{180x + x^2 - 180 - x}{x} = 190 \)

\( \implies 180x + x^2 - 180 - x = 190x \)

\( \implies x^2 - 11x - 180 = 0 \)

\( \implies (x - 20)(x + 9) = 0 \)

\( \implies x = 20 \).
Number of toys is 20.

Question. Nine times the side of one square exceeds a perimeter of a second square by one metre and six times the area of the second square exceeds twenty-nine times the area of the first by one square metre. Find the side of each square.
Answer: Let side of first square be \( x \) and second be \( y \).
\( 9x = 4y + 1 \implies y = \frac{9x - 1}{4} \).
\( 6y^2 = 29x^2 + 1 \).

\( \implies 6(\frac{9x - 1}{4})^2 = 29x^2 + 1 \)

\( \implies 6(\frac{81x^2 - 18x + 1}{16}) = 29x^2 + 1 \)

\( \implies 3(81x^2 - 18x + 1) = 8(29x^2 + 1) \)

\( \implies 243x^2 - 54x + 3 = 232x^2 + 8 \)

\( \implies 11x^2 - 54x - 5 = 0 \)

\( \implies 11x^2 - 55x + x - 5 = 0 \)

\( \implies (11x + 1)(x - 5) = 0 \)

\( \implies x = 5 \).
\( y = \frac{9(5) - 1}{4} = 11 \).
Sides are 5 m and 11 m.

Question. One-fourth of a herd of camels was seen in a 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: Let total number of camels be \( x \).
\( \frac{x}{4} + 2\sqrt{x} + 15 = x \)

\( \implies \frac{3x}{4} - 2\sqrt{x} - 15 = 0 \)
Let \( \sqrt{x} = a \). \( 3a^2 - 8a - 60 = 0 \).

\( \implies 3a^2 - 18a + 10a - 60 = 0 \)

\( \implies 3a(a - 6) + 10(a - 6) = 0 \)

\( \implies a = 6 \).
Total camels \( x = a^2 = 36 \).

Question. One pipe can fill a cistern in \( (x + 2) \) hours and the other pipe can fill the same cistern in \( (x + 7) \) hours. If both the pipes, when opened together take 6 hours to fill the empty cistern, find the value of x.
Answer: \( \frac{1}{x + 2} + \frac{1}{x + 7} = \frac{1}{6} \)

\( \implies \frac{x + 7 + x + 2}{(x + 2)(x + 7)} = \frac{1}{6} \)

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

\( \implies 12x + 54 = x^2 + 9x + 14 \)

\( \implies x^2 - 3x - 40 = 0 \)

\( \implies (x - 8)(x + 5) = 0 \)

\( \implies x = 8 \).

Question. One pipe can fill a tank in \( (x - 2) \) hours and the other pipe can empty the full tank in \( (x + 2) \) hours. If the tank is empty and both the pipes are opened together, the tank is filled completely in 24 hours. Find how much time will the second pipe take to empty the tank?
Answer: \( \frac{1}{x - 2} - \frac{1}{x + 2} = \frac{1}{24} \)

\( \implies \frac{x + 2 - (x - 2)}{(x - 2)(x + 2)} = \frac{1}{24} \)

\( \implies \frac{4}{x^2 - 4} = \frac{1}{24} \)

\( \implies x^2 - 4 = 96 \)

\( \implies x^2 = 100 \)

\( \implies x = 10 \).
Time for second pipe \( = x + 2 = 12 \) hours.