CBSE Class 10 Mathematics Quadratic Equations VBQs Set O

Passage Based Questions

During the battle of Mahabharta, Arjun carried some arrows for fighting with Bheeshma. With half of the arrows, he cut down the arrows thrown by Bheeshma on him and with six other arrows he killed the rath driver of Bheeshma. With one arrow each, he knocked down respectively the rath, flag and the bow of Bheeshma. Finally, with one more than four times the square root of total arrows, he laid Bheeshma unconscious on an arrow bed. Based on above information answer the following questions :

Question. Find the total number of arrows Arjun had.
Answer: Sol. (i) Suppose Arjun had \( x \) arrows.
Number of arrows used to cut arrows of Bheeshma = \( \frac{x}{2} \)
Number of arrows used to kill the rath driver = 6
Number of arrows used to knock down rath, flag and bow = \( 1 + 1 + 1 = 3 \)
Number of arrows used to lay Bheeshma unconscious = \( 4\sqrt{x} + 1 \)
According to question we have
\( \frac{x}{2} + 6 + 3 + 4\sqrt{x} + 1 = x \)
\( \Rightarrow \frac{x}{2} + 10 + 4\sqrt{x} = x \)
\( \Rightarrow x + 20 + 8\sqrt{x} = 2x \)
\( \Rightarrow x - 8\sqrt{x} - 20 = 0 \)
Putting \( \sqrt{x} = y \), the above equation becomes
\( y^2 - 8y - 20 = 0 \)
\( \Rightarrow y^2 - 10y + 2y - 20 = 0 \)
\( \Rightarrow (y - 10) (y + 2) = 0 \)
\( \Rightarrow y = 10 \) or \( y = - 2 \)
\( \Rightarrow y = 10 \) [\( \because y \) cannot be negative]
\( \Rightarrow \sqrt{x} = 10 \Rightarrow x = 100 \)
Hence, Arjun had 100 arrows.

Question. Find the number of arrows that Arjun used to lay unconsicious Bheeshma.
Answer: Required no. of arrows = \( 4\sqrt{x} + 1 \)
\( = 4\sqrt{100} + 1 \)
\( = 40 + 1 = 41 \).

In the courtyard at the back of the house, Sunita prepared a small beautiful garden. The courtyard is of square shaped whose each side is 44 m. At the centre, she prepared a square flower bed leaving a gravel path all around it. One day a friend of her visited and praised the garden a lot. She also asked for the cost of laying the flower bed and gravelling the path, then sunita told her that the total cost of laying the flower bed and gravelling the path at Rs 2.75 and Rs 1.50 per square metre, respectively, is Rs 4904. Using the above information find :

Question. The width of the path for gravelling.
Answer: Sol. Let the width of the gravel path be \( x \) metres. Then, each side of the square flower bed is \( (44 - 2x) \) metres.
Area of the square field = \( 44 \times 44 = 1936 \text{ m}^2 \)
Area of the flower bed = \( (44 - 2x)^2 \text{ m}^2 \)
Area of gravel path = Area of square field – Area of flower bed
\( = 1936 - (44 - 2x)^2 = 1936 - (1936 - 176x + 4x^2) = (176x - 4x^2) \text{ m}^2 \)
Cost of laying the flower bed = \( (44 - 2x)^2 \times \frac{275}{100} = \frac{11}{4}(44 - 2x)^2 = 11(22 - x)^2 \)
Cost of gravelling the path = \( (176x - 4x^2) \times \frac{150}{100} = 6(44x - x^2) \)
It is given that the total cost is Rs 4904
\( \therefore 11(22 - x)^2 + 6(44x - x^2) = 4904 \)
\( \Rightarrow 11[484 - 44x + x^2] + 264x - 6x^2 = 4904 \)
\( \Rightarrow 5324 - 484x + 11x^2 + 264x - 6x^2 = 4904 \)
\( \Rightarrow 5x^2 - 220x + 420 = 0 \)
\( \Rightarrow x^2 - 44x + 84 = 0 \)
\( \Rightarrow x^2 - 42x - 2x + 84 = 0 \)
\( \Rightarrow x(x - 42) - 2(x - 42) = 0 \)
\( \Rightarrow (x - 2) (x - 42) = 0 \)
\( \Rightarrow x = 2 \) or \( x = 42 \)
By \( x \neq 42 \), as the side of the square is 44 m.
Therefore, \( x = 2 \). Hence, the width of the gravel path is 2.

Question. The cost of laying the flower bed.
Answer: Cost of laying flower bed = Area of flower bed \( \times \) rate per m\(^2\).
Area of flower bed = \( [44 - 2 \times 2]^2 = (40)^2 = 1600 \text{ m}^2 \).
Hence, cost of laying flower bed = \( 1600 \times 2.75 = Rs 4400 \).

Question. The cost of gravelling the path.
Answer: Cost of gravelling the path = Area of path \( \times \) Cost per m\(^2\)
Area of path = Area of square field – Area of flower bed
\( = [1936 - 1600] \text{ m}^2 = 336 \text{ m}^2 \)
Total cost of gravelling the path = \( 336 \times 1.50 = Rs 504 \).

Priyanka had her birthday next week. Some of her friends plan to arrange a surprise party for her. The budget for food was Rs 2400. But at the end moment 4 of her friends refused to come in the party. Due to this, the cost of food for each friend went up by Rs 50. Based on the above information, answer the following questions :

Question. How many friends planned the party?
Answer: Sol. Let the total number of friends be \( x \).
\( \therefore \) Share of each friend = Rs \( \frac{2400}{x} \)
When 4 friends refused to join, then share of each friend = Rs \( \frac{2400}{x - 4} \)
According to the question, \( \frac{2400}{x - 4} - \frac{2400}{x} = 50 \)
\( \Rightarrow \frac{2400x - 2400(x - 4)}{x(x - 4)} = 50 \)
\( \Rightarrow \frac{9600}{x^2 - 4x} = 50 \)
\( \Rightarrow x^2 - 4x = 192 \)
\( \Rightarrow x^2 - 4x - 192 = 0 \)
\( \Rightarrow x^2 - 16x + 12x - 192 = 0 \)
\( \Rightarrow x(x - 16) + 12(x - 16) = 0 \)
\( \Rightarrow (x - 16)(x + 12) = 0 \)
\( x = 16 \) or \( - 12 \)
[\( x = - 12 \), rejected as number of friends cannot be negative]
Total number of friends who planned for the party were 16.

Question. How many friends actually joined the party?
Answer: Number of friends who actually joined the party were \( 16 - 4 = 12 \).

Question. What was the final contribution each of them did?
Answer: Finaly contribution by each of them is Rs \( \frac{2400}{12} = Rs 200 \).

An industry produces a certain number of toys in a day. On a particular day, the cost of production of each toy was 9 less than twice the number of toys produced on that day. The total cost of production on that day was Rs 143. Based on the given information, answer the following questions :

Question. Find the number of toys produced in the industry on that day.
Answer: Sol. (i) Let the number of toys be \( n \)
According to the question, Cost of each toy = \( 2n - 9 \)
Total cost of toys = \( n(2n - 9) \)
But Total cost = Rs 143
\( \therefore n(2n - 9) = 143 \)
\( \Rightarrow 2n^2 - 9n - 143 = 0 \)
\( \Rightarrow n = \frac{9 \pm \sqrt{81 + 1144}}{4} = \frac{9 \pm \sqrt{1225}}{4} = \frac{9 \pm 35}{4} \)
\( = \frac{44}{4} \) or \( -\frac{26}{4} \)
\( \Rightarrow n = 11 \). Since the number of toys cannot be negative or fraction, 11 toys were produced on that day.

Question. What is the cost of each toy?
Answer: Cost of each toy = Rs \( (2 \times 11 - 9) = Rs 13 \).

In a society, there is a big swimming pool. It has three pipes with uniform flow to fill the swimming pool. If first two pipes operate simultaneously then they fill the pool in same time during which the pool is filled by the third pipe alone. If second pipe is operated along it fills the pool five hour faster than the first pipe and four hours slower than the third pipe. Pool is closed for monthly maintenance. Based on the above information, answer the following questions :

Question. Find the time required by each pipe to fill the pool separately.
Answer: Sol. Let \( x \) be the number of hours required by the second pipe alone to fill the pool. Then, the first pipe takes \( (x + 5) \) hours, while the third pipe takes \( (x - 4) \) hours to fill the pool. So, the parts of the pool filled by the first, second and third pipes in one hour are \( \frac{1}{x+5}, \frac{1}{x}, \frac{1}{x-4} \) respectively.
Let the time taken by the first and second pipes to fill the pool simultaneously by \( t \) hours then, the third pipe also takes the same time to fill the pool.
\( \therefore (\frac{1}{x+5} + \frac{1}{x})t = (\frac{1}{x-4})t \)
\( \Rightarrow \frac{1}{x+5} + \frac{1}{x} = \frac{1}{x-4} \)
\( \Rightarrow \frac{x + x + 5}{x(x + 5)} = \frac{1}{x - 4} \)
\( \Rightarrow (2x + 5) (x - 4) = x^2 + 5x \)
\( \Rightarrow 2x^2 - 8x + 5x - 20 = x^2 + 5x \)
\( \Rightarrow x^2 - 8x - 20 = 0 \)
\( \Rightarrow x^2 - 10x + 2x - 20 = 0 \)
\( \Rightarrow (x - 10) (x + 2) = 0 \)
\( \Rightarrow x = 10 \) or \( x = - 2 \)
But time cannot be negative, So, \( x = 10 \).
Hence, the time required to fill the pool by first pipe is 15 hrs, second pipe is 10 hrs and third pipe is 6 hrs.

Question. If all the three pipes are opened simultaneously then in how much time the pool be filled?
Answer: Sol. Parts of pool filled by three pipes in 1 hour = \( \frac{1}{15} + \frac{1}{10} + \frac{1}{6} = \frac{2+3+5}{30} = \frac{10}{30} = \frac{1}{3} \).
So, the pool will be filled in 3 hours if all three pipes are opened simultaneously.

Self- Assessment

Question. In a flight of 6000 km, an aircraft was slowed down due to bad weather. The average speed for the trip was reduced by 400 km/hr and the time of the flight was increased by 30 minutes. Find the original duration of the flight.
Answer: \( 2 \frac{1}{2} \) hrs.

Question. A train travels a distance of 480 km at uniform speed. If the speed had been 8 km/hr less, it would have taken 3 hours more to cover the same distance. Formulate the quadratic equation in terms of the speed of the train.
Answer: \( x^2 - 8x - 1280 = 0 \).

Question. The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if \( x \) denotes the smaller integer. 
Answer: \( x^2 + x - 306 = 0 \).

Question. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be the product of the numbers of toys produced per day and 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. If \( x \) denotes the number of toys produced that day, form the quadratic equation to find \( x \). 
Answer: \( x^2 - 55x + 750 = 0 \).

Question. The height of a right-triangle is 7 cm less than the base. If the hypotenuse is 13 cm form the quadratic equation to find the base of the triangle. 
Answer: \( x^2 - 7x - 60 = 0 \).

Question. Solve : \( \frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x} \). 
Answer: \( x = 3 \) and \( \frac{4}{3} \).

Question. Solve : \( x + \frac{1}{x} = 3 \), where \( x \neq 0 \). 
Answer: \( x = \frac{3 \pm \sqrt{5}}{2} \).

Question. Determine the roots of the equation \( 2x^2 - 6x + 3 = 0 \). 
Answer: \( x = \frac{3 \pm \sqrt{6}}{2} \).

Question. Find the value of \( k \) for which the equation \( kx(x - 2) + 6 = 0 \) has real and equal roots. 
Answer: \( k = 6 \).

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 the passenger train, find the speeds of the two trains. 
Answer: The speed of the passenger train is 33 km/hr and the speed of the express train is 44 km / hr.

Question. The sum of the reciprocals of Rehman’s ages 3 years ago and 5 years hence is \( \frac{1}{3} \). Find his present age. 
Answer: 7 years.

Question. A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. If it is possible to do so, at what distance from the gates should the pole be erected ? 
Answer: 5 metres and 12 metres.

Question. Is it possible to design a rectangular mango grove whose length is twice its breadth and area is 800 m\(^2\) ? If so, find the length and the breadth. 
Answer: Breadth = 20 m and length = 40 m.

Question. Two water taps together take \( 9 \frac{3}{8} \) hours to fill a tank. If the tap with the larger diameter takes 10 hours lesser than the tap with the smaller diameter, then find the time in which each tap can separately fill the tap. 
Answer: The smaller tap takes 25 hours to fill the tank and the larger one takes 15 hours to do so.

Question. In a class test, the sum of the marks obtained by Shefali in Mathematics and English was 30. Had she secured 2 more marks in Mathematics and 3 less in English then the product of the marks in both the tests would have been 210. Find the marks obtained by her in the two subjects separately.
Answer: The marks obtained in Mathematics are either 13 or 12 and in English are 17 or 18.