Maharashtra Board Class 10 Maths Chapter 2 Quadratic Equations Set 2.6 Solutions

Question 1. Product of Pragati's age 2 years ago and years hence is 84. Find her present age. Solution: Let the present age of Pragati be x years.
\( \therefore \) 2 years ago,
Age of Pragati = \( (x - 2) \) years
After 3 years,
Age of Pragati = \( (x + 3) \) years
According to the given condition,
\( (x - 2) (x + 3) = 84 \)
\( \therefore \) \( x(x + 3) - 2(x + 3) = 84 \)
\( \therefore \) \( x^2 + 3x - 2x - 6 = 84 \)
\( \therefore \) \( x^2 + x - 6 - 84 = 0 \)
\( \therefore \) \( x^2 + x - 90 = 0 \)
\( x^2 + 10x - 9x - 90 = 0 \)
\( \therefore \) \( x(x + 10) - 9(x + 10) = 0 \)
\( \therefore \) \( (x + 10)(x - 9) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x + 10 = 0 \) or \( x - 9 = 0 \)
\( \therefore \) \( x = -10 \) or \( x = 9 \)
But, age cannot be negative.
\( \therefore \) \( x = 9 \)
\( \therefore \) Present age of Pragati is 9 years.
Answer: The present age of Pragati is 9 years.
In simple words: We set up a quadratic equation based on Pragati's age in the past and future, and then solved it to find her current age, rejecting the negative solution.

🎯 Exam Tip: Remember to always consider the practical implications of your mathematical solutions, especially for real-world problems like age, which cannot be negative.

Question 2. The sum of squares of two consecutive even natural numbers is 244; find the numbers. Solution: Let the first even natural number be x.
\( \therefore \) the next consecutive even natural number will be \( (x + 2) \).
According to the given condition,
\( x^2 + (x + 2)^2 = 244 \)
\( \therefore \) \( x^2 + x^2 + 4x + 4 = 244 \)
\( \therefore \) \( 2x^2 + 4x + 4 - 244 = 0 \)
\( \therefore \) \( 2x^2 + 4x - 240 = 0 \)
\( \therefore \) \( x^2 + 2x - 120 = 0 \) ...[Dividing both sides by 2]
\( \therefore \) \( x^2 + 12x - 10x - 120 = 0 \)
\( \therefore \) \( x(x + 12) - 10 (x + 12) = 0 \)
\( \therefore \) \( (x + 12) (x - 10) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x + 12 = 0 \) or \( x - 10 = 0 \)
\( \therefore \) \( x = -12 \) or \( x = 10 \)
But, natural number cannot be negative.
\( \therefore \) \( x = 10 \) and \( x + 2 = 10 + 2 = 12 \)
\( \therefore \) The two consecutive even natural numbers are 10 and 12.
Answer: The two consecutive even natural numbers are 10 and 12.
In simple words: By representing two consecutive even natural numbers with 'x' and 'x+2', we formed and solved a quadratic equation based on the sum of their squares, finding the positive integer solution.

🎯 Exam Tip: When dealing with "natural numbers" or "even natural numbers," remember that solutions must be positive integers, and filter out any negative or non-integer results.

Question 3. In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक फ्लोचार्ट को दर्शाता है जो एक नारंगी के बगीचे में पेड़ों की संख्या ज्ञात करने के चरणों को बताता है। इसमें कॉलम में पेड़ों की संख्या (x), पंक्ति में पेड़ों की संख्या (x+5), कुल पेड़ों की संख्या, द्विघात समीकरण बनाना, x का मान ज्ञात करना और फिर से कॉलम व पंक्ति में पेड़ों की संख्या ज्ञात करने के चरण शामिल हैं।
Solution:
i. Number of trees in a column is x.
ii. Number of trees in a row = x + 5
iii. Total number of trees = \( x \times (x + 5) \)
iv. According to the given condition,
\( x(x + 5) = 150 \)
\( \therefore \) \( x^2 + 5x = 150 \)
\( \therefore \) \( x^2 + 5x - 150 = 0 \)
v. \( x^2 + 15x - 10x - 150 = 0 \)
\( \therefore \) \( x(x + 15) - 10(x + 15) = 0 \)
\( \therefore \) \( (x + 15)(x - 10) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x + 15 = 0 \) or \( x - 10 = 0 \)
\( \therefore \) \( x = -15 \) or \( x = 10 \)
But, number of trees cannot be negative.
\( \therefore \) \( x = 10 \)
vi. Number of trees in a column is 10.
vii. Number of trees in a row = \( x + 5 = 10 + 5 = 15 \)
\( \therefore \) Number of trees in a row is 15.
Answer: The number of trees in each column is 10, and in each row is 15.
In simple words: We defined the number of trees in columns as 'x' and in rows as 'x+5', formulated a quadratic equation from the total tree count, solved for 'x', and then determined the trees in rows and columns, discarding the negative solution.

🎯 Exam Tip: Clearly define your variables (e.g., number of trees in a column) and translate the word problem into a correct mathematical equation. Practical quantities like 'number of trees' must always be positive.

Question 4. Vivek is older than Kishor by 5 years. The Find their present ages is Find their Present ages
\( \frac{1}{x} + \frac{1}{x+5} = \frac{1}{6} \)
Solution:
Let the present age of Kishor be x.
\( \therefore \) Present age of Vivek = \( (x + 5) \) years
According to the given condition,
\[ \frac{1}{x} + \frac{1}{x+5} = \frac{1}{6} \]
\( \therefore \) \( \frac{x+5+x}{x(x+5)} = \frac{1}{6} \)
\( \therefore \) \( \frac{2x+5}{x(x+5)} = \frac{1}{6} \)
\( \therefore \) \( 6(2x + 5) = x(x + 5) \)
\( \therefore \) \( 12x + 30 = x^2 + 5x \)
\( \therefore \) \( x^2 + 5x - 12x - 30 = 0 \)
\( \therefore \) \( x^2 - 7x - 30 = 0 \)
\( \therefore \) \( x^2 - 10x + 3x - 30 = 0 \)
\( \therefore \) \( x(x - 10) + 3(x - 10) = 0 \)
\( \therefore \) \( (x - 10)(x + 3) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x - 10 = 0 \) or \( x + 3 = 0 \)
\( \therefore \) \( x = 10 \) or \( x = - 3 \)
But, age cannot be negative.
\( \therefore \) \( x = 10 \) and \( x + 5 = 10 + 5 = 15 \)
\( \therefore \) Present ages of Kishor and Vivek are 10 years and 15 years respectively.
Answer: The present ages of Kishor and Vivek are 10 years and 15 years respectively.
In simple words: We set up an equation relating the reciprocals of Kishor's and Vivek's ages, solved the resulting quadratic equation, and found their current ages after discarding the invalid negative solution.

🎯 Exam Tip: Pay close attention to how age differences are translated into algebraic expressions. For fractions involving variables, always remember to check for denominators becoming zero, although in age problems, negative age rules often handle this.

Question 5. Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test. Solution: Let the score of Suyash in the first test be x.
\( \therefore \) Score in the second test = \( x + 10 \) According to the given condition,
\( 5(x + 10) = x^2 \)
\( \therefore \) \( 5x + 50 = x^2 \)
\( \therefore \) \( x^2 - 5x - 50 = 0 \)
\( \therefore \) \( x^2 - 10x + 5x - 50 = 0 \)
\( \therefore \) \( x(x - 10) + 5(x - 10) = 0 \)
\( \therefore \) \( (x - 10) (x + 5) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x - 10 = 0 \) or \( x + 5 = 0 \)
\( \therefore \) \( x = 10 \) or \( x = - 5 \)
But, score cannot be negative.
\( \therefore \) \( x = 10 \)
\( \therefore \) The score of Suyash in the first test is 10.
Answer: The score of Suyash in the first test is 10.
In simple words: We used 'x' for the first test score to set up a quadratic equation based on the given conditions and solved for 'x', choosing the positive result as scores cannot be negative.

🎯 Exam Tip: Define variables clearly for each unknown quantity. In real-world scenarios like scores, ensure your final answer is a positive and reasonable value.

Question 6. 'Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is Rs. 600, find production cost of one pot and number of pots he makes per day. Solution: Let Mr. Kasam make x number of pots on daily basis.
Production cost of each pot = Rs. \( (10x + 40) \)
According to the given condition,
\( x(10x + 40) = 600 \)
\( \therefore \) \( 10x^2 + 40x = 600 \)
\( \therefore \) \( 10x^2 + 40x - 600 = 0 \)
\( \therefore \) \( x^2 + 4x - 60 = 0 \) ...[Dividing both sides by 10]
\( \therefore \) \( x^2 + 10x - 6x - 60 = 0 \)
\( \therefore \) \( x(x + 10) - 6(x + 10) = 0 \)
\( \therefore \) \( (x + 10) (x - 6) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x + 10 = 0 \) or \( x - 6 = 0 \)
\( \therefore \) \( x = - 10 \) or \( x = 6 \)
But, number of pots cannot be negative.
\( \therefore \) \( x = 6 \)
\( \therefore \) Production cost of each pot = \( (10 \times 6 + 40) \)
= Rs. \( [(10 \times 6) + 40] \)
= \( (60 + 40) = \) Rs. 100
Production cost of one pot is Rs. 100 and the number of pots Mr. Kasam makes per day is 6.
Answer: The production cost of one pot is Rs. 100, and Mr. Kasam makes 6 pots per day.
In simple words: We formed a quadratic equation by relating the number of pots made daily to their production cost and total daily cost, solving for the number of pots and then calculating the cost per pot, excluding the negative result.

🎯 Exam Tip: Carefully set up expressions for 'cost per item' and 'total cost', ensuring all parts of the word problem are correctly translated into your equation before solving.

Question 7. Pratik takes 8 hours to travel 36 km downstream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current. Solution: Let the speed of water current be x km/hr. Speed of boat is 12 km/hr. \( (x < 12) \)
In upstream, speed of the water current decreases the speed of the boat and it is the opposite in downstream.
\( \therefore \) speed of the boat in upstream = \( (12 - x) \) km/hr and speed of the boat in downstream = \( (12 + x) \) km/hr.
Now, Time \( = \frac{\text{Distance}}{\text{Speed}} \)
Time required to cover 36 km upstream \( = \frac{36}{12-x} \) hrs
Time required to cover 36 km downstream \( = \frac{36}{12+x} \) hrs
According to the given condition,
\[ \frac{36}{12-x} + \frac{36}{12+x} = 8 \]
\( \therefore \) \( 36 \left( \frac{1}{12-x} + \frac{1}{12+x} \right) = 8 \)
\( \therefore \) \( \frac{1}{12-x} + \frac{1}{12+x} = \frac{8}{36} \)
\( \therefore \) \( \frac{(12+x)+(12-x)}{(12-x)(12+x)} = \frac{2}{9} \)
\( \therefore \) \( \frac{24}{144-x^2} = \frac{2}{9} \) ...\( [(a+b)(a - b) = a^2 - b^2] \)
\( \therefore \) \( 24 \times 9 = 2 (144 - x^2) \)
\( \therefore \) \( 216 = 288 - 2x^2 \)
\( \therefore \) \( 2x^2 = 288 - 216 \)
\( \therefore \) \( 2x^2 = 72 \)
\( \therefore \) \( x^2 = 36 \)
\( \therefore \) \( x = \pm 6 \) ... [Taking square root of both sides]
But, speed cannot be negative.
\( x = 6 \)
\( \therefore \) The speed of water current is 6 km/hr.
Answer: The speed of water current is 6 km/hr.
In simple words: By using the formulas for upstream and downstream speeds and the total time taken, we set up and solved a quadratic equation to find the positive speed of the water current.

🎯 Exam Tip: Always define speeds for still water, upstream, and downstream clearly. Remember that time is distance divided by speed, and negative speeds are not physically possible in such contexts.

Question 8. Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone. Solution: Let Nishu take x days to complete the work alone.
\( \therefore \) Total work done by Nishu in 1 day \( = \frac{1}{x} \)
Also, Pintu takes \( (x + 6) \) days to complete the work alone.
\( \therefore \) Total work done by Pintu in 1 day \( = \frac{1}{x+6} \)
\( \therefore \) Total work done by both in 1 day \( = \left( \frac{1}{x} + \frac{1}{x+6} \right) \)
But, both take 4 days to complete the work together.
\( \therefore \) Total work done by both in 1 day \( = \frac{1}{4} \)
According to the given condition,
\[ \frac{1}{x} + \frac{1}{x+6} = \frac{1}{4} \]
\( \therefore \) \( \frac{x+6+x}{x(x+6)} = \frac{1}{4} \)
\( \therefore \) \( \frac{2x+6}{x(x+6)} = \frac{1}{4} \)
\( \therefore \) \( 4(2x + 6) = x(x + 6) \)
\( \therefore \) \( 8x + 24 = x^2 + 6x \)
\( \therefore \) \( x^2 + 6x - 8x - 24 = 0 \)
\( \therefore \) \( x^2 - 2x - 24 = 0 \)
\( \therefore \) \( x^2 - 6x + 4x - 24 = 0 \)
\( \therefore \) \( x(x - 6) + 4(x - 6) = 0 \)
\( \therefore \) \( (x - 6) (x + 4) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x - 6 = 0 \) or \( x + 4 = 0 \)
\( \therefore \) \( x = 6 \) or \( x = -4 \)
But, number of days cannot be negative,
\( \therefore \) \( x = 6 \) and \( x + 6 = 6 + 6 = 12 \)
\( \therefore \) Number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.
Answer: Nishu takes 6 days and Pintu takes 12 days to complete the work alone.
In simple words: We calculated the individual work rates of Nishu and Pintu, formed a quadratic equation based on their combined work rate, and solved it to determine the number of days each would take alone, ignoring the negative result.

🎯 Exam Tip: For "work done" problems, always express the work done per unit of time (e.g., 1 day) as the reciprocal of the total time taken. Remember that time cannot be negative.

Question 9. If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and divisor. Solution: Let the natural number be x.
\( \therefore \) Divisor = x, Quotient = \( 5x + 6 \)
Also, Dividend = 460 and Remainder = 1
Dividend = Divisor \( \times \) Quotient + Remainder
\( \therefore \) \( 460 = x \times (5x+6) + 1 \)
\( \therefore \) \( 460 = 5x^2 + 6x + 1 \)
\( \therefore \) \( 5x^2 + 6x + 1 - 460 = 0 \)
\( \therefore \) \( 5x^2 + 6x - 459 = 0 \)
\( 5x -459 \)
\( = 5x -9 \times 51 \)
\( = -45, +51 \)
\( \therefore \) \( 5x^2 - 45x + 51x - 459 = 0 \)
\( \therefore \) \( 5x(x - 9) + 51(x - 9) = 0 \)
\( \therefore \) \( (x - 9) (5x + 51) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( x - 9 = 0 \) or \( 5x + 51 = 0 \)
\( \therefore \) \( x = 9 \) or \( x = \frac{-51}{5} \)
But, natural number cannot be negative,
\( \therefore \) \( x = 9 \)
\( \therefore \) Quotient = \( 5x + 6 = 5(9) + 6 = 45 + 6 = 51 \)
\( \therefore \) Quotient is 51 and Divisor is 9.
Answer: The divisor is 9 and the quotient is 51.
In simple words: Using the division algorithm (Dividend = Divisor \( \times \) Quotient + Remainder), we formed a quadratic equation with the divisor 'x' and solved it, finding the valid natural number as the divisor and subsequently the quotient.

🎯 Exam Tip: Remember the division algorithm and correctly translate the relationships between divisor, quotient, dividend, and remainder into a quadratic equation. Natural numbers must be positive integers.

Question 10. In the given fig. []ABCD is a trapezium, AB || CD and its area is 33 cm2. From the information given in the figure find the lengths of all sides of the []ABCD. Fill in the empty boxes to get the solution.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समलंब चतुर्भुज ABCD को दर्शाता है जिसमें AB, CD के समानांतर है। इसमें भुजाओं AB, AD, BC और CD को x के व्यंजकों (x, x-4, x-2, 2x+1) के रूप में दर्शाया गया है। इसमें बिंदु M, CD पर इस प्रकार है कि AM, CD पर लंब है और इस प्रकार चतुर्भुज की ऊंचाई को दर्शाता है।
Solution:
\( \therefore \) ABCD is a trapezium. AB || CD
Area of trapezium \( = \frac{1}{2} \times \) (Sum of parallel sides) \( \times \) Height
\( \therefore \) A(ABCD) \( = \frac{1}{2} \times \) (AB + CD) \( \times \) AM
\( \therefore \) \( 33 = \frac{1}{2} (x + 2x + 1)x(x-4) \)
\( \therefore \) \( 66 = (3x + 1)x(x-4) \)
\( \therefore \) \( 66 = 3x^2 - 12x + x - 4 \)
\( \therefore \) \( 3x^2 - 11x - 70 = 0 \)
\( \therefore \) \( 3x^2 - 21x + 10x - 70 = 0 \)
\( \therefore \) \( 3x(x - 7) + 10(x - 7) = 0 \)
\( \therefore \) \( (3x + 10) (x - 7) = 0 \)
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get
\( \therefore \) \( (3x + 10) = 0 \) or \( (x - 7) = 0 \)
\( \therefore \) \( x = \frac{-10}{3} \) or \( x = 7 \)
But, length is never negative.
\( \therefore \) \( x \ne \frac{-10}{3} \)
\( \therefore \) \( x = 7 \)
AB \( = x = 7 \)cm
CD \( = 2x + 1 = 2(7) + 1 = 15 \) cm
AD \( = \) BC \( = x - 2 = 7 - 2 = 5 \) cm
Answer: The lengths of the sides are AB = 7 cm, CD = 15 cm, AD = 5 cm, and BC = 5 cm.
In simple words: We used the formula for the area of a trapezium with the given expressions for sides and height in terms of 'x', formed a quadratic equation, and solved for 'x' to find the lengths of all sides, rejecting the negative value.

🎯 Exam Tip: When working with geometric figures, ensure you correctly identify the corresponding algebraic expressions for sides and height. Always remember that lengths must be positive, and discard any negative solutions for 'x'.