Class 10 Maths Part 1 Practice Set 2.6 Solutions | Quadratic Equations Maharashtra Board
Welcome to our comprehensive, step-by-step Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1. In this post, we will tackle Chapter 2: Quadratic Equations (Practice Set 2.6). Word problems involving quadratic equations often appear in the SSC Board Exams for 3 to 4 marks. Mastering these real-life applications is crucial for maximizing your score. Below, you will find detailed calculations, key formulas, and exam tips to help you easily understand and solve every question perfectly!
Question 1
Product of Pragati's age 2 years ago and 3 years hence is 84. Find her present age.
Given:
- Let the present age of Pragati be \( x \) years.
- 2 years ago, Age of Pragati = \( (x - 2) \) years.
- After 3 years, Age of Pragati = \( (x + 3) \) years.
Calculation:
According to the given condition,
\[ (x - 2)(x + 3) = 84 \] \[ x(x + 3) - 2(x + 3) = 84 \] \[ x^2 + 3x - 2x - 6 = 84 \] \[ x^2 + x - 6 - 84 = 0 \] \[ x^2 + x - 90 = 0 \]Now, factorize the quadratic equation by splitting the middle term:
\[ x^2 + 10x - 9x - 90 = 0 \] \[ x(x + 10) - 9(x + 10) = 0 \] \[ (x + 10)(x - 9) = 0 \]By using the property, if the product of two numbers is zero, then at least one of them is zero, we get:
\[ x + 10 = 0 \quad \text{or} \quad x - 9 = 0 \] \[ x = -10 \quad \text{or} \quad x = 9 \]Since age cannot be negative, \( x \neq -10 \).
Therefore, \( x = 9 \).
Final Answer:
The present age of Pragati is 9 years.
Question 2
The sum of squares of two consecutive even natural numbers is 244; find the numbers.
Given:
- Let the first even natural number be \( x \).
- The next consecutive even natural number will be \( (x + 2) \).
Calculation:
According to the given condition,
\[ x^2 + (x + 2)^2 = 244 \] \[ x^2 + x^2 + 4x + 4 = 244 \] \[ 2x^2 + 4x + 4 - 244 = 0 \] \[ 2x^2 + 4x - 240 = 0 \]Dividing both sides by 2 to simplify the equation:
\[ x^2 + 2x - 120 = 0 \]Factorizing the equation:
\[ x^2 + 12x - 10x - 120 = 0 \] \[ x(x + 12) - 10(x + 12) = 0 \] \[ (x + 12)(x - 10) = 0 \] \[ x + 12 = 0 \quad \text{or} \quad x - 10 = 0 \] \[ x = -12 \quad \text{or} \quad x = 10 \]But a natural number cannot be negative. Therefore, \( x \neq -12 \).
So, \( x = 10 \).
The next consecutive even natural number is \( x + 2 = 10 + 2 = 12 \).
Final Answer:
The two consecutive even natural numbers are 10 and 12.
Exam Tip: Whenever a question mentions "consecutive even" or "consecutive odd" numbers, always assume them as \( x \) and \( x+2 \).
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.
Step-by-Step Flow Chart Solution:
- No. of trees in a column: Let it be \( x \).
- No. of trees in a row: \( x + 5 \) (Since it is 5 more than the column).
- Total no. of trees: \( x \times (x + 5) \).
- Frame quadratic equation:
According to the given condition, total trees = 150. \[ x(x + 5) = 150 \] \[ x^2 + 5x = 150 \] \[ x^2 + 5x - 150 = 0 \] - Find \( x \):
\[ x^2 + 15x - 10x - 150 = 0 \]
\[ x(x + 15) - 10(x + 15) = 0 \]
\[ (x + 15)(x - 10) = 0 \]
\[ x = -15 \quad \text{or} \quad x = 10 \]
Since the number of trees cannot be negative, \( x = 10 \). - No. of trees in a column: 10.
- No. of trees in a row: \( x + 5 = 10 + 5 = 15 \).
Final Answer:
The number of trees in a column is 10, and the number of trees in a row is 15.
Question 4
Vivek is older than Kishor by 5 years. The sum of the reciprocals of their ages is \( \frac{1}{6} \). Find their present ages.
Given:
- Let the present age of Kishor be \( x \) years.
- Present age of Vivek = \( (x + 5) \) years.
Calculation:
According to the given condition,
\[ \frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6} \] \[ \frac{(x + 5) + x}{x(x + 5)} = \frac{1}{6} \] \[ \frac{2x + 5}{x^2 + 5x} = \frac{1}{6} \]By cross-multiplying:
\[ 6(2x + 5) = x^2 + 5x \] \[ 12x + 30 = x^2 + 5x \] \[ x^2 + 5x - 12x - 30 = 0 \] \[ x^2 - 7x - 30 = 0 \]Factorizing the equation:
\[ x^2 - 10x + 3x - 30 = 0 \] \[ x(x - 10) + 3(x - 10) = 0 \] \[ (x - 10)(x + 3) = 0 \] \[ x - 10 = 0 \quad \text{or} \quad x + 3 = 0 \] \[ x = 10 \quad \text{or} \quad x = -3 \]But age cannot be negative. Therefore, \( x = 10 \).
Age of Vivek = \( x + 5 = 10 + 5 = 15 \).
Final Answer:
The present ages of Kishor and Vivek are 10 years and 15 years respectively.
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.
Given:
- Let the score of Suyash in the first test be \( x \).
- Score in the second test = \( (x + 10) \).
Calculation:
According to the given condition,
\[ 5(x + 10) = x^2 \] \[ 5x + 50 = x^2 \] \[ x^2 - 5x - 50 = 0 \]Factorizing the equation:
\[ x^2 - 10x + 5x - 50 = 0 \] \[ x(x - 10) + 5(x - 10) = 0 \] \[ (x - 10)(x + 5) = 0 \] \[ x - 10 = 0 \quad \text{or} \quad x + 5 = 0 \] \[ x = 10 \quad \text{or} \quad x = -5 \]Since a test score cannot be negative, \( x = 10 \).
Final Answer:
The score of Suyash in the first test is 10.
Question 6
Mr. Kasam runs a small business of making earthen pots. He makes a certain number of pots on a 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 ₹600, find production cost of one pot and number of pots he makes per day.
Given:
- Let Mr. Kasam make \( x \) number of pots on a daily basis.
- Production cost of each pot = ₹\( (10x + 40) \).
Calculation:
According to the given condition (Total Cost = Number of Pots × Cost per Pot):
\[ x(10x + 40) = 600 \] \[ 10x^2 + 40x = 600 \] \[ 10x^2 + 40x - 600 = 0 \]Dividing both sides by 10:
\[ x^2 + 4x - 60 = 0 \]Factorizing the equation:
\[ x^2 + 10x - 6x - 60 = 0 \] \[ x(x + 10) - 6(x + 10) = 0 \] \[ (x + 10)(x - 6) = 0 \] \[ x = -10 \quad \text{or} \quad x = 6 \]But the number of pots cannot be negative, so \( x = 6 \).
Now, substitute \( x \) to find the production cost of each pot:
\[ \text{Cost} = 10(6) + 40 = 60 + 40 = 100 \]Final Answer:
The number of pots Mr. Kasam makes per day is 6, and the production cost of one pot is ₹100.
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.
Key Formula:
\( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)
Upstream Speed = \( (\text{Speed of boat in still water} - \text{Speed of stream}) \)
Downstream Speed = \( (\text{Speed of boat in still water} + \text{Speed of stream}) \)
Given:
- Let the speed of water current be \( x \) km/hr.
- Speed of boat in still water is 12 km/hr. (\( x < 12 \))
- Speed of the boat in upstream = \( (12 - x) \) km/hr.
- Speed of the boat in downstream = \( (12 + x) \) km/hr.
Calculation:
- 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, total time taken is 8 hours:
\[ \frac{36}{12 - x} + \frac{36}{12 + x} = 8 \] \[ 36 \left( \frac{1}{12 - x} + \frac{1}{12 + x} \right) = 8 \] \[ \frac{1}{12 - x} + \frac{1}{12 + x} = \frac{8}{36} = \frac{2}{9} \]Taking LCM on the left side:
\[ \frac{(12 + x) + (12 - x)}{(12 - x)(12 + x)} = \frac{2}{9} \] \[ \frac{24}{144 - x^2} = \frac{2}{9} \]By cross-multiplying:
\[ 24 \times 9 = 2(144 - x^2) \] \[ 216 = 288 - 2x^2 \] \[ 2x^2 = 288 - 216 \] \[ 2x^2 = 72 \] \[ x^2 = 36 \] \[ x = \pm 6 \]Since speed cannot be negative, \( x = 6 \).
Final Answer:
The speed of the water current is 6 km/hr.
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?
Given:
- Let Nishu take \( x \) days to complete the work alone.
- Work done by Nishu in 1 day = \( \frac{1}{x} \).
- Pintu takes \( (x + 6) \) days to complete the work alone.
- Work done by Pintu in 1 day = \( \frac{1}{x + 6} \).
- Together they finish the work in 4 days. Therefore, work done by both in 1 day = \( \frac{1}{4} \).
Calculation:
According to the given condition,
\[ \frac{1}{x} + \frac{1}{x + 6} = \frac{1}{4} \] \[ \frac{(x + 6) + x}{x(x + 6)} = \frac{1}{4} \] \[ \frac{2x + 6}{x^2 + 6x} = \frac{1}{4} \]By cross-multiplying:
\[ 4(2x + 6) = x^2 + 6x \] \[ 8x + 24 = x^2 + 6x \] \[ x^2 + 6x - 8x - 24 = 0 \] \[ x^2 - 2x - 24 = 0 \]Factorizing the equation:
\[ x^2 - 6x + 4x - 24 = 0 \] \[ x(x - 6) + 4(x - 6) = 0 \] \[ (x - 6)(x + 4) = 0 \] \[ x = 6 \quad \text{or} \quad x = -4 \]Since the number of days cannot be negative, \( x = 6 \).
Number of days taken by Pintu = \( x + 6 = 6 + 6 = 12 \).
Final Answer:
The number of days taken by Nishu and Pintu to complete the work alone is 6 days and 12 days respectively.
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.
Key Formula:
\( \text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder} \)
Given:
- Let the divisor (natural number) be \( x \).
- Quotient = \( 5x + 6 \).
- Dividend = 460.
- Remainder = 1.
Calculation:
Using the division algorithm formula:
\[ 460 = x(5x + 6) + 1 \] \[ 460 = 5x^2 + 6x + 1 \] \[ 5x^2 + 6x + 1 - 460 = 0 \] \[ 5x^2 + 6x - 459 = 0 \]Factorizing the equation (Product needed = \( 5 \times -459 = -2295 \). Split 6 into +51 and -45):
\[ 5x^2 - 45x + 51x - 459 = 0 \] \[ 5x(x - 9) + 51(x - 9) = 0 \] \[ (x - 9)(5x + 51) = 0 \] \[ x = 9 \quad \text{or} \quad x = \frac{-51}{5} \]Since a natural number cannot be negative or a fraction, \( x = 9 \).
Now, calculate the Quotient:
\[ \text{Quotient} = 5x + 6 = 5(9) + 6 = 45 + 6 = 51 \]Final Answer:
The Divisor is 9 and the Quotient is 51.
Question 10
In the given figure, 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 trapezium ABCD. Fill in the empty boxes to get the solution.
Given:
- \( \Box ABCD \) is a trapezium. AB || CD.
- Area of trapezium = 33 cm².
- From figure: \( AB = x \), \( CD = 2x + 1 \), \( AD = BC = x - 2 \), and height \( AM = x - 4 \).
Calculation (Fill in the boxes):
Area of trapezium = \( \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height} \)
\[ \text{A}(\Box ABCD) = \frac{1}{2} \times (AB + CD) \times \boxed{AM} \] \[ 33 = \frac{1}{2} (x + 2x + 1) \times \boxed{(x - 4)} \] \[ 66 = (3x + 1) \times \boxed{(x - 4)} \] \[ 66 = 3x^2 - 12x + x - 4 \] \[ 3x^2 - \boxed{11x} - \boxed{70} = 0 \]Factorizing:
\[ 3x^2 - 21x + 10x - 70 = 0 \] \[ 3x(x - 7) + 10(x - 7) = 0 \] \[ (3x + 10)(x - 7) = 0 \]By using the property of zero products:
\[ (3x + 10) = 0 \quad \text{or} \quad \boxed{(x - 7)} = 0 \] \[ x = \frac{-10}{3} \quad \text{or} \quad x = \boxed{7} \]But length is never negative, so \( x \neq \frac{-10}{3} \).
\[ x = \boxed{7} \]Finding the sides:
- \( AB = x = 7 \text{ cm} \)
- \( CD = \boxed{2x + 1} = 2(7) + 1 = 15 \text{ cm} \)
- \( AD = BC = \boxed{x - 2} = 7 - 2 = 5 \text{ cm} \)
Final Answer:
The lengths of the sides of the trapezium are AB = 7 cm, CD = 15 cm, and AD = BC = 5 cm.