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Maharashtra Board Class 10 Maths Part 1 Chapter 2 Quadratic Equations Problem Set 2 Solutions

Welcome to our comprehensive guide for Maharashtra Board Class 10 Maths Part 1. In this post, we provide detailed, step-by-step Textbook Exercise Solutions for Chapter 2 Quadratic Equations, specifically focusing on Problem Set 2. These solutions are designed to help you understand core concepts like finding roots, evaluating the discriminant, and solving real-world word problems. We have formatted this guide with clear steps and exam tips to help you maximize your board exam scores. Let's dive in!

Question 1. Choose the correct answers for the following questions.

i. Which one is the quadratic equation?

  • (A) \( \frac{5}{x} - 3 = x^2 \)
  • (B) \( x(x+5) = 2 \)
  • (C) \( n - 1 = 2n \)
  • (D) \( \frac{1}{x^2}(x+2) = x \)

Answer: (B)
Explanation: Expanding option (B) gives \( x^2 + 5x - 2 = 0 \), where the maximum index of the variable is 2.

ii. Out of the following equations, which one is not a quadratic equation?

  • (A) \( x^2 + 4x = 11 + x^2 \)
  • (B) \( x = 4x \)
  • (C) \( 5x^2 = 90 \)
  • (D) \( 2x - x^2 = x^2 + 5 \)

Answer: (A)
Explanation: In option (A), the \( x^2 \) terms cancel out, leaving \( 4x = 11 \), which is a linear equation.

iii. The roots of \( x^2 + kx + k = 0 \) are real and equal, find k.

  • (A) 0
  • (B) 4
  • (C) 0 or 4
  • (D) 2

Answer: (C)
Explanation: For real and equal roots, \( \Delta = 0 \). Here, \( b^2 - 4ac = k^2 - 4(1)(k) = 0 \Rightarrow k(k - 4) = 0 \Rightarrow k = 0 \) or \( k = 4 \).

iv. For \( \sqrt{2}x^2 - 5x + \sqrt{2} = 0 \), find the value of the discriminant.

  • (A) -5
  • (B) 17
  • (C) \( \sqrt{2} \)
  • (D) \( \sqrt{2} - 5 \)

Answer: (B)
Explanation: \( \Delta = b^2 - 4ac = (-5)^2 - 4(\sqrt{2})(\sqrt{2}) = 25 - 8 = 17 \).

v. Which of the following quadratic equations has roots 3, 5?

  • (A) \( x^2 - 15x + 8 = 0 \)
  • (B) \( x^2 - 8x + 15 = 0 \)
  • (C) \( x^2 + 3x + 5 = 0 \)
  • (D) \( x^2 + 8x - 15 = 0 \)

Answer: (B)
Explanation: Equation is \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \Rightarrow x^2 - (3+5)x + (3)(5) = 0 \Rightarrow x^2 - 8x + 15 = 0 \).

vi. Out of the following equations, find the equation having the sum of its roots -5.

  • (A) \( 3x^2 - 15x + 3 = 0 \)
  • (B) \( x^2 - 5x + 3 = 0 \)
  • (C) \( x^2 + 3x - 5 = 0 \)
  • (D) \( 3x^2 + 15x + 3 = 0 \)

Answer: (D)
Explanation: Sum of roots \( = \frac{-b}{a} \). In option (D), \( \frac{-15}{3} = -5 \).

vii. \( \sqrt{5}m^2 - \sqrt{5}m + \sqrt{5} = 0 \). Which of the following statements is true for this given equation?

  • (A) Real and unequal roots
  • (B) Real and equal roots
  • (C) Roots are not real
  • (D) Three roots

Answer: (C)
Explanation: \( \Delta = (-\sqrt{5})^2 - 4(\sqrt{5})(\sqrt{5}) = 5 - 20 = -15 \). Since \( \Delta < 0 \), roots are not real.

viii. One of the roots of equation \( x^2 + mx - 5 = 0 \) is 2; find m.

  • (A) -2
  • (B) \( -\frac{1}{2} \)
  • (C) \( \frac{1}{2} \)
  • (D) 2

Answer: (C)
Explanation: Substitute \( x = 2 \): \( 2^2 + m(2) - 5 = 0 \Rightarrow 4 + 2m - 5 = 0 \Rightarrow 2m = 1 \Rightarrow m = \frac{1}{2} \).

Question 2. Which of the following equations is quadratic?

i. \( x^2 + 2x + 11 = 0 \)

Solution:
Here, \( x \) is the only variable, and the maximum index of the variable is 2.
Comparing with standard form, \( a = 1, b = 2, c = 11 \) are real numbers and \( a \neq 0 \).
Conclusion: The given equation is a quadratic equation.

ii. \( x^2 - 2x + 5 = x^2 \)

Solution:
Simplifying the equation:
\[ x^2 - x^2 - 2x + 5 = 0 \] \[ -2x + 5 = 0 \]
Here, \( x \) is the only variable, but the maximum index of the variable is 1 (not 2).
Conclusion: The given equation is not a quadratic equation.

iii. \( (x + 2)^2 = 2x^2 \)

Solution:
Simplifying the equation:
\[ x^2 + 4x + 4 = 2x^2 \] \[ 2x^2 - x^2 - 4x - 4 = 0 \] \[ x^2 - 4x - 4 = 0 \]
Here, \( x \) is the only variable, and the maximum index of the variable is 2.
Comparing with standard form, \( a = 1, b = -4, c = -4 \) are real numbers and \( a \neq 0 \).
Conclusion: The given equation is a quadratic equation.

Question 3. Find the value of discriminant for each of the following equations.

Key Formula: The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is denoted by \( \Delta \) (Delta) and is calculated as \( \Delta = b^2 - 4ac \).

i. \( 2y^2 - y + 2 = 0 \)

Solution:
Comparing with \( ay^2 + by + c = 0 \):
\( a = 2, b = -1, c = 2 \)
\[ b^2 - 4ac = (-1)^2 - 4(2)(2) \] \[ b^2 - 4ac = 1 - 16 = -15 \]

ii. \( 5m^2 - m = 0 \)

Solution:
Writing in standard form: \( 5m^2 - m + 0 = 0 \)
Comparing with \( am^2 + bm + c = 0 \):
\( a = 5, b = -1, c = 0 \)
\[ b^2 - 4ac = (-1)^2 - 4(5)(0) \] \[ b^2 - 4ac = 1 - 0 = 1 \]

iii. \( \sqrt{5}x^2 - x - \sqrt{5} = 0 \)

Solution:
Comparing with \( ax^2 + bx + c = 0 \):
\( a = \sqrt{5}, b = -1, c = -\sqrt{5} \)
\[ b^2 - 4ac = (-1)^2 - 4(\sqrt{5})(-\sqrt{5}) \] \[ b^2 - 4ac = 1 - (-20) = 1 + 20 = 21 \]

Question 4. One of the roots of quadratic equation \( 2x^2 + kx - 2 = 0 \) is -2, find k.

Solution:
Given that \( -2 \) is one of the roots of the equation \( 2x^2 + kx - 2 = 0 \).
Substitute \( x = -2 \) into the given equation:
\[ 2(-2)^2 + k(-2) - 2 = 0 \] \[ 2(4) - 2k - 2 = 0 \] \[ 8 - 2k - 2 = 0 \] \[ 6 - 2k = 0 \] \[ 2k = 6 \] \[ k = \frac{6}{2} \] Final Answer: \( k = 3 \)

Question 5. Two roots of quadratic equations are given; frame the equation.

Key Formula: If \( \alpha \) and \( \beta \) are the roots, the required quadratic equation is: \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \).

i. 10 and -10

Solution:
Let \( \alpha = 10 \) and \( \beta = -10 \).
Sum of roots: \( \alpha + \beta = 10 + (-10) = 0 \)
Product of roots: \( \alpha\beta = 10 \times (-10) = -100 \)
Required equation:
\[ x^2 - (0)x + (-100) = 0 \] Final Answer: \( x^2 - 100 = 0 \)

ii. \( 1 - 3\sqrt{5} \) and \( 1 + 3\sqrt{5} \)

Solution:
Let \( \alpha = 1 - 3\sqrt{5} \) and \( \beta = 1 + 3\sqrt{5} \).
Sum of roots: \( \alpha + \beta = 1 - 3\sqrt{5} + 1 + 3\sqrt{5} = 2 \)
Product of roots: \( \alpha\beta = (1 - 3\sqrt{5})(1 + 3\sqrt{5}) \)
\[ = (1)^2 - (3\sqrt{5})^2 = 1 - 45 = -44 \]
Required equation:
\[ x^2 - (2)x + (-44) = 0 \] Final Answer: \( x^2 - 2x - 44 = 0 \)

iii. 0 and 7

Solution:
Let \( \alpha = 0 \) and \( \beta = 7 \).
Sum of roots: \( \alpha + \beta = 0 + 7 = 7 \)
Product of roots: \( \alpha\beta = 0 \times 7 = 0 \)
Required equation:
\[ x^2 - (7)x + 0 = 0 \] Final Answer: \( x^2 - 7x = 0 \)

Question 6. Determine the nature of roots for each of the quadratic equations.

Exam Tip: Examiners look for the conclusion statement linked to the value of \( \Delta \):
If \( \Delta > 0 \), roots are real and unequal.
If \( \Delta = 0 \), roots are real and equal.
If \( \Delta < 0 \), roots are not real.

i. \( 3x^2 - 5x + 7 = 0 \)

Solution:
\( a = 3, b = -5, c = 7 \)
\[ \Delta = b^2 - 4ac = (-5)^2 - 4(3)(7) \] \[ = 25 - 84 = -59 \]
Since \( \Delta < 0 \), the roots of the given quadratic equation are not real.

ii. \( \sqrt{3}x^2 + \sqrt{2}x - 2\sqrt{3} = 0 \)

Solution:
\( a = \sqrt{3}, b = \sqrt{2}, c = -2\sqrt{3} \)
\[ \Delta = b^2 - 4ac = (\sqrt{2})^2 - 4(\sqrt{3})(-2\sqrt{3}) \] \[ = 2 - (-24) = 2 + 24 = 26 \]
Since \( \Delta > 0 \), the roots of the given quadratic equation are real and unequal.

iii. \( m^2 - 2m + 1 = 0 \)

Solution:
\( a = 1, b = -2, c = 1 \)
\[ \Delta = b^2 - 4ac = (-2)^2 - 4(1)(1) \] \[ = 4 - 4 = 0 \]
Since \( \Delta = 0 \), the roots of the given quadratic equation are real and equal.

Question 7. Solve the following quadratic equations.

i. \( \frac{1}{x+5} = \frac{1}{x^2} \)

Solution:
Cross multiplying:
\[ x^2 = x + 5 \] \[ x^2 - x - 5 = 0 \]
Comparing with standard form: \( a = 1, b = -1, c = -5 \)
\[ b^2 - 4ac = (-1)^2 - 4(1)(-5) = 1 + 20 = 21 \]
Using formula method:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] \[ x = \frac{-(-1) \pm \sqrt{21}}{2(1)} \] \[ x = \frac{1 \pm \sqrt{21}}{2} \] Final Answer: The roots are \( \frac{1 + \sqrt{21}}{2} \) and \( \frac{1 - \sqrt{21}}{2} \).

ii. \( x^2 - \frac{3x}{10} - \frac{1}{10} = 0 \)

Solution:
Multiplying both sides by 10:
\[ 10x^2 - 3x - 1 = 0 \]
Splitting the middle term:
\[ 10x^2 - 5x + 2x - 1 = 0 \] \[ 5x(2x - 1) + 1(2x - 1) = 0 \] \[ (2x - 1)(5x + 1) = 0 \]
Either \( 2x - 1 = 0 \) or \( 5x + 1 = 0 \)
\( x = \frac{1}{2} \) or \( x = -\frac{1}{5} \)
Final Answer: The roots are \( \frac{1}{2} \) and \( -\frac{1}{5} \).

iii. \( (2x+3)^2 = 25 \)

Solution:
Taking the square root of both sides:
\[ 2x + 3 = \pm 5 \]
Case 1: \( 2x + 3 = 5 \Rightarrow 2x = 2 \Rightarrow x = 1 \)
Case 2: \( 2x + 3 = -5 \Rightarrow 2x = -8 \Rightarrow x = -4 \)
Final Answer: The roots are 1 and -4.

iv. \( m^2 + 5m + 5 = 0 \)

Solution:
\( a = 1, b = 5, c = 5 \)
\[ b^2 - 4ac = (5)^2 - 4(1)(5) = 25 - 20 = 5 \]
Using formula method:
\[ m = \frac{-5 \pm \sqrt{5}}{2(1)} = \frac{-5 \pm \sqrt{5}}{2} \]
Final Answer: The roots are \( \frac{-5 + \sqrt{5}}{2} \) and \( \frac{-5 - \sqrt{5}}{2} \).

v. \( 5m^2 + 2m + 1 = 0 \)

Solution:
\( a = 5, b = 2, c = 1 \)
\[ b^2 - 4ac = (2)^2 - 4(5)(1) = 4 - 20 = -16 \]
Since \( b^2 - 4ac < 0 \), roots of the given equation are not real.

vi. \( x^2 - 4x - 3 = 0 \)

Solution:
\( a = 1, b = -4, c = -3 \)
\[ b^2 - 4ac = (-4)^2 - 4(1)(-3) = 16 + 12 = 28 \]
Using formula method:
\[ x = \frac{-(-4) \pm \sqrt{28}}{2(1)} \] \[ x = \frac{4 \pm \sqrt{4 \times 7}}{2} \] \[ x = \frac{4 \pm 2\sqrt{7}}{2} = 2 \pm \sqrt{7} \]
Final Answer: The roots are \( 2 + \sqrt{7} \) and \( 2 - \sqrt{7} \).

Question 8. Find m, if \( (m - 12)x^2 + 2(m - 12)x + 2 = 0 \) has real and equal roots.

Solution:
Given equation: \( (m - 12)x^2 + 2(m - 12)x + 2 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \):
\( a = (m - 12) \), \( b = 2(m - 12) \), \( c = 2 \)
\[ \Delta = b^2 - 4ac = [2(m - 12)]^2 - 4(m - 12)(2) \] \[ = 4(m - 12)^2 - 8(m - 12) \] Taking \( 4(m - 12) \) common:
\[ = 4(m - 12)[(m - 12) - 2] \] \[ \Delta = 4(m - 12)(m - 14) \]
Since the roots are real and equal, \( \Delta = 0 \):
\[ 4(m - 12)(m - 14) = 0 \]
\[ m - 12 = 0 \text{ or } m - 14 = 0 \]
\[ m = 12 \text{ or } m = 14 \]
But, if \( m = 12 \), the quadratic coefficient \( (m - 12) \) becomes zero, and it is no longer a quadratic equation. Therefore, \( m \neq 12 \).
Final Answer: \( m = 14 \)

Question 9. The sum of two roots of a quadratic equation is 5 and the sum of their cubes is 35. Find the equation.

Solution:
Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation.
According to the given conditions:
\( \alpha + \beta = 5 \) ... (i)
\( \alpha^3 + \beta^3 = 35 \) ... (ii)

Using the algebraic identity:
\[ (\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha\beta(\alpha + \beta) \]
Substituting the given values:
\[ (5)^3 = 35 + 3\alpha\beta(5) \] \[ 125 = 35 + 15\alpha\beta \] \[ 125 - 35 = 15\alpha\beta \] \[ 90 = 15\alpha\beta \] \[ \alpha\beta = \frac{90}{15} = 6 \]
The required quadratic equation is \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \).
Final Answer: \( x^2 - 5x + 6 = 0 \)

Question 10. Find a quadratic equation such that its roots are the square of the sum of the roots and the square of the difference of the roots of equation \( 2x^2 + 2(p + q)x + p^2 + q^2 = 0 \).

Solution:
Given equation: \( 2x^2 + 2(p + q)x + (p^2 + q^2) = 0 \)
Here, \( a = 2, b = 2(p + q), c = p^2 + q^2 \).
Let \( \alpha \) and \( \beta \) be the roots.
Sum of roots: \( \alpha + \beta = \frac{-b}{a} = \frac{-2(p + q)}{2} = -(p + q) \)
Product of roots: \( \alpha\beta = \frac{c}{a} = \frac{p^2 + q^2}{2} \)
We need to find \( (\alpha - \beta)^2 \):
\[ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \] \[ = [-(p + q)]^2 - 4\left(\frac{p^2 + q^2}{2}\right) \] \[ = (p^2 + 2pq + q^2) - 2(p^2 + q^2) \] \[ = p^2 + 2pq + q^2 - 2p^2 - 2q^2 \] \[ = -p^2 + 2pq - q^2 = -(p^2 - 2pq + q^2) = -(p - q)^2 \]
The new roots are \( R_1 = (\alpha + \beta)^2 \) and \( R_2 = (\alpha - \beta)^2 \).
Let's find their Sum and Product:
Sum of new roots (\( R_1 + R_2 \)):
\[ = [-(p + q)]^2 + [-(p - q)^2] = (p + q)^2 - (p - q)^2 \] \[ = (p^2 + 2pq + q^2) - (p^2 - 2pq + q^2) = 4pq \]
Product of new roots (\( R_1 \times R_2 \)):
\[ = (\alpha + \beta)^2 (\alpha - \beta)^2 = (p + q)^2 \times [-(p - q)^2] \] \[ = - [(p + q)(p - q)]^2 = - (p^2 - q^2)^2 \]
Required Equation: \( x^2 - (\text{Sum})x + (\text{Product}) = 0 \)
Final Answer: \( x^2 - 4pq x - (p^2 - q^2)^2 = 0 \)

Question 11. Mukund possesses ₹50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.

Solution:
Given:
Let the amount Sagar possesses be ₹ \( x \).
Then, the amount Mukund possesses = ₹ \( (x + 50) \).
According to the given condition:
\[ x(x + 50) = 15000 \] Calculation:
\[ x^2 + 50x - 15000 = 0 \] Splitting the middle term (factors of -15000 that add up to 50 are 150 and -100):
\[ x^2 + 150x - 100x - 15000 = 0 \] \[ x(x + 150) - 100(x + 150) = 0 \] \[ (x + 150)(x - 100) = 0 \]
Either \( x + 150 = 0 \Rightarrow x = -150 \) OR \( x - 100 = 0 \Rightarrow x = 100 \).
Since an amount of money cannot be negative, \( x = 100 \).
Mukund's amount = \( x + 50 = 100 + 50 = 150 \).
Final Answer: The amount possessed by Sagar is ₹ 100 and by Mukund is ₹ 150.

Question 12. The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.

Solution:
Given:
Let the greater number be \( x \) and the smaller number be \( y \).
According to the given conditions:
\( x^2 - y^2 = 120 \) ... (i)
\( y^2 = 2x \) ... (ii)

Calculation:
Substituting equation (ii) into equation (i):
\[ x^2 - 2x = 120 \] \[ x^2 - 2x - 120 = 0 \] \[ x^2 - 12x + 10x - 120 = 0 \] \[ x(x - 12) + 10(x - 12) = 0 \] \[ (x - 12)(x + 10) = 0 \]
So, \( x = 12 \) or \( x = -10 \).
If \( x = -10 \), then \( y^2 = 2(-10) = -20 \). Since the square of a real number cannot be negative, \( x = -10 \) is discarded.
Therefore, \( x = 12 \).
Now, find \( y \):
\[ y^2 = 2(12) = 24 \] \[ y = \pm \sqrt{24} = \pm 2\sqrt{6} \]
Final Answer: The greater number is 12 and the smaller number is \( \sqrt{24} \) or \( -\sqrt{24} \).

Question 13. Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.

Solution:
Given:
Let the original number of students be \( x \).
Total oranges = 540.
Number of oranges each student gets initially = \( \frac{540}{x} \).
If there were 30 more students, the total number of students = \( x + 30 \).
Number of oranges each student would get = \( \frac{540}{x + 30} \).

Calculation:
According to the given condition, the new amount is 3 less than the old amount:
\[ \frac{540}{x} - \frac{540}{x + 30} = 3 \]
Dividing both sides by 3:
\[ \frac{180}{x} - \frac{180}{x + 30} = 1 \]
\[ 180 \left( \frac{x + 30 - x}{x(x + 30)} \right) = 1 \]
\[ 180 \left( \frac{30}{x^2 + 30x} \right) = 1 \]
\[ 5400 = x^2 + 30x \]
\[ x^2 + 30x - 5400 = 0 \]
Factoring (factors of -5400 that sum to 30 are 90 and -60):
\[ x^2 + 90x - 60x - 5400 = 0 \] \[ x(x + 90) - 60(x + 90) = 0 \] \[ (x + 90)(x - 60) = 0 \]
\( x = -90 \) or \( x = 60 \).
Since the number of students cannot be negative, \( x = 60 \).
Final Answer: The original number of students is 60.

Question 14. Mr. Dinesh owns a rectangular agricultural farm at village Talvel. The length of the farm is 10 metre more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is \( \frac{1}{3} \) of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and side of the pond.

Solution:
Given:
Let the breadth of the farm be \( x \) m.
Length of the farm = \( (2x + 10) \) m.
Area of rectangular farm = Length × Breadth = \( x(2x + 10) = 2x^2 + 10x \) sq. m.
Side of the square pond = \( \frac{x}{3} \) m.
Area of the pond = \( \left(\frac{x}{3}\right)^2 = \frac{x^2}{9} \) sq. m.

Calculation:
Area of farm = 20 × Area of pond
\[ 2x^2 + 10x = 20 \left( \frac{x^2}{9} \right) \]
Dividing both sides by 2:
\[ x^2 + 5x = \frac{10x^2}{9} \]
Multiplying by 9 to clear the fraction:
\[ 9x^2 + 45x = 10x^2 \] \[ 10x^2 - 9x^2 - 45x = 0 \] \[ x^2 - 45x = 0 \] \[ x(x - 45) = 0 \]
\( x = 0 \) or \( x = 45 \).
Since the breadth of the farm cannot be zero, \( x = 45 \) m.
Breadth = 45 m.
Length = \( 2(45) + 10 = 90 + 10 = 100 \) m.
Side of pond = \( \frac{45}{3} = 15 \) m.
Final Answer: The length of the farm is 100 m, the breadth is 45 m, and the side of the pond is 15 m.

Question 15. A tank fills completely in 2 hours if both the taps are open. If only one of the taps is open at the given time, the smaller tap takes 3 hours more than the larger one to fill the tank. How much time does each tap take to fill the tank completely?

Solution:
Given:
Let the larger tap take \( x \) hours to fill the tank.
Part of tank filled by larger tap in 1 hour = \( \frac{1}{x} \).
The smaller tap takes \( (x + 3) \) hours to fill the tank.
Part of tank filled by smaller tap in 1 hour = \( \frac{1}{x + 3} \).
Both taps together fill the tank in 2 hours. Therefore, part of tank filled by both taps in 1 hour = \( \frac{1}{2} \).

Calculation:
\[ \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2} \] \[ \frac{x + 3 + x}{x(x + 3)} = \frac{1}{2} \] \[ \frac{2x + 3}{x^2 + 3x} = \frac{1}{2} \] Cross multiplying:
\[ 2(2x + 3) = x^2 + 3x \] \[ 4x + 6 = x^2 + 3x \] \[ x^2 + 3x - 4x - 6 = 0 \] \[ x^2 - x - 6 = 0 \] \[ x^2 - 3x + 2x - 6 = 0 \] \[ x(x - 3) + 2(x - 3) = 0 \] \[ (x - 3)(x + 2) = 0 \]
\( x = 3 \) or \( x = -2 \).
Since time cannot be negative, \( x = 3 \).
Larger tap time = \( x = 3 \) hours.
Smaller tap time = \( x + 3 = 3 + 3 = 6 \) hours.
Final Answer: The larger tap takes 3 hours and the smaller tap takes 6 hours to fill the tank completely.

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