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Maharashtra Board Class 10 Maths Part 1 Practice Set 3.3 Solutions

Welcome to the comprehensive Textbook Exercise Solutions for Maharashtra Board Class 10 Maths Part 1 (Algebra). In this post, we provide complete, step-by-step answers for Chapter 3: Arithmetic Progression, Practice Set 3.3. Designed to help you prepare effectively for your SSC board exams, these solutions break down complex numericals, highlight key formulas, and provide clear calculation steps to ensure you score full marks!

Practice Set 3.3 Solutions

Question 1

First term and common difference of an A.P. are 6 and 3 respectively; find \( S_{27} \).

Given:

First term \( (a) = 6 \)
Common difference \( (d) = 3 \)
\( S_{27} = ? \)

Key Formula: \[ S_{n} = \frac{n}{2}[2a + (n-1)d] \]

Calculation:

Substitute the given values into the formula:
\( S_{27} = \frac{27}{2}[2(6) + (27 - 1)3] \)
\( S_{27} = \frac{27}{2}[12 + (26 \times 3)] \)
\( S_{27} = \frac{27}{2}[12 + 78] \)
\( S_{27} = \frac{27}{2} \times 90 \)
\( S_{27} = 27 \times 45 \)
\( S_{27} = 1215 \)

Final Answer: The value of \( S_{27} \) is 1215.

Question 2

Find the sum of the first 123 even natural numbers.

Solution:

The sequence of even natural numbers is: 2, 4, 6, 8, ...
This sequence forms an Arithmetic Progression (A.P.).

Given:

First term \( (a) = 2 \)
Common difference \( (d) = 4 - 2 = 2 \)
Number of terms \( (n) = 123 \)

Calculation:

\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{123} = \frac{123}{2}[2(2) + (123 - 1)2] \)
\( S_{123} = \frac{123}{2}[4 + 122(2)] \)
\( S_{123} = \frac{123}{2}[4 + 244] \)
\( S_{123} = \frac{123}{2} \times 248 \)
\( S_{123} = 123 \times 124 \)
\( S_{123} = 15252 \)

Final Answer: The sum of the first 123 even natural numbers is 15252.

Question 3

Find the sum of all even numbers between 1 and 350.

Solution:

The even numbers strictly between 1 and 350 are: 2, 4, 6, ..., 348.
This sequence forms an A.P.

Given:

First term \( (a) = 2 \)
Common difference \( (d) = 4 - 2 = 2 \)
Last term \( (t_{n}) = 348 \)

Exam Tip: First, find the total number of terms \( (n) \) using the \( t_{n} \) formula before calculating the sum.

Step 1: Find \( n \)

\( t_{n} = a + (n-1)d \)
\( 348 = 2 + (n-1)2 \)
\( 348 - 2 = (n-1)2 \)
\( 346 = (n-1)2 \)
\( n - 1 = \frac{346}{2} \)
\( n - 1 = 173 \)
\( n = 173 + 1 = 174 \)

Step 2: Calculate Sum \( (S_{n}) \)

\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{174} = \frac{174}{2}[2(2) + (174 - 1)2] \)
\( S_{174} = 87[4 + 173 \times 2] \)
\( S_{174} = 87[4 + 346] \)
\( S_{174} = 87 \times 350 \)
\( S_{174} = 30450 \)

Final Answer: The sum of all even numbers between 1 and 350 is 30450.

Question 4

In an A.P., the 19^th term is 52 and the 38^th term is 128. Find the sum of the first 56 terms.

Given:

\( t_{19} = 52 \)
\( t_{38} = 128 \)

Calculation:

Let \( a \) be the first term and \( d \) be the common difference.

Using the formula \( t_{n} = a + (n-1)d \):
\( t_{19} = a + (19 - 1)d \)
\( 52 = a + 18d \quad \text{... (i)} \)

Similarly, for the 38^th term:
\( t_{38} = a + (38 - 1)d \)
\( 128 = a + 37d \quad \text{... (ii)} \)

Adding equations (i) and (ii):

\( (a + 18d) + (a + 37d) = 52 + 128 \)
\( 2a + 55d = 180 \quad \text{... (iii)} \)

Finding \( S_{56} \):

\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{56} = \frac{56}{2}[2a + (56 - 1)d] \)
\( S_{56} = 28(2a + 55d) \)
Substituting the value from equation (iii):
\( S_{56} = 28 \times 180 \)
\( S_{56} = 5040 \)

Final Answer: The sum of the first 56 terms is 5040.

Question 5

Complete the following activity to find the sum of natural numbers between 1 to 140 which are divisible by 4.

Solution (Activity Completed):

From 1 to 140, natural numbers divisible by 4 are:
\( 4, 8, 12, \dots, 136 \)

How many numbers? \( n = \mathbf{34} \)
\( n = \mathbf{34}, \ a = \mathbf{4}, \ d = \mathbf{4} \)

Step 1: Finding \( n \)
\( t_{n} = a + (n-1)d \)
\( 136 = \mathbf{4} + (n-1) \times \mathbf{4} \)
\( n = \mathbf{34} \)

Step 2: Finding Sum \( S_{n} \)
\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{\mathbf{34}} = \frac{\mathbf{34}}{2}[2 \times 4 + (34 - 1)4] \)
\( S_{34} = 17[8 + 132] = \mathbf{2380} \)

Final Answer: The sum of numbers from 1 to 140 which are divisible by 4 is 2380.

Question 6

Sum of the first 55 terms in an A.P. is 3300. Find its 28^th term.

Given:

\( S_{55} = 3300 \)

Calculation:

\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{55} = \frac{55}{2}[2a + (55 - 1)d] \)
\( 3300 = \frac{55}{2}[2a + 54d] \)
Taking 2 common from the bracket:
\( 3300 = \frac{55}{2} \times 2(a + 27d) \)
\( 3300 = 55(a + 27d) \)
\( a + 27d = \frac{3300}{55} \)
\( a + 27d = 60 \quad \text{... (i)} \)

Finding the 28^th term:

\( t_{n} = a + (n-1)d \)
\( t_{28} = a + (28 - 1)d \)
\( t_{28} = a + 27d \)
From equation (i):
\( t_{28} = 60 \)

Final Answer: The 28^th term of the A.P. is 60.

Question 7

In an A.P., the sum of three consecutive terms is 27 and their product is 504. Find the terms.
(Assume that three consecutive terms in A.P. are \( a - d \), \( a \), and \( a + d \).)

Solution:

Let the three consecutive terms in an A.P. be \( a - d \), \( a \), and \( a + d \).

First Condition (Sum is 27):

\( (a - d) + a + (a + d) = 27 \)
\( 3a = 27 \)
\( a = \frac{27}{3} \)
\( a = 9 \quad \text{... (i)} \)

Second Condition (Product is 504):

\( (a - d) \times a \times (a + d) = 504 \)
\( a(a^{2} - d^{2}) = 504 \)
Substitute \( a = 9 \) from (i):
\( 9(9^{2} - d^{2}) = 504 \)
\( 9(81 - d^{2}) = 504 \)
\( 81 - d^{2} = \frac{504}{9} \)
\( 81 - d^{2} = 56 \)
\( d^{2} = 81 - 56 \)
\( d^{2} = 25 \)
Taking square root on both sides:
\( d = \pm 5 \)

Case 1: When \( a = 9 \) and \( d = 5 \)

\( a - d = 9 - 5 = 4 \)
\( a = 9 \)
\( a + d = 9 + 5 = 14 \)

Case 2: When \( a = 9 \) and \( d = -5 \)

\( a - d = 9 - (-5) = 9 + 5 = 14 \)
\( a = 9 \)
\( a + d = 9 - 5 = 4 \)

Final Answer: The three consecutive terms are 4, 9, 14 or 14, 9, 4.

Question 8

Find four consecutive terms in an A.P. whose sum is 12 and the sum of the 3^rd and 4^th term is 14.
(Assume the four consecutive terms in A.P. are \( a - d \), \( a \), \( a + d \), and \( a + 2d \).)

Solution:

Let the four consecutive terms be \( a - d \), \( a \), \( a + d \), and \( a + 2d \).

First Condition (Sum is 12):

\( (a - d) + a + (a + d) + (a + 2d) = 12 \)
\( 4a + 2d = 12 \)
Dividing by 2:
\( 2a + d = 6 \quad \text{... (i)} \)

Second Condition (Sum of 3^rd and 4^th terms is 14):

\( (a + d) + (a + 2d) = 14 \)
\( 2a + 3d = 14 \quad \text{... (ii)} \)

Subtracting equation (i) from (ii):

\( (2a + 3d) - (2a + d) = 14 - 6 \)
\( 2d = 8 \)
\( d = \frac{8}{2} = 4 \)

Substitute \( d = 4 \) in equation (i):

\( 2a + 4 = 6 \)
\( 2a = 6 - 4 = 2 \)
\( a = \frac{2}{2} = 1 \)

Finding the terms:

\( a - d = 1 - 4 = -3 \)
\( a = 1 \)
\( a + d = 1 + 4 = 5 \)
\( a + 2d = 1 + 2(4) = 1 + 8 = 9 \)

Final Answer: The four consecutive terms are -3, 1, 5, and 9.

Question 9

If the 9^th term of an A.P. is zero, then show that the 29^th term is twice the 19^th term.

To Prove: \( t_{29} = 2 \times t_{19} \)

Proof:

Let \( a \) be the first term and \( d \) be the common difference.

Given: \( t_{9} = 0 \)
\( t_{n} = a + (n-1)d \)
\( t_{9} = a + (9 - 1)d \)
\( 0 = a + 8d \)
\( a = -8d \quad \text{... (i)} \)

Finding the 19^th term:
\( t_{19} = a + (19 - 1)d \)
\( t_{19} = a + 18d \)
Substitute \( a = -8d \) from (i):
\( t_{19} = -8d + 18d \)
\( t_{19} = 10d \quad \text{... (ii)} \)

Finding the 29^th term:
\( t_{29} = a + (29 - 1)d \)
\( t_{29} = a + 28d \)
Substitute \( a = -8d \) from (i):
\( t_{29} = -8d + 28d \)
\( t_{29} = 20d \)

Conclusion:
We can write \( 20d \) as \( 2(10d) \).
From equation (ii), we know \( 10d = t_{19} \).
Therefore, \( t_{29} = 2(t_{19}) \).

Hence Proved: The 29^th term is twice the 19^th term.

Additional Question (Textbook Pg. No. 71)

Find the sum of all odd numbers from 1 to 150.

Solution:

Odd numbers from 1 to 150 are: 1, 3, 5, 7, ..., 149.
Here, the difference between any two consecutive terms is 2, hence it forms an A.P.

Given:

First term \( (a) = 1 \)
Common difference \( (d) = 2 \)
Last term \( (t_{n}) = 149 \)

Step 1: Find the number of terms \( (n) \)

\( t_{n} = a + (n-1)d \)
\( 149 = 1 + (n-1)2 \)
\( 149 - 1 = (n-1)2 \)
\( \frac{148}{2} = n - 1 \)
\( 74 = n - 1 \)
\( n = 75 \)

Step 2: Find the sum \( (S_{75}) \)

Memory Trick: If both the first and last terms are known, you can use the shortcut formula: \( S_{n} = \frac{n}{2}(t_{1} + t_{n}) \).

Method I (Using standard formula):
\( S_{n} = \frac{n}{2}[2a + (n-1)d] \)
\( S_{75} = \frac{75}{2}[2(1) + (75 - 1)2] \)
\( S_{75} = \frac{75}{2}[2 + 148] \)
\( S_{75} = \frac{75}{2}[150] \)
\( S_{75} = 75 \times 75 = 5625 \)

Method II (Using shortcut formula):
\( S_{n} = \frac{n}{2}(t_{1} + t_{n}) \)
\( S_{75} = \frac{75}{2}(1 + 149) \)
\( S_{75} = \frac{75}{2}(150) \)
\( S_{75} = 75 \times 75 = 5625 \)

Final Answer: The sum of all odd numbers from 1 to 150 is 5625.

Class 10 Maths Part 1 Chapter 3 Arithmetic Progression Practice Set 3.2 Solutions (Maharashtra Board)

Welcome to our comprehensive guide for Maharashtra Board Class 10 Maths Part 1. In this post, we provide complete, step-by-step Textbook Exercise Solutions for Chapter 3: Arithmetic Progression, specifically focusing on Practice Set 3.2. These verified solutions will help you master sequence calculations, score maximum marks in your board exams, and easily find the nth term of any sequence.

Question 1: Write the correct number in the given boxes from the following A.P.

Given A.P.: \( 1, 8, 15, 22, \dots \)[cite: 1]

Calculation:
- Here \( a = 1 \), \( t_1 = 1 \), \( t_2 = 8 \), \( t_3 = 15 \)[cite: 2]
- \( t_2 - t_1 = 8 - 1 = 7 \)[cite: 2]
- \( t_3 - t_2 = 15 - 8 = 7 \)[cite: 2]
Final Answer: \( d = 7 \)[cite: 2]
Given A.P.: \( 3, 6, 9, 12, \dots \)[cite: 1]

Calculation:
- Here \( t_1 = 3 \), \( t_2 = 6 \), \( t_3 = 9 \), \( t_4 = 12 \)[cite: 2]
- \( t_2 - t_1 = 6 - 3 = 3 \) (Note: Derived from sequence pattern)
- \( t_3 - t_2 = 9 - 6 = 3 \)[cite: 2]
Final Answer: \( d = 3 \)[cite: 2]
Given A.P.: \( -3, -8, -13, -18, \dots \)[cite: 1]

Calculation:
- Here \( t_1 = -3 \), \( t_2 = -8 \), \( t_3 = -13 \), \( t_4 = -18 \)[cite: 2]
- \( t_2 - t_1 = -8 - (-3) = -5 \)[cite: 2]
- \( t_3 - t_2 = -13 - (-8) = -5 \)[cite: 2]
Final Answer: \( d = -5 \)[cite: 2]
Given A.P.: \( 70, 60, 50, 40, \dots \)[cite: 1]

Calculation:
- Here \( t_1 = 70 \), \( t_2 = 60 \), \( t_3 = 50 \)[cite: 2]
- \( a = 70 \)[cite: 2]
- \( t_2 - t_1 = 60 - 70 = -10 \)
Final Answer: \( d = -10 \)[cite: 2]

Question 2: Decide whether following sequence is an A.P., if so find the 20th term of the progression.

Given Sequence: \( -12, -5, 2, 9, 16, 23, 30, \dots \)[cite: 2]

Calculation (Checking for A.P.):

Here, \( t_1 = -12 \), \( t_2 = -5 \), \( t_3 = 2 \), \( t_4 = 9 \)[cite: 2]
\( t_2 - t_1 = -5 - (-12) = -5 + 12 = 7 \)[cite: 2]
\( t_3 - t_2 = 2 - (-5) = 2 + 5 = 7 \)[cite: 2]
\( t_4 - t_3 = 9 - 2 = 7 \)[cite: 2]
\( t_2 - t_1 = t_3 - t_2 = \dots = 7 = d = \text{constant} \)[cite: 2]

The difference between two consecutive terms is constant. Therefore, the given sequence is an Arithmetic Progression (A.P.).[cite: 3]

Key Formula: \( t_n = a + (n-1)d \)[cite: 3]

Calculation (Finding 20th term):

Substitute \( a = -12 \), \( d = 7 \), and \( n = 20 \)[cite: 3]
\( t_{20} = -12 + (20 - 1)7 \)[cite: 3]
\( t_{20} = -12 + 19 \times 7 \)[cite: 3]
\( t_{20} = -12 + 133 \)[cite: 3]
\( t_{20} = 121 \)[cite: 3]

Final Answer: The 20th term of the given A.P. is 121.[cite: 3]

Question 3: Given Arithmetic Progression is 12, 16, 20, 24, ... Find the 24th term of this progression.

Given: A.P. is \( 12, 16, 20, 24, \dots \)[cite: 3]

First term \( a = 12 \)[cite: 3]
Common difference \( d = 16 - 12 = 4 \)[cite: 3]

Formula: \( t_n = a + (n-1)d \)[cite: 3]

Calculation:

\( t_{24} = 12 + (24 - 1)4 \)[cite: 3]
\( t_{24} = 12 + 23 \times 4 \)[cite: 3]
\( t_{24} = 12 + 92 \)[cite: 3]
\( t_{24} = 104 \)[cite: 3]

Final Answer: The 24th term of the given A.P. is 104.[cite: 3]

Question 4: Find the 19th term of the following A.P. 7, 13, 19, 25, ...

Given: A.P. is \( 7, 13, 19, 25, \dots \)[cite: 3]

First term \( a = 7 \)[cite: 3]
Common difference \( d = 13 - 7 = 6 \)[cite: 3]

Formula: \( t_n = a + (n-1)d \)[cite: 3]

Calculation:

\( t_{19} = 7 + (19 - 1)6 \)[cite: 3]
\( t_{19} = 7 + 18 \times 6 \)[cite: 3]
\( t_{19} = 7 + 108 \)[cite: 3]
\( t_{19} = 115 \)[cite: 4]

Final Answer: The 19th term of the given A.P. is 115.[cite: 4]

Question 5: Find the 27th term of the following A.P. 9, 4, -1, -6, -11, ...

Given: A.P. is \( 9, 4, -1, -6, -11, \dots \)[cite: 4]

First term \( a = 9 \)[cite: 4]
Common difference \( d = 4 - 9 = -5 \)[cite: 4]

Formula: \( t_n = a + (n-1)d \)[cite: 4]

Calculation:

\( t_{27} = 9 + (27 - 1)(-5) \)[cite: 4]
\( t_{27} = 9 + 26 \times (-5) \)[cite: 4]
\( t_{27} = 9 - 130 \)[cite: 4]
\( t_{27} = -121 \)[cite: 4]

Final Answer: The 27th term of the given A.P. is -121.[cite: 4]

Question 6: Find how many three-digit natural numbers are divisible by 5.

Given:

The three-digit natural numbers divisible by 5 are \( 100, 105, 110, \dots, 995 \).[cite: 4]
This sequence is an A.P.[cite: 4]
First term \( a = 100 \)[cite: 4]
Common difference \( d = 105 - 100 = 5 \)[cite: 4]
Last term \( t_n = 995 \)[cite: 4]

Formula: \( t_n = a + (n-1)d \)[cite: 4]

Calculation:

\( 995 = 100 + (n - 1)5 \)[cite: 4]
\( 995 - 100 = (n - 1)5 \)[cite: 4]
\( 895 = (n - 1)5 \)[cite: 4]
\( n - 1 = \frac{895}{5} \)[cite: 4]
\( n - 1 = 179 \)[cite: 5]
\( n = 179 + 1 = 180 \)[cite: 5]

Final Answer: There are 180 three-digit natural numbers which are divisible by 5.[cite: 5]

Exam Tip: Whenever asked for "how many numbers", form the A.P., identify the first and last term, and solve for \( n \) using the \( t_n \) formula.

Question 7: The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find the 41st term of that A.P.

Given:

\( t_{11} = 16 \)[cite: 5]
\( t_{21} = 29 \)[cite: 5]

Formula: \( t_n = a + (n-1)d \)[cite: 5]

Calculation:

For the 11th term: \( t_{11} = a + (11 - 1)d \) → \( 16 = a + 10d \). Let this be Equation (i): \( a + 10d = 16 \).[cite: 5]
For the 21st term: \( t_{21} = a + (21 - 1)d \) → \( 29 = a + 20d \). Let this be Equation (ii): \( a + 20d = 29 \).[cite: 5]
Subtracting equation (i) from (ii):
\( (a + 20d) - (a + 10d) = 29 - 16 \)
\( 10d = 13 \)[cite: 6]
\( d = \frac{13}{10} \)[cite: 6]
Substituting \( d = \frac{13}{10} \) in equation (i):
\( a + 10(\frac{13}{10}) = 16 \)[cite: 6]
\( a + 13 = 16 \)[cite: 6]
\( a = 16 - 13 = 3 \)[cite: 6]
Now, to find the 41st term:
\( t_{41} = 3 + (41 - 1)(\frac{13}{10}) \)[cite: 6]
\( t_{41} = 3 + 40 \times \frac{13}{10} \)[cite: 6]
\( t_{41} = 3 + 52 \)[cite: 6]
\( t_{41} = 55 \)[cite: 6]

Final Answer: The 41st term of the A.P. is 55.[cite: 6]

Question 8: 11, 8, 5, 2, ... In this A.P. which term is number -151?

Given:

The A.P. is \( 11, 8, 5, 2, \dots \)[cite: 6]
\( a = 11 \)[cite: 6]
\( d = 8 - 11 = -3 \)[cite: 6]
Let the nth term be -151, so \( t_n = -151 \)[cite: 6]

Formula: \( t_n = a + (n-1)d \)[cite: 6]

Calculation:

\( -151 = 11 + (n - 1)(-3) \)[cite: 6]
\( -151 - 11 = (n - 1)(-3) \)[cite: 6]
\( -162 = (n - 1)(-3) \)[cite: 6]
\( n - 1 = \frac{-162}{-3} \)[cite: 6, 7]
\( n - 1 = 54 \)[cite: 7]
\( n = 54 + 1 = 55 \)[cite: 7]

Final Answer: The 55th term of the given A.P. is -151.[cite: 7]

Question 9: In the natural numbers from 10 to 250, how many are divisible by 4?

Given:

The natural numbers from 10 to 250 divisible by 4 form the sequence: \( 12, 16, 20, \dots, 248 \).[cite: 7]
This sequence is an A.P.[cite: 7]
\( a = 12 \)[cite: 7]
\( d = 16 - 12 = 4 \)[cite: 7]
Let the number of terms be \( n \), so \( t_n = 248 \).[cite: 7]

Formula: \( t_n = a + (n-1)d \)[cite: 7]

Calculation:

\( 248 = 12 + (n - 1)4 \)[cite: 7]
\( 248 - 12 = (n - 1)4 \)[cite: 7]
\( 236 = (n - 1)4 \)[cite: 7]
\( n - 1 = \frac{236}{4} \)[cite: 7]
\( n - 1 = 59 \)[cite: 7]
\( n = 59 + 1 = 60 \)[cite: 7]

Final Answer: There are 60 natural numbers from 10 to 250 which are divisible by 4.[cite: 7]

Question 10: In an A.P., the 17th term is 7 more than its 10th term. Find the common difference.

Given Condition: \( t_{17} = t_{10} + 7 \)[cite: 8]

Formula: \( t_n = a + (n-1)d \)[cite: 8]

Calculation:

\( a + (17 - 1)d = a + (10 - 1)d + 7 \)[cite: 8]
\( a + 16d = a + 9d + 7 \)[cite: 8]
\( a + 16d - a - 9d = 7 \)[cite: 8]
\( 7d = 7 \)[cite: 8]
\( d = \frac{7}{7} = 1 \)[cite: 8]

Final Answer: The common difference is 1.[cite: 8]

Application Question 1: Kabir's Height

Problem Statement: Kabir's mother keeps a record of his height on each birthday. When he was one year old, his height was 70 cm, at 2 years he was 80 cm tall and 3 years he was 90 cm tall. His aunt Meera was studying in the 10th class. She said, "it seems like Kabir's height grows in Arithmetic Progression". Assuming this, she calculated how tall Kabir will be at the age of 15 years when he is in 10th. You too assume that Kabir grows in A.P. and find out his height at the age of 15 years.[cite: 8]

Given:

Height at 1 year = 70 cm (\( a = 70 \))[cite: 8]
Height at 2 years = 80 cm[cite: 8]
Height at 3 years = 90 cm[cite: 8]
The heights form an A.P. Common difference \( d = 80 - 70 = 10 \).[cite: 8]
We need to find his height at age 15 (\( t_{15} \)).[cite: 8]

Formula: \( t_n = a + (n-1)d \)[cite: 8]

Calculation:

\( t_{15} = 70 + (15 - 1)10 \)[cite: 8]
\( t_{15} = 70 + 14 \times 10 = 70 + 140 \)[cite: 8]
\( t_{15} = 210 \)[cite: 9]

Final Answer: The height of Kabir at the age of 15 years will be 210 cm.[cite: 9]

Application Question 2: Testing Sequence Terms

Problem Statement: Is 5, 8, 11, 14, ... an A.P.? If so then what will be the 100th term? Check whether 92 is in this A.P.? Is number 61 in this A.P.?[cite: 9]

Part i: Is it an A.P.?

Given: \( 5, 8, 11, 14, \dots \)[cite: 9]
\( t_1 = 5 \), \( t_2 = 8 \), \( t_3 = 11 \), \( t_4 = 14 \)[cite: 9]
\( t_2 - t_1 = 8 - 5 = 3 \)[cite: 9]
\( t_3 - t_2 = 11 - 8 = 3 \)[cite: 9]
\( t_4 - t_3 = 14 - 11 = 3 \)[cite: 9]

The difference between two consecutive terms is constant (\( d = 3 \)). Therefore, the given sequence is an A.P.[cite: 9]

Part ii: What will be the 100th term?

\( t_{100} = 5 + (100 - 1)3 \) (using \( a = 5 \), \( d = 3 \))[cite: 9]
\( t_{100} = 5 + 99 \times 3 \)[cite: 9]
\( t_{100} = 5 + 297 \)[cite: 9]
\( t_{100} = 302 \)[cite: 9]

The 100th term is 302.[cite: 9]

Part iii: Check whether 92 is in this A.P.

Let \( t_n = 92 \)[cite: 9]
\( 92 = 5 + (n - 1)3 \)[cite: 9]
\( 92 = 5 + 3n - 3 \)[cite: 9]
\( 92 = 2 + 3n \)[cite: 9]
\( 90 = 3n \)[cite: 9]
\( n = \frac{90}{3} = 30 \)[cite: 9, 10]

Since \( n \) is a natural number, 92 is the 30th term of the given A.P.[cite: 10]

Part iv: Check whether 61 is in this A.P.

Let \( t_n = 61 \)[cite: 10]
\( 61 = 5 + (n - 1)3 \)[cite: 10]
\( 61 = 5 + 3n - 3 \)[cite: 10]
\( 61 = 2 + 3n \)[cite: 10]
\( 61 - 2 = 3n \)[cite: 10]
\( 59 = 3n \)[cite: 10]
\( n = \frac{59}{3} \)[cite: 10]

But, \( n \) must be a natural number, and 59 is not divisible by 3. Therefore, 61 is not in the given A.P.[cite: 10]

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.