Maharashtra Board Class 6 Maths Chapter 10 Equations Set 27 Solutions

Equations Class 6 Maths Chapter 10 Practice Set 27 Solutions Maharashtra Board

Std 6 Maths Practice Set 27 Solutions Answers

 

Question 1. Rewrite the following using a letter:
(i) The sum of a certain number and 3.
(ii) The difference is obtained by subtracting 11 from another number.
(iii) The product of 15 and another number.
(iv) Four times a number is 24.
Answer:
(i) Let the number be \(x\).
\( \therefore x + 3 \) represents the sum of a certain number \(x\) and 3.
(ii) Let the number be \(x\).
\( \therefore x - 11 \) represents the number obtained by subtracting 11 from another number \(x\).
(iii) Let the number be \(x\).
\( \therefore 15x \) represents the product of 15 and another number \(x\).
(iv) Let the number be \(x\).
\( \therefore 4x = 24 \) represents four times the number \(x\).
In simple words: Algebraic expressions use letters (variables) to represent unknown numbers, making it easier to write mathematical statements concisely.

🎯 Exam Tip: Always state what your chosen letter (variable) represents at the beginning of word problems for clarity and correctness.

 

Question 2. Find out which operation must be done on both sides of these equations in order to solve them:
1. \(x + 9 = 11\)
2. \(x - 4 = 9\)
3. \(8x = 24\)
4. \( \frac{x}{6} = 3 \)
Answer:
1. Subtract 9 from both sides.
2. Add 4 to both sides.
3. Divide both sides by 8.
4. Multiply both sides by 6.
In simple words: To solve an equation, you perform the inverse operation on both sides to isolate the variable. For example, to undo addition, you subtract.

🎯 Exam Tip: Remember the principle of balance: whatever operation you perform on one side of an equation, you must perform the exact same operation on the other side to maintain equality.

 

Question 3. Given below are some equations and the values of the variables. Are these values the solutions to those equations?

No.	Equation	Value of the Variable	Solution (Yes/No)
i.	\(y - 3 = 11\)	\(y = 3\)	No
ii.	\(17 = n + 7\)	\(n = 10\)	Yes
iii.	\(30 = 5x\)	\(x = 6\)	Yes
iv.	\( \frac{m}{2} = 14 \)	\(m = 7\)	No

Answer:
(i) \(y - 3 = 11\)
\( \therefore y - 3 + 3 = 11 + 3 \)
(Adding 3 to both sides)
\( \therefore y + 0 = 14 \)
\( \therefore y = 14 \)
Since the value \(y = 3\) does not satisfy the equation (\(3 - 3 \ne 11\)), it is not a solution.

(ii) \(17 = n + 7\)
\( \therefore 17 - 7 = n + 7 - 7 \)
(Subtracting 7 from both sides)
\( \therefore 17 + (-7) = n + 7 - 7 \)
\( \therefore 10 = n \)
\( \therefore n = 10 \)
Since the value \(n = 10\) satisfies the equation (\(17 = 10 + 7\)), it is a solution.

(iii) \(30 = 5x\)
\( \therefore \frac{30}{5} = \frac{5x}{5} \)
(Dividing both sides by 5)
\( \therefore 6 = 1x \)
\( \therefore 6 = x \)
\( \therefore x = 6 \)
Since the value \(x = 6\) satisfies the equation (\(30 = 5 \times 6\)), it is a solution.

(iv) \( \frac{m}{2} = 14 \)
\( \therefore \frac{m}{2} \times 2 = 14 \times 2 \)
(Multiplying both sides by 2)
\( \frac{m \times 2}{2 \times 1} = 28 \)
\( \therefore m = 28 \)
Since the value \(m = 7\) does not satisfy the equation (\( \frac{7}{2} \ne 14 \)), it is not a solution.
In simple words: To check if a given value is a solution to an equation, substitute the value into the equation. If both sides of the equation are equal, then the value is a solution.

🎯 Exam Tip: Always show your substitution and the resulting calculation clearly when verifying a solution to ensure full credit.

 

Question 4. Solve the following equations:
(i) \(y - 5 = 1\)
(ii) \(8 = t + 5\)
(iii) \(4x = 52\)
(iv) \(19 = m - 4\)
(v) \( \frac{p}{4} = 9 \)
(vi) \(x + 10 = 5\)
(vii) \(m - 5 = -12\)
(viii) \(p + 4 = -1\)
Answer:
(i) \(y - 5 = 1\)
\( \therefore y - 5 + 5 = 1 + 5 \)
(Adding 5 to both sides)
\( \therefore y + 0 = 6 \)
\( \therefore y = 6 \)

(ii) \(8 = t + 5\)
\( \therefore 8 - 5 = t + 5 - 5 \)
(Subtracting 5 from both sides)
\( \therefore 8 + (-5) = t + 0 \)
\( \therefore 3 = t \)
\( \therefore t = 3 \)

(iii) \(4x = 52\)
\( \therefore \frac{4x}{4} = \frac{52}{4} \)
(Dividing both sides by 4)
\( \therefore 1x = 13 \)
\( \therefore x = 13 \)

(iv) \(19 = m - 4\)
\( \therefore 19 + 4 = m - 4 + 4 \)
(Adding 4 to both sides)
\( \therefore 23 = m + 0 \)
\( \therefore m = 23 \)

(v) \( \frac{p}{4} = 9 \)
\( \therefore \frac{p}{4} \times 4 = 9 \times 4 \) (Multiplying both sides by 4)
\( \therefore \frac{p \times 4}{4 \times 1} = 36 \)
\( \therefore 1p = 36 \)
\( \therefore p = 36 \)

(vi) \(x + 10 = 5\)
\( \therefore x + 10 - 10 = 5 - 10 \)
(Subtracting 10 from both sides)
\( \therefore x + 0 = 5 + (-10) \)
\( \therefore x = -5 \)

(vii) \(m - 5 = -12\)
\( \therefore m - 5 + 5 = -12 + 5 \)
(Adding 5 to both sides)
\( \therefore m + 0 = -7 \)
\( \therefore m = -7 \)

(viii) \(p + 4 = -1\)
\( \therefore p + 4 - 4 = -1 - 4 \)
(Subtracting 4 from both sides)
\( \therefore p + 0 = (-1) + (-4) \)
\( \therefore P = -5 \)
In simple words: Solving equations involves using inverse operations to isolate the variable, step by step, ensuring the equation remains balanced.

🎯 Exam Tip: For each step in solving an equation, clearly write down the operation performed on both sides (e.g., "Adding 5 to both sides") to demonstrate your understanding and reduce errors.

 

Question 5. Write the given information as an equation and find its solution:
(i) Haraba owns some sheep. After selling 34 of them in the market, he still has 176 sheep. How many sheep did Haraba have at first?
(ii) Sakshi prepared some jam at home and filled it in bottles. After giving away 7 of the bottles to her friends she still has 12 for herself. How many bottles had she made in all? If she filled 250g of jam in each bottle, what was the total weight of the jam she made?
(iii) Archana bought some kilograms of wheat. She requires 12 kg per month and she got enough wheat milled for 3 months. After that, she had 14 kg left. How much wheat had Archana bought altogether?
Answer:
(i) Let the number of sheep before selling be \(x\).
Equation: \(x - 34 = 176\)
Solution:
\( \therefore x - 34 + 34 = 176 + 34 \) (Adding 34 to both sides)
\( \therefore x + 0 = 210 \)
\( \therefore x = 210 \)
The number of sheep with Haraba before selling is 210.

(ii) Let the total number of bottles be \(x\).
Equation: \(x - 7 = 12\)
Solution:
\( \therefore x - 7 + 7 = 12 + 7 \) (Adding 7 to both sides)
\( \therefore x + 0 = 19 \)
\( \therefore x = 19 \)
Weight of jam in each bottle = 250g
\( \therefore \) Total weight of jam = \(19 \times 250\)g = 4750g = \( \frac{4750}{1000} \) kg = 4.75 kg
\( \therefore \) The total number of bottles of jam made by Sakshi is 19, and the total weight of jam made is 4.75 kg.

(iii) Let the total wheat bought by Archana be \(x\) kg.
Wheat used in 1 month = 12 kg
\( \therefore \) Wheat used in 3 months = \(3 \times 12 = 36\) kg
Equation: \(x - 36 = 14\)
Solution:
\( \therefore x - 36 + 36 = 14 + 36 \) (Adding 36 to both sides)
\( \therefore x + 0 = 50 \)
\( \therefore x = 50 \)
\( \therefore \) The total amount of wheat bought by Archana was 50 kg.
In simple words: Word problems can be solved by first converting the given information into a mathematical equation with a variable representing the unknown quantity, and then solving that equation.

🎯 Exam Tip: When solving word problems, read carefully to identify the unknown, define a variable, form the correct equation, and always write your final answer with appropriate units and context.

Std 6 Maths Digest