Samacheer Kalvi Class 6 Maths Solutions Term 1 Chapter 2 Introduction to Algebra Exercise 2.2

 

Question 1. Find in the blanks.
(i) The algebraic statement of 'f' decreased by 5 is __________
(ii) The algebraic statement of 's' divided by 5 is __________
(iii) The verbal statement of '2m – 10' is __________
(iv) If A's age is 'n' years now, 7 years ago A's age was __________
(v) If p - 5 = 12, then 'p' is __________
Answer:
(i) \( f - 5 \). When a number is "decreased by" another, it means subtraction.
(ii) \( \frac { s }{ 5 } \). Dividing 's' by 5 can be written as a fraction.
(iii) 10 less than 2 times m (or) Take away 10 from the product of 2 and m. This describes the order of operations.
(iv) \( n - 7 \). To find someone's age in the past, you subtract the number of past years from their current age.
(v) 17. To solve \( p - 5 = 12 \), we add 5 to both sides, so \( p = 12 + 5 = 17 \). We are finding the number that was 5 less than 12.
In simple words: We fill in the blanks using algebra or basic arithmetic. For an unknown like 'f', 's', 'n', or 'p', we use letters to represent them and write down the operation.

🎯 Exam Tip: Remember to express operations like "decreased by" as subtraction, "divided by" as division, and "times" as multiplication. For equations, perform the inverse operation to find the unknown value.

 

Question 2. Say True or False.
(i) 10 more to three times 'c' is '3c + 13'.
(ii) If the cost of 10 rice bags is 't', then the cost of 1 rice bag is \( \frac { t }{ 10 } \).
(iii) The statements 'x' divided by 3 and 3 divided by 'x' are the same.
(iv) The product of 'q' and 20 is '20q'.
(v) 7 less to 7 times 'y' is '7 - 7y'.
Answer:
(i) False. "10 more to three times 'c'" means \( 3c + 10 \), not \( 3c + 13 \). The '13' is incorrect.
(ii) True. To find the cost of one item when you know the cost of many, you divide the total cost by the number of items.
(iii) False. Division is not commutative, meaning the order matters. \( \frac{x}{3} \) is different from \( \frac{3}{x} \).
(iv) True. The product means multiplication, and in algebra, we write a number followed by a variable (like 20q) to show multiplication.
(v) False. "7 less to 7 times 'y'" means \( 7y - 7 \), not \( 7 - 7y \). The order of subtraction is very important.
In simple words: We check if each statement correctly translates English words into algebraic expressions. Pay close attention to keywords like "more than," "divided by," and "less than" because the order matters.

🎯 Exam Tip: Always be careful with the order of operations, especially in subtraction and division, as changing the order can change the entire meaning of the expression.

 

Question 3. Express the following verbal statement to an algebraic statement.
(i) 100 more than 't'
(ii) 4 times 'q'
(iii) 8 is reduced by 'y'
(iv) 56 added to 2 times 'x'
(v) 4 less to 9 times of 'y'
Answer:
(i) \( t + 100 \). "More than" indicates addition.
(ii) \( 4q \). "Times" indicates multiplication, and we usually write the number first.
(iii) \( 8 - y \). "Reduced by" means subtraction, where the amount reduced is taken from the original number.
(iv) \( 2x + 56 \). "2 times 'x'" is \( 2x \), and "added to" means we add 56 to this product.
(v) \( 9y - 4 \). "9 times of 'y'" is \( 9y \), and "4 less to" means we subtract 4 from this product.
In simple words: We change word problems into math symbols. "More than" means add, "times" means multiply, and "reduced by" or "less than" means subtract. The key is to put the parts in the correct order.

🎯 Exam Tip: Break down complex verbal statements into smaller parts. For example, "56 added to 2 times 'x'" should first be interpreted as "2 times 'x'" (\( 2x \)) and then "56 added to" (\( + 56 \)).

 

Question 4. Express the following algebraic statement to a verbal statement.
(i) \( x \div 3 \)
(ii) \( 5n - 12 \)
(iii) \( 11 + 10x \)
(iv) \( 70s \)
Answer:
(i) \( x \) divided by 3. This is the most direct way to express division.
(ii) 12 less to 5 times \( n \). This means 12 is subtracted from the product of 5 and \( n \).
(iii) 11 added to 10 times \( x \). We add 11 to the product of 10 and \( x \).
(iv) 7 times \( s \). When a number is next to a variable, it means multiplication.
In simple words: We turn math expressions back into words. For \( x \div 3 \), we say "x divided by 3". For \( 5n - 12 \), we say "12 less than 5 times n".

🎯 Exam Tip: Use clear and concise language. For subtraction, phrases like "less than" or "taken away from" correctly indicate the order of operations.

 

Question 5. The teacher asked two students to write the algebraic statement for the verbal statement "8 more than a number”. Vetri wrote '8 + x' but Maran wrote '8x'. Who gave the correct answer?
Answer: Let the number be \( x \). The verbal statement "8 more than a number" means we add 8 to that number. So, the correct algebraic statement is \( x + 8 \) or \( 8 + x \). This means Vetri gave the correct answer. Maran's answer, \( 8x \), means "8 times a number," which is different from "8 more than a number."
In simple words: Vetri is right. "8 more than a number" means you add 8 to the number (\( x + 8 \)). Maran's answer, \( 8x \), means to multiply the number by 8, which is not what the question asked.

🎯 Exam Tip: Distinguish carefully between "more than" (addition) and "times" (multiplication) when converting verbal statements to algebraic expressions. This is a common point of confusion.

 

Question 6. Answer the following questions.
(i) If 'n' takes the value 3 then find the value of 'n + 10'?
(ii) If 'g' is equal to 300 what is the value of 'g – 1' and 'g + 1'?
(iii) What is the value of 's', if '2s – 6' gives 30?
Answer:
(i) If \( n = 3 \), then \( n + 10 = 3 + 10 = 13 \). We simply substitute the value of \( n \) into the expression.
(ii) If \( g = 300 \):
\( g - 1 = 300 - 1 = 299 \).
\( g + 1 = 300 + 1 = 301 \). We replace \( g \) with 300 and perform the operations.
(iii) To find the value of \( s \) if \( 2s - 6 = 30 \):
Add 6 to both sides: \( 2s = 30 + 6 \)
\( 2s = 36 \)
Divide by 2: \( s = \frac{36}{2} \)
\( s = 18 \). We solve the equation by isolating \( s \) using inverse operations.
In simple words: For (i) and (ii), we put the given number into the letter's place and do the math. For (iii), we work backwards from the equation to find the value of the letter 's'.

🎯 Exam Tip: When solving equations, remember to perform the same operation on both sides to keep the equation balanced and isolate the variable.

 

Question 7. Complete the table and find the value of 'k' for which 'k/3' gives 5.

Answer:To find the value of \( k \) for which \( \frac{k}{3} = 5 \):
Multiply both sides by 3: \( k = 5 \times 3 \)
\( k = 15 \). The table helps us see the pattern, and then we solve the equation to find the specific value of \( k \).
In simple words: We fill in the table by dividing each 'K' value by 3. Then, to find 'k' when \( \frac{k}{3} \) is 5, we multiply 5 by 3 to get \( k = 15 \).

🎯 Exam Tip: When completing tables, look for the rule or pattern that connects the values. For equations, use inverse operations to solve for the unknown variable.

 

Question 8. The value of 'y' in y + 7 = 13 is
(a) y = 5
(b) y = 6
(c) y = 7
(d) y = 8
Answer: (b) y = 6
In simple words: To find 'y', we take 7 away from both sides of the equation. So, \( y = 13 - 7 \), which means \( y = 6 \).

🎯 Exam Tip: To solve simple linear equations, use the inverse operation: if a number is added, subtract it from both sides; if multiplied, divide both sides.

 

Question 9. 6 less to 'n' gives 8 is represented as
(a) \( n - 6 = 8 \)
(b) \( 6 - n = 8 \)
(c) \( 8 - n = 6 \)
(d) \( n - 8 = 6 \)
Answer: (a) \( n - 6 = 8 \)
In simple words: "6 less to 'n'" means we subtract 6 from 'n'. "Gives 8" means the result is 8. So, it should be \( n - 6 = 8 \).

🎯 Exam Tip: The phrase "less to" or "less than" implies subtraction from the preceding term. For example, "A less than B" is \( B - A \).

 

Question 10. The value of 'c' for which \( \frac{3c}{4} \) gives 18 is
(a) c = 15
(b) c = 21
(c) c = 24
(d) c = 27
Answer: (c) c = 24
In simple words: First, we multiply both sides by 4 to get \( 3c = 72 \). Then, we divide by 3 to find \( c = 24 \).

🎯 Exam Tip: When solving equations with fractions, multiply both sides by the denominator to clear the fraction first. Then proceed to isolate the variable.