AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.1

AP State Syllabus AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.1 Textbook Questions and Answers.

AP State Syllabus 6th Class Maths Solutions 7th Lesson Introduction to Algebra Ex 7.1

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns.
(i) A pattern of letter ‘T’
(ii) A pattern of letter ‘E’
(iii) A pattern of letter ‘Z’
Answer:

Letter Matchsticks required for
1 2 3 4 5 n
i) T 2 2 × 2 3 × 2 4 × 2 5 × 2 2n
ii) E 4 2 × 4 3 × 4 4 × 4 5 × 4 4n
iii) Z 3 2 × 3 3 × 3 4 × 3 5 × 3 3n

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.1

Question 2.
Make a rule between the number of blades required and the number of fans (say n) in a hall.
Answer:
Number of blades for one fan = 3 = 1 × 3
Number of blades for two fans = 3 + 3 = 2 × 3
Number of blades for three fans = 3 + 3 + 3 = 3 × 3
Number of blades for four fans = 3 + 3 + 3 + 3 = 4 × 3
………………………………………
Number of blades for n fans = 3 + 3 + …. (n times) = 3 × n
Rule of number of blades required for n fans = 3.n

Question 3.
The cost of one pen is Rs. 7, then what is the rule for the cost of ‘n’ pens ?
Answer:
Cost of one pen = Rs. 7 = 1 × 7
Cost of two pens = Rs. 7 + Rs. 7 = 2 × 7
Cost of three pens = Rs. 7 + Rs. 7 + Rs. 7 = 3 × 7
……………………………………..
Cost of n pens = Rs. 7 + Rs. 7 + Rs. 7 ….. n times = 7 × n
Rule for cost n pens = 7.n.

Question 4.
The rule for purchase of books is that the cost of q books is Rs. 25q, then find the price of one book.
Answer:
Given, rule for cost of q books = Rs. 25q
In this rule q is variable i.e., q = 1, 2, 3,….
To get the cost of one book, put q = 1 in rule 25q
∴ Cost of one book = 25 (q) = 25(1) = Rs. 25

AP Board 6th Class Maths Solutions Chapter 7 Introduction to Algebra Ex 7.1

Question 5.
Harshini says that she has 5 biscuits more than Padma has. How can you express the relationship using the variable ‘y’?
Answer:
Given, Harshini has 5 biscuits more than Padma.
Let number of biscuits Padma has = y
Number of biscuits Harshini has = 5 more than Padma
∴ Number of biscuits Harshini has = y + 5

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