AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions

Well-designed AP 6th Class Maths Guide Chapter 11 Algebra Exercise 11.1 offers step-by-step explanations to help students understand problem-solving strategies.

Algebra Class 6 Exercise 11.1 Solutions – 6th Class Maths 11.1 Exercise Solutions

Question 1.
Find the rule which gives the number of matchsticks required to make the following matchstick patterns. Use a variable to write the rule.
a) A pattern of letter T as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 1
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 2
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 5
For n = 1, the number of matchsticks required = 2 × 1 = 2
n = 2, the number of matchsticks required = 2 × 2 = 4
n = 3, the number of matchsticks required = 2 × 3 = 6
∴ The rule = 2 × n = 2 n, where n is number of Ts.

b) A pattern of letter Z as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 3
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 4
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 6
For n = 1, the number of matchsticks required = 3 × 1 = 3
n = 2, the number of matchsticks required = 3 × 2 = 6
n = 3, the number of matchsticks required = 3 × 3 = 9
∴ The rule = 3 × n = 3n, where n is number of Zs .

c) A pattern of letter U as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 7
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 8
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 9
For n = 1, the number of matchsticks required = 3 × 1 = 3
n = 2, the number of matchsticks required = 3 × 2 = 6
n = 3, the number of matchsticks required = 3 × 3 = 9
∴ The rule = 3 × n = 3n, where n is number of Us .

d) A pattern of letter V as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 10
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 11
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 12
For n = 1, the number of matchsticks required = 2 × 1 = 2
n = 2, the number of matchsticks required = 2 × 2 = 4
n = 3, the number of matchsticks required = 2 × 3 = 6
n = 4, the number of matchsticks required = 2 × 4 = 8
∴ The rule = 2 × n = 2n, where n is number of Vs.

e) A pattern of letter E as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 13
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 14
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 15
For n = 1, the number of matchsticks required = 5 × 1 = 5
n = 2, the number of matchsticks required = 5 × 2 = 10
n = 3, the number of matchsticks required = 5 × 3 = 15
∴ The rule = 5 × n = 5n, where n is number of Es.

f) A pattern of letter S as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 16
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 17
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 18
For n = 1, the number of matchsticks required = 5 × 1 = 5
n = 2, the number of matchsticks required = 5 × 2 = 10
n = 3, the number of matchsticks required = 5 × 3 = 15
∴ The rule = 5 × n = 5n, where n is number of Ss.

g) A pattern of letter A as AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 19
Solution:
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 20
Number of matchsticks required to make the pattern AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 21
For n = 1, the number of matchsticks required = 6 × 1 = 6
n = 2, the number of matchsticks required = 6 × 2 = 12
n = 3, the number of matchsticks required = 6 × 3 = 18
∴ The rule = 6 × n = 6n, where n is number of As.

AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Q. 1 (given above) give us the same rule as that given by L. Which are these ? Why does this happen ?
Solution:
Rule for the following letters.

Letter Rule
L 2n
V 2n
T 2n
Letter Rule
C 3n
F 3n
U 3n

From the above tables,
i) We obseerve that the rule is same of L,V and T i.e. 2n as they are required only two matchsticks.
ii) Letters C, F and U have the same rule i.e. 3n, as they required only 3 matchsticks.

Question 3.
Cadets are marching in a parade. There are 5 cadets in a row. What is the rule which gives the number of cadets, given the number of rows? (Use n for the number of rows.)
Solution:
Number of cadets in a row = 5
Number of rows = n
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 22
Number of cadets = n
For n = 1 is 5 × 1 = 5
n = 2 is 5 × 2 = 10
n = 3 is 5 × 3 = 15
∴ Rule is 5n, where n represents the number of rows.

Question 4.
If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes.)
Solution:
Number of boxes = b
Number of mangoes in a box = 50
Number of unangoes
For n – 1 is 50 × 1 = 50
For n = 2 is 50 × 2 = 100
For n = 3 is 50 × 3 = 150
∴ Rule is 50 b, where b represents the number of boxes.

Question 5.
The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students.)
Solution:
Number of students = s
Number of pencils ditributed per student = 5
Number of pencils required
For n = 1 is 5 × 1 = 5
For n =2 is 5 × 2 = 10
For n = 3 is 5 × 3 = 15
∴ Rule is 5s, where s represents the number of students.

AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions

Question 6.
A bird files 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes.)
Solution:
Distance covered in 1 minute = 1 km
The flying time = t
Distance covered
For n = 1 is 1 × 1 = 1 km
For n = 2 is 1× 2 = 2 km
For n = 3 is 1 × 3 = 3 km
∴ Rule is 1 × t km, where t represents, the flying time.

Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder. She has 9 dots in a row. How many dots will her Rangoll have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Solution:
Number of rows = 9
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 23
Number of dots in a row drawn by Radha = 8
∴ The number of dots required
For r = 1 is 9 × 1 = 9
For r = 2 is 9 × 2 = 18
For r = 3 is 9 × 3 = 27
∴ Rule is 9r, where r represents the number of rows.
If r = 8 (i.e. 8 rows), the number of dots = 9 × 8 = 72
r = 10 (i.e. 10 rows), the number of dots = 9 × 10 = 90

Question 8.
Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age ? Take Radha’s age to be x years.
Solution:
Radha’s age = x years
Given that Leela’s age = Radha’s age – 4 years
= x years – 4 years
= (x – 4) years

Question 9.
Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution:
Given that the number of laddus given away = l
Number of laddus left = 5
∴ Number of laddus made by mother = l + 5

Question 10.
Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box ?
Solution:
Given that, the number of oranges in smaller box = x
∴ Number of oranges in bigger box = 2 (number of oranges in small box) + Number of oranges remain outside)
So, the number of oranges in bigger box = 2x + 10

Question 11.
a) Look at the following matchstick pattern of squares (Figure). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint : If you remove the vertical stick at the end, you will get a pattern of Cs.)
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 24
Solution:
Let n be the number of squares.
∴ Number of matchsticks required.
For n = 1 = 3n + 1 = 3 × 1 + 1 = 4
For n = 2 = 3n + 1 = 3 × 2 + 1 = 7
For n = 3 = 3n + 1 = 3 × 3 + 1 = 10
For n = 4 = 3n + 1 = 3 × 4 + 1 = 13
For n, 3 × n + 1 = 3n + 1
∴ Rule is 3n + 1, where n represents the number of squares.

b) Following figure gives a matchstick pattern of triangles. As in Exercise 11 (a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.
AP 6th Class Maths 11th Chapter Algebra Exercise 11.1 Solutions Img 25
Solution:
Let n be the number of triangles.
∴ Number of matchsticks required.
For n = 1 = 2 × 1 = 1
For n = 2 = 2 × 2 = 1
For n = 3 = 2 × 3 + 1
For n = 4 = 2 × 4 + 1
For n, 2 × n + 1 = 2n + 1
∴ Rule is 2n + 1, where n represents the number of matchsticks.

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