NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1 are the part of NCERT Solutions for Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.1.



Ex 11.1 Class 6 Maths 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 T
(b) A pattern of letter Z as Z
(c) A pattern of letter U as U
(d) A pattern of letter V as V
(e) A pattern of letter E as E
(f) A pattern of letter S as S
(g) A pattern of letter A as A

Solution:

Number of matchsticks required to make one T is 2.

Number of matchsticks required to make two T is 4.

Number of matchsticks required to make three T is 6.

Rule is 2n where n is number of Ts.

Because
For n = 1, 2n = 2 × n = 2 × 1 = 2
For n = 2, 2n = 2 × n = 2 × 2 = 4
For n = 3, 2n = 2 × n = 2 × 3 = 6


Number of matchsticks required to make one Z is 3.

Number of matchsticks required to make two Z is 6.

Number of matchsticks required to make three Z is 9.

Rule is 3n where n is number of Zs.

Because
For n = 1, 3n = 3 × n = 3 × 1 = 3
For n = 2, 3n = 3 × n = 3 × 2 = 6
For n = 3, 3n = 3 × n = 3 × 3 = 9

Number of matchsticks required to make one U is 3.

Number of matchsticks required to make two U is 6.

Number of matchsticks required to make three U is 9.

Number of matchsticks required to make four U is 12.

Rule is 3n where n is number of Us.

Because
For n = 1, 3n = 3 × n = 3 × 1 = 3
For n = 2, 3n = 3 × n = 3 × 2 = 6
For n = 3, 3n = 3 × n = 3 × 3 = 9
For n = 4, 3n = 3 × n = 3 × 4 = 12

Number of matchsticks required to make one V is 2.

Number of matchsticks required to make two V is 4.

Number of matchsticks required to make three V is 6.

Number of matchsticks required to make four V is 8.

Rule is 2n where n is number of Vs.

Because
For n = 1, 2n = 2 × n = 2 × 1 = 2
For n = 2, 2n = 2 × n = 2 × 2 = 4
For n = 3, 2n = 2 × n = 2 × 3 = 6
For n = 4, 2n = 2 × n = 2 × 4 = 8

Number of matchsticks required to make one E is 5.

Number of matchsticks required to make two E is 10.

Number of matchsticks required to make three E is 15.

Rule is 5n where n is number of Es.

Because
For n = 1, 5n = 5 × n = 5 × 1 = 5
For n = 2, 5n = 5 × n = 5 × 2 = 10
For n = 3, 5n = 5 × n = 5 × 3 = 15

Number of matchsticks required to make one S is 5.

Number of matchsticks required to make two S is 10.

Number of matchsticks required to make three S is 15.

Rule is 5n where n is number of S.

Because
For n = 1, 5n = 5 × n = 5 × 1 = 5
For n = 2, 5n = 5 × n = 5 × 2 = 10
For n = 3, 5n = 5 × n = 5 × 3 = 15

Number of matchsticks required to make one A is 6.

Number of matchsticks required to make two A is 12.

Number of matchsticks required to make three A is 18.

Rule is 6n where n is number of As.

Because
For n = 1, 6n = 6 × n = 6 × 1 = 6
For n = 2, 6n = 6 × n = 6 × 2 = 12
For n = 3, 6n = 6 × n = 6 × 3 = 18

 

Ex 11.1 Class 6 Maths 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:
Rules for the following letters are as follows:
For L it is 2n.
For C it is 3n.
For V it is 2n.
For F it is 3n.
For T it is 2n.
For U it is 3n.
We observe that the rule is the same for L, V and T, i.e., 2n, as they required only 2 matchsticks.
Letters C, F and U have the same rule, i.e., 3n, as they require only 3 matchsticks.

 

Ex 11.1 Class 6 Maths 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

Number of cadets
For n = 1 is 5 × n = 5 × 1 = 5
For n = 2 is 5 × n = 5 × 2 = 10
For n = 3 is 5 × n = 5 × 3 = 15
Rule is 5n where n is the number of rows.

 

Ex 11.1 Class 6 Maths 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 mangoes in a box = 50
Number of boxes = b
Number of mangoes,
For b = 1 is 50 × b = 50 × 1 = 50
For b = 2 is 50 × b = 50 × 2 = 100
For b = 3 is 50 × b = 50 × 3 = 150
Rule is 50b where b represents the number of boxes.

 

Ex 11.1 Class 6 Maths 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 pencils distributed per student = 5
Number of students = s
Number of pencils required
For s = 1 is 5 × s = 5 × 1 = 5
For s = 2 is 5 × s = 5 × 2 = 10
For s = 3 is 5 × s = 5 × 3 = 15
Rule is 5s where s represents the number of students.

 

Ex 11.1 Class 6 Maths Question 6.

A bird flies 1 kilometre in one minute. Can you express the distance covered by the bird in terms of is 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 t = 1 is 1 × t = 1 × 1 = 1 km
For t = 2 is 1 × t = 1 × 2 = 2 km
For t = 3 is 1 × t = 1 × 3 = 3 km
Rule is 1.t km where t represents the flying time.

 

Ex 11.1 Class 6 Maths 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 rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?

Solution:
Number of rows = r
Number of dots in a row drawn by Radha = 9
The number of dots required
For r = 1 is 9 × r = 9 × 1 = 9
For r = 2 is 9 × r = 9 × 2 = 18
For r = 3 is 9 × r = 9 × 3 = 27
Rule is 9r where r represents the number of rows.
For r = 8, the number of dots = 9 × 8 = 72
For r = 10, the number of dots = 9 × 10 = 90

 

Ex 11.1 Class 6 Maths 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:
Given: Radha’s age = x years
Given that Leela’s age = Radha’s age – 4 years
= x years – 4 years
= (x – 4) years

 

Ex 11.1 Class 6 Maths 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: The number of laddus given away = l
Number of laddus left = 5
Number of laddus made by mother = l + 5

 

Ex 11.1 Class 6 Maths 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: The number of oranges in smaller box = x
Number of oranges in larger box = 2(number of oranges in small box) + (Number of oranges remain outside)
So, the number of oranges in larger box = 2x + 10

 

Ex 11.1 Class 6 Maths Question 11.

(a) Look at the following matchstick pattern of square. 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.)

(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.

Solution:
(a) Let n be the number of squares.
Number of matchsticks required
For n = 1 is 3 × n + 1 = 3n + 1 = 4
For n = 2 is 3 × n + 1 = 3n + 1 = 7
For n = 3 is 3 × n + 1 = 3n + 1 = 10
For n = 4 is 3 × n + 1 = 3n + 1 = 13
Rule is 3n + 1 where n represents the number of squares.

(b) Let n be the number of triangles.
Number of matchsticks required
For n = 1 is 2n + 1 = 3
For n = 2 is 2n + 1 = 5
For n = 3 is 2n + 1 = 7
For n = 4 is 2n + 1 = 9
Rule is 2n + 1 where n represents the number of triangles.



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