**Chapter
11 Algebra**

**Ex 11.1 Class 6 Maths **

**Question 1.
Find the rule which gives the number of matchsticks required to make the
following matchsticks 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 matchstieks required to make the pattern of T

For n = 1 is 2 x n

For n = 2 is 2 x n

For n = 3 is x n

Solution:

Number of matchstieks required to make the pattern of T

For n = 1 is 2 x n

For n = 2 is 2 x n

For n = 3 is x n

**∴**

**Rule is 2n where n is number of Ts.**

Number of matchstieks required to make the pattern of Z.

For n = 1 is 3 x n

For n = 2 is 3 x n

For n = 3 is 3 x n

**∴**** Rule is 3n
where n is number of Zs.**

Number of matchstieks required to make the pattern U.

For n = 1 is 3 x n

For n = 2 is 3 x n

For n = 3 is 3 x n

For n = 4 is 3 x n

**∴ Rule is 3n where n is number of
Us.**

Number of matchstieks required

For n = 1 is 2 x n

For n = 2 is 2 x n

For n = 3 is 2 x n

For n = 4 is 2 x n

**∴ Rule is 2n where n is number of
Vs.**

Number of matchstieks required

For n = 1 is 5 x n

For n = 2 is 5 x n

For n = 3 is 5 x n

**∴ Rule is 5n where n is number of
Es.**

Number of matchstieks required

For n = 1 is 5 x n

For n = 2 is 5 x n

For n = 3 is 5 x n

**∴ Rule is 5n where n is number of
Ss.**

Number of matchstieks required

For n = 1 is 6 x n

For n = 2 is 6 x n

For n = 3 is 6 x n

**∴ Rule is 6n where n is number of
As.**

**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 Ql. (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

For L it is 2n

For C it is 2n

For V it is 2n

For F it is 3n

For T it is 3n

For U it is 3n

We observe that the rule is same of L, V and T as they required only 2 matchsticks.

Letters C, F and U have the same rule, i.e., 3n as they require only 3 sticks.

Solution:

Rule for the following letters

For L it is 2n

For C it is 2n

For V it is 2n

For F it is 3n

For T it is 3n

For U it is 3n

We observe that the rule is same of L, V and T as they required only 2 matchsticks.

Letters C, F and U have the same rule, i.e., 3n as they require only 3 sticks.

**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 x n

For n = 2 is 5 x n

For n = 3 is 5 x n

Solution:

Number of cadets in a row = 5

Number of rows = n

Number of cadets

For n = 1 is 5 x n

For n = 2 is 5 x n

For n = 3 is 5 x n

**∴**

**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 boxes = b

Number of mangoes in a box = 50

Number of mangoes,

For n = 1 is 50 x b

For n = 2 is 50 x b

For n = 3 is 50 x b

Solution:

Number of boxes = b

Number of mangoes in a box = 50

Number of mangoes,

For n = 1 is 50 x b

For n = 2 is 50 x b

For n = 3 is 50 x b

**∴**

**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? (Uses for the number of students.)**

Solution:

Number of students = s

Number of pencils distributed per students = 5

Number of pencils required

For n = 1 is 5 x s

For n = 2 is 5 x s

For n = 3 is 5 x s

Solution:

Number of students = s

Number of pencils distributed per students = 5

Number of pencils required

For n = 1 is 5 x s

For n = 2 is 5 x s

For n = 3 is 5 x s

**∴**

**Rule is 5s where s represents the number of students.**

**Ex 11.1 Class
6 Maths **

**Question 6.
A bird flies 1 kilometer 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 n = 1 is 1 x t km
For n = 2 is 1 x t km
For n = 3 is 1 x t 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 a 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 = 8

Solution:

Number of rows = r

Number of dots in a row drawn by Radha = 8

**∴**

**The number of dots required**

For r = 1 is 8 x r

For r = 2 is 8 x r

For r = 3 is 8 x r

For r = 1 is 8 x r

For r = 2 is 8 x r

For r = 3 is 8 x r

**∴**

**Rule is 8r where r represents the number of rows.**

For r = 8, the number of dots = 8 x 8 = 64

For r = 10, the number of dots = 8 x 10 = 80

For r = 8, the number of dots = 8 x 8 = 64

For r = 10, the number of dots = 8 x 10 = 80

**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:

Radha’s age = x yeas.

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

Solution:

Given that 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 that, the number of oranges in smaller box = x

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

So, the number of oranges in bigger 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.

(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.

Solution:

(a) Let n be the number of squares.

**∴**

**Number of matchsticks required**

For n = 1 is 3 x n + l = 3n + 1 = 4

For n = 2is 3 x n + l = 3n + 1 = 7

For n = 3is 3 x n + l = 3n + 1 = 10

For n = 4is 3 x n + l = 3n + 1 = 13

For n = 1 is 3 x n + l = 3n + 1 = 4

For n = 2is 3 x n + l = 3n + 1 = 7

For n = 3is 3 x n + l = 3n + 1 = 10

For n = 4is 3 x n + l = 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

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 matchsticks.**

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