HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1

Haryana State Board HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 6th Class Maths Solutions Chapter 11 Algebra Exercise 11.1

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.
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (1)
Solution:
(a)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (2)
To make one T, we use two matchsticks as shown in Fig.(i)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (2) - 1
Thus, number of matchsticks required = 2n; n is a variable taking values 1, 2, 3, 4 ………..

(b) To make one Z, we use three matchsticks as shown in Fig. 11.2(f)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (3)
Thus, number of matchsticks required = 3n; n is a variable taking values 1, 2, 3, ……………

HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1

(c) To make one U, we use three matchsticks as shown in Fig. (i)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (4)
Thus, number of matchsticks required = 3n; n is a variable taking values 1, 2, 3, 4,

(d) To make one V, we use two matchsticks as shown in Fig. (i)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (5)
Thus, number of matchsticks required = 2n; n is a variable taking values 1, 2, 3, 4, 5, ………….

(e) To make one E, we use five matchsticks as shown in Fig. (i)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (6)
Thus, number of matchsticks required = 5n; n is a variable taking values 1, 2, 3, …………

(f) To make one S, we use five matchsticks as shown in Fig. (i)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (7)
Thus, number of matchsticks required = 5n; n is a variable taking values 1, 2, 3, 4, 5

(g) To make one A, we use six rnatchsticks as shown in Fig.(i)
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (8)
Thus, number of matchsticks required = 6n; n is a variable taking values 1, 2, 3, 4, 5, …………….

Question 2.
We already know the rule for the pattern of letters L, C and F. Some of the letters from Question 1 (given back) give us the same rule as that given by L. Which are these ? Why does this happen ?
Solution:
(a) and (d) give us the same rule as that given by L. Because to make one T and one V, we require two matchsticks.

HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1

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:
The number of cadets will depend on the number of rows. If there is 1 row, there will be 5 cadets. If there are 2 rows, there will be 2 x 5 or 10 cadets and so on. If there are ‘n’ rows, there will be n x 5 or 5n cadets, here n is a variable which stands for the number of rows and takes values 1, 2, 3, 4 ……..

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 ‘6’ for the number of boxes.)
Solution:
Number of mangoes in one box = 50
Number of mangoes in two boxes = 50 x 2 = 100
Number of mangoes in three boxes
= 50 x 3 = 150 and so on.
Number of mangoes in ‘b’ boxes = 50 x b = 50b
where ‘b’ is the no. 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 pencils needed for one student = 5
Number of pencils needed for two students = 5 x 2 = 10
Number of pencils needed for three students
= 5 x 3 = 15 and so on.
∴ Number of pencils needed for ‘s’ students = 5 x s = 50s.
where ‘s’ is the number of students.

Question 6.
A bird flies 1 kilometre 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.)
Sol. Distance covered by the bird in one minute = 1 km
Distance covered by the bird in two minutes = 1 x 2 = 2 km
Distance covered by the bird in three minutes
= 1 x 3 = 3 Ion and so on.
∴ Distance covered by the bird in‘t’ minutes 1 x t = t km.
where ‘t’ is the flying time in minutes.

HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1

Question 7.
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder as in (Fig. 11.8). She has 8 dots in a row. How many dots will her Rangoli have for V rows ? How many dots are there if there are 8 rows ? If there are 10 rows ?
Solution:
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (9)
Number of dots in one row = 8
Number of dots in two rows = 8 x 2 = 16
Number of dots in three rows = 8 x 3 = 24 and so on.
Number of dots in rows = 8 x r = 8r
Now, number of dots in 8 rows = 8 x 8 = 64
and number of dots in 10 rows = 8 x 10 = 80

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 V years.
Solution:
∵ Leela is 4 years younger than Radha.
∴ Leela’s age will be 4 years less than Radha’s age.
If Radha’s age = x years, then Leela’s age = (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:
Number of laddus given to guests and family members = l
Number of laddus still remain = 5
Total number of laddus she made = 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:
Number of oranges in one small box = x
Number of oranges in two small boxes = 2x
Number of oranges remain outside = 10
∴ Number of oranges in the larger box = 2x + 10.

Question 11.
(a) Look at the following matchstick pattern of squares (Fig. 11.9).
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (10)
The squares are not separate. Two neighbouring squares have a common matchstick. Observe the pattern 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.)
Solution:
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (10) - 1
Thus, number of matchsticks required = 3x + 1, where x is a variable and represents the number of squares.

HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1

(b) Fig. 11.10 gives a matchstick pattern of triangles. As in Exercise 11(a) above, find the general rule that gives the no. of matchsticks in terms of the number of triangles.
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (11)
Solution:
HBSE 6th Class Maths Solutions Chapter 11 Algebra Ex 11.1 (12)
Thus number of matchsticks required = 2x + 1, where x is a variable and represents the number of triangles.

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