NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

Mathematics (Maths) Class 6

Class 6 : NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

The document NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6

Exercise 11.1

Q1. 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  NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans:  Pattern of letter NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
So,
For 1T, Number for Matchsticks = 2
For 2T, Number for Matchsticks = 4
For 3T, Number for Matchsticks = 6
So, we write
Number of matchsticks = 2n
where, n = Number of T

(b) A pattern of letter Z as NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: Pattern of letter  NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
So,
For 1Z, Number of matchsticks = 3
For 2Z, Number of matchsticks = 6
For 3Z, Number of matchsticks = 9
So, we write,
Number of matchsticks = 3n
where n = Number of Z

(c) A pattern of letter U as NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: Pattern of letter NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev 
So,
For 1U, Number of matchsticks = 3
Similarly
For 2U, Number of matchsticks = 6
For 3U, Number of matchsticks = 9
So, we write,
Number of matchsticks = 3n
where n = Number of U

(d) A pattern of letter V as NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: Pattern of letter NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
So,
For 1V, Number of matchsticks = 2
Similarly
For 2V, Number of matchsticks = 4
For 4V, Number of matchsticks = 6
So, we write,
Number of matchsticks = 2n
where n = Number of V 

(e) A pattern of letter E as NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: Pattern of letter NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
So,
For one E, Number of matchsticks = 5
For 2E, Number of matchsticks = 10
For 3E, Number of matchsticks = 15
So, we write,
Number of matchsticks = 5n
where, n = Number of E

(f) A pattern of letter S as NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: Pattern of letter NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev 
So,
For one S, Number of matchsticks = 5
For 2S, Number of matchsticks = 10
For 3S, Number of matchsticks = 15
So, we write,
Number of matchsticks = 5n
where, n = Number of S

(g) A pattern of letter A as NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: Pattern of letter NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev 
So,
For 1A, Number of matchsticks = 6
For 2A, Number of matchsticks = 12
For 3A, Number of matchsticks = 18
So, we write
Number of matchsticks = 6n
where n = Number of A


Q2. 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?
Ans: The letter ‘T’ and ‘V’ has pattern 2n since 2 matchsticks are used in all these letters.


Q3. 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.)
Ans: Number of rows = n
Cadets in each row = 5
Therefore, total number of cadets = 5n


Q4. 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.)
Ans: Number of boxes = b
Number of mangoes in each box = 50
Therefore, total number of mangoes = 50b


Q5. 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.)
Ans: Number of students = s
Number of pencils to each student = 5
Therefore, the total number of pencils needed are = 5s


Q6. 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.)
Ans: Time taken by bird = t minutes
Speed of bird = 1 km per minute
Therefore, Distance covered by bird = speed x time = 1 x t = t km


Q7. 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?

NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

Ans: Number of dots in each row = 8 dots
Number of rows = r
Therefore, number of dots = 8r
When there are 8 rows, then number of dots = 8 x 8 = 64 dots
When there are 10 rows, then number of dots = 8 x 10 = 80 dots


Q8. 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.
Ans: Given that,
Leela is 4 years younger than Radha
So,
Leela's Age = Radha's age - 4
Let Radha's age = x years
So, Leela's age = x - 4 years
where x = Radha's age

Q9. 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?
Ans: Number of laddus gave away = l
Number of laddus remaining = 5
Total number of laddus = (l + 5)


Q10. 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?
Ans: Number of oranges in one box = x
Number of boxes = 2
Therefore, total number of oranges in boxes = 2x
Remaining oranges = 10

Thus, number of oranges = 2x + 10


Q11: (a) Look at the following matchstick pattern of squares (Fig 11.6). 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

NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

in terms of the number of squares. (Hint: If you remove vertical stick at the end, you will get a pattern of Cs)
Ans: We may observe that in the given matchstick pattern, the number of matchsticks are 4, 7, 10 and 13, which is 1 more than the thrice of the number of squares in the pattern

Therefore the pattern is 3x + 1, where x is the number of squares

(b) Fig 11.7 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.

NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans: We may observe that in the given matchstick pattern, the number of matchsticks is 3, 5, 7 and 9 which is 1 more than twice the number of triangles in the pattern.
Therefore the pattern is 2x + 1, where x is the number of triangles.


Exercise 11.2

Q1. The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Ans: Side of equilateral triangle = l

Therefore, Perimeter of equilateral triangle = 3 x side = 3l


Q2. The Side of a regular hexagon (see the given figure) is denoted by l. Express the perimeter of the hexagon using l.

(Hint: A regular hexagon has all its six sides equal in length.)

NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

Ans: Side of hexagon = l
Therefore, Perimeter of Hexagon = 6 x side = 6l


Q3. A cube is a three-dimensional figure as shown in the given figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube.

NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

Ans: Length of one edge of cube = l
Number of edges in a cube = 12
Therefore, total length = 12 x l = 12l


Q4. The diameter of a circle is a line that joins two points on the circle and also passed through the centre of the circle. (In the adjoining figure AB is the diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius(r).

NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev

Ans: Since the length of diameter is double the length of the radius.
Therefore, d = 2r


Q5. To find the sum of three numbers 14, 27 and 13, we can have two ways:
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus, (14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on whole numbers, in a general way, by using variables a, b and c.
Ans: For any three whole numbers a, b and c
(a + b) + c = a + (b + c)


Exercise 11.3

Q1. Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication.
(Hint: Three possible expressions are 5 + (8 − 7), 5 − (8 − 7), (5 × 8) + 7; make the other expressions.)
Ans: 
(a) (8 x 5) – 7
(b) (8 + 5) – 7
(c) (8 x 7) – 5
(d) (8 + 7) – 5
(e) 5 x (7 + 8)
(f) 5 + (7 x 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)


Q2. Which out of the following are expressions with numbers only?
(a) y + 3 
(b) (7 × 20) − 8z
(c) 5 (21 − 7) + 7 × 2 
(d) 5
(e) 3x 
(f) 5 − 5n
(g) (7 × 20) − (5 × 10) − 45 + p
Ans: (c) and (d) are the expressions with numbers only.


Q3. Identify the operations (addition, subtraction, division, multiplication) in the following expressions and tell how the expressions have been formed.

(a) z + 1, z − 1, y + 17, y − 17
Ans:
z +1→ Addition    
z -1→ Subtraction
y + 17→ Addition      
y - 17→ Subtraction

(b) NCERT Solutions(Part - 1) - Algebra Class 6 Notes | EduRev
Ans:
17y → Multiplication
y/17 → Subtraction

5z → Multiplication

(c) 2y + 17, 2y − 17
Ans:
2y + 17→Multiplication and Addition
2y - 17→ Multiplication and Subtraction

(d) 7m, − 7m + 3, − 7m − 3
Ans:
7m → Multiplication
-7m +3 → Multiplication and Addition
7m -3 → Multiplication and Subtraction


Q4. Give expressions for the following cases.

(a) 7 added to p
Ans: p + 7

(b) 7 subtracted from p
Ans: p - 7

(c) p multiplied by 7
Ans: 7p

(d) p divided by 7
Ans: p/7

(e) 7 subtracted from − m
Ans: -m- 7

(f) − p multiplied by 5
Ans: -5 p

(g) − p divided by 5
Ans: -p/5  

(h) p multiplied by − 5
Ans: -5p

Q5. Give expressions in the following cases.

(a) 11 added to 2m
Ans: 2m + 11

(b) 11 subtracted from 2m
Ans: 2m - 11

(c) 5 times y to which 3 is added
Ans: 5y + 3

(d) 5 times y from which 3 is subtracted
Ans: 5y - 3

(e) y is multiplied by − 8
Ans: -8y

(f) y is multiplied by − 8 and then 5 is added to the result
Ans: -8y + 5

(g) y is multiplied by 5 and the result is subtracted from 16
Ans: 16 - 5y

(h) y is multiplied by − 5 and the result is added to 16
Ans: -5y + 16

Q6. (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
Ans: (t + 4), (t – 4), 4t, (t / 4), (4 / t), (4 – t), (4 + t) are the expressions using t and 4

(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Ans: 2y + 7, 2y – 7, 7y + 2,… are the expression using y, 2 and 7

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