Kseeb Class 6 Mathematics Chapter 11 Algebra Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students by providing Solutions for KSEEB Class 6 Mathematics chapter 11 Algebra. Here students can easily find all the solutions for Algebra Exercise 11.1, 11.2, 11.3, 114 and 11.5 Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 11 solutions. Here all the solutions are based on Karnataka State Board latest syllabus.
Algebra Exercise 11.1 Solutions :
(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
(b) A pattern of letter Z
(c) A pattern of letter U
(d) A pattern of letter V
(e) A pattern of letter E
(f) A pattern of letter S
(g) A pattern of letter A
Solution :
(a) Number of matchsticks required is 2.
Rule which matchstick pattern is 2 x n.
(b) Number of matchsticks required is 3.
Rule which matchstick pattern is 3 x n.
(c) Number of matchsticks required is 3 .
Rule which matchstick pattern is 3 x n.
(d) Number of matchsticks required is 2.
Rule which matchstick pattern is 2 x n.
(e) Number of matchsticks required is 5.
Rule which matchstick pattern is 5 x n.
(f) Number of matchsticks required is 5.
Rule which matchstick pattern is 5 x n.
(g) Number of matchsticks required is 6.
Rule which matchstick pattern is 6 x n.
(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 :
Let n be given number of rows.
Since, 5 cadets in each row.
Thus, 5n is the rule for the number of cadets, for a given number of rows.
(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 :
There are 50 mangoes in a box.
We uses ‘b’ for boxes.
Total number of mangoes in terms of the number of boxes = 50 x b
(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 :
The teacher distributes 5 pencils per student.
We uses ‘s’ for number of students.
Pencils are needed, given the number of students = 5 x s
(6) A bird flies 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 :
A bird flies 1 kilometre in one minute.
We uses ‘t’ for flying time in minutes.
Distance covered by the bird in terms of its flying time in minutes = t km
(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 :
Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots) with chalk powder.
She has 9 dots in a row.
We uses’ r’ for dots in a row.
Pattern is 8 x r.
Are there if there are 8 rows
= 8 x 8
= 64
Are there if there are 8 rows
= 8 x 10
= 80
(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 :
Leela is Radha’s younger sister. Leela is 4 years younger than Radha.
Leela’s age in terms of Radha’s age,
We use Radha’s age to be x years.
Leela’s age in terms of Radha’s age is (x – 4) years
(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 :
She gives some laddus to guests and family members; still 5 laddus remain.
We use number of laddus mother gave away is “l”.
laddus she make = l + 5
(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 :
When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside.
The number of oranges in a small box are taken to be x.
Number of oranges in the larger box = 2x + 10
Algebra Exercise 11.2 solutions :
(2) The side of a regular hexagon (Fig 11.10) is denoted by l. Express the perimeter of the hexagon using l.
Solution :
The side of a regular hexagon is denoted by “l”.
The perimeter of the hexagon = 6 x l
(3) A cube is a three-dimensional figure as shown in Fig 11.11. 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.
Solution :
We can solve this problem like as follows:-
The length of an edge of the cube is given by “l”.
Formula for the total length of the edges of a cube = 12 x l
(4) The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.12) AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).
Solution :
AB is a diameter of the circle; C is its centre.
The diameter of the circle (d) in terms of its radius (r) = d = 2r
Algebra Exercise 11.3 Solutions :
(2) 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
Solution :
(a) y + 3
= No
(b) (7 × 20) – 8z
= No
(c) 5 (21 – 7) + 7 × 2
= Yes
(d) 5
= Yes
(e) 3x
= No
(f) 5 – 5n
= No
(g) (7 × 20) – (5 × 10) – 45 + p
= No
(3) Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed
(a) z +1, z – 1, y + 17, y – 17
(b) 17y, y /17, 5 z
(c) 2y + 17, 2 y – 17
(d) 7 m, – 7 m + 3, – 7 m
Solution :
(a) z +1, z – 1, y + 17, y – 17
The operations used as follows:
Addition, subtraction, addition, subtraction
(b) 17y, y /17, 5 z
The operations used as follows:
Multiplication, division, multiplication
(c) 2y + 17, 2 y – 17
The operations used as follows:
Multiplication and addition, multiplication and subtraction
(d) 7 m, – 7 m + 3, – 7 m
The operations used as follows:
Multiplication, multiplication and addition, multiplication and subtraction
(4) Give expressions for the following cases.
(a) 7 added to p
(b) 7 subtracted from p
(c) p multiplied by 7
(d) p divided by 7
(e) 7 subtracted from – m
(f) – p multiplied by 5
(g) – p divided by 5
(h) p multiplied by – 5
Solution :
(a) = p + 7
(b) = 7 – p
(c) = 7p
(d) = p/7
(e) = m – 7
(f) = (– 5 x p)
(g) = (- p / 5)
(h) = (– 5 x p)
(5) Give expressions in the following cases.
(a) 11 added to 2 m
(b) 11 subtracted from 2 m
(c) 5 times y to which 3 is added
(d) 5 times y from which 3 is subtracted
(e) y is multiplied by – 8
(f) y is multiplied by – 8 and then 5 is added to the result
(g) y is multiplied by 5 and the result is subtracted from 16
(h) y is multiplied by – 5 and the result is added to 16.
Solution :
(a) = (2 x m) + 11
(b) = 2m – 11
(c) (5 x y) + 3
(d) (5 x y) – 3
(e) = (– 8 x y)
(f) = (– 8 x y) + 5
(g) = 16 – (5 x y)
(h) = (– 5 x y) + 16
(6) – (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
Solution :
The expressions formed from the given numbers stratifying the given condition are, t + 4, t – 4, 4 – t, 4 + t, 4t, t/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.
Solution :
The expressions formed from the given numbers stratifying the given condition are – 2y + 7, 2y – 7, 7y + 2, 7y – 2, y/2 + 7, y/2 – 7…..
Algebra Exercise 11.4 Solutions :
(1) Answer the following :
(a) Take Sarita’s present age to be y years
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
Solution :
(i) Sarita’s present age to be “y” years.
age 5 years from now = y + 5 years.
(ii) Age 3 years back = y – 3 years.
(iii) Sarita’s grandfather is 6 times her age.
Age of her grandfather = (6 x y) years
(iv) Grandmother is 2 years younger than grandfather
Grandmother’s age = (6 x y) – 2 years
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age.
Father’s age = (3 x y) + 5 years
(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?
Solution :
The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall.
The length, if the breadth is b meters = (3b – 4) metres.
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
Solution :
The length and the breadth of the box in terms of the height
Length in terms of the height = (5 x h) cm.
Breadth in terms of the height = (5 x h)– 10 cm
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
Solution :
Meena, Beena and Leena are climbing the steps to the hill top.
Beena is 8 steps ahead and Leena 7 steps behind.
Total number of steps to the hill top is 10 less than 4 times what Meena has reached.
Total number of steps using “s” = (4 x s) – 10
Beena position = s + 8
Meena position = s – 7
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Solution :
A bus travels at “v”km per hour.
The bus has travelled 5 hours, Beespur is still 20 km away.
The distance from Daspur to Beespur =
Expression of distance from Daspur to Beespur = (5v + 20) km
(2) Change the following statements using expressions into statements in ordinary language.
(a) A notebook costs ₹p. A book costs ₹3 p.
(b) Tony puts q marbles on the table. He has 8 q marbles in his box.
(c) Our class has n students. The school has 20 n students.
(d) Jaggu is z years old. His uncle is 4 z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution :
(a) A book costs three times the cost of a notebook.
(b) Tony’s box contains 8 times the marbles on the table.
(c) Total number of students in the school is 20 times that of our class.
(d) Jaggu’s uncle is 4 times older than Jaggu and Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.
Algebra Exercise 11.5 Solutions :
(1) State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 7
(b) (t – 7) > 5
(c) 4/2 = 2
(d) (7 × 3) – 13 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
Solution :
(a) 17 = x + 7
= x is the variable
(b) (t – 7) > 5
= t is the variable
(e) 5 × 4 – 8 = 2x
= x is the variable
(f) x – 2 = 0
= x is the variable
(g) 2m < 30
= m is the variable
(h) 2n + 1 = 11
(2) Complete the entries in the third column of the table.
| Sl.No | Equation | Value of variable | Equation satisfied Yes/ No |
| (a) | 10y = 80 | y = 10 | |
| (b) | 10y = 80 | y = 8 | |
| (c) | 10y = 80 | y = 5 | |
| (d) | 4l = 20 | l = 20 | |
| (e) | 4l = 20 | l = 80 | |
| (f) | 4l = 20 | l = 5 | |
| (g) | b + 5 = 9 | b = 5 | |
| (h) | b + 5 = 9 | b = 9 | |
| (i) | b + 5 = 9 | b = 4 | |
| (j) | h – 8 = 5 | h = 13 | |
| (k) | h – 8 = 5 | h = 8 | |
| (l) | h – 8 = 5 | h = 0 | |
| (m) | p +3 = 1 | p = 3 | |
| (n) | p +3 = 1 | p = 1 | |
| (o) | p +3 = 1 | p = 0 | |
| (p) | p +3 = 1 | p = -1 | |
| (q) | p +3 = 1 | p = -2 |
Solution :
| S.NO | Equation | Value of variable | Equation satisfied Yes/ No |
| (a) | 10y = 80 | y = 10 | No |
| (b) | 10y = 80 | y = 8 | Yes |
| (c) | 10y = 80 | y = 5 | No |
| (d) | 4l = 20 | l = 20 | No |
| (e) | 4l = 20 | l = 80 | No |
| (f) | 4l = 20 | l = 5 | Yes |
| (g) | b + 5 = 9 | b = 5 | No |
| (h) | b + 5 = 9 | b = 9 | No |
| (i) | b + 5 = 9 | b = 4 | Yes |
| (j) | h – 8 = 5 | h = 13 | Yes |
| (k) | h – 8 = 5 | h = 8 | No |
| (l) | h – 8 = 5 | h = 0 | No |
| (m) | p +3 = 1 | p = 3 | No |
| (n) | p +3 = 1 | p = 1 | No |
| (o) | p +3 = 1 | p = 0 | No |
| (p) | p +3 = 1 | p = -1 | No |
| (q) | p +3 = 1 | p = -2 | Yes |
(4) – (a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16.
| M | 1 2 3 4 5 6 7 8 9 10 _ _ _ _ |
| M + 10 | _ _ _ _ _ _ _ _ _ _ |
Solution :
m 12 3 4 5 67
m + 10, 11 , 12 , 13 , 14 , 15 , 16 , 17
Equation m + 10 = 16.
We use m = 6 which satisfy the Equation m + 10 = 16.
6 + 10 = 16
(b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35.
| T | 3 4 5 6 7 8 9 10 11 _ _ _ _ _ |
| 5t | _ _ _ _ _ _ _ _ _ _ |
Solution :
t345 6 7 8 9
5t , 15 , 20 , 25 , 30 , 35 , 40 , 45
5t = 35
We use t = 7 which satisfy the Equation 5t = 35
5 x 7 = 35
(c) Complete the table and find the solution of the equation z/3 = 4 using the table.
| Z | 8 9 10 11 12 13 14 15 16 _ _ _ |
| z/3 | 2 ~2/3 3 3~1/3 _ _ _ _ _ _ _ |
Solution :
Equation z/3 =4
We use z = 12 which satisfy the Equation z/3 =4
12 /3 = 4
(d) Complete the table and find the solution to the equation m – 7 = 3.
| M | 5 6 7 8 9 10 11 12 13 _ _ |
| M -7 | 2 ~2/3 3 3~1/3 _ _ _ _ _ _ _ |
Solution :
m56 7 8 9 10 11
m – 7 , -2 , -1 , 0 , 1 , 2 , 3 , 4
Equation m – 7 = 3
We use m = 10 which satisfy the Equation m – 7 = 3
m – 7 = 3 = 10 – 3 = 7
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