NCERT Solutions For Class 6 Maths Chapter 11 Algebra
Here at MentorAtHome, We provide NCERT Solutions For Class 6 Maths Chapter 11 Algebra. NCERT solutions help students to improve their performance, and attain better results. In NCERT Solutions Class 6 Maths Chapter 11 there is 5 exercise.
NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercises 11.1
Class 6th Maths Chapter 11 Solutions ex 11.1
Q.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
Ans:-
(a)
∴ The rule is 2n where n is the number of Ts.
(b)
∴ The rule is 3n where n is the number of Zs.
(c)
∴ The rule is 3n where n is the number of Us.
(d)
∴ The rule is 2n where n is the number of Vs.
(e)
∴ The rule is 5n where n is the number of Es.
(f)
∴ The rule is 5n where n is the number of Ss.
(g)
∴ The rule is 6n where n is the number of As
Q.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?
Ans:-
The rule for the following letters
For L it is 2n
For T it is 2n
For V it is 2n
For C it is 3n
For F it is 3n
For U it is 3n
We observe that the rule is the same for 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.
Q.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)
Ans:-
Number of cadets in a row = 5
Let the number of rows = n
Total number of cadets = number of cadets in a row × number of rows
5 x n = 5n
Q.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.)
Ans:-
Let the number of boxes = b
Number of mangoes in a box = 50
Total number of mangoes = number of mangoes in a box × number of boxes
50 x b = 50b
Q.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.)
Ans:-
Let the number of students = s
Number of pencils distributed per student = 5
Total number of pencils = number of pencils given to each student × number of students
5 x s = 5s
Q.6. A bird flies 1 kilometer in one minute. Can you express the distance covered by the birds in terms of their flying time in minutes? (Use t for flying time in minutes.)
Ans:-
Distance covered in one minute = 1 km.
Let the flying time = t
Distance covered in.t minutes = Distance covered in one minute × Flying time
= 1 × t = t km
Q.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?
Ans:-
Let the number of rows = r
Number of dots in a row drawn by Radha = 9
Total number of dots in r rows = Number of dots in a row × number of rows
9 x r = 9r
Number of dots in 8 rows = 8 × 9 = 72
Number of dots in 10 rows = 10 × 9 = 90
Q.8. Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write about Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Ans:-
Let Radha’s age = x yeas.
Given that Leela’s age = 4 years younger than Radha
= x years – 4 years
= (x – 4) years
Q.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?
Ans:-
Given that the number of laddus gave = l
Number of laddus left = 5
Total number of laddus made by mother = l + 5
Q.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 is taken to be x, what is the number of oranges in the larger box?
Ans:-
Let the number of oranges in smaller box = x
Number of oranges in two small boxes = 2x
Number of oranges remained = 10
Number of oranges in a large box
= (number of oranges in two small boxes + number of oranges that remained)
= 2x + 10
Q.11. (a) Look at the following matchstick pattern of squares (Fig 11.6). The squares are not separate. Two neighboring 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) 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.
Ans:-
(a) Let n be the number of squares.
∴ Number of matchsticks required
3n + 1
For n = 1 is
3 x 1 + l = 3 + 1 = 4
For n = 2 is
3 x 2 + l = 6 + 1 = 7
For n = 3 is
3 x 3 + l = 9 + 1 = 10
For n = 4 is
3 x 4 + l = 12 + 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
2n + 1
For n = 1 is
2 x 1 + 1 = 2 + 1 = 3
For n = 2 is
2 x 2 + 1 4 + 1 = 5
For n = 3 is
2 x 3 + 1 6 + 1 = 7
For n = 4 is
2 x 4 + 1 8 + 1 = 9
∴ Rule is 2n + 1 where n represents the number of matchsticks.
NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercises 11.2
NCERT Class 6 Maths Chapter 11 Solutions ex 11.2
Q.1. 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
Perimeter = l + l + l = 3l
Q.2. The side of the regular hexagon (Fig 11.10) is denoted by l. Express the perimeter of the hexagon using l.
(Hint: A regular hexagon has all its six sides equal in length.)
Ans:-
each side of a hexagon = l
∴ Perimeter of the regular hexagon = l + l + l + l + l + l
= 6 x l = 6 l units
Q.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.
Ans:-
We know that a cube has 12 edges and 6 identical faces.
Length of an edge of the cube = l
Number of edges = 12
Total length of the edges = 12 x l = 12l
Q.4. The diameter of a circle is a line which joins two points on the circle and also passes through the center of the circle. (In the adjoining figure (Fig 11.2) AB is a diameter of a circle; C is its center.) Express the diameter of the circle (d) in terms of its radius (r).
Ans:-
Let diameter = d
radius = r
∴ d = 2 x radius = 2 x r = 2r
So, diameter = 2r.
Q.5. 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 41and then add 13 to it to get the total sum of 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)
Ans:-
For any three whole numbers a, b and c
(a + b) + c = a + (b + c)
NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercises 11.3
Ch 11 Class 6 Maths Solutions ex 11.3
Q.1. Makeup 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.
Ans:-
(i) 8 x (5 + 7)
(ii) 5 + (8 – 7)
(iii) 8 + (5 x 7)
(iv) 7 – (8 – 5)
(v) 7 x (8 + 5)
(vi) 5 x (8 + 7)
(vii) 8 + (5 + 7)
(viii) 7 + (8 – 5)
(ix) (5 x 7) – 8
(x) 7 + (8 x 5)
Q.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
Ans:-
(c) and (d) are the expressions with numbers only.
Q.3. Identify the operations (addition, subtraction, division, multiplication) informing the following expressions and tell how the expressions have been formed.
(a) z + 1, z – 1, y + 17, y – 17
(b) 17y, y / 17, 5z
(c) 2y + 17, 2y – 17
(d) 7m, -7m + 3, -7m – 3
Ans:-
(a) z + 1 = Addition = 1 is added to z
z – 1 = Subtraction = 1 is subtracted from z
y + 17 = Addition = 17 is added to y
y – 17 = Subtraction = 17 is subtracted from y
(b) 17y = Multiplication = y is multiplied by 17
y / 17 = Divisio = y is divided by 17
5z = Multiplication = z is multiplied by 5
(c) 2y + 17 = Multiplication and addition = y is multiplied by 2 and 17 is added to the result
2y – 17 = Multiplication and subtraction = y is multiplied by 2 and 17 is subtracted from the result
(d) 7m = multiplication = m is multiplied by 7
-7m + 3 = Multiplication, and addition = m is multiplied by -7, and 3 is added to the result
-7m – 3 = Multiplication, and subtraction = m is multiplied by -7, and 3 is subtracted from the result
Q.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
Ans:-
(a) (p + 7) 7 is added to p
(b) (p – 7) 7 subtracted from p
(c) (7p) p multiplied by 7
(d) (p / 7) p divided by 7
(e) (-m – 7) 7 subtracted from –m
(f) (-5p) –p multiplied by 5
(g) (–p / 5) –p divided by 5
(h) (-5p) p multiplied by -5
Q.5. Give expressions in the following cases.
(a) 11 added to 2m
(b) 11 subtracted from 2m
(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.
Ans:-
(a) 2m + 11
(b) 2m – 11
(e) 5y + 3
(d) 5y – 3
(e) -8y
(f) -8y+5
(g) 16 – 5y
(h) -5y + 16
Q.6. (a) Form expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(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:-
(a) the expressions using t and 4
(i) (t + 4)
(ii) (t – 4)
(iii) 4t
(iv) (t / 4)
(b) the expression using y, 2 and 7
(i) 2y + 7
(ii) 2y – 7
(iii) 7y + 2
(iv) 7y -2
NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercises 11.4
NCERT Class 6 Maths Chapter 11 solutions Ex 11.4
Q.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 two 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?
(b) The length of a rectangular hall is 4 meters less than three times the breadth of the hall. What is the length, if the breadth is b meters?
(c) A rectangular box has a height of 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 height.
(d) Meena, Beena, and Reena are climbing the steps to the hilltop. Meena is at step s, Beena is 8 steps ahead, and Leena is 7 steps behind. Where are Beena and Meena? The total number of steps to the hilltop is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has traveled 5 hours, Beespur is still 20 km away. What is the distance from Daspur to Beespur? Express it using v.
Ans:-
(a)
(i) Sarita’s present age = y
Sarita’s age aftyer 5 years = Sarita’s present age + 5
= (y + 5) years
(ii) Sarita’s age Before 3 years = Sarita’s present age – 3
= (y – 3) years
(iii) Grandfather’s age = 6 × Sarita’s present age
= (6 x y) = 6y years
(iv) Grandmother’s age = granfather’s present age – 2
= (6 x y + 2) = (6y -2) years
(v) Father’s age = 5 + 3 × Sarita’s present age
= (5 + 3 x y) = (5 + 3y) years
(b)
Let the length of the rectangular hall = L
L = 3 x b – 4
∴ length = (3b – 4) metre
(c) Height of the rectangular box = h
Length = 5 × Breadth
L = (5 x h)
Length = 5h cm
Breadth = 5 × length – 10
b = (5 x h – 10)
Breadth = (5h – 10) cm
(d) The step at meena = s
Beena step = (s + 8)
Leena stap = (s – 7)
Total steps = (4 × step at which Meena is) – 10)
=(4 x s – 10)
= (4s – 10)
(e) Speed = v km / hr
Distance travelled in 5 hours = 5 × v
= 5v km
Total distance travelled between Daspur to Beespur = (5v + 20) km
Q.2. Change the following statements using expressions into statements in ordinary language.
(For example, Given Salim scores r runs in a cricket match, Nalin scores (r + 15) runs. In ordinary
language – Nalin scores 15 runs more than Salim.)
(a) A notebook costs ₹ p. A book costs ₹ 3p
(b) Tony put 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 4z 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
Ans:-
(a) A book costs 3 times the cost of a notebook.
(b) Tony has 8 times the number of marbles put on the table by him.
(c) The school has 20 times the number of students in a class.
(d) Jaggu’s uncle’s age is 4 times his age and his aunt’s age is 3 years less than the age of his uncle.
(e) Number of dots in a row is 5 times the number of rows.
Q.3. (a) Given Munnu’s age to be x years, can you guess what (x – 2) may show?
Can you guess what (x + 4) may show? What (3x + 7) may show?
(b) Given Sara’s age today to be y years. Think of her age in the future or in the past.
What will the following expression indicate? Y + 7, y – 3, [y + (4 x 1/2)], [y – (2 x 1 / 2)]
(c) Given n students in the class like football, what may 2n shows? What may n / 2 show?
Ans:-
(a) Given that Mannu’s age = x years.
(x – 2) years may be the age of her younger brother or sister.
(x + 4) years show the age of her elder brother or sister.
(3x + 7) years may be the age of her father, mother, or uncle.
(b) Sara’s age represents by = y
Shows her future age. = y + 7
Shows her past age. = y – 3
[y + (4 x 1 / 2)] show her future age i.e., the age after z
four and half years.
[y – (2 x 1 / 2)] shows her past age i.e., the age before two and half years.
(c) Number of students who like football = n
twice the number of football players who may like to play cricket. = 2n
n / 2 = half of the number of football 2 players may like to play football.
NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercises 11.5
Ch 11 Class 6 Maths Solutions Ex 11.5
Q.1. State which of the following are equations (with a variable). Give a reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 17
(b) (t – 7) > 5
(c) 4 / 2 = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 × 2) + p
(k) 20 = 5y
(l) 3q/ 2 < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 × 5
(o) 7 – x = 5
Ans:-
(a) 17 = x + 17 is an equation with variable x
(b) (t – 7) > 5 is an inequality equation
(c) 4 / 2 = 2 is No, it’s a numerical equation
(d) (7 x 3) – 19 = 8 is No, it’s a numerical equation
(e) 5 x 4 – 8 = 2x is An equation with variable x
(f) x – 2 = 0 is An equation with variable x
(g) 2m < 30 is An inequality equation
(h) 2n + 1 = 11 is An equation with variable n
(i) 7 = (11 x 5) – (12 x 4) is No, it’s a numerical equation
(j) 7 = (11 x 2) + p is An equation with variable p
(k) 20 = 5y is An equation with variable y
(l) 3q /2 < 5 is An inequality equation
(m) z + 12 > 24 is An inequality equation
(n) 20 – (10 – 5) = 3 x 5 is No, it’s a numerical equation
(o) 7 – x = 5 is An equation with variable x
Q.2. Complete the entries in the third column of the table.
| S.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 |
| 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 |
Q.3. Pick out the solution from the values given in the bracket next to each equation.
Show that the other values do not satisfy the equation.
(a) 5m = 60 (10, 5, 12, 15)
(b) n + 12 (12, 8, 20, 0)
(c) p – 5 = 5 (0, 10, 5, – 5)
(d) q / 2 = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4, -4, 8, 0)
(f) x + 4 = 2 (-2, 0, 2, 4)
Ans:-
(a) 5m = 60
Put m = 10
5m = 5 × 10 = 50
m = 10 is a solution for this equation
Put m = 5
5m = 5 × 5 = 25
m = 5 is not a solution for this equation
Put m = 12
5m = 5 × 12 = 60
m = 12 is a solution for this equation
∴ Equation satisfied
Put m = 15
5m = 5 × 15 = 75
m = 15 is not a solution for this equation
(b) n + 12 = 20
Put n = 12
n + 12 = 12 + 12 = 24
n = 12 is not a solution for this equation
Put n = 8
n + 12 = 8 + 12 = 20
n = 8 is a solution for this equation
∴ Equation satisfied
Put n = 20
n + 12 = 20 + 12 = 32
n = 20 is not a solution for this equation
Put n = 0
n + 12 = 0 + 12 = 12
n = 0 is not a solution for this equation
(c) p – 5 = 5
Put p = 0
p – 5 = 0 – 5 = -5
p = 0 is not a solution for this equation
Put p = 10
p – 5 = 10 – 5 = 5
p = 10 is a solution for this equation
∴ Equation satisfied
Put p = 5
p – 5 = 5 – 5 = 0
p = 5 is not a solution for this equation
Put p = -5
p – 5 = -5 – 5 = -10
p = -5 is not a solution for this equation
(d) q / 2 = 7
Put q = 7
q / 2 = 7 / 2
q = 7 is not a solution for this equation
Put q = 2
q / 2 = 2 / 2 = 1
q = 2 is not a solution for this equation
Put q = 10
q / 2 = 10 / 2 = 5
q = 10 is not a solution for this equation
Put q = 14
q / 2 = 14 / 2 = 7
q = 14 is a solution for this equation
∴ Equation satisfied
(e) r – 4 = 0
Put r = 4
r – 4 = 4 – 4 = 0
r = 4 is a solution for this equation
∴ Equation satisfied
Put r = – 4
r – 4 = – 4 – 4 = – 8
r = – 4 is not a solution for this equation
Put r = 8
r – 4 = 8 – 4 = 4
r = 8 is not a solution for this equation
Put r = 0
r – 4 = 0 – 4 = – 4
r = 0 is not a solution for this equation
(f) x + 4 = 2
Put x = – 2
x + 4 = – 2 + 4 = 2
x = – 2 is a solution for this equation
∴ Equation satisfied
Put x = 0
x + 4 = 0 + 4 = 4
x = 0 is not a solution for this equation
Put x = 2
x + 4 = 2 + 4 = 6
x = 2 is not a solution for this equati
Put x = 4
x + 4 = 4 + 4 = 8
x = 4 is not a solution for this equati
Q.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 | — | — | — | — | — | — | — | — | — | — |
(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 |
| 5t | — | — | — | — | — | — | — | — |
(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 |  | 3 |  | — | — | — | — | — | — |
(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 | — | — | — | — | — | — | — | — | — |
Ans:-
(a) For m + 10 = 16 the table is represented as below
| m | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| m+10 | 1+10 = 11 | 2+10 = 12 | 3+10 = 13 | 4+10 = 14 | 5+10 = 15 | 6+10 = 16 | 7+10 = 17 | 8+10 = 18 | 9+10 = 19 | 10+10 = 20 |
So, m – 6 is the solution of the equation.
(b) For 5t = 35 the table is represented as below
| t | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5t | 3×5 = 15 | 4×5 = 20 | 5×5 = 25 | 6×5 = 30 | 7×5 = 35 | 8×5 = 40 | 9×5 = 45 | 10×5 = 50 |
So, t = 7 is the solution of the equation.
(c) For z / 3 = 4 the table is represented as below
| z | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |
| z/3 | 8/3 = | 9/3 = 3 | 10/3 = | 11/3 = | 12/3 = 4 | 13/3 = | 14/3 = | 15/3 = 5 | 16/3 = |
So, z = 12 is the solution of the equation.
(d) For m – 7 = 3 the table is represented as below
| m | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |
| m – 7 | 5-7 = -2 | 6-7 = -1 | 7-7 = 0 | 8-7 = 1 | 9-7 = 2 | 10-7 = 3 | 11-7 = 4 | 12-5 = 7 | 13-7 = 6 |
So, m = 10 is the solution of the equation.
Q.5. Solve the following riddles, you may yourself construct such riddles. Who am I?
(i) Go round a square
Counting every corner
Thrice and no more!
Add the count to me
To get exactly thirty-four!
(ii) For each day of the week
Make an account from me
If you make no mistake
You will get twenty-three!
(iii) I am a special number
Take away from me a six!
A whole cricket team
You will still be able to fix it!
(iv) Tell me who I am
I shall give a pretty clue!
You will get me back
If you take me out of twenty-two!
Ans:-
(i) There are 4 corners in a square.
Thrice the number of corners in the square = 3 × 4 = 12
x + 12 = 34
x = 34 – 12 = 22
22 + 12 = 34
So, I am 22.
(ii) Let I am ‘x’.
We know that there are 7 days a week.
x + 7 = 23
x = 23 – 7 = 16
16 + 7 = 23
∴ x = 16
Thus I am 16.
(iii) Let the special number be x and there are 11 players in cricket team.
∴ Special Number -6 = 11
∴ x – 6 = 11
x = 11 + 6 = 17
By inspection, we get
17 – 6 = 11
∴ x = 17
Thus I am 17.
(iv) Suppose I am ‘x’.
22 – x = x
22 = x + x
2x = 22
x =22/2 =11
∴ x = 11
Thus I am 11.
| Class 6 Science NCERT Notes | Class 6 Complete Study Material |
| NCERT Solutions for Class 6 Science | Class 6 Maths Chapter 6 Solutions |
NCERT Solutions For Class 6 Maths Chapter 11 algebra are based on NCERT Books. You can download ncert book for class 6 maths chapter 11. Ch 11 class 6 maths solutions are designed by our subject expert team.
What We Learn?
- We looked at patterns of making letters and other shapes using matchsticks. We learnt how to write the general relation between the number of matchsticks required for repeating a given shape. The number of times a given shape is repeated varies; it takes on values 1,2,3,… . It is a variable, denoted by some letter like n.A variable takes on different values, its value is not fixed. The length of a square can have any value. It is a variable. But the number of angles of a triangle has a fixed value. It is not a variable.
- We may use any letter n, l, m, p, x, y, z, etc. to show a variable.
- A variable allows us to express relations in any practical situation.
- Variables are numbers, although their value is not fixed. We can do the operations of addition, subtraction, multiplication and division on them just as in the case of fixed numbers. Using different operations we can form expressions with variables like x – 3, x3, 2n, 5m, 3p, 2y + 3, 3l – 5, etc.
- Variables allow us to express many common rules in both geometry and arithmetic in a general way. For example, the rule that the sum of two numbers remains the same if the order in which the numbers are taken is reversed can be expressed as a + b = b + a. Here, the variables a and b stand for any number, 1, 32, 1000 – 7,– 20, etc.
- An equation is a condition on a variable. It is expressed by saying that an expression with a variable is equal to a fixed number, e.g. x – 3 = 10.
- An equation has two sides, LHS and RHS, between them is the equal (=) sign.
- The LHS of an equation is equal to its RHS only for a definite value of the variable in the equation. We say that this definite value of the variable satisfies the equation. This value itself is called the solution of the equation.
- For getting the solution of an equation, one method is the trial and error method. In this method, we give some value to the variable and check whether it satisfies the equation. We go on giving this way different values to the variable` until we find the right value which satisfies the equation.
