## NCERT Solutions for Class 6 Maths Chapter - 11: Algebra## Exercise 11.1
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter T requires two numbers of matchsticks. For n =1 Number of matchsticks required to make the pattern = 2 For n = 2, Number of matchsticks required to make the pattern = 4 Similarly, the following numbers of matchsticks are required for different number of n. It is shown in the below table:
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter Z requires three numbers of matchsticks. For n = 1 Number of matchsticks required to make the pattern = 3 For n = 2, Number of matchsticks required to make the pattern = 6 Similarly, the following numbers of matchsticks are required for different number of n. It is shown in the below table:
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter U requires three numbers of matchsticks. For n = 1 Number of matchsticks required to make the pattern = 3 For n = 2, Number of matchsticks required to make the pattern = 6 Similarly, the following numbers of matchsticks are required for different number of n. It is shown in the below table:
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter V requires two numbers of matchsticks. For n =1 Number of matchsticks required to make the pattern = 2 For n = 2, Number of matchsticks required to make the pattern = 4
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter E requires five numbers of matchsticks. For n =1 Number of matchsticks required to make the pattern = 5 For n = 2, Number of matchsticks required to make the pattern = 10
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter S requires five numbers of matchsticks. For n =1 Number of matchsticks required to make the pattern = 5 For n = 2, Number of matchsticks required to make the pattern = 10
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n Letter A requires six numbers of matchsticks. For n =1 Number of matchsticks required to make the pattern = 6 For n = 2, Number of matchsticks required to make the pattern = 12
Thus,
Similarly, letter T and V from the options (a) and (d) also requires two numbers of matchsticks. It is shown in the below table:
Thus,
Let the number of rows be n. For n = 1, Number of cadets = 5 For n = 2, Number of cadets = 10 Similarly, for different values of n, the numbers of cadets are shown in the below table:
Let the number of boxes be b. For b = 1, Number of mangoes = 50 For b = 2, Number of mangoes = 100 Similarly, for different values of b, the numbers of mangoes are shown in the below table:
The pencils distributed per student = 5 For s = 1, Number of pencils distributed = 5 For s = 2, Number of pencils distributed = 10 Similarly, for different values of s, the numbers of pencils distributed are shown in the below table:
Distance travelled by bird in 1 minute = 1 kilometer For t = 1, Distance = 1 km For t = 2, Distance = 2 km
Let the number of rows be r. For r = 1, Number of dots = 9 For r = 2, Number of cadets = 18 Similarly, for different values of r, the numbers of dots are shown in the below table:
If there are 8 rows, Number of dots = 9r Number of dots = 9 � 8
Thus, there are 72 dots in 8 rows. If there are 10 rows, Number of dots = 9r Number of dots = 9 � 10
Thus, there are 90 dots in 10 rows.
Leela is gour year younger than Radha. Thus, leela's age = (x - 4) years
Total laddus made = Laddus given + 5 Total laddus made = l + 5
Number of oranges in larger box = 2 � Number of oranges in smaller box + 10 Number of oranges in larger box = 2X + 10
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n n = Number of squares The above pattern is drawn using different number of matchsticks. For n = 1 Number of matchsticks required to make the pattern = 4 For n = 2, Number of matchsticks required to make the pattern = 7 For n = 3 Number of matchsticks required to make the pattern = 10 Thus, the pattern forms triplet the number of the squares and one extra matchstick.
Thus,
Where, n = 1, 2, 3, 4, 5, 6, 7, 8, 9 ... n n = Number of triangles The above pattern is drawn using different number of matchsticks. For n = 1 Number of matchsticks required to make the pattern = 3 For n = 2, Number of matchsticks required to make the pattern = 5 For n = 3 Number of matchsticks required to make the pattern = 7 Thus, the pattern forms double the number of the triangles and one extra matchstick.
Thus,
## Exercise 11.2
Perimeter of a triangle = Sum of all its three sides An equilateral triangle has all its three sides equal = l Perimeter of an equilateral triangle = l + l + l Perimeter of an equilateral triangle = 3l
Perimeter of a hexagon = Sum of all its six sides The hexagon has all its six sides equal = l Perimeter of the hexagon = l + l + l + l + l + l Perimeter of the hexagon = 6l
A cube has twelve edges or twelve sides Perimeter of a cube = Sum of all its twelve sides The cube has all its twelve sides equal = l Perimeter of the cube = l + l + l + l + l + l + l + l + l + l + l + l Perimeter of the cube = 12l
CA = CP It is because both are the radius of the circle. The centre point equally divides the diameter into two equal parts. Thus, diameter is equal to twice the radius of the circle. d = 2 r
(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.
## Exercise 11.3
The possible expressions using the three numbers, 5, 7, and 8 are as follows: - 5 + (8 - 7)
- 5 + (8 + 7)
- 5 � (8 - 7)
- 5 - (8 - 7)
- 5 - (8 + 7)
- 5 � (8 + 7)
- 8 - (5 + 7)
- 8 - (5 - 7)
- 8 � (5 + 7)
- 8 + (5 + 7)
- 8 + (5 - 7)
- 8 � (5 - 7)
- 7 - (8 - 5)
- 7 - (8 + 5)
- 7 + (8 + 5)
- 7 + (8 - 5)
- 7 � (8 - 5)
- 7 � (8 + 5)
Let's discuss each expression individually.
It has a variable '
It has a variable '
It has no variable.
It has no variable.
It has a variable '
It has a variable '
It has a variable ' Thus, option
(a) z +1, z - 1, y + 17, y - 17
Z is increased by 1.
Z is subtracted by 1
Y is increased by 17.
Y is subtracted by 17.
Y is multiplied by 17.
Y is divided by 17.
Y is multiplied by 5.
Y is multiplied by 2 and then increased by 17.
Y is multiplied by 2 and then decreased by 17.
M is multiplied by 7.
Y is multiplied by -7 and then increased by 3.
Y is multiplied by -7 and then decreased by 3.
(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
(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.
- t - 4
- t + 4
- 4 - t
- 4 + t
- t/4
- 4t
- 4/t
- 2y - 7
- 2y + 7
- y/2 - 7
- y - 7/2
- 7y - 2
- 7y + 2
- y/7 - 2
## Exercise 11.4
(a) Take Sarita's present age to be y years
Age of Sarita 5 years from now = current age + 5 years = y + 5
Age of Sarita 3 years back = current age - 3 years = y - 3
Age of her grandfather = 6 time her age (Times means multiplication) Age of her grandfather = 6y
Age of her grandfather = 6 time her age (Times means multiplication) Age of her grandfather = 6y Age of her grandmother = 2 years younger than grandfather Age of her grandmother = Age of grandfather - 2 = 6y - 2
Age of her father = 3 times Sarita's age + 5 years Age of her father = 3y + 5
Length of the rectangular hall = 4 meters less than 3 times the breadth Length of the rectangular hall = 3 times he breadth - 4 = (3b - 4) meters Thus, the length of the rectangular hall is
Breadth = (5h - 10) cm Height = h cm
Length of the box = 5 times the height Length of the box = 5h cm Breadth of the box = 10 cm less than the length = Length of the box - 10 = 5h - 10 Thus, the dimensions of the box are: Length = 5h cm Breadth = (5h - 10) cm Height = h cm
Leena = s- 7 Total number of steps to the hill top = 4s - 10
Steps of Beena = 8 steps ahead of Meena Steps of Beena = steps of Meena + 8 = s + 8 Steps of Leena = 8 steps behind of Meena Steps of Leena = steps of Meena - 7 = s - 7 Total number of steps to the hill top = 10 less than 4 times of Meena = 4 time Meena steps - 10 = 4s - 10
The distance travelled by the bus in 5 hours = 5v km Total distance between Daspur to Beespur = distance travelled in 5 hours + 20 km Total distance between Daspur to Beespur = (5v + 20) km
Cost of a notebook = p Thus, it signifies that cost of book is 3 times p, i.e., 3 times the cost of a notebook.
Marbles in the Tony's box = 8q Thus, it signifies that ton's box has the marbles equal to 8 times the marbles on the table.
Total number of students in the school = 20n Thus, it signifies that school has 20 times the students in the class.
Age of his uncle = 4z Thus, the age of uncle is 4 times the age of Jaggu. Age of her aunt = 4z - 3 Age of her aunt = Uncle's age - 3 Thus, the age of Jaggu's aunt is 3 times less than the age of his uncle.
Total number of dots in a row = 5 Total number of dots in an arrangement = Total number of rows x 5 Thus, thetotal number of dots in an arrangement is five times the number of rows. 3.
X - 2 = 2 years less than the age of Munnu X - 2 =Munnu's younger brother/sister X + 4 = 4 years older than the age of Munnu X + 4 = Elder sister/brother 3X + 7 = 7 added to 3 times the age of Munnu Thus, 3X + 7 can be the age of Munnu's mother/father.
Y + 7 = Seven years after the current age
Y - 3 = Three years before the current age
y + 4 ½ = Four and a half years after the current age
y - 2 ½ =Two and a half years before the current age
2n may be any other game like basketball, cricket, etc. Total number of students who like basketball = 2n Thus, total numbers of students who like other game (for example, basketball) are two times the number of students who like football. ## Exercise 11.5
Yes, it is an equation with a variable x. Variable:
No, it is an equation with any variable. An equation always has an equal sign (=) sign between them. Variable:
No, it is not an equation with any variable. Variable:
No, it is not an equation with any variable. Variable:
Yes, it is an equation with a variable x. Variable:
Yes, it is an equation with a variable x. Variable:
No, it is an equation with any variable. An equation always has an equal sign (=) sign between them. Variable:
Yes, it is an equation with a variable n. Variable:
No, it is not an equation with any variable. Variable:
Yes, it is an equation with a variable p. Variable:
Yes, it is an equation with a variable y. Variable:
No, it is an equation with any variable. An equation always has an equal sign (=) sign between them. Variable:
No, it is an equation with any variable. An equation always has an equal sign (=) sign between them. Variable:
No, it is not an equation with any variable. Variable:
Yes, it is an equation with a variable x. Variable:
Answer:
For y = 10, LHS = 10(10) LHS = 100 100 is not equal to 80 Thus, the above equation is not satisfied
10y = 80 Y = 80/10 Y = 8 Thus, the value of 'y' for which the equation is satisfied is 8.
Y = 80/10 Y = 8 Thus, the value of 'y' for which the equation is satisfied is 8.
LHS = 10(5) LHS = 50 50 is not equal to 80 Thus, the above equation is not satisfied
10y = 80 Y = 80/10 Y = 8 Thus, the value of 'y' for which the equation is satisfied is 8.
LHS = 4(20) LHS = 80 80 is not equal to 20 Thus, the above equation is not satisfied
4l = 20 L = 20/4 L = 5 Thus, the value of 'l' for which the equation is satisfied is 5.
LHS = 4(80) LHS = 320 320 is not equal to 20 Thus, the above equation is not satisfied
4l = 20 L = 20/4 L = 5 Thus, the value of 'l' for which the equation is satisfied is 5.
L = 20/4 L = 5 Thus, the value of 'l' for which the equation is satisfied is 5.
LHS = 5 + 5 LHS = 10 10 is not equal to 9 Thus, the above equation is not satisfied
b + 5 = 9 b = 9 - 5 b = 4 Thus, the value of 'b' for which the equation is satisfied is 4.
LHS = 9 + 5 LHS = 14 14 is not equal to 9 Thus, the above equation is not satisfied
b + 5 = 9 b = 9 - 5 b = 4 Thus, the value of 'b' for which the equation is satisfied is 4.
b = 9 - 5 b = 4 Thus, the value of 'b' for which the equation is satisfied is 4.
h = 5 + 8 h = 13 Thus, the value of 'h' for which the equation is satisfied is 13.
LHS = 8 - 8 LHS = 0 0 is not equal to 5 Thus, the above equation is not satisfied
h - 8 = 5 h = 5 + 8 h = 13 Thus, the value of 'h' for which the equation is satisfied is 13.
LHS = 0 - 8 LHS = -8 -8 is not equal to 5 Thus, the above equation is not satisfied
h - 8 = 5 h = 5 + 8 h = 13 Thus, the value of 'h' for which the equation is satisfied is 13.
LHS = 3 + 3 LHS = 6 6 is not equal to 1 Thus, the above equation is not satisfied
p + 3 = 1 p = 1 - 3 p = -2 Thus, the value of 'p' for which the equation is satisfied is -2.
LHS = 1 + 3 LHS = 4 4 is not equal to 1 Thus, the above equation is not satisfied
p + 3 = 1 p = 1 - 3 p = -2 Thus, the value of 'p' for which the equation is satisfied is -2.
LHS = 0 + 3 LHS = 3 3 is not equal to 1 Thus, the above equation is not satisfied
p + 3 = 1 p = 1 - 3 p = -2 Thus, the value of 'p' for which the equation is satisfied is -2.
LHS = -1 + 3 LHS = 2 2 is not equal to 1 Thus, the above equation is not satisfied
p + 3 = 1 p = 1 - 3 p = -2 Thus, the value of 'p' for which the equation is satisfied is -2.
p = 1 - 3 p = -2 Thus, the value of 'p' for which the equation is satisfied is -2.
LHS = 5 (10) LHS = 50 50 is not equal to 60 Hence, the value for m = 10 is not satisfied.
LHS = 5 (5) LHS = 25 25 is not equal to 60 Hence, the value for m = 5 is not satisfied.
LHS = 5 (12) LHS = 60 60 is equal to 60 Hence, the value for
LHS = 5 (15) LHS = 75 75 is not equal to 60 Hence, the value for m = 15 is not satisfied.
LHS = 12 + 12 LHS = 24 24 is not equal to 20 Hence, the value for n = 12 is not satisfied.
LHS = 8 + 12 LHS = 20 20 is equal to 20 Hence, the value for
LHS = 20 + 12 LHS = 32 32 is not equal to 20 Hence, the value for n = 20 is not satisfied.
LHS = 0 + 12 LHS = 12 12 is not equal to 20 Hence, the value for n = 0 is not satisfied.
LHS = 0 - 5 LHS = -5 -5 is not equal to 5 Hence, the value for p = 0 is not satisfied.
LHS = 10 - 5 LHS = 5 5 is equal to 5 Hence, the value for
LHS = 5 - 5 LHS = 0 0 is not equal to 5 Hence, the value for p = 5 is not satisfied.
LHS = -5 - 5 LHS = -10 -10 is not equal to 5 Hence, the value for p = -5 is not satisfied.
LHS = 7/2 LHS = 3.5 3.5 is not equal to 7 Hence, the value for q = 7 is not satisfied.
LHS = 2/2 LHS = 1 1 is not equal to 7 Hence, the value for q = 2 is not satisfied.
LHS = 10/2 LHS = 5 5 is not equal to 7 Hence, the value for q = 10 is not satisfied.
LHS = 14/2 LHS = 7 7 is equal to 7 Hence, the value for
LHS = 4 - 4 LHS = 0 0 is equal to 0 Hence, the value for
LHS = -4 - 4 LHS = -8 -8 is not equal to 0 Hence, the value for r = -4 is not satisfied.
LHS = 8 - 4 LHS = 4 4 is not equal to 0 Hence, the value for r = 8 is not satisfied.
LHS = 0 - 4 LHS = -4 -4 is not equal to 0 Hence, the value for r = 0 is not satisfied.
LHS = -2 + 4 LHS = 2 2 is equal to 2 Hence, the value for
LHS = 0 + 4 LHS = 4 4 is not equal to 2 Hence, the value for x = 0 is not satisfied.
LHS = 2 + 4 LHS = 6 6 is not equal to 2 Hence, the value for x = 2 is not satisfied.
LHS = 4 + 4 LHS = 8 8 is not equal to 2 Hence, the value for x = 4 is not satisfied. 4.
m + 10 = 16 Let's find the solution for different values of m.
At
5t = 35 Let's find the solution for different values of t.
At
z/3 = 4 Let's find the solution for different values of z.
At
m - 7 = 3 Let's find the solution for different values of m.
At
Who am I?
Let the riddles be x. Thrice every corner = 3 (4) = 12 Add the count to get thirty four. Thus, the equation can be written as: 12 + x = 34 X =34 - 12
Count from me signifies addition Let the riddles be X. For every day of a week, the count can be written as X + 7. Thus, the equation can be written as: X + 7 = 23 X = 23 - 7 X = 16
Take from a special number signifies subtraction. Six taken from a number = X - 6 Total number of players in a cricket team = 11 Thus, the equation can be written as: X - 6 = 11 X = 11 + 6 X = 17
You will get me back means twice. It represents a double sided count. Thus, the equation can be written as: 2X = 22 X = 22/2 X = 11 Next TopicNCERT Solutions Class 10 Maths |