NCERT Solutions for Class 6 Maths Chapter 11 Algebra

NCERT Solutions for Class 6 Maths Chapter 11 Algebra, are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 11 Algebra.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 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.

(a) A pattern of letter T as

(b) A pattern of letter Z as 

(c) A pattern of letter U as 

(d) A pattern of letter V as 

(e) A pattern of letter E as 

(f) A pattern of letter S as 

(g) A pattern of letter A as 

Solution.

(a) Number of matchsticks required = 2n

(b) Number of matchsticks required = 3n

(c) Number of matchsticks required = 3n

(d) Number of matchsticks required = 2n

(e) Number of matchsticks required = 5n

(f) Number of matchsticks required = 5n

(g) Number of matchsticks required = 6

Question 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?

Solution.

These letters are T and V. This happens since the number>of matchsticks require,d in each of them is 2.

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 = 5n.

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 b for the number of boxes.)

Solution.

Total number of mangoes = 50b.

Question 5.

The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use for the number of students.)

Solution.

Number of pencils needed = 5s

Question 6.

A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in tertns of its flying time in minutes? (Use t for flying time in minutes.)

Solution.

Yes! / kilometers

The bird flies in one minute = 1 kilometer

The bird flies in / minutes = 1 x t kilometers

= kilometers

Question 7.

Radha is drawing a dot Rangoli (a beautified 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.

∵ Number of dots in 1 row = 9

∴ Number of dots in r rows = 9 x r=9r

Number of dots in 8 rows = 9 x 8 = 72

Number of dots in 10 row = 9 x 10 = 90

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 x years.

Solution.

Yes! we can write Leela’s age in terms of Radha’s age.

Age of Radha = x years

∵ Leela is 4 years younger than Radha.

∴ Age of Leela = (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 away to guests and family members = l

Number of laddus remained = 5

∴ Number of laddus she made = 1 + 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 is taken to be x, what is the number of oranges in the larger box?

Solution.

Let the number of oranges in a smaller box box

∴ Number of oranges in two smaller boxes = 2x

Number of oranges remained outside = 10

∴ Number of oranges in the larger box = 2x+ 10

Question 11.

(a) Look at the following matchstick pattern of squares (figure). The squares are not separate. Two neighboring 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.)



(b) Figure gives a matchstick pattern of triangles. Av in Exercise 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles.





Solution.

(a)

 Rule: Number of matchsticks required = 3x + I

where x is the number of squares.



(b)



Rule: Number of matchsticks required = 2x + 1,

where x is the number of triangles.

NCERT Solutions for Class 6 Maths Chapter 11 Alge
bra Ex 11.2

Question 1.

The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.

Solution.

Perimeter (p) of the equilateral triangle with side l = Sum of the lengths of sides of the equilateral triangle = 1 + 1 + 1 = 31.

Question 2.

The side of a regular hexagon (Fig.) is denoted by l. Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides equal in length.)



Solution.

Perimeter (p) of the regular hexagon with side

= Sum of the lengths of all sides of the regular hexagon

=l+l+l+l+l+l

= 6 l.

Question 3.

A cube is a three-dimensional figure as shown in Fig. 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.

Total length (L) of the edges of a cube

= Sum of the lengths of all (12) edges of the cube

= l+l+l+l+l+l+l+l+l+l+l+l

=12 l

Question 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.) AB is a diameter of the circle; C is its center.) Express the diameter of the circle (d) in terms of its radius (r).



Solution.

AB = AC + CB

⇒ d=r+r ⇒ d=2r

Question 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 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 the 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.

Solution.

Let a, b and c be three variables, each of which can take any numerical value. Then.

(a + b) + c = a + (b + c)

This property is known as the associativity of the addition of numbers.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.3

Question 1.

Make up as many expressions with numbers (no variables) as you conform with 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 + (8­7), 5 – (8-7), 5 x 8 + 7; make the other expressions.)

Solution.

The possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 + 8) + 7, 5 + (8 + 7),

5 x 8 + 7, 5 x 7 + 8, 5 x 8-7, 5 x 7-8,

5 x (8 – 7), 5 x (8 + 7), 8 x (7 – 5), 8 x (7 + 5) etc.

Question 2.

Which out of the following are expressions with numbers only ?

(a) y + 3

(b) 7 x 20- 8z

(c) 5(21 -7)+ 7 x 2

(d) 5

(e) 3x

(f) 5 – 5n

(g) 7 x 20 – 5 x 10 – 45 + p.

Solution.

(c), (d).

Question 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, (cfrac { y }{ 17 } ), 5z

(c) 2y + 17, 2y – 17

(d) 7m, -7m + 3, -7m- 3.

Solution.

(a) Addition, subtraction, addition, subtraction

(b) Multiplication, division, multiplication.

(c) Multiplication and addition, multiplication and subtraction.

(d) Multiplication, multiplication and addition,multiplication and subtraction.

Question 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) p-1

(c) 7p

(d) (frac { p }{ 7 } )

(e) -m-1

(f) -5p

(g) – (frac { p }{ 5 } )

(h) – 5p.

Question 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 v 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

(i) y is multiplied by 5 and the result is subtracted from 16

(j) y is multiplied by-5 and the result is added to 16.

Solution.

(a) 2m +11

(b) 2m – 11

(c) 5y + 3

(d) 5v – 3

(e)-8v

(f)– 8y + 5

(g) 16 – 5y

(h) – 5y + 16.

Question 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.

Solution.

(a) t + 4, t – 4,4t, (frac { t }{ 4 } ) , (frac { 4 }{ t } )  , 4 -1,4 +1 4 t

(b) 2y + 7, 2y – 7, 7y + 2

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.4

Question 1.

Answer the following :

(a) Take Sarita’s present age to he 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’sfather’s age is 5years 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 3 times the breadth of the hall. What is the length, if the breadth is b meters?

(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.

(d) Meena, Beena, and Leena are climbing the steps to the hilltop. Meena is at step s, Beena is 8 steps ahead and Leena7 steps behind
. Where are Beena and Leena? 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.

(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.



Solution.

(a) (y + 5) years

(ii) (y – 3) years

(iii) 6y years

(iv) (6y – 2) years

(v) (3y + 5) years

(b) Length (l) = (3b – 4) meters

(c) Length of the box = 5h cm

Breadth of the box = (5h – 10) cm

(d) Beena is at step (s + 8)

Leena is at step (s – 7)

Total number of steps = 4s – 10.

(e) Speed of the bus = v km/hr

Distance travelled in 5 hours = 5v km.

∴Total distance = (5v + 20) km

Question 2.

Change the following statements using expressions into statements in the 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 Rs p. A book costs Rs 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 20n 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.

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) The total number of students in the school in 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 in an arrangement is 5 times the number of rows.

Question 3.

(a) Given Munnu ’s age to be x years, can you guess what (x – 2) may show?

(Hint: Think of Munnu’s younger brother.) 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(frac { 1 }{ 2 } ) ,y-2(frac { 1 }{ 2 } ).

(c) Given n students in the class like football, what may 2n show? What may — show?(Hint: Think of games other than football).

Solution.

(a) (x – 2) may show the age is Munnu’s younger brother. (a + 4) may show the age of Munnu’s elder brother. (3a+ 7) may show the age of Munnu’s grand mother.

(b) y + l indicates her age 7 years from now.

y – 3 indicates her age 3 years back.

y+ 4(frac { 1 }{ 2 } ) indicates her age 4(frac { 1 }{ 2 } ) years from now

y-2(frac { 1 }{ 2 } ) indicates her age 2(frac { 1 }{ 2 } ) years back.

(c) 2n may show the number of students who like cricket.

(frac { n }{ 2 } )may show the number of students who like hockey.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 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) (frac { 4 }{ 2 } ) =2

(d) (7 x 3)-19 = 8

(e) 5 x 4 – 8 =2x

(f) x – 2 =0

(g) 2m<30

(h) 2n+1=11

(i) 7=(11 x 5) -(12x 4)

(j) 7=(11×2)+p

(k) 20=5y

(l) latex]cfrac { 3q }{ 2 } [/latex]<5

(m) z+12>24

(n) 20-(10-5)=3×5

(o) 7-x =5

Solution.



Question 2.

Complete the entries in the third column of the table.





Solution.

(a) Yes

(b) Yes

(c) No

(d) No

(e) No

(f) Yes

(g) No

(h) No

(i)  Yes

(j) Yes

(k) No

(l) No

(m) No

(n) No

(o) No

(p) No

(q) Yes

Question 3.

Pick out the solution from the values given bracket next to each equation. Show that the values do not satisfy the equation.

(a) 5m – 60 (10, 5, 12, 15)

(b) n + 12 = 20 (12, 8, 20, 0)

(c) p-5 = 5 (0, 10, 5,-5)

(d) (cfrac { q }{ 2 } ) = 7 (7, 2, 10, 14)

(e) r-4- 0 (4, – 4. 8, 0)

(f) x + 4 – . y (-2, 0, 2. 4)

Solution.

(a) Solution is 12

For m = 10

L.H.S. = 5 x 10 = 50 which is not equal to R.H.S.

For m = 5

L.H.S. = 5 x 5 = 25 which is not equal to R.H.S.

For m = 15

L.H.S. = 5 x 15 = 75 which is not equal to R.H.S.



(b) Solution is 8

For n = 12

L.H.S. = 12+12

= 24 which is not equal to R.H.S.

For n = 20

L.H.S. = 20+12

= 32 which is not equal to R.H.S.

For n =0

L.H.S. = 0+12

= 12 which is not equal to R.H.S.

(c) Solution is 10

For p = 0

L.H.S. = 0 – 5 = – 5 which is not equal to R.H.S.

For p = 5

L.H.S. = 5-5 = 0 which is not equal to R.H.S.

For p = – 5

L.H.S. = – 5 – 5 = – 10 which is not equal to R.H.S.

(d) Solution is 14

For g = 7

L.H.S. = (frac { q }{ 2 } ) which is not equal to R.H.S.

For q = 2

L.H.S. = (frac { 2 }{ 2 } ) = 1 which is not equal to R.H.S.

For q = 10

L.H.S. = (frac { 10 }{ 2 } ) = 5 which is not equal to R.H.S.

(e) Solution is 4

For r = – 4

L.H.S. = -4-4

= – 8 which is not equal to R.H.S.

For r = 8

L.H.S. = 8-4 = 4 which is not equal to R.H.S.

For r = 0

L.H.S. = 0 – 4 = – 4 which is not equal to R.H.S.

(f) Solution is – 2

For x = 0

L.H.S. = 0 + 4 = 4 which is not equal to R.H.S. For x = 2

L.H.S. = 2 + 4 = 6 which is not equal to R.H.S.

For x : = 4

L.H.S. = 4 + 4 = 8 which is not equal to R.H.S.

Question 4.

(a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16



(b) Complete the table and by inspection of the table find the solution to the equation 5t = 35



(c) Complete the table and find using the table the solution of the equation z/3 = 4



(d) Complete the table and find the solution to the equation m-7=3



Solution.

Question 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!

(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!

Solution.

(i) Let the required number be x.

Number of corners of the square = 4

Number obtained on counting every comer thrice = 4 x 3=12

According to the question, x + 12 = 34 ⇒ x = 34 -12 = 22

Hence, I am 22.

(ii) Let the required number be x.

Number of players in a cricket team = 11

According to the question.

x – 6 = 11  ⇒ x =11+6=17

Hence, lam 17.

(iii) Let the required number be v.

Number of days of a week = 7

According to the question.

x + 7 = 23    ⇒ x = 23-7 =16

Hence, lam 16.

(iv) Let the required number be x.

According to the question.

22 – x = x ⇒ x + x = 22

2x = 22    ⇒ x = (cfrac { 10 }{ 2 } ) = 11

Hence, lam 11.

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