NCERT Solutions for Class 6 Maths Exercise 11.5

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NCERT Solutions for Class 6 Maths Exercise 11.5

NCERT Solutions for Class 6 Maths Algebra

Class –VI Mathematics
(Ex. 11.5)
Question 1.State which of the following are equations (with a variable). Given reason for your answer. Identify the variable from the equations with a variable.

(a) {tex}17 = x + 7{/tex}

(b) {tex}\left( {t – 7} \right) > 5{/tex}

(c) {tex}\frac{4}{2} = 2{/tex}

(d) {tex}\left( {7 \times 3} \right) – 19 = 8{/tex}

(e) {tex}5 \times 4 – 8 = 2x{/tex}

(f) {tex}x – 2 = 0{/tex}

(g) {tex}2m < 30{/tex}

(h) {tex}2n + 1 = 11{/tex}

(i) {tex}7 = \left( {11 \times 5} \right) – \left( {12 \times 4} \right){/tex}

(j) {tex}7 = \left( {11 \times 2} \right) + p{/tex}

(k) {tex}20 = 5y{/tex}

(l) {tex}\frac{{3q}}{2} < 5{/tex}

(m) {tex}z + 12 > 24{/tex}

(n) {tex}20 – \left( {10 – 5} \right) = 3 \times 5{/tex}

(o) {tex}7 – x = 5{/tex}

Answer:

(a) It is an equation of variable as both the sides are equal. The variable is {tex}x.{/tex}

(b) It is not an equation as L.H.S. is greater than R.H.S.

(c) It is an equation with no variable. But it is a false equation.

(d) It is an equation with no variable. But it is a false equation.

(e) It is an equation of variable as both the sides are equal. The variable is {tex}x.{/tex}

(f) ) It is an equation of variable {tex}x.{/tex}

(g) It is not an equation as L.H.S. is less than R.H.S.

(h) It is an equation of variable as both the sides are equal. The variable is {tex}n.{/tex}

(i) It is an equation with no variable as its both sides are equal.

(j) It is an equation of variable {tex}p.{/tex}

(k) It is an equation of variable {tex}y.{/tex}

(l) It is not an equation as L.H.S. is less than R.H.S.

(m) It is not an equation as L.H.S. is greater than R.H.S.

(n) It is an equation with no variable.

(o) It is an equation of variable {tex}x.{/tex}


NCERT Solutions for Class 6 Maths Exercise 11.5

Question 2.Complete the entries of the third column of the table:
S. No.EquationValue of variableEquation satisfied Yes/No
(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)

(n)

(o)

(p)

(q)

{tex}10y = 80{/tex}

{tex}10y = 80{/tex}

{tex}10y = 80{/tex}

{tex}4l = 20{/tex}

{tex}4l = 20{/tex}

{tex}4l = 20{/tex}

{tex}b + 5 = 9{/tex}

{tex}b + 5 = 9{/tex}

{tex}b + 5 = 9{/tex}

{tex}h – 8 = 5{/tex}

{tex}h – 8 = 5{/tex}

{tex}h – 8 = 5{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}y = 10{/tex}

{tex}y = 8{/tex}

{tex}y = 5{/tex}

{tex}l = 20{/tex}

{tex}l = 80{/tex}

{tex}l = 5{/tex}

{tex}b = 5{/tex}

{tex}b = 9{/tex}

{tex}b = 4{/tex}

{tex}h = 13{/tex}

{tex}h = 8{/tex}

{tex}h = 0{/tex}

{tex}p = 3{/tex}

{tex}p = 1{/tex}

{tex}p = 0{/tex}

{tex}p =- 1{/tex}

{tex}p =- 2{/tex}

Answer:

S. No.EquationValue of variableEquation satisfied Yes/NoSolution of L.H.S.
(a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k)

(l)

(m)

(n)

(o)

(p)

(q)

{tex}10y = 80{/tex}

{tex}10y = 80{/tex}

{tex}10y = 80{/tex}

{tex}4l = 20{/tex}

{tex}4l = 20{/tex}

{tex}4l = 20{/tex}

{tex}b + 5 = 9{/tex}

{tex}b + 5 = 9{/tex}

{tex}b + 5 = 9{/tex}

{tex}h – 8 = 5{/tex}

{tex}h – 8 = 5{/tex}

{tex}h – 8 = 5{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}p + 3 = 1{/tex}

{tex}y = 10{/tex}

{tex}y = 8{/tex}

{tex}y = 5{/tex}

{tex}l = 20{/tex}

{tex}l = 80{/tex}

{tex}l = 5{/tex}

{tex}b = 5{/tex}

{tex}b = 9{/tex}

{tex}b = 4{/tex}

{tex}h = 13{/tex}

{tex}h = 8{/tex}

{tex}h = 0{/tex}

{tex}p = 3{/tex}

{tex}p = 1{/tex}

{tex}p = 0{/tex}

{tex}p =- 1{/tex}

{tex}p =- 2{/tex}

No

Yes

No

No

No

Yes

No

Yes

Yes

Yes

No

No

No

No

No

No

Yes

10 x 10 = 100

10 x 8 = 80

10 x 5 = 50

4 x 20 = 80

4 x 80 = 320

4 x 5 = 20

5 + 5 = 10

9 + 5 = 14

4 + 5 = 9

13 – 8 = 5

8 – 8 = 0

0 – 8 = –8

3 + 3 = 6

1 + 3 = 4

0 + 3 = 3

–1 + 3 = 2

–2 + 3 = 1


NCERT Solutions for Class 6 Maths Exercise 11.5

Question 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) {tex}5m = 60{/tex} (10, 5, 12, 15)

(b) {tex}n + 12 = 20{/tex} (12, 8, 20, 0)

(c) {tex}p – 5 = 5{/tex} (0, 10, 5, –5)

(d) {tex}\frac{q}{2} = 7{/tex} (7, 2, 10, 14)

(e) {tex}r – 4 = 0{/tex} (4, –4, 8, 0)

(f) {tex}x + 4 = 2{/tex} (–2, 0, 2, 4)

Answer:

(a) {tex}5m = 60{/tex}

Putting the given values in L.H.S.,

5 x 10 = 50, 5 x 5 = 25

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S.{tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}m = 10{/tex} is not the solution. {tex}\therefore {/tex}{tex}m = 5{/tex} is not the solution.

5 x 12 = 60, 5 x 15 = 75

{tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}m = 12{/tex} is a solution. {tex}\therefore {/tex}{tex}m = 15{/tex} is not the solution.

(b) {tex}n + 12 = 20{/tex}

Putting the given values in L.H.S.,

12 + 12 = 24, 8 + 12 = 20

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S.

{tex}\therefore {/tex}{tex}n = 12{/tex} is not the solution. {tex}\therefore {/tex}{tex}n = 8{/tex} is a solution.

20 + 12 = 32, 0 + 12 = 12

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}n = 20{/tex} is not the solution. {tex}\therefore {/tex}{tex}n = 0{/tex} is not the solution.

(c) {tex}p – 5 = 5{/tex}

Putting the given values in L.H.S.,

0 – 5 = –5, 10 – 5 = 5

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S.

{tex}\therefore {/tex}{tex}p = 0{/tex} is not the solution. {tex}\therefore {/tex}{tex}p = 10{/tex} is a solution.

5 – 5 = 0, –5 – 5 = –10

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}p = 5{/tex} is not the solution. {tex}\therefore {/tex}{tex}p = – 5{/tex} is not the solution.

(d) {tex}\frac{q}{2} = 7{/tex}

Putting the given values in L.H.S.,

{tex}\frac{7}{2}{/tex} {tex}\frac{2}{2} = 1{/tex}

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}q = 7{/tex} is not the solution. {tex}\therefore {/tex}{tex}q = 2{/tex} is not the solution.

{tex}\frac{{10}}{2} = 5{/tex} {tex}\frac{{14}}{2} = 7{/tex}

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S.

{tex}\therefore {/tex}{tex}q = 10{/tex} is not the solution. {tex}\therefore {/tex}{tex}q = 14{/tex} is a solution.

(e) {tex}r – 4 = 0{/tex}

Putting the given values in L.H.S.,

4 – 4 = 0, –4 – 4 = –8

{tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}r = 4{/tex} is a solution. {tex}\therefore {/tex}{tex}r = – 4{/tex} is not the solution.

8 – 4 = 4, 0 – 4 = –4

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}r = 8{/tex} is not the solution. {tex}\therefore {/tex}{tex}r = 0{/tex} is not the solution.

(f) {tex}x + 4 = 2{/tex}

Putting the given values in L.H.S.,

–2 + 4 = 2, 0 + 4 = 4

{tex}\because {/tex} L.H.S. {tex} = {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}x = – 2{/tex} is a solution. {tex}\therefore {/tex}{tex}x = 0{/tex} is not the solution.

2 + 4 = 6, 4 + 4 = 8

{tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S. {tex}\because {/tex} L.H.S. {tex} \ne {/tex} R.H.S.

{tex}\therefore {/tex}{tex}x = 2{/tex} is not the solution. {tex}\therefore {/tex}{tex}x = 4{/tex} is not the solution.


NCERT Solutions for Class 6 Maths Exercise 11.5

Question 4.

(a) Complete the table and by inspection of the table find the solution to the equation

{tex}m + 10 = 16.{/tex}

{tex}m{/tex}12345678910
{tex}m + 10{/tex}

(b) Complete the table and by inspection of the table find the solution to the equation {tex}5t = 35{/tex}.

{tex}t{/tex}34567891011
{tex}5t{/tex}

(c)Complete the table and by inspection of the table find the solution to the equation {tex}\frac{z}{3} = 4.{/tex}

{tex}z{/tex}8910111213141516
{tex}\frac{z}{3}{/tex}{tex}2\frac{2}{3}{/tex}3{tex}3\frac{1}{3}{/tex}

(d)Complete the table and by inspection of the table find the solution to the equation {tex}m – 7 = 3.{/tex}

{tex}m{/tex}5678910111213
{tex}m – 7{/tex}

Answer:

(a)

{tex}m{/tex}12345678910111213
{tex}m + 10{/tex}11121314151617181920212223

{tex}\therefore {/tex} {tex}m = 6{/tex} is the solution.{tex}\because {/tex} At {tex}m = 6,{/tex}{tex}m + 10 = 16{/tex}

(b)

{tex}t{/tex}345678910111213141516
{tex}5t{/tex}1520253035404550556065707580

{tex}\therefore {/tex} {tex}t = 7{/tex} is the solution.{tex}\because {/tex} At {tex}t = 7,{/tex}{tex}5t = 35{/tex}

(c)

{tex}z{/tex}891011121314151617181920
{tex}\frac{z}{3}{/tex}{tex}2\frac{2}{3}{/tex}3{tex}3\frac{1}{3}{/tex}{tex}3\frac{2}{3}{/tex}4{tex}4\frac{1}{3}{/tex}{tex}4\frac{2}{3}{/tex}5{tex}5\frac{1}{3}{/tex}{tex}5\frac{2}{3}{/tex}6{tex}6\frac{1}{3}{/tex}{tex}6\frac{2}{3}{/tex}

{tex}\therefore {/tex} {tex}z = 12{/tex} is the solution.{tex}\because {/tex} At {tex}z = 12,{/tex}{tex}\frac{z}{3} = 4{/tex}

(d)

{tex}m{/tex}56789101112131415
{tex}m – 7{/tex}-2-1012345678

{tex}\therefore {/tex} {tex}m = 10{/tex} is the solution.{tex}\because {/tex} At {tex}m = 10,{/tex}{tex}m – 7 = 3{/tex}

NCERT Solutions for Class 6 Maths Exercise 11.5

NCERT Solutions Class 6 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 6 Maths includes text book solutions from Class 6 Maths Book . NCERT Solutions for CBSE Class 6 Maths have total 14 chapters. 6 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 6 solutions PDF and Maths ncert class 6 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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