NCERT Solutions for Class 7 Maths Exercise 12.3

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NCERT Solutions for Class 7 Maths Algebraic Expressions

Class –VII Mathematics (Ex. 12.3)

Question 1.If {tex}m = 2,{/tex} find the value of:

(i) {tex}m – 2{/tex}

(ii) {tex}3m – 5{/tex}

(iii) {tex}9 – 5m{/tex}

(iv) {tex}3{m^2} – 2m – 7{/tex}

(v) {tex}\frac{{5m}}{2} – 4{/tex}

Answer:

(i) {tex}m – 2{/tex} = {tex}2 – 2{/tex} [Putting {tex}m = 2{/tex}]

= 0

(ii) {tex}3m – 5{/tex} = {tex}3 \times 2 – 5{/tex} [Putting {tex}m = 2{/tex}]

= 6 – 5 = 1

(iii) {tex}9 – 5m{/tex} = 9 – 5 x 2 [Putting {tex}m = 2{/tex}]

= 9 – 10 = {tex} – 1{/tex}

(iv) {tex}3{m^2} – 2m – 7{/tex} = {tex}3{\left( 2 \right)^2} – 2\left( 2 \right) – 7{/tex} [Putting {tex}m = 2{/tex}]

= 3 x 4 – 2 x 2 – 7 = 12 – 4 – 7

= 12 – 11 = 1

(v) {tex}\frac{{5m}}{2} – 4{/tex} = {tex}\frac{{5 \times 2}}{2} – 4{/tex}

[Putting {tex}m = 2{/tex}]

= 5 – 4 = 1

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 2.If {tex}p = – 2,{/tex} find the value of:

(i) {tex}4p + 7{/tex}

(ii) {tex} – 3{p^2} + 4p + 7{/tex}

(iii) {tex} – 2{p^3} – 3{p^2} + 4p + 7{/tex}

Answer:

(i) {tex}4p + 7{/tex} = {tex}4\left( { – 2} \right) + 7{/tex}

[Putting {tex}p = – 2{/tex}]

= {tex} – 8 + 7{/tex} = {tex} – 1{/tex}

(ii) {tex} – 3{p^2} + 4p + 7{/tex} = {tex} – 3{\left( { – 2} \right)^2} + 4\left( { – 2} \right) + 7{/tex} [Putting {tex}p = – 2{/tex}]

= {tex} – 3 \times 4 – 8 + 7{/tex} = {tex} – 12 – 8 + 7{/tex}

= {tex} – 20 + 7{/tex} = {tex} – 13{/tex}

(iii) {tex} – 2{p^3} – 3{p^2} + 4p + 7{/tex} = {tex} – 2{\left( { – 2} \right)^3} – 3{\left( { – 2} \right)^2} + 4\left( { – 2} \right) + 7{/tex} [Putting {tex}p = – 2{/tex}]

= = {tex} – 2 \times \left( { – 8} \right) – 3 \times 4 – 8 + 7{/tex} = {tex}16 – 12 – 8 + 7{/tex}

= {tex} – 20 + 23{/tex} = {tex}3{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 3.Find the value of the following expressions, when {tex}x = – 1:{/tex}

(i) {tex}2x – 7{/tex}

(ii) {tex} – x + 2{/tex}

(iii) {tex}{x^2} + 2x + 1{/tex}

(iv) {tex}2{x^2} – x – 2{/tex}

Answer:

(i) {tex}2x – 7{/tex} = {tex}2\left( { – 1} \right) – 7{/tex}

[Putting {tex}x = – 1{/tex}]

= {tex} – 2 – 7{/tex} = {tex} – 9{/tex}

(ii) {tex} – x + 2{/tex} = {tex} – \left( { – 1} \right) + 2{/tex}

[Putting {tex}x = – 1{/tex}]

= 1 + 2 = 3

(iii) {tex}{x^2} + 2x + 1{/tex} = {tex}{\left( { – 1} \right)^2} + 2\left( { – 1} \right) + 1{/tex} [Putting {tex}x = – 1{/tex}]

= 1 – 2 + 1 = 2 – 2 = 0

(iv) {tex}2{x^2} – x – 2{/tex} = {tex}2{\left( { – 1} \right)^2} – \left( { – 1} \right) – 2{/tex} [Putting {tex}x = – 1{/tex}]

= 2 x 1 + 1 – 2 = 2 + 1 – 2

= 3 – 2

= 1

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 4.If {tex}a = 2,b = – 2,{/tex} find the value of:

(i) {tex}{a^2} + {b^2}{/tex}

(ii) {tex}{a^2} + ab + {b^2}{/tex}

(iii) {tex}{a^2} – {b^2}{/tex}

Answer:

(i) {tex}{a^2} + {b^2}{/tex} = {tex}{\left( 2 \right)^2} + {\left( { – 2} \right)^2}{/tex} [Putting {tex}a = 2,b = – 2{/tex}]

= 4 + 4 = 8

(ii) {tex}{a^2} + ab + {b^2}{/tex} = {tex}{\left( 2 \right)^2} + \left( 2 \right)\left( { – 2} \right) + {\left( { – 2} \right)^2}{/tex} [Putting {tex}a = 2,b = – 2{/tex}]

= 4 – 4 + 4 = 4

(iii) {tex}{a^2} – {b^2}{/tex} = {tex}{\left( 2 \right)^2} – {\left( { – 2} \right)^2}{/tex} [Putting {tex}a = 2,b = – 2{/tex}]

= 4 – 4 = 0

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 5.When {tex}a = 0,b = – 1,{/tex} find the value of the given expressions:

(i) {tex}2a + 2b{/tex}

(ii) {tex}2{a^2} + {b^2} + 1{/tex}

(iii) {tex}2{a^2}b + 2a{b^2} + ab{/tex}

(iv) {tex}{a^2} + ab + 2{/tex}

Answer:

(i) {tex}2a + 2b{/tex} = {tex}2\left( 0 \right) + 2\left( { – 1} \right){/tex}

[Putting {tex}a = 0,b = – 1{/tex}]

= 0 – 2 = {tex} – 2{/tex}

(ii) {tex}2{a^2} + {b^2} + 1{/tex} = {tex}2{\left( 0 \right)^2} + {\left( { – 1} \right)^2} + 1{/tex} [Putting {tex}a = 0,b = – 1{/tex}]

= 2 x 0 + 1 + 1 = 0 + 2 = 2

(iii) {tex}2{a^2}b + 2a{b^2} + ab{/tex} = {tex}2{\left( 0 \right)^2}\left( { – 1} \right) + 2\left( 0 \right){\left( { – 1} \right)^2} + \left( 0 \right)\left( { – 1} \right){/tex} [Putting {tex}a = 0,b = – 1{/tex}]

= 0 + 0 + 0 = 0

(iv) {tex}{a^2} + ab + 2{/tex} = {tex}{\left( 0 \right)^2} + \left( 0 \right)\left( { – 1} \right) + 2{/tex} [Putting {tex}a = 0,b = – 1{/tex}]

= 0 + 0 + 2 = 2

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 6.Simplify the expressions and find the value if {tex}x{/tex} is equal to 2:

(i) {tex}x + 7 + 4\left( {x – 5} \right){/tex}

(ii) {tex}3\left( {x + 2} \right) + 5x – 7{/tex}

(iii) {tex}6x + 5\left( {x – 2} \right){/tex}

(iv) {tex}4\left( {2x – 1} \right) + 3x + 11{/tex}

Answer:

(i) {tex}x + 7 + 4\left( {x – 5} \right){/tex} = {tex}x + 7 + 4x – 20{/tex} = {tex}x + 4x + 7 – 20{/tex}

= {tex}5x – 13{/tex} = {tex}5 \times 2 – 13{/tex} [Putting {tex}x = 2{/tex}]

= {tex}10 – 13{/tex} = {tex} – 3{/tex}

(ii) {tex}3\left( {x + 2} \right) + 5x – 7{/tex} = {tex}3x + 6 + 5x – 7{/tex} = {tex}3x + 5x + 6 – 7{/tex}

= {tex}8x – 1{/tex} = 8 x 2 – 1 [Putting {tex}x = – 1{/tex}]

= 16 – 1 = 15

(iii) {tex}6x + 5\left( {x – 2} \right){/tex} = {tex}6x + 5x – 10{/tex} = {tex}11x – 10{/tex}

= 11 x 2 – 10 [Putting {tex}x = – 1{/tex}]

= 22 – 10 = 12

(iv) {tex}4\left( {2x – 1} \right) + 3x + 11{/tex} = {tex}8x – 4 + 3x + 11{/tex} = {tex}8x + 3x – 4 + 11{/tex}

= {tex}11x + 7{/tex} = 11 x 2 + 7 [Putting {tex}x = – 1{/tex}]

= 22 + 7 = 29

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 7.Simplify these expressions and find their values if {tex}x = 3,a = – 1,b = – 2:{/tex}

(i) {tex}3x – 5 – x + 9{/tex}

(ii) {tex}2 – 8x + 4x + 4{/tex}

(iii) {tex}3a + 5 – 8a + 1{/tex}

(iv) {tex}10 – 3b – 4 – 5b{/tex}

(v) {tex}2a – 2b – 4 – 5 + a{/tex}

Answer:

(i) {tex}3x – 5 – x + 9{/tex} = {tex}3x – x – 5 + 9{/tex} = {tex}2x + 4{/tex}

= {tex}2 \times 3 + 4{/tex} [Putting {tex}x = 3{/tex}]

= 6 + 4 = 10

(ii) {tex}2 – 8x + 4x + 4{/tex} = {tex} – 8x + 4x + 2 + 4{/tex} = {tex} – 4x + 6{/tex}

= {tex} – 4 \times 3 + 6{/tex} [Putting {tex}x = 3{/tex}]

= {tex} – 12 + 6 = – 12{/tex}

(iii) {tex}3a + 5 – 8a + 1{/tex} = {tex}3a – 8a + 5 + 1{/tex} = {tex} – 5a + 6{/tex}

= {tex} – 5\left( { – 1} \right) + 6{/tex} [Putting {tex}a = – 1{/tex}]

= 5 + 6 = 11

(iv) {tex}10 – 3b – 4 – 5b{/tex} = {tex} – 3b – 5b + 10 – 4{/tex} = {tex} – 8b + 6{/tex}

= {tex} – 8\left( { – 2} \right) + 6{/tex} [Putting {tex}b = – 2{/tex}]

= 16 + 6 = 22

(v) {tex}2a – 2b – 4 – 5 + a{/tex} = {tex}2a + a – 2b – 4 – 5{/tex}

= {tex}3a – 2b – 9{/tex} = {tex}3\left( { – 1} \right) – 2\left( { – 2} \right) – 9{/tex} [Putting {tex}a = – 1{/tex}, {tex}b = – 2{/tex}]

= {tex} – 3 + 4 – 9{/tex} = {tex} – 8{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 8.

(i) If {tex}z = 10,{/tex} find the value of {tex}{z^3} – 3\left( {z – 10} \right).{/tex}

(ii) If {tex}p = – 10,{/tex} find the value of {tex}{p^2} – 2p – 100.{/tex}

Answer:

(i) {tex}{z^3} – 3\left( {z – 10} \right){/tex} = {tex}{\left( {10} \right)^3} – 3\left( {10 – 10} \right){/tex} [Putting {tex}z = 10{/tex}]

= 1000 – 3 x 0 = 1000 – 0 = 1000

(ii) {tex}{p^2} – 2p – 100{/tex} = {tex}{\left( { – 10} \right)^2} – 2\left( { – 10} \right) – 100{/tex} [Putting {tex}p = – 10{/tex}]

= {tex}100 + 20 – 100{/tex} = 20

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 9.What should be the value of {tex}a{/tex} if the value of {tex}2{x^2} + x – a{/tex} equals to 5, when {tex}x = 0{/tex} ?

Answer:

Given: {tex}2{x^2} + x – a = 5{/tex}

{tex} \Rightarrow {/tex} {tex}2{\left( 0 \right)^2} + 0 – a = 5{/tex} [Putting {tex}x = 0{/tex}]

{tex} \Rightarrow {/tex} {tex}0 + 0 – a = 5{/tex} {tex} \Rightarrow {/tex} {tex}a = – 5{/tex}

Hence, the value of {tex}a{/tex} is {tex} – 5.{/tex}

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NCERT Solutions for Class 7 Maths Exercise 12.3

Question 10.Simplify the expression and find its value when {tex}a = 5{/tex} and {tex}b = – 3:{/tex}{tex}2\left( {{a^2} + ab} \right) + 3 – ab{/tex}

Answer:

Given: {tex}2\left( {{a^2} + ab} \right) + 3 – ab{/tex}

{tex} \Rightarrow {/tex} {tex}2{a^2} + 2ab + 3 – ab{/tex} {tex} \Rightarrow {/tex} {tex}2{a^2} + 2ab – ab + 3{/tex}

{tex} \Rightarrow {/tex} {tex}2{a^2} + ab + 3{/tex}

{tex} \Rightarrow {/tex} {tex}2{\left( 5 \right)^2} + \left( 5 \right)\left( { – 3} \right) + 3{/tex} [Putting {tex}a = 5{/tex}, {tex}b = – 3{/tex}]

{tex} \Rightarrow {/tex} 2 x 25 – 15 + 3

{tex} \Rightarrow {/tex} 50 – 15 + 3

{tex} \Rightarrow {/tex} 38

NCERT Solutions for Class 7 Maths Exercise 12.3

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