Selina Solutions Concise Mathematics Class 6 Chapter 20 Substitution Exercise 20(A) are well structured by highly experienced faculty having vast knowledge in the education industry. Solutions contain steps to be followed in evaluating the problems relying on algebraic expressions, which is the main concept discussed under this exercise. Practicing solved examples present before the exercise problems help students, in analyzing the question that would appear in the final exam. Students who aspire to obtain proficiency in Mathematics can access Selina Solutions Concise Mathematics Class 6 Chapter 20 Substitution Exercise 20(A) PDF, from the links which are present below.

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### Access Selina Solutions Concise Mathematics Class 6 Chapter 20 Substitution Exercise 20(A)

#### Exercise 20(A)

**1. Fill in the following blanks, when:**

**x = 3, y = 6, z = 18, a = 2, b = 8, c = 32 and d = 0.**

**(i) x + y = â€¦â€¦â€¦.**

**(ii) y â€“ x = â€¦â€¦â€¦.**

**(iii) y / x = â€¦â€¦â€¦..**

**(iv) c Ã· b = â€¦â€¦â€¦..**

**(v) z Ã· x = â€¦â€¦â€¦..**

**Solution:**

(i) x + y = â€¦â€¦..

The value of x + y is calculated as shown below

x + y = 3 + 6

= 9

âˆ´ x + y = 9

(ii) y â€“ x = â€¦â€¦â€¦.

The value of y â€“ x is calculated as shown below

y â€“ x = 6 â€“ 3

= 3

âˆ´ y â€“ x = 3

(iii) y / x = â€¦â€¦â€¦..

The value of y / x is calculated as shown below

y / x= 6 / 3

= 2

âˆ´ y / x = 2

(iv) c Ã· b = â€¦â€¦â€¦

The value of c Ã· b is calculated as shown below

c Ã· b = 32 Ã· 8

32 / 8 = 4

âˆ´ c Ã· b = 4

(v) z Ã· x = â€¦â€¦â€¦.

The value of z Ã· x is calculated as shown below

z Ã· x = 18 Ã· 3

= 6

âˆ´ z Ã· x = 6

**2. Find the value of:**

**(i) p + 2q + 3r, when p = 1, q = 5 and r = 2**

**(ii) 2a + 4b + 5c, when a = 5, b = 10 and c = 20**

**(iii) 3a â€“ 2b, when a = 8 and b = 10**

**(iv) 5x + 3y â€“ 6z, when x = 3, y = 5 and z = 4**

**(v) 2p â€“ 3q + 4r â€“ 8s, when p = 10, q = 8, r = 6 and s = 2**

**Solution:**

(i) p + 2q + 3r, when p = 1, q = 5 and r = 2

The value of p + 2q + 3r is calculated as shown below

p + 2q + 3r = 1 + 2 Ã— 5 + 3 Ã— 2

= 1 + 10 + 6

= 17

Therefore, p + 2q + 3r = 17

(ii) 2a + 4b + 5c, when a = 5, b = 10 and c = 20

The value of 2a + 4b + 5c is calculated as shown below

2a + 4b + 5c = 2 Ã— 5 + 4 Ã— 10 + 5 Ã— 20

= 10 + 40 + 100

= 150

Therefore, 2a + 4b + 5c = 150

(iii) 3a â€“ 2b, when a = 8 and b = 10

The value of 3a â€“ 2b is calculated as shown below

3a â€“ 2b = 3 Ã— 8 â€“ 2 Ã— 10

= 24 â€“ 20

= 4

Therefore, 3a â€“ 2b = 4

(iv) 5x + 3y â€“ 6z, when x = 3, y = 5 and z = 4

The value of 5x + 3y â€“ 6z is calculated as shown below

5x + 3y â€“ 6z = 5 Ã— 3 + 3 Ã— 5 â€“ 6 Ã— 4

= 15 + 15 â€“ 24

= 30 â€“ 24

= 6

Therefore, 5x + 3y â€“ 6z = 6

(v) 2p â€“ 3q + 4r â€“ 8s, when p = 10, q = 8, r = 6 and s = 2

The value of 2p â€“ 3q + 4r â€“ 8s is calculated as shown below

2p â€“ 3q + 4r â€“ 8s = 2 Ã— 10 â€“ 3 Ã— 8 + 4 Ã— 6 â€“ 8 Ã— 2

= 20 â€“ 24 + 24 â€“ 16

= 4

Therefore, 2p â€“ 3q + 4r â€“ 8s = 4

**3. Find the value of:**

**(i) 4pq Ã— 2r, when p = 5, q = 3 and r = 1 / 2**

**(ii) yx / z, when x = 8, y = 4 and z = 16**

**(iii) (a + b â€“ c) / 2a, when a = 5, b = 7 and c = 2**

**Solution:**

(i) 4pq Ã— 2r, when p = 5, q = 3 and r = 1 / 2

The value of 4pq Ã— 2r is calculated as below

4pq Ã— 2r = 4 Ã— 5 Ã— 3 Ã— 2 Ã— (1 / 2)

= 4 Ã— 5 Ã— 3

= 60

âˆ´ 4pq Ã— 2r = 60

(ii) yx / z, when x = 8, y = 4 and z = 16

The value of yx / z is calculated as below

yx / z = (4 Ã— 8) / 16

= 32 / 16

= 2

âˆ´ yx / z = 2

(iii) (a + b â€“ c) / 2a, when a = 5, b = 7 and c = 2

The value of (a + b â€“ c) / 2a is calculated as below

(a + b â€“ c) / 2a = (5 + 7 â€“ 2) / (2 Ã— 5)

= 10 / 10

= 1

**4. If a = 3, b = 0, c = 2 and d = 1, find the value of:**

**(i) 3a + 2b â€“ 6c + 4d **

**(ii) 6a â€“ 3b â€“ 4c â€“ 2d**

**(iii) ab â€“ bc + cd â€“ da**

**(iv) abc â€“ bcd + cda**

**(v) a ^{2} + 2b^{2} â€“ 3c^{2}**

**Solution:**

(i) 3a + 2b â€“ 6c + 4d

The value of 3a + 2b â€“ 6c + 4d is calculated as shown below

3a + 2b â€“ 6c + 4d = 3 Ã— 3 + 2 Ã— 0 â€“ 6 Ã— 2 + 4 Ã— 1

On further calculation, we get

= 9 + 0 â€“ 12 + 4

= 9 â€“ 12 + 4

= 13 â€“ 12

= 1

Therefore, 3a + 2b â€“ 6c + 4d = 1

(ii) 6a â€“ 3b â€“ 4c â€“ 2d

The value of 6a â€“ 3b â€“ 4c â€“ 2d is calculated as shown below

6a â€“ 3b â€“ 4c â€“ 2d = 6 Ã— 3 â€“ 3 Ã— 0 â€“ 4 Ã— 2 â€“ 2 Ã— 1

On further calculation, we get

= 18 â€“ 0 â€“ 8 â€“ 2

= 18 â€“ 10

= 8

Therefore, 6a â€“ 3b â€“ 4c â€“ 2d = 8

(iii) ab â€“ bc + cd â€“ da

The value of ab â€“ bc + cd â€“ da is calculated as shown below

ab â€“ bc + cd â€“ da = 3 Ã— 0 â€“ 0 Ã— 2 + 2 Ã— 1 â€“ 1 Ã— 3

On further calculation, we get

= 0 â€“ 0 + 2 â€“ 3

= 2 â€“ 3

= – 1

Therefore, ab â€“ bc + cd â€“ da = – 1

(iv) abc â€“ bcd + cda

The value of abc â€“ bcd + cda is calculated as shown below

abc â€“ bcd + cda = 3 Ã— 0 Ã— 2 â€“ 0 Ã— 2 Ã— 1 + 2 Ã— 1 Ã— 3

On further calculation, we get

= 0 â€“ 0 + 6

= 6

Therefore, abc â€“ bcd + cda = 6

(v) a^{2} + 2b^{2} â€“ 3c^{2}

The value of a^{2} + 2b^{2} â€“ 3c^{2} is calculated as shown below

a^{2} + 2b^{2} â€“ 3c^{2} = (3)^{2} + 2 Ã— (0)^{2} â€“ 3 Ã— (2)^{2}

On further calculation, we get

= 9 + 0 â€“ 12

= 9 â€“ 12

= – 3

Therefore, a^{2} + 2b^{2} â€“ 3c^{2} = – 3

**5. Find the value of 5x ^{2} â€“ 3x + 2, when x = 2**

**Solution:**

The value of 5x^{2} â€“ 3x + 2 when x = 2 is calculated as below

5x^{2} â€“ 3x + 2 = 5 Ã— (2)^{2} â€“ 3 Ã— (2) + 2

On simplification, we get

= 5 Ã— 4 â€“ 3 Ã— 2 + 2

= 20 â€“ 6 + 2

= 22 â€“ 6

= 16

Hence, the value of 5x^{2} â€“ 3x + 2 when x = 2 is 16

**6. Find the value of 3x ^{3} â€“ 4x^{2} + 5x â€“ 6, when x = – 1**

**Solution:**

The value of 3x^{3} â€“ 4x^{2} + 5x â€“ 6 when x = -1 is calculated as below

3x^{3} â€“ 4x^{2} + 5x â€“ 6 = 3 Ã— (- 1)^{3} â€“ 4 Ã— (- 1)^{2} + 5 Ã— (- 1) â€“ 6

On simplification, we get

= – 3 â€“ 4 â€“ 5 â€“ 6

= – 18

Hence, the value of 3x^{3} â€“ 4x^{2} + 5x â€“ 6 when x = – 1 is â€“ 18

**7. Show that the value of x ^{3} â€“ 8x^{2} + 12x â€“ 5 is zero, when x = 1**

Solution:

The value of x^{3} â€“ 8x^{2} + 12x â€“ 5 = 0 when x = 1 is calculated as below

x^{3} â€“ 8x^{2} + 12x â€“ 5 = (1)^{3} â€“ 8 Ã— (1)^{2} + 12 Ã— (1) â€“ 5

On simplification, we get

= 1 â€“ 8 Ã— 1 + 12 Ã— 1 â€“ 5

= 1 â€“ 8 + 12 â€“ 5

= 0

The value of x^{3} â€“ 8x^{2} + 12x â€“ 5 = 0 when x = 1

Hence, proved

**8. State true and false:**

**(i) The value of x + 5 = 6, when x = 1**

**(ii) The value of 2x â€“ 3 = 1, when x = 0**

**(iii) (2x â€“ 4) / (x + 1) = -1, when x = 1**

**Solution:**

(i) The value of x + 5 = 6, when x = 1

The value of x + 5 = 6 for x = 1 is calculated as below

x + 5 = 6

Adding the value of x = 1, we get

1 + 5 = 6

6 = 6

Therefore, the given statement is true

(ii) The value of 2x â€“ 3 = 1, when x = 0

The value of 2x â€“ 3 = 1 for x = 0 is calculated as below

2x â€“ 3 = 1

Adding the value of x = 0, we get

2(0) â€“ 3 = 1

0 â€“ 3 = 1

– 3 = 1

Therefore, the given statement is false

(iii) (2x â€“ 4) / (x + 1) = -1, when x = 1

The value of (2x â€“ 4) / (x + 1) = -1 for x = 1 is calculated as below

(2x â€“ 4) / (x + 1) = -1

Adding x = 1, we get

2(1) â€“ 4 / (1 + 1) = – 1

– 2 / 2 = – 1

– 1 = – 1

Therefore, the given statement is true

**9. If x = 2, y = 5 and z = 4, find the value of each of the following:**

**(i) x / 2x ^{2}**

**(ii) xz / yz**

**(iii) z ^{x}**

**(iv) y ^{x}**

**(v) x ^{2}y^{2}z^{2} / xz**

**Solution:**

(i) x / 2x^{2}

The value of x / 2x^{2} for x = 2, y = 5 and z = 4 is calculated as below

x / 2x^{2}

Now, adding x = 2, y = 5 and z = 4, we get

x / 2x^{2} = 2 / 2(2)^{2}

On calculation, we get

= 2 / 8

= 1 / 4

(ii) xz / yz

The value of xz / yz for x = 2, y = 5 and z = 4 is calculated as below

xz / yz

Now, adding x = 2, y = 5 and z = 4, we get

xz / yz = (2) (4) / (5) (4)

On calculation, we get

= 8 / 20

= 2 / 5

(iii) z^{x}

The value of z^{x} for x = 2, y = 5 and z = 4 is calculated as below

Now, adding x = 2 and z = 4, we get

z^{x} = (4)^{2}

We get

= 4 Ã— 4

= 16

(iv) y^{x}

The value of y^{x} for x = 2, y = 5 and z = 4 is calculated as below

Now, adding x = 2 and y = 5, we get

y^{x }= (5)^{2}

We get,

= 5 Ã— 5

= 25

(v) x^{2}y^{2}z^{2} / xz

The value of x^{2}y^{2}z^{2} / xz for x = 2, y = 5 and z = 4 is calculated as below

Now, adding x = 2, y = 5 and z = 4, we get

x^{2}y^{2}z^{2} / xz = (2)^{2} Ã— (5)^{2} Ã— (4)^{2} / (2 Ã— 4)

We get,

= 2^{2-1} Ã— 5^{2} Ã— 4^{2-1}

= 2 Ã— 5 Ã— 5 Ã— 4

= 200

**10. If a = 3, find the values of a ^{2} and 2^{a}**

**Solution:**

The value of a^{2} and 2^{a} for a = 3 is calculated as below

a^{2} = 3^{2}

= 3 Ã— 3

= 9

2^{a} = 2^{3}

= 2 Ã— 2 Ã— 2

= 8

Hence, the values of a^{2} = 9 and 2^{a} = 8

**11. If m = 2, find the difference between the values of 4m ^{3} and 3m^{4}.**

**Solution:**

The difference between the values of 4m^{3} and 3m^{4} for m = 2 is calculated as below

4m^{3} = 4 Ã— (2)^{3}

= 4 Ã— 2 Ã— 2 Ã— 2

We get,

= 32

3m^{4} = 3 Ã— (2)^{4}

= 3 Ã— 2 Ã— 2 Ã— 2 Ã— 2

We get,

= 48

Therefore, the difference of 4m^{3} and 3m^{4} is calculated as,

3m^{4} â€“ 4m^{3} = 48 â€“ 32

= 16

Hence, the difference between the given values is 16

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