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 = ……….
(ii) y – x = ……….
(iii) y / x = ………..
(iv) c ÷ b = ………..
(v) z ÷ x = ………..
Solution:
(i) x + y = ……..
The value of x + y is calculated as shown below
x + y = 3 + 6
= 9
∴ x + y = 9
(ii) y – x = ……….
The value of y – x is calculated as shown below
y – x = 6 – 3
= 3
∴ y – x = 3
(iii) y / x = ………..
The value of y / x is calculated as shown below
y / x= 6 / 3
= 2
∴ y / x = 2
(iv) c ÷ b = ………
The value of c ÷ b is calculated as shown below
c ÷ b = 32 ÷ 8
32 / 8 = 4
∴ c ÷ b = 4
(v) z ÷ x = ……….
The value of z ÷ x is calculated as shown below
z ÷ x = 18 ÷ 3
= 6
∴ z ÷ 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 – 2b, when a = 8 and b = 10
(iv) 5x + 3y – 6z, when x = 3, y = 5 and z = 4
(v) 2p – 3q + 4r – 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 × 5 + 3 × 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 × 5 + 4 × 10 + 5 × 20
= 10 + 40 + 100
= 150
Therefore, 2a + 4b + 5c = 150
(iii) 3a – 2b, when a = 8 and b = 10
The value of 3a – 2b is calculated as shown below
3a – 2b = 3 × 8 – 2 × 10
= 24 – 20
= 4
Therefore, 3a – 2b = 4
(iv) 5x + 3y – 6z, when x = 3, y = 5 and z = 4
The value of 5x + 3y – 6z is calculated as shown below
5x + 3y – 6z = 5 × 3 + 3 × 5 – 6 × 4
= 15 + 15 – 24
= 30 – 24
= 6
Therefore, 5x + 3y – 6z = 6
(v) 2p – 3q + 4r – 8s, when p = 10, q = 8, r = 6 and s = 2
The value of 2p – 3q + 4r – 8s is calculated as shown below
2p – 3q + 4r – 8s = 2 × 10 – 3 × 8 + 4 × 6 – 8 × 2
= 20 – 24 + 24 – 16
= 4
Therefore, 2p – 3q + 4r – 8s = 4
3. Find the value of:
(i) 4pq × 2r, when p = 5, q = 3 and r = 1 / 2
(ii) yx / z, when x = 8, y = 4 and z = 16
(iii) (a + b – c) / 2a, when a = 5, b = 7 and c = 2
Solution:
(i) 4pq × 2r, when p = 5, q = 3 and r = 1 / 2
The value of 4pq × 2r is calculated as below
4pq × 2r = 4 × 5 × 3 × 2 × (1 / 2)
= 4 × 5 × 3
= 60
∴ 4pq × 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 × 8) / 16
= 32 / 16
= 2
∴ yx / z = 2
(iii) (a + b – c) / 2a, when a = 5, b = 7 and c = 2
The value of (a + b – c) / 2a is calculated as below
(a + b – c) / 2a = (5 + 7 – 2) / (2 × 5)
= 10 / 10
= 1
4. If a = 3, b = 0, c = 2 and d = 1, find the value of:
(i) 3a + 2b – 6c + 4d
(ii) 6a – 3b – 4c – 2d
(iii) ab – bc + cd – da
(iv) abc – bcd + cda
(v) a2 + 2b2 – 3c2
Solution:
(i) 3a + 2b – 6c + 4d
The value of 3a + 2b – 6c + 4d is calculated as shown below
3a + 2b – 6c + 4d = 3 × 3 + 2 × 0 – 6 × 2 + 4 × 1
On further calculation, we get
= 9 + 0 – 12 + 4
= 9 – 12 + 4
= 13 – 12
= 1
Therefore, 3a + 2b – 6c + 4d = 1
(ii) 6a – 3b – 4c – 2d
The value of 6a – 3b – 4c – 2d is calculated as shown below
6a – 3b – 4c – 2d = 6 × 3 – 3 × 0 – 4 × 2 – 2 × 1
On further calculation, we get
= 18 – 0 – 8 – 2
= 18 – 10
= 8
Therefore, 6a – 3b – 4c – 2d = 8
(iii) ab – bc + cd – da
The value of ab – bc + cd – da is calculated as shown below
ab – bc + cd – da = 3 × 0 – 0 × 2 + 2 × 1 – 1 × 3
On further calculation, we get
= 0 – 0 + 2 – 3
= 2 – 3
= – 1
Therefore, ab – bc + cd – da = – 1
(iv) abc – bcd + cda
The value of abc – bcd + cda is calculated as shown below
abc – bcd + cda = 3 × 0 × 2 – 0 × 2 × 1 + 2 × 1 × 3
On further calculation, we get
= 0 – 0 + 6
= 6
Therefore, abc – bcd + cda = 6
(v) a2 + 2b2 – 3c2
The value of a2 + 2b2 – 3c2 is calculated as shown below
a2 + 2b2 – 3c2 = (3)2 + 2 × (0)2 – 3 × (2)2
On further calculation, we get
= 9 + 0 – 12
= 9 – 12
= – 3
Therefore, a2 + 2b2 – 3c2 = – 3
5. Find the value of 5x2 – 3x + 2, when x = 2
Solution:
The value of 5x2 – 3x + 2 when x = 2 is calculated as below
5x2 – 3x + 2 = 5 × (2)2 – 3 × (2) + 2
On simplification, we get
= 5 × 4 – 3 × 2 + 2
= 20 – 6 + 2
= 22 – 6
= 16
Hence, the value of 5x2 – 3x + 2 when x = 2 is 16
6. Find the value of 3x3 – 4x2 + 5x – 6, when x = – 1
Solution:
The value of 3x3 – 4x2 + 5x – 6 when x = -1 is calculated as below
3x3 – 4x2 + 5x – 6 = 3 × (- 1)3 – 4 × (- 1)2 + 5 × (- 1) – 6
On simplification, we get
= – 3 – 4 – 5 – 6
= – 18
Hence, the value of 3x3 – 4x2 + 5x – 6 when x = – 1 is – 18
7. Show that the value of x3 – 8x2 + 12x – 5 is zero, when x = 1
Solution:
The value of x3 – 8x2 + 12x – 5 = 0 when x = 1 is calculated as below
x3 – 8x2 + 12x – 5 = (1)3 – 8 × (1)2 + 12 × (1) – 5
On simplification, we get
= 1 – 8 × 1 + 12 × 1 – 5
= 1 – 8 + 12 – 5
= 0
The value of x3 – 8x2 + 12x – 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 – 3 = 1, when x = 0
(iii) (2x – 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 – 3 = 1, when x = 0
The value of 2x – 3 = 1 for x = 0 is calculated as below
2x – 3 = 1
Adding the value of x = 0, we get
2(0) – 3 = 1
0 – 3 = 1
– 3 = 1
Therefore, the given statement is false
(iii) (2x – 4) / (x + 1) = -1, when x = 1
The value of (2x – 4) / (x + 1) = -1 for x = 1 is calculated as below
(2x – 4) / (x + 1) = -1
Adding x = 1, we get
2(1) – 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 / 2x2
(ii) xz / yz
(iii) zx
(iv) yx
(v) x2y2z2 / xz
Solution:
(i) x / 2x2
The value of x / 2x2 for x = 2, y = 5 and z = 4 is calculated as below
x / 2x2
Now, adding x = 2, y = 5 and z = 4, we get
x / 2x2 = 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) zx
The value of zx for x = 2, y = 5 and z = 4 is calculated as below
Now, adding x = 2 and z = 4, we get
zx = (4)2
We get
= 4 × 4
= 16
(iv) yx
The value of yx for x = 2, y = 5 and z = 4 is calculated as below
Now, adding x = 2 and y = 5, we get
yx = (5)2
We get,
= 5 × 5
= 25
(v) x2y2z2 / xz
The value of x2y2z2 / xz for x = 2, y = 5 and z = 4 is calculated as below
Now, adding x = 2, y = 5 and z = 4, we get
x2y2z2 / xz = (2)2 × (5)2 × (4)2 / (2 × 4)
We get,
= 22-1 × 52 × 42-1
= 2 × 5 × 5 × 4
= 200
10. If a = 3, find the values of a2 and 2a
Solution:
The value of a2 and 2a for a = 3 is calculated as below
a2 = 32
= 3 × 3
= 9
2a = 23
= 2 × 2 × 2
= 8
Hence, the values of a2 = 9 and 2a = 8
11. If m = 2, find the difference between the values of 4m3 and 3m4.
Solution:
The difference between the values of 4m3 and 3m4 for m = 2 is calculated as below
4m3 = 4 × (2)3
= 4 × 2 × 2 × 2
We get,
= 32
3m4 = 3 × (2)4
= 3 × 2 × 2 × 2 × 2
We get,
= 48
Therefore, the difference of 4m3 and 3m4 is calculated as,
3m4 – 4m3 = 48 – 32
= 16
Hence, the difference between the given values is 16
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