 # Substitution

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• Q1

If p = 1, q= 2 and r = 3, then the value of
2p2q – 6pq + pq2r is

Marks:1

4.

##### Explanation:
Putting p = 1, q = 2 and r = 3 in 2p2q – 6pq + pq2r, we get
2p2q – 6pq + pq2r = 2(1)2(2) – 6(1)(2) + (1)(2)2(3)
= 4 – 12 + 12
= 4.
• Q2

If a = 0, b= 2 and c = 1, then the value of
2a2b – 3bc2 + 4b2 is

Marks:1

10.

##### Explanation:
Putting a = 0, b = 2 and c = 1 in 2a2b – 3bc2 + 4b2, we get
2(0)2(2) – 3(2)(1)2 + 4(2)2
= 0 – 6 + 16 = 10.
• Q3

If a = 3, b = 2 and k = 7, What is the value of 5a3 - 2b2 - k - 5?

Marks:1

115

##### Explanation:

5a3 – 2b2 – k – 5
Putting the values of a , b and k in the given expression, we get
= 5 × 3 × 3 × 3 – 2 × 2 × 2 – 7 – 5
= 135 – 8 – 12
= 135 – 20
= 115

• Q4

If x = 1 and y = 2, then the value of x2y + x2y2 – xy2 is

Marks:1

2

##### Explanation:

Putting x = 1 and y = 2 in x2y + x2y2 – xy2, we get

(1)2(2) + (1)2(2)2 – (1)(2)2

= 2 + 4 – 4

= 2