Substitution
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Q1
If p = 1, q= 2 and r = 3, then the value of
2p^{2}q – 6pq + pq^{2}r isMarks:1Answer:
4.
Explanation:
Putting p = 1, q = 2 and r = 3 in 2p^{2}q – 6pq + pq^{2}r, we get
2p^{2}q – 6pq + pq^{2}r = 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
2a^{2}b – 3bc^{2} + 4b^{2} isMarks:1Answer:
10.
Explanation:
Putting a = 0, b = 2 and c = 1 in 2a^{2}b – 3bc^{2} + 4b^{2}, 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 5a^{3}  2b^{2}  k  5?
Marks:1Answer:
115
Explanation:
5a^{3} – 2b^{2} – 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 x^{2}y + x^{2}y^{2} – xy^{2} is
Marks:1Answer:
2
Explanation:
Putting x = 1 and y = 2 in x^{2}y + x^{2}y^{2} – xy^{2}, we get
(1)^{2}(2) + (1)^{2}(2)^{2} – (1)(2)^{2}
= 2 + 4 – 4
= 2

Q5
If the value of t = 4, then the value of 5t + 4 is
Marks:1Answer:
24.
Explanation:
The value of 5t + 4 is given by: 5(4) + 4 = 20 + 4 = 24, as t = 4.