SSC CGL 16 Aug 2015 Shift 1 Paper

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Question : 102 of 200

Marks: +1, -0

If p = 99 then, the value of p(

p^2

+ 3p + 3) is :

989898
988899
999999
998889

Solution:

p = 99 (Given)
Expression = p (

p^2

+ 3p + 3)
=

p^3 + 3p^2

+ 3p
=

p^3 + 3p^2

+ 3p + 1 – 1
=

(p + 1)^3

– 1
=

(99 + 1)^3

– 1 =

(100)^3

– 1
= 1000000 – 1 = 999999

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