question_answer

Santosh has Rs. \[({{x}^{3}}-3{{x}^{2}}+4x+50).\]He want to buy chocolates each of cost Rs.\[(x-3).\] After buying maximum number of chocolates with his money, how much money is left with him?

A)  Rs. 50             

B)  Rs. 40

C)         Rs. 62             

D)         Rs. 20

Correct Answer: C

Solution :

Amount of money Santosh has = Rs. \[({{x}^{3}}-3{{x}^{2}}+4x+50)\] Cost of each chocolate = Rs.\[(x-3)\] By long division method, we have \[\therefore \]\[{{x}^{3}}-3{{x}^{2}}+4x+50=(x-3)({{x}^{2}}+4)+62\] So, Santosh bought\[({{x}^{2}}+4)\]chocolates and Rs. 62 left with him.