NIMCET 2015 Question Paper with solutions

Question : 58 of 120

Marks: +1, -0

In an examination there are 100 questions divided into 3 parts A, B, C and each part should contain at least one question. Each question in parts A, B and C carry 1, 2 and 3 marks respectively. Part A is for at least 60% of the total marks and part B should contain 23 questions. How many questions must be set in part C ?

1
2
3
cannot be determined

Validate

Solution:

Suppose, the number of questions in group A, B and C be x, y, and z, respectively.
Then,

x+y+z=100

(given)
Total marks would be

x+2y+3z

Given,

y=23

Total marks from section B = 46
Different possible values for z are 1,2, 3....
∴ Corresponding values for x are 76, 75, 74...
(Since, total questions are 100)
when

z=1,x=76,y=23

Total marks from groups A, B and C are 76, 46, 3 respectively.
Percentage marks from

A=76/{76+46+3}

=75/125 > 60%

when

z=2,x=75,y=23

Percentage marks from

A=75/{75+46+6}

=75/127 < 60%

For all other values, when z increases, x decreases and contribution of marks from A keeps decreasing.
∴ There is only 1 possible value for questions from group C.

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