PE 2 - Mathematical Reasoning | LR - Mathematical Reasoning

Answer: Option C

Explanation :

Let
x = Number of correct answers
y = Number of wrong answers
z = Number of unattempted questions.

Given, x + y + z = 100    …(1)
6x – 2y – z = 321    …(2)
(1) + (2) : 
⇒ 7x + y = 421
⇒ y = 421 – 7x

We know y ≥ 0
Hence, 421 – 7x ≥ 0
⇒ x ≤ 421/7
⇒ x ≤ 60.14

∴ Highest value of x can be 60, for which y = 1 and z = 39
Next possible value of x = 59, for which y = 8 and z = 33
Next possible value of x = 58, for which y = 15 and z = 27

For a decrease of 1 in x, z decreases by 6. 

∴ Possible values of z = 39, 33, 27, 21, 15, 9, and 3 i.e., 7 values.
⇒ Corresponding values of x = 60, 59, 58, 57, 56, 55 and 54.
⇒ Corresponding values of y = 1, 8, 15, 22, 29, 36 and 43.

∴ There are 7 ways of attempting the exam.

Hence, option (c).

Answer: 97

Explanation :

Consider the solution to first question of this set.

We need to maximize (x + y) and hence minimize z.

The least value of z is 3, hence maximum value of (x + y) = 100 – 3 = 97.

Hence, 97.

Answer: 36

Explanation :

Consider the solution to first question of this set.

The number of correct questions can only be 55 (since it has to be a multiple of 11).

∴ Number of wrong questions will be 36.

Hence, 36.

Answer: 33

Explanation :

Consider the solution to first question of this set.

We need to maximize z.

The highest value of z is 33.

Hence, 33.