The number of ways in which Varun could have attempted the exam is _____.
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 = 7x - 421
We know y ≥ 0 Hence, 7x - 421 ≥ 0 ⇒ x ≥ 421/7 ⇒ x ≥ 60.14
∴ Least value of x can be 61, for which y = 6 and z = 33. Next possible value of x = 62, for which y = 13 and z = 25. Next possible value of x = 63, for which y = 20 and z = 17. Next possible value of x = 64, for which y = 27 and z = 9. Next possible value of x = 65, for which y = 34 and z = 1.
∴ There are 5 ways of attempting the exam.
Hence, option (b).
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