CAT 2003 Slot 2 — QA Question 10
Number TheoryEasy
If 13x + 1 < 2z, and z + 3 = 5y2, then
Answer & solution
- A
x is necessarily less than y.
- B
x is necessarily greater than y.
- C
x is necessarily equal to y.
None of the above is necessarily true.
Solution
We have,
13x + 1 < 2z … (i)
z + 3 = 5y2 …(ii)
From (i) and (ii), we get,
13x + 1 < 2(5y2 – 3)
∴ 13x + 1 < 10y2 – 6
∴ x < (10y2 – 7)/13
For y = 1, we get x < 3/13
∴ x < y
For y = 2, we get x < 33/13
∴ x could be greater than or less than y.
∴ None of the given options are necessarily true.
Hence, option (d).