CAT 2023 QA Question and Solution
Question 13
A lab experiment measures the number of organisms at 8 am every day. Starting with 2 organisms on the first day, the number of organisms on any day is equal to 3 more than twice the number on the previous day. If the number of organisms on the \( n \)-th day exceeds one million, then the lowest possible value of \( n \) is:
A. 16
B. 17
C. 18
D. 19
Answer: D. 19
Detailed Solution:
Let the number of organisms on day 1 be 2.
On day 2, the number of organisms = \( 2 \times 2 + 3 = 7 \).
On day 3, the number of organisms = \( 2 \times 7 + 3 = 17 \).
On day 4, the number of organisms = \( 2 \times 17 + 3 = 37 \), and so on.
The number of organisms on day \( n \) can be represented as: \( 2^{(n-1)} + 3 \times (2^{(n-1)} – 1) \).
For the number to exceed one million, we solve for \( n \) such that:
\( 5 \times 2^{(n-1)} – 3 \geq 10^6 \).
Solving, \( 2^{(n-1)} \geq 200,000 \Rightarrow n – 1 \geq 18 \Rightarrow n = 19 \).
