CAT 2023 Question Paper | Quant Slot 3

CAT 2023 Quant was dominated by Arithmetic followed by Algebra. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 7 : Let \(n\) be any natural number such that \(5^{n-1} \lt 3^{n+1}\). Then, the least integer value of \(m\) that satisfies \(3^{n+1} \lt 2^{n+m}\) for each such \(n\), is

Video Explanation

Explanatory Answer

\( 5 ^ { n - 1 } < 3 ^ { n + 1 } \)
\( 5 ^ { n - 1 } < 9 \times 3 ^ { n - 1 } \)
\( 1 < 9 \times \left( \frac { 3 } { 5 } \right) ^ { n - 1 } \)
\( ( 0.6 ) ^ { n - 1 } > \frac { 1 } { 9 } = 0.111 \)
\( 0.6 ^ { 4 } = 0.1296 ; 0.6 ^ { 5 } = 0.07776 \)
\( \therefore n - 1 \leq 4 \)
\( n \leq 5 \)
\( 1 \leq n \leq 5 \)
\( 3 ^ { n + 1 } < 2 ^ { n + m } \)
\( 3 ^ { n } \times 3 < 2 ^ { n } \times 2 ^ { m } \)
\( ( 1.5 ) ^ { n } \times 3 < 2 ^ { m } \)
We see that m increases as n increases. So we just need to check the minimum value of m when n reaches its maximum value.
\( ( 1.5 ) ^ { 5 } \times 3 < 2 ^ { m } \)
\( 2 ^ { m } > 22.78125 \)
\( 2 ^ { 5 } = 32 \)
\( \therefore m \geq 5 \)
The minimum value of m is 5.

The question is " Let \(n\) be any natural number such that \(5^{n-1} \lt 3^{n+1}\). Then, the least integer value of \(m\) that satisfies \(3^{n+1} \lt 2^{n+m}\) for each such \(n\), is "

Hence, the answer is '5'

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