CAT 2021 Question Paper | Quant Slot 3

CAT 2021 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 20 : If \(n\) is a positive integer such that \((\sqrt[7]{10})(\sqrt[7]{10})^{2} \ldots(\sqrt[7]{10})^{n}>999\), then the smallest value of \(n\) is

Video Explanation

Explanatory Answer

\( ( \sqrt [ 7 ] { 10 } ) ( \sqrt [ 7 ] { 10 } ) ^ { 2 } ( \sqrt [ 7 ] { 10 } ) ^ { 3 } ( \sqrt [ 7 ] { 10 } ) ^ { 4 } ( \sqrt [ 7 ] { 10 } ) ^ { 5 } \ldots \ldots ( \sqrt [ 7 ] { 10 } ) ^ { n } > 999 \)
999 can be approximated as 1000 as we are dealing with powers of 10, 1000 can be written as 10³.
\( ( \sqrt [ 7 ] { 10 } ) ( \sqrt [ 7 ] { 10 } ) ^ { 2 } ( \sqrt [ 7 ] { 10 } ) ^ { 3 } ( \sqrt [ 7 ] { 10 } ) ^ { 4 } ( \sqrt [ 7 ] { 10 } ) ^ { 5 } \ldots \ldots ( \sqrt [ 7 ] { 10 } ) ^ { n } > 10 ^ { 3 } \)
The bases are the same in the product so the powers will get added up till n.
So,
\( ( 10 ) ^ { \frac { 1 } { 7 } } \cdot \frac { n ( n + 1 ) } { 2 } > 10 ^ { 3 } \)
Comparing the bases on both the sides of the inequality,
\( \frac { n ( n + 1 ) } { 14 } > 3 \)
n(n+1) > 42
Can n be 6?
If n was 6, then 6×7 = 42
Which just works, because we approximated 999 as 1000 and so without the approximation the right hand side would be smaller than 42.
Hence 6 satisfies the condition.

The question is " If \(n\) is a positive integer such that \((\sqrt[7]{10})(\sqrt[7]{10})^{2} \ldots(\sqrt[7]{10})^{n}>999\), then the smallest value of \(n\) is "

Hence, the answer is '6'