CAT 2023 Question Paper | Quant Slot 2

CAT Questions

CAT 2023 Quant was dominated by Algebra followed by Arithmetic. 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 : Let both the series $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3 \ldots$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $a_5=b_9, a_{19}=b_{19}$ and $b_2=0$, then $a_{11}$ equals

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Video Explanation

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Explanatory Answer

Let the common differences of the series $a _ { 1 } , a _ { 2 } , a _ { 3 } , \ldots$ and $b _ { 1 } , b _ { 2 } , b _ { 3 }$ be p and q respectively.
We are told that p and q are prime numbers.
$a _ { 5 } = b _ { 9 }$
$a _ { 19 } = b _ { 19 }$
$a _ { 19 } - a _ { 5 } = b _ { 19 } - b _ { 9 }$
$14 \times p = 10 \times q$
$\frac { p } { q } = \frac { 10 } { 14 } = \frac { 5 } { 7 }$
Since $p \& q$ are primes & $\frac { p } { q } = \frac { 5 } { 7 }$, we have $p = 5 \& q = 7$
$a _ { 11 } = a _ { 5 } + 6 p$
$a _ { 11 } = b _ { 9 } + 6 p$
$a _ { 11 } = b _ { 2 } + 7 q + 6 p$
$a _ { 11 } = 0 + 7 \times 7 + 6 \times 5$
$a _ { 11 } = 79$

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The question is " Let both the series $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3 \ldots$ be in arithmetic progression such that the common differences of both the series are prime numbers. If $a_5=b_9, a_{19}=b_{19}$ and $b_2=0$, then $a_{11}$ equals "

The answer is '79'

Choice A is the correct answer.

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