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CAT 2023 Question Paper | Quant Slot 1

CAT Previous Year Paper | CAT Quant Questions | Question 1

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 1 : If \(x\) and \(y\) are positive real numbers such that \(\log _x\left(x^2+12\right)=4\) and \(3 \log _y x=1\), then \(x+y\) equals

  1. 20
  2. 11
  3. 68
  4. 10

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Explanatory Answer

\( \left( x ^ { 2 } + 12 \right) = 4 \)
This means, \( x ^ { 2 } + 12 = x ^ { 4 } \)
Let k = x2
k + 12 = k2
k2 – k – 12 = 0
k2 – 4k + 3k – 12 = 0
k(k – 4) + 3(k – 4) = 0
k = 4 or k = -3
But since k = x2, k is always non-negative.
∴ k = 4
∴ x2 = 4
Since x is the base of the log function, it can should always be positive.
∴ x = 2
\( 3 x = 1 \)
\( x ^ { 3 } = 1 \)
\( x ^ { 3 } = y ^ { 1 } \)
y = x3 = 23 = 8
∴ x + y = 2 + 8 = 10


The question is " If \(x\) and \(y\) are positive real numbers such that \(\log _x\left(x^2+12\right)=4\) and \(3 \log _y x=1\), then \(x+y\) equals "

Hence, the answer is '10'

Choice D is the correct answer.

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