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CAT 2023 Question Paper | Quant Slot 1

CAT Previous Year Paper | CAT Quant Questions | Question 6

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 6 : Let \(\alpha\) and \(\beta\) be the two distinct roots of the equation \(2 x^2-6 x+k=0\), such that \((\alpha+\beta)\) and \(\alpha \beta\) are the distinct roots of the equation \(x^2+p x+p=0\). Then, the value of \(8(k-p)\) is


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Explanatory Answer

\( \alpha \) and \( \beta \) are the distinct roots of the equation \( 2 x ^ { 2 } - 6 x + k = 0 \).
Therefore,
\( \alpha + \beta = \frac { - ( - 6 ) } { 2 } = 3 \)
\( \alpha \cdot \beta = \frac { k } { 2 } \)
We know that \( ( \alpha + \beta ) \) and \( ( \alpha , \beta ) \) are the roots of the equation: \( x ^ { 2 } + p x + p = 0 \)
\( ( \alpha + \beta ) + ( \alpha \cdot \beta ) = 3 + \frac { k } { 2 } = \frac { - ( p ) } { 1 } = - p \)
\( ( \alpha + \beta ) \cdot ( \alpha \cdot \beta ) = 3 \frac { k } { 2 } = \frac { p } { 1 } = p \)
\( k = \frac { 2 } { 3 } p \)
\( 3 + \frac { k } { 2 } = 3 + \frac { p } { 3 } = - p \)
\( \frac { - 4 p } { 3 } = 3 \)
\( p = - \frac { 9 } { 4 } \)
\( k = \frac { 2 } { 3 } p = - \frac { 3 } { 2 } \)
8(k - p) = 8\( \left( - \frac { 3 } { 2 } + \frac { 9 } { 4 } \right) = 8 \times \frac { 3 } { 4 } = 6 \)


The answer is '6'

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