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 17 : In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is
Let BP = x, PQ = y and QC = z.
Since \( \operatorname { Ar } ( A B P ) , \operatorname { Ar } ( A P Q ) \& A r ( A Q C D ) \) are in \( G P \) and \( \operatorname { Ar } ( A Q C D ) = 4 \times \operatorname { Ar } ( A B P ) \)
\( \operatorname { Ar } ( A B P ) : \operatorname { Ar } ( A P Q ) : \operatorname { Ar } ( A Q C D ) = 1 : 2 : 4 \)
\( \frac { 9 x } { 2 } : \frac { 9 y } { 2 } : 27 + \frac { 9 z } { 2 } = 1 : 2 : 4 \)
\( x : y : 6 + z = 1 : 2 : 4 \)
\( y = 2 x \)
\( 6 + z = 4 x \)
x + y + z = 6
x + 2x + 4x – 6 = 6
x = \( \frac { 12 } { 7 } \)
z = 4x – 6 = \( \frac { 48 - 42 } { 7 } = \frac { 6 } { 7 } \)
∴ x = 2z
Therefore, \( x : y : z = 2 : 4 : 1 \)
BP : PQ : QC = 2 : 4 : 1
The question is " In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is "
Choice C is the correct answer.
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