CAT 2024 Question Paper | Quant Slot 3

CAT Previous Year Paper | CAT Quant Questions | Question 4

CAT 2024 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 4 A circular plot of land is divided into two regions by a chord of length \(10 \sqrt{3}\) meters such that the chord subtends an angle of \(120^{\circ}\) at the center. Then, the area, in square meters, of the smaller region is

  1. \(25\left(\frac{4 \pi}{3}+\sqrt{3}\right)\)
  2. \(20\left(\frac{4 \pi}{3}-\sqrt{3}\right)\)
  3. \(25\left(\frac{4 \pi}{3}-\sqrt{3}\right)\)
  4. \(20\left(\frac{4 \pi}{3}+\sqrt{3}\right)\)

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Explanatory Answer

\[ O A=O B=\text { Radii; } \angle A O B=120^{\circ} \] So, \[ \angle OAB=\angle OBA=30^{\circ} \] Consider \(O M \perp A B\);
In right triangle \(\triangle OMB\),
\[ \begin{array}{l} \cos 30^{\circ}=\frac{M B}{O B} \\ \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{O B} \\ \Rightarrow O B=10 \end{array} \]
\[ \begin{array}{l} \sin 30^{\circ}=\frac{O M}{O B} \\ \frac{1}{2}=\frac{O M}{10} \\ \Rightarrow O M=5 \end{array} \] Area of the smaller region \(=\) Area of sector - Area of \(\triangle OMB\) \[ \begin{array}{l} =\frac{120}{360} \times \pi \times 10^2-\frac{1}{2} \times 10 \sqrt{3} \times 5 \\ =25\left(\frac{4 \pi}{3}-\sqrt{3}\right) \end{array} \]


The question is "A circular plot of land is divided into two regions by a chord of length \(10 \sqrt{3}\) meters such that the chord subtends an angle of \(120^{\circ}\) at the center. Then, the area, in square meters, of the smaller region is"

Hence, the answer is '\(25\left(\frac{4 \pi}{3}-\sqrt{3}\right)\)'

Choice C is the correct answer.

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