CAT 2025 Slot 2 QA Q21 Geometry

Question 21 of the Quantitative Ability (QA) section, CAT 2025 Slot 2, with CAT's official answer and a worked solution.

In a \( \triangle A B C \), points D and E are on the sides BC and AC , respectively. BE and AD intersect at point T such that \( \mathrm { AD } : A T = 4 : 3 \), and \( \mathrm { BE } : \mathrm { BT } = 5 : 4 \). Point F lies on AC such that DF is parallel to BE . Then, \( \mathrm { BD } : \mathrm { CD } \) is
  1. A
    \( 9 : 4 \)
  2. B
    \( 15 : 4 \)
  3. C
    \( 11 : 4 \)
  4. D
    \( 7 : 4 \)

Answer: (C) 11 : 4 Geometry MCQ

Solution

  1. \(AT : TD = 3 : 1\) and \(BT : TE = 4 : 1\). Use mass points with masses \(a, b, c\) at \(A, B, C\).
  2. On \(AD\): \(a \cdot 3 = (b + c) \cdot 1\), so \(b + c = 3a\). On \(BE\): \(b \cdot 4 = (a + c) \cdot 1\), so \(a + c = 4b\).
  3. Subtracting: \(b - a = 3a - 4b \Rightarrow b = \frac{4a}{5}\), \(c = \frac{11a}{5}\).
  4. \(BD : DC = c : b = 11 : 4\).

Shortcut. Mass points make cevian-ratio questions two equations; the point F is not needed.

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