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Day 119: SSC CGL Tier II Quant

2 questions. Try each one first, then open its answer.

  1. Question 1

    Triangles. Asked once: 15 Nov 2020 Tier II

    16
    In Δ\DeltaABC, ∠A=90∘\angle A = 90^\circ, AD is the bisector of ∠A\angle A meeting BC at D, and DE ⊥\perp AC at E. If AB = 10 cm and AC = 15 cm, then the length of DE, in cm, is:

    Tap an option to check your answer.

  2. Question 2

    Triangles. Asked once: 15 Nov 2020 Tier II

    41
    In Δ\DeltaABC, ∠C=90∘\angle C = 90^\circ. Points P and Q are on the sides AC and BC, respectively, such that AP : PC = BQ : QC = 1 : 2. Then, AQ2+BP2AB2\frac{AQ^2 + BP^2}{AB^2} is equal to:

    Tap an option to check your answer.