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CAT 1995 — QA Question 10

Number TheoryEasy
Passage / Data

Direction: Answer the questions based on the following information.
Four sisters — Suvarna, Tara, Uma and Vibha are playing a game such that the loser doubles the money of each of the other players from her share. They played four games and each sister lost one game in alphabetical order. At the end of fourth game, each sister had Rs.32.

The value of 553+453552-55×45+452 is

Answer & solution

  • 100

  • B

    105

  • C

    125

  • D

    75

Solution

The given expression is of the form [x3+y3][x2-xy+y2].

We know, x3 + y3 = (x + y)(x2 - xy + y2).

Thus, required value = (x + y) = (55 + 45) = 100.

CAT 1995 QA Q10: The value of 55 3 + 45 3 55 2 - 55 × 45 + 45 2 is — Solution | TheCATExam