Edition 23

Edition 12

1. Suppose you have a 12-hour digital clock where the number representing the hour is always the same as the number representing the minute. That is, the clock can only show times like 8:08, 9:09, 10:10, etc. What is the smallest time difference between two such times?

  • (A) 101 minutes
  • (B) 61 minutes
  • (C) 60 minutes
  • (D) 49 minutes
  • (E) 11 minutes

2. Let Г be the circle of radius 1 around 0 in the complex plane and z0 be a fixed point on Г. Then the number of ordered pairs (z1, z2) of points on Г such that z0 + z1 + z2 = 0 is

  • (A) 0
  • (B) 1
  • (C) 2
  • (D) 3
  • (E) ꝏ

3. The maximum of the areas of the isosceles triangles lying between the curve y = ex and the x axis, with base on the positive x axis, is

  • (A) e
  • (B) 1/e
  • (C) 1
  • (D) 1/2
  • (E) 2

7 comments

  1. Problem #2) Solution :
    B D=x cm;;
    C o s
    Or (A B ^2+ B D ^2-A D ^2)/(2 * A B* B D)=(A C ^2+C D ^2-A D ^2)/(2* A C * C D);
    Or ( 1 7^2 +x ^2 -1 5 ^2/(2* 17 * x)=(1 7^2 + 4 ^2-1 5 ^2)/(2 * 17* 4);
    Or (64+ x^2)/x= (80)/4=20;
    Or 64+ x^2=20 *x; or x=4 or 16 ;
    Admissible value of x=16B D= x cm=16 cm

    1. Solution of problem1.
      First we will measure 3 conis each side.
      Case1 – if balance is eqall we will measure two out of 3 coins putting one on each side, thus we will get the heavier one.
      Case 2 – if one side is heavier them we will take 2 coins of the heavier side and measure by putting one each side, thus we can find the heavier one.

  2. Solution of problem1.
    First we will measure 3 conis each side.
    Case1 – if balance is eqall we will measure two out of 3 coins putting one on each side, thus we will get the heavier one.
    Case 2 – if one side is heavier them we will take 2 coins of the heavier side and measure by putting one each side, thus we can find the heavier one.

  3. Question 1 solution :
    (c)
    As 10 houses have less than 6rooms, they are to be excluded.
    Given,
    4 houses have more than 8 rooms .
    Therefore, number of houses having either less than 6 or greater than 8 rooms = 10 + 4 = 14.
    The remaining houses, that is 11 houses fulfill the above mentioned criteria.

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