Edition 23

Edition 10

1. The number of triplets (p, q, r) of integers such that p < q < r and p, q, r are sides of a triangle with perimeter 21 is

  • (A) 7
  • (B) 9
  • (C) 8
  • (D) 11
  • (E) 12

2. The angles of a convex pentagon are in A.P. Then, the minimum possible value of the smallest angle is

  • (A) 36°
  • (B) 45°
  • (C) 30°
  • (D) 75°
  • (E) 54°

3. Let A be the set of all points (x, y) such that |x| + |y| ≤ 1. Given a point K in the plane, let CK be the point in A which is closest to K. Then the points K for which CK = (1, 0) are

  • (A) All points K = (x, y) with x ≥ 1;
  • (B) All points K = (x, y) with x ≥ 1 and y < 0;
  • (C) All points K = (x, y) with x ≥ 1 and y = 0;
  • (D) All points K = (x, y) with xy + 1 and x ≥ 1 − y;
  • (E) All points K = (x, y) with x ≥ 0 and y = 0;

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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