Edition 23

Edition 18

1. In an examination, a student answered 15 questions correctly and secured 40 marks. If there were two types of questions (2 marks and 4 marks questions), how many questions of 2 marks did he answer correctly?

  • (A) 5
  • (B) 10
  • (C) 20
  • (D) 40
  • (E) None of these

2. T is an obtuse angled triangle. Two of its sides are 7 cm and 13 cm. How many possibilities exist for T such that the third side has on integral measure?

  • (A) 8
  • (B) 7
  • (C) 9
  • (D) 15
  • (E) 10

3. Let θ1, θ2, θ3, . . . , θ13 be real numbers and let A be the average of the complex numbers e1, e2, e3, . . . , e13, where i = √−1. As the values of θ’s vary over all 13-tuples of real numbers, the minimum value attained by |A| is

  • (A) −7
  • (B) −8
  • (C) −3
  • (D) −4
  • (E) 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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