Edition 23

Edition 6

1. The Fibonacci sequence is given by 1, 1, 2, 3, 5, 8, 13, 21, . . . Which one of the ten digits is the last to appear in the unit’s position of a number in the Fibonacci sequence?

  • (A) 0
  • (B) 4
  • (C) 6
  • (D) 7
  • (E) 9

2. Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?

  • (A) 21
  • (B) 60
  • (C) 119
  • (D) 180
  • (E) 231

3. The sides of a triangle with positive area have lengths 4, 6 and x. The sides of a second triangle with positive area have lengths 4, 6 and y. What is the smallest positive number that is not a possible value of |xy|?

  • (A) 2
  • (B) 4
  • (C) 6
  • (D) 8
  • (E) 10

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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