 # 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

1. DEB JYOTI MITRA says:

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. Math1089 says:

Thank you sir for another solution

1. arka bhattacharya says:

2. Rajan kumar says:

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. Rajan kumar says:

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.

1. Math1089 says:

Thank you so much Sir

3. arka bhattacharya says:

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.