Edition 14
1. 50 × 50 × 50 × ··· (where there are a hundred 50s) is how many times 100 × 100 × 100 × ··· (where there are fifty 100s)?
- (A) 25 × 25 × 25 × ··· (where there are fifty 25s)
- (B) 4 × 4 × 4 × ··· (where there are fifty 4s)
- (C) 2 × 2 × 2 × ··· (where there are fifty 2s)
- (D) 1 time
- (E) None of these
2. In the above figure, ABC is a right-angled triangle, right angled at B and AD is the external bisector of angle A of triangle ABC. If AC = 17 cm and BC = 8 cm, then AD equals
- (A) 15 cm
- (B) 23 cm
- (C) 60 cm
- (D) 40 cm
- (E) 15√17 cm
3. Let θ1, θ2, θ3, . . . , θ13 be real numbers and let A be the average of the complex numbers eiθ1, eiθ2, eiθ3, . . . , eiθ13, where i = √−1. As the values of θ’s vary over all 13-tuples of real numbers, the maximum value attained by |A| is
- (A) 7
- (B) 8
- (C) 3
- (D) 4
- (E) 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
Thank you sir for another solution
Please explain this answer.
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.
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.
Thank you so much Sir
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.