Edition 11
1. Let P, Q and R be three sets such that P ∩ Q ∩ R = ∅ and the number of elements in each of P ∆ Q, Q ∆ R and R ∆ P equals 100. Then the number of elements in P ∪ Q ∪ R equals
- (A) 150
- (B) 300
- (C) 230
- (D) 210
- (E) 249
2. The ratio of the area of a triangle XYZ to the area of the triangle whose sides are equal to the medians of the triangle XYZ is
- (A) 2:1
- (B) 3:1
- (C) 4:3
- (D) 3:2
- (E) 1:4
3. Let x and y be positive numbers. Which of the following always implies xy ≥ yx?
- (A) y ≥ e ≥ x
- (B) e ≥ y ≥ x or x ≥ y ≥ e
- (C) e ≥ x ≥ y or y ≥ x ≥ e
- (D) x ≥ e ≥ y
- (E) x/2 ≥ e ≥ y/3
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.