Edition 23

Edition 10

1. The number of triplets (p, q, r) of integers such that p < q < r and p, q, r are sides of a triangle with perimeter 21 is

• (A) 7
• (B) 9
• (C) 8
• (D) 11
• (E) 12

2. The angles of a convex pentagon are in A.P. Then, the minimum possible value of the smallest angle is

• (A) 36°
• (B) 45°
• (C) 30°
• (D) 75°
• (E) 54°

3. Let A be the set of all points (x, y) such that |x| + |y| ≤ 1. Given a point K in the plane, let CK be the point in A which is closest to K. Then the points K for which CK = (1, 0) are

• (A) All points K = (x, y) with x ≥ 1;
• (B) All points K = (x, y) with x ≥ 1 and y < 0;
• (C) All points K = (x, y) with x ≥ 1 and y = 0;
• (D) All points K = (x, y) with xy + 1 and x ≥ 1 − y;
• (E) All points K = (x, y) with x ≥ 0 and y = 0;

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. Thank you sir for another solution

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.

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.