Edition 5
1. If none of the faces of a tetrahedron is an isosceles triangle, what is the minimum number of edges no two of which have the same length?
- (A) 3
- (B) 4
- (C) 5
- (D) 6
- (E) 8
2. Consider the following two statements about two triangles:
- P: They have equal area, and two sides of one are equal to two sides of the other.
- Q: They are congruent.
Which of the following statements is true?
- (A) P is necessary and sufficient for Q.
- (B) P is necessary but not sufficient for Q.
- (C) P is sufficient but not necessary for Q.
- (D) P is neither necessary nor sufficient for Q.
- (E) None of these
3. In the sequence 1, 4, 8, 10, 16, 21, 25, 30 and 43, how many blocks of consecutive terms have sums divisible by 11?
- (A) 1
- (B) 2
- (C) 5
- (D) 6
- (E) None of these
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.