Edition 17
1. On a street there are 25 houses —10 of the houses have fewer than six rooms, 10 of the houses have more than seven rooms, and 4 houses have more than eight rooms. What is the total number of houses that are either six, seven or eight rooms?
- (A) 5
- (B) 9
- (C) 11
- (D) 14
- (E) 15
2. ABCD is a trapezium, in which AD and BC are parallel. If the four sides AB, BC, CD and DA are respectively 9 cm, 12 cm, 15 cm and 20 cm, then the magnitude of the sum of the squares of the two diagonals is
- (A) 638
- (B) 786
- (C) 838
- (D) 648
- (E) None of these
3. You have a piece of land close to a river, running straight. You are required to cut off a rectangular portion of the land, with the river forming one of the sides of the rectangle so, your fence will have three sides to it. You only have 60 meters of fencing. The maximum area that you can enclose is
- (A) 450 square meters
- (B) 460 square meters
- (C) 440 square meters
- (D) 445 square meters
- (E) 455 square meters
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.