Edition 19
1. Let P be a prime number greater than 37. Then the largest number that will always divide (P – 1) × (P + 1) is
- (A) 32
- (B) 16
- (C) 8
- (D) 4
- (E) 24
2. In the given figure, AL is perpendicular to AB and AL || BK || CI || DG || EF. Also AE || KG || JF and AL = AB = 2BK = BC = CI = 4CD = 3IJ. Then the ratio of the areas of the quadrilaterals ABKL and IJKG is
- (A) 3 : 1
- (B) 36 : 19
- (C) 19 : 12
- (D) 22 : 13
- (E) None of these
3. How many functions are there from the set {1, 2, . . . , k} to the set {1, 2, . . . , n}?
- (A) kn
- (B) nk
- (C) nk − 1
- (D) kn − 1
- (E) kn
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.