Edition 20
1. When a certain natural number is multiplied by 7, the product obtained consists only of fives. What is the least value of such a natural number?
- (A) 78465
- (B) 79365
- (C) 755
- (D) 7965
- (E) None of these
2. In the given figure, PQ = QR = RS = ST = TU = UV = VP. The value of ∠SPT is
- (A) 30°
- (B) 26°
- (C) 20°
- (D) 15°
- (E) None of these
3. Let R be the set of real numbers. A continuous function f : R → R satisfies f(1) = 1, f(2) = 4, f(3) = 9 and f(4) = 16. Suppose f is a polynomial, then which of the following are possible values of its degree?
- (A) 5
- (B) 3
- (C) 4
- (D) 2
- (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.