# 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 : RR 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

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

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1. 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.

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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.

Liked by 1 person

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.

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