# Edition 23

## Edition 16

1. There are nine coins that are identical in appearance. One weighs more than the others, which have equal weight. With a balance scale to determine the coin that is heavier in only two weighings, how many coins on each side of the balance scale would you weigh first?

• (A) 1 vs. 1
• (B) 2 vs. 2
• (C) 3 vs. 3
• (D) 4 vs. 4
• (E) none of these

2. The sum of the circumferences of two circles which touch each other externally is 176 cm. What is the ratio of the radius of the larger circle to that of the smaller circle, if the sum of squares of the radii of the circles is 400 cm2?

• (A) 7 : 8
• (B) 8 : 7
• (C) 3 : 4
• (D) 4 : 3
• (E) 1 : 2

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. Which of the following value(s) must be in the range of f?

• (A) 5
• (B) 7
• (C) 16
• (D) 25
• (E) All 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

1. Thank you sir for another solution

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.

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.

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.