 # Edition 23

## Edition 13

1. A square ABCD is inscribed in a quarter circle, where B is on the circumference of the circle and D is the centre of the circle. What is the length of diagonal AC of the square if the circle’s radius is 5?

• (A) 3
• (B) 4
• (C) 5
• (D) 6
• (E) The length cannot be determined.

2. In how many ways we can choose three distinct natural numbers from the set {1, 2, 3, . . . ,20} such that their product is divisible by 4?

• (A) 795
• (B) 810
• (C) 855
• (D) 1665
• (E) 954

3. Let M be a 3×3 matrix with all entries being 0 or 1. Then, all possible values for det (M) are

• (A) {0, ±1}
• (B) {0, ±1, ±2}
• (C) {0, ±1, ±3}
• (D) {0, ±1, ±2, ±3}
• (E) {0, ±2, ±3}

1. DEB JYOTI MITRA says:

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. Math1089 says:

Thank you sir for another solution

1. arka bhattacharya says:

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.

1. Math1089 says:

Thank you so much Sir

3. arka bhattacharya says:

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.