1. 10

2. 15

3. 12

4. 20

**ANSWERS WITH SOLUTIONS:**

(1) 3

(2) 4

(3) 2 dog : hare = 3*3 leaps of hare :5*1

9:5

(4) 4

(5) Let the quantity of the wine in the cask originally be x litres.

Then, quantity of wine left in cask after 4 operations = [x(1-8/X)4]litres

= [x(1-8/X)4] = 16/81

x = 24

(6) 1

(7) Speed of the first train = 120/10 = 12 m/sec.

Speed of the second train =120/15= 8 m/sec.

Relative speed = (12 + 8) = 20 m/sec.

Therefore Required time =(120 + 120)/20 = 12 sec.

(8) 1 Let Arun’s weight by X kg.

According to Arun, 65 < X < 72

According to Arun’s brother, 60 < X < 70.

According to Arun’s mother, X <= 68

The values satisfying all the above conditions are 66, 67 and 68.

Required average = 66 + 67 + 68/3 = 201/3 = 67 kg.

(9) 4

L.C.M. of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

(10) 3