question is incomplete
10 Camels = 68000
A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m
CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
a) who started with small amount of money?
b) who started with greatest amount of money?
c) what amount did B have?
status after 3 games
A-40
B-40
c-80
d-16
status after 2 games
A-20
B-20
C-40
D-96
status after 1 game
A-10
B-10
C-108
D-48
status after 0 games
A-5
B-93
C-54
D-24
so answers are
a)A
b)B
c)93
Put switch 1 on and leave it on for 2 minutes, then switch it off.
Put switch 2 on and leave it on, then walk into the room.
If the light is on, the answer is switch 2.
If the light is off but the bulb is warm when you feel it, the answer is switch 1.
If the light is off and the bulb is cold when you feel it, the answer is switch 3.
32,7
The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.
Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?
CPNCBZ
How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.
area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2
first fill 3quart pail and pour in 5quart pail.fill again 3
quart pail and fill remaining 2 quarts in 5 quart
pail.hence 1 quart remains in 3quart pail.empty 5quart pail
and pour 1 quart from 3 quart pail. also add another 3
quart from 3 quart pail. hence 1 + 3 quarts = 4 quarts.
B. 3 : 2
Sunday
Monday
Tuesday
2a; 2a+2; 2a+4; 2a+6; 2a+8; 2a+10
t_2 + t_6 = 24 => 2a+2 + 2a + 10 = 24 => a = 3
therefore t_4 = 2(3) + 6 = 12