The answer cannot be determined as there is a particular formula where the consecutive numbers start.
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 ?
Relative speed = 60 + 90 = 150 km/hr.
= 150 x 5/18 = 125/3 m/sec.
Distance covered = 1.10 + 0.9 = 2 km = 2000 m.
Required time = 2000 x 3/125 = 48 sec.
I have been teaching some students for 1 year.
x = 5/3y
x – 4 = y + 4
x – 8 = y
x = 5/3 * (x-8)
5/3x – 40/3 = x
2/3x = 40/3
2x = 40
x = 20
y = 20 – 8
y = 12
x + 4 = 24
y – 4 = 8
Ratio is 24:8 , so 3:1
2, 4, 12, 48, 240, (…..)
C)1440
2*2=4
4*3=12
12*4=48
48*5=240
240*6=1440
Slice the cake
9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec
Average = (1+2+3+4+5+6+7+8+9)/3
= 45/3 = 15
Every side will contain sum of 15.
1st side contains 8 and 7.
2nd side contains 9 and 6.
3rd side contains 1,2,3,4 and 5.
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.
13.5