Wb games Interview Questions, Process, and Tips

8400

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 ?

under pressure a person more effort to his work

b

xy+2y-x=6
xy+2y-x-2=6-2
y(x+2)-1(x+2)=4
(y-1)(x+2)=4
so c is 4

Triveni Sangam of the river
Ganga, Yamuna, and Saraswati at Allahabad.

8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8

2100

when know that 13 is factor of 39 so it will also not be able to divide it differently. so the remainder will be same 20 but as we know that 20 is greater then 13 so we can divide it once again and in the casE of 13 the remainder will be 20-13=7

The Answer is 3

1/221

on the earth

let,first num be x and second num be y
hcf =264
lcm=44
given equation
x/2=44 i.e x=88
product of 2 numbers is equal to lcm*hcf
88*y=44*264
y=132

1+0^3=1
1+1^3=2
2+2^3=10
10+3^3=37
37+4^3=101
101+5^3=226
1,2,10,37,101,226……..