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9
b
41, 43, 47, 53, 61, 71, 73, 81
81
no
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 ?
Yes answer is 18 days
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
Friends,
The Answer given by ‘Gaurav Sharma’ is correct and
the approach (bottom to top) suggested by ‘Shailesh’ is
good. But with minor correction we can arrive the solution
using this approach:
After 5th loot, No. of breads left = 3
after 4th loot, no. of breads left = (3+0.5)x2 = 7
after 3rd loot, no. of breads left = (7+0.5)x2 = 15
after 2nd loot, no. of breads left = (15+0.5)x2 = 31
after 1st loot, no. of breads left = (31+0.5)x2 = 63
So, before 1st loot, no. of breads left = (63+0.5)x2 = 127
237.6 Kmh
let x be sum.
SI of 18% for 1 year= x*18/100=0.18x;
SI of 18% for 2 years is 0.36x;
SI of 12% for 1 year= x*12/100=0.12x;
SI of 12% for 2 years is 0.24x;
Given, 0.36x-0.24x=840
0.12x=840
x=840/0.12=7000.
First nos series is 7,9,11,?
ie odd number siries ie 7,9,11,13
Second number series is 16,15,14
ie 1 less the previous number 16,15,14,13
Ans —-series is 7,16,9,15,11,14,13,13