35 km/h = 10 m/s
Speed = Distance/Time
Distance= (100+150)m = 250m
Speed = 10 m/s
Time= 250/10 = 25s
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
d
360
6.40 am
China
B.25
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 ?
The phrase is usually ‘like looking for a needle in a haystack’. It describes a task that is virtually impossible because you would have to search a huge area
The answer for this question is very simple: it is called Ceaser Cipher.
It consists of replacing a character by another located in (current character position + (key-1)), where key = how many positions you want to skip.
For Example: VICTORY -> YLFWRUB with the key = 3.
Then SUCCESS -> QXFFHQQ
Note: it should work like a circle(Z+1 = A).
Assume there are 52 weeks in one year.
Since he supposed to have a new order for every two weeks, he
needs 52/2 = 26 orders to break the office record.
Now after 28 weeks,he has got only 28/2 -6 = 8 orders.
Hence,he needs 26-8 = 18 new orders in the remaining 24= 52-28 weeks to break the office record.
Compute 24 orders/18 weeks = 4/3 orders/week ,
we see that averagely he has a new order for
every 4/3 weeks in the remaining weeks to break the office record.
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