wake up guys and see modulo % can return only integer value.
So answer will be 1 and all other answers are wrong. You can
run your equation in C code and check the output.
The probability will be
white ball = 5/9
red ball = 4/9
The answer can be 5/9 and 4/9 or 1
3hour 45 min = 13500 sec
In 1 sec he covers = 12 m
In 13500 sec he covers = 12×13500 m = (12×13500)/1000 km = 162 km
Ans : 162 km
Number of students behind Aruna in rank = (46 – 12) = 34. So, Arun is 35th from the last
1728
2*3*5*7*11*13*17*19=9699690
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
10C3 = 120
Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE
– 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
0
Quick
1:2