4
Alcohol content from
2 l of A (62.5%) : 1.35L
4L of B (87.5%) : 3.50L
Total Acohol content in 6 L : 4.85L or 80.8%
E. Copper
A Roman was born the first day of the 35th year before Christ means before the start of year 0, 35 years has been past and died the first day of the 35th year after Christ means 35th year is about to start at his death so only 34 years has been completed so total years = 69
5/9
45
1>=y>x => y belongs to (-infinity,1]
x belongs to (-infinity,y)
if both x and y are negitive z can be greater than zero and
can be greater than y(1 and 4 are true)
if y equals 1 then x can be equal to z(2 is true)
therefore y=z is not true for any value of x and y
(1) There exists a smaller natural number.
(2) There exists a largest natural number.
(3) Between two natural numbers, there is always a natural number.
Which of the above statement is/are correct?
B
– 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
(10×1000) / (2×3.1416×1.75) = 909 times (about)
L*I/3(I-H)
16