area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2
B
C
Put switch 1 on and leave it on for 2 minutes, then switch it off.
Put switch 2 on and leave it on, then walk into the room.
If the light is on, the answer is switch 2.
If the light is off but the bulb is warm when you feel it, the answer is switch 1.
If the light is off and the bulb is cold when you feel it, the answer is switch 3.
Ans is 25
600
900
Each PAIR of stations means a PAIR of tickets (A to B and B to A)
2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46
(N * X) + (X * (X – 1)) = 23
factoring ___ X (N + X – 1) = 23
23 is a prime number with only two factors, 1 and 23
so N = 23 and X = 1
13 and 15
Put a= 4, b = 2 in the equation and multiplying by 2/2 then you will get same value in right hand side. It mean a is 2a which mean b<a
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
– 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
23.5 days