S=D/T
S=624KM/6.5HRS
S=96KM/HR
C. 4
5/9
The answer probably lies in finding the in-centre of the
traingle. Bisect all the angles of the triangles and the
point where these angle bisectors meet gives u the point P
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? 28 m
211
A rude person who doesn’t respect me
X+x+2+x+3=42
3x+6=42
3x=42-6=36
X=12
X+2=14
X+3=15
Middle number 14
24
Fuel to go = x + (x/4)=5x/4
Fuel to come = x
now,
x+(5x/4) = 4.5
9x/4 = 4.5
9x = 18
x = 2 (Fuel to comeup)
Fuel to go will be:
2+(2/4) = 2 + (0.5) = 2.5
999,919 = 991 × 1,009
Number of cats= 991
Number of mice killed by each cat=1,009
M
M. L
L
Petroleum
– 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
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
2100