I think it’s 32 not 37
– 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
exactly
bogus, counterfeit, fake, false, illegitimate, sham, unreal
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
3
B.25
3 groups
9
9/25
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
26
w, l=2w , area of square =8^2 = 64
area of rectangle = l*b
so w*2w = 8*64
w=16
length = 32