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Circular Track sis of 11 km
Speed of Mens are
4 , 5.5 , 8 km/hr
Time at Which they are at starting Points again
11/4 , 11/5 , 11/8
11/4 , 2.2 , 11/8
We need to find LCM of these
to find at what time they Meet again at starting point
LCM 11 * 2 = 22
After 22 Hrs they will meet at starting point
Relative speed = 60 + 90 = 150 km/hr.
= 150 x 5/18 = 125/3 m/sec.
Distance covered = 1.10 + 0.9 = 2 km = 2000 m.
Required time = 2000 x 3/125 = 48 sec.
Bus started at 8:00
it travelled with 18mph…
distance of destination is 27 miles…
we know velocity=(distance)/time
i.e., time=27/18hours=3/2hours=90 min…
i.e., the bus reached destination at 9:30
the bus stayed for 30 min…
so the bus started return journey at 10:00
now the bus returned with velocity 18 + (18/2)=27mph
the taken to the bus to travel = 27/27=1 hour = 60 min…
so bus would be returned on 11:00…
Use 3pt. Formula
( (x-x1)/(x2-x1) )=( (y-y1)/(y2-y1) )
We get 150
2*3*5*7*11*13*17*19=9699690
30%
suppose the area is 100 and it was increased by 69%, then
the area is 169 which it indicates the side of the square
is 13.. which means 30% increase in its side.
9999991 (6k+_1) form
Calculation:
⇒ If 1000 divided by 112, the remainder is 104. ⇒ 112 – 104 = 8 ⇒ If 8 is added to 1000 it will become the smallest four-digit number and a multiple of 112. ⇒ 1000 + 8 = 1008 ∴ The required result will be 1008.
24000
Answer is 400.(14^2=196, 16^2=256, 18^2=324, 20^2=400.)
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