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it is based on the right shift operation on the perticular digit
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…
2637
4815
OR
2817
4635
600
copy cat
373
21
20
=> 3s = d
=> 5(s-9) = d
therefore
5s-45=3s
2s = 45
s=22.5
hence
d= 67.5
LCM=48
Then add remainder 3
ans =51
X-340 = X x 4 x 8/ 100
100X – 34000 = 32X
68X = 34000
X = 34000/68
X= 500.
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
Take one screw out from the rest of the three tyres and put the screws in the fourth tyre. Now all tyres will be held by three screws each.
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
66 seconds
20000
6
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
121
31yrs
7/5 * 5/q = 7/10
35/5q = 7/10
q=10