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.
23
to find the root of f(x) = 0;
say the work is w and let no of days taken by b is ‘x’ which
we have to calculate
so work done by a in one day is w/6
work done by b in one day is w/x
a and b together can do work in 4 days ie=(w/6)+(w/x)=(w/4)
solving equation x=12
so no of days taken by b=12
1min = 60 seconds
6 min=6*60=360seconds
4 sec=10
360sec=360*10 /4=900
The local value of 7 in the number is at 10000 so 7×10000 and face of 7 is 7
So 70000 -7 so the answer is C 69993
1267
d
HCF= 2
LCM = 2*5*7 = 70
56.25 % is the right answer.
area of square = side ^ 2
at first, it was 100% , means 1
then, a side was enlarged to 25%, means now, 125% means new side is 1.25
taking square will give 1.5625
1.5625 – 1 = 0.5625
so, 56.25% increment in area.
14
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
mausi
E non of those
I would design a voice integrated clock that reads out the time aloud whenever people ask for time
3000