n(S) = 2100
n(E) = No. of favourable ways = 100C1 + 100C3 + ... 100 C99 = 2100-1 = 299
[? nC1 + nC3 + nC5 + ........................... = 2n-1 ]
? P(E)= n(E)/n(S) = 299/2100 = 1/2
Note : The given case can be generalised as "If a unbiased coin is tossed 'n' times, then the chance that the head will present itself an odd number of times is 1/2. "
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
NA
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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