Consider from 2nd June 2010 to 2nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1 + 1 + 2 = 4 so 2nd June 2010 must be 4 days back from Sunday and that day is Wednesday.
From Zeller's Formula:
f = k + [13 x m - 1/ 5 ] + D + [D/4] + [C/4] - 2 x C.
In this case k = 2 (since 2nd June)
Month m = 4 (As march = 1, April = 2, May = 3, June = 4 )
D is the last two digit of year here D = 10 (As year is 2010)
C is 1 st two digit of century here C = 20 (As year is 2010)
f = 2 + [13 x 4 - 1 / 5] + 10 + [10/ 4] + [ 20/4] - 2 x 20.
f = 2 + [51/5] + 10 +[2.5] + [5] - 40.
f = 2 + 10 + 10 + 2 + 5 - 40 = -11
This - ve value of f can be made positive by adding multiple of 7
So f = - 11 + 14 = 3
When divided by 7 we will get remainder 3, hence number of odd days is 3,
So 2nd June 2010 is 3 days more than Monday, i.e Wednesday.
? a / b = 2/3 and 2 / a = 1/2
? a/b x 2/a = 2/3 x 1/2
? 2/b = 1/3
? b = 6
0.6 x a = 0.09 x b
? a/b = 0.09 / 0.06 = 9 / 60 = 3 / 20 = 3 : 20
Let a/3 = b/8 = K
Then, a = 3K, b = 8K
? (a + 3) / (b + 8) = (3K + 3) / (8K + 8)
= [3( K + 1)] / [8(K+1)] = 3/8
? ( a + 3 ) : (b + 8) = 3 : 8
Series pattern +15, + 30, +60, +120
? Missing term = 49 + 60 = 109
777777 / 11 = 70707.
NA
According to the question,
1728 = 1331(1 + R/100)3
1728/1331 = (1 + R/100)3
? (12/11)3 = (1 + R/100)3
? 1 + R/100 = 12/11
? R/100 = (12/11) - 1 = 1/11
? R = 100/11= 9.09%
Here, M + N = 10 and MN = 20
According to the question,
(1/M) + (1/N) = (M + N) / MN = 10/20 = 1/2
Time = distance / speed
=110 / {60 x 18/5} = 6.6 sec.
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