Quantitative Aptitude - Arithmetic Ability Questions

Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E = event of getting at most two heads.

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.

P(E) =n(E)/n(S)=7/8.

L.C.M. of 6, 9, 15 and 18 is 90.

Let required number be 90k + 4, which is multiple of 7.

Least value of k for which (90k + 4) is divisible by 7 is k = 4.

=>Required number = (90 x 4) + 4   = 364.

We know that,

Odd days --> days more than complete weeks

Number of odd days in 400/800/1200/1600/2000 years are 0.

Hence, the number of odd days in first 1600 years are 0.

Number of odd days in 300 years = 1

Number of odd days in 49 years = (12 x 2 + 37 x 1) = 61 days = 5 odd days

Total number of odd days in 1949 years = 1 + 5 = 6 odd days

Now look at the year 1950

Jan 26 = 26 days = 3 weeks + 5 days = 5 odd days

Total number of odd days = 6 + 5 = 11 => 4 odd days

Odd days :-

0 = sunday ;

1 = monday ;

2 = tuesday ;

3 = wednesday ;

4 = thursday ;

5 = friday ;

6 = saturday

Therefore, Jan 26th 1950 was Thursday.