Quantitative Aptitude - Arithmetic Ability Questions

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.

(A+B+C) do 1 work in 10 days.

So (A+B+C)'s 1 day work=1/10 and as they work together for 4 days so workdone by them in 4 days=4/10=2/5

Remaining work=1-2/5=3/5

(B+C) take 10 more days to complete 3/5 work. So( B+C)'s 1 day work=3/50

Now A'S 1 day work=(A+B+C)'s 1 day work - (B+C)'s 1 day work=1/10-3/50=1/25

A does 1/25 work in in 1 day

Therefore 1 work in 25 days.

The given number series is 3, 11, 80, 291, 803

$$\begin{gathered} 3 \\ 3 + 23 = 3 + 8 = 11 \\ 11 + 43 = 11 + 64 = 75 (not 80) \\ 75 + 63 = 75 + 216 = 291 \\ 291 + 83 = 291 + 512 = 803 \end{gathered}$$

The given number series is 5, 8, 40, 90, 664

5

5 - 1 x 2 = 8

8 - 3 x 4 = 20 (NOT 40)

20 - 5 x 6 = 90

90 - 7 x 8 = 664