Aptitude and Reasoning Questions

After day 1, A finishes 1/9 of the work.

After day 2, B finishes 1/7 more of the total work. (1/9) + (1/7) is finished.

After day 3, C finishes 1/5 more of total work. Total finished is 143/315.

So, after day 6, total work finished is 286/315.

Now remaining work =  29 /315

On day 7, A will work again 

Work will be completed on day 7 when A is working. He must finish 29/315 of total remaining work.

Since he takes 9 days to finish the total task, he will need 261/315 of the day. 

Total days required is 6 + (261/315) days.

Area of the square = 12×diagonal2=  1/2 * 3.8 * 3.8 = 7.22 sq.m

ja ke mo pe    =>  at a frog's leap          --------(i)

mo la ki so     =>  take a leapahead       --------(ii)

re bo na ja     =>  insects are frog's diet  --------(iii)

From (i) and (ii),   a leap  =>  ki mo       --------(iv)

From (i) and (iii),  frog's   =>   ja           ---------(v)

From (i), (iv) and (v),  'at' => pe

Let C.P =Rs. x

Then as given, (753 - x ) = (x - 455)

=>2x = 1208 => x = 604

 There fore S.P= 150% of 604 => Rs. 150100×604= Rs.906

Given,
The ratio of the doctors to the nurses is 11 : 8
Number of nurses = 8/19 x 209 = 88.

K x 3/8 = 141 => K = 376

376 x 32.08/100 = 120

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be 8C2 games.Since,each pairing of teams will occur three times,the answer will be triple.

Therfore, 8C2*3=84

The logic behind the sequence given in the question is +4, +4, +6, +6
Therefore, the sequence which follows the same logic in the given options is DHLRX.