CAT Questions

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Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed = (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
so
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec

x = 31.4208 km/hr and 2x = 62.8416 km/hr.

work done=total number of person x number of days;
half of work done = 140 x 66;
For half of remaining work 25 extra men are added.
Therefore, total men for half work remaining = 140 + 25 = 165;
Let 2nd half work will be completed in K days;
140 x 66 = 165 x K
K = 122;
Hence, extra days => 122-120 = 2days.

Let p be the speed of man in kmph

According to the given data in the question,

Distance travelled by bus in 10 min with 20 kmph == Distance travelled by man in 8 min with (20 + p) kmph in opposite direction

=> 20 x 10/60 = 8/60 (20 + p)

=> 200 = 160 + 8p

=> p = 40/8 = 5 kmph.

Excretion is best described as the removal of Metabolic wastes from a cell.

Let the number be n. Then,

50% of n - 35% of n = 24

50/100 n - 35/100 n = 24

n = (24 x 100)/15 = 160.

Protein digestion occurs in the stomach and duodenum in which 3 main enzymes:

1. Pepsin secreted by the stomach 

2. Trypsin and

3. Chymotrypsin secreted by the pancreas

Break down food proteins into polypeptides that are then broken down by various exopeptidases and dipeptidases into amino acids.

Given X can do in 10 days 

=> 1 day work of X = 1/10

Y can do in 15 days 

=> 1 day work of Y = 1/15

1day work of (X + Y) = 1/10 + 1/15 = 1/6

Given they are hired for 5 days

 => 5 days work of (X + Y) = 5 x 1/6 = 5/6

 Therefore, Unfinished work = 1 - 5/6 = 1/6 

Walking at 3/4th of usual rate implies that time taken would be 4/3th of the usual time. In other words, the time taken is 1/3rd more than his usual time

so 1/3rd of the usual time = 15min
or usual time = 3 x 15 = 45min = 45/60 hrs = 3/4 hrs.