AIEEE Questions

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The distance to be covered = Sum of their lengths = 200 + 300 = 500 m.

Relative speed = 72 -36 = 36 kmph = 36 x 5/18 = 10 mps.

Time required = d/s = 500/10 = 50 sec.

Downstream speed = 72km/8hrs  = 9 kmph
upstream speed = 84km/12hrs = 7 kmph
speed of boat = avg of downstream and upstream speeds
speed of boat = (9+7)/2kmph = 8 kmph.
current speed = half of the difference of downstream and upstream speeds
currend speed = (9-7)/2kmph = 1 kmph

First send 17 and 7. After 7 minutes, 7 is at finish and 17 needs 10 more minutes to complete his travel. now send 13. After 10 minutes 17 reaches finish and 13 needs 3 more minutes to complete. now send 3. both 3 and 13 reach destination at the same time after 3 minutes. hence the time taken is 7+10+3=20 minutes.

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Vol of silver = πR2h  = 66 C.C  

Given D= 1mm => R = D/2 = 1/20 cm  

=> 227×1202×h = 66

= 66 x 400 x (7/22) = 8400 cm = 84 mtrs.

Given ratio of initial mixture of milk and water in Q = 5 : 3

Let the initial quantity of mixture in vessel Q = 8x

Let quantity of Milk = 5x and

Let quantity of water = 3x

According to the question,

5x + 10 x 353x + 10 x 25 = 85

=> 25x + 30 = 24x + 32

=> x = 2

Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.

Let there are L men

job requires LK man hours.

job completed in first 4 hrs = L x 4 = 4L

job completed in next 4 hrs = 4 x L/2 = 2L

job completed in next 4 hrs = 4 x L/4 = L

job completed in last 4 hrs = 4 x L/8 = L/2

4L + 2L + L + L/2 = KL

K = 7+1/2 = 7.5 hours.

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