CAT Questions

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Distance d = 1200km
let S be the speed

he walks 15 hours a day(i.e 24 - 9)
so totally he walks for 70 x 15 = 1050hrs.

S = 1200/1050 => 120/105 = 24/21 => 8/7kmph

given 1 1/2 of previous speed
so 3/2 * 8/7= 24/14 = 12/7

New speed = 12/7kmph

Now he rests 10 hrs a day that means he walks 14 hrs a day.

time = 840 x 7 /12 => 490 hrs
=> 490/14 = 35 days

So he will take 35 days to cover 840 km.

The day of the week repeats every 7 days.

Given today is Friday. Again Friday is repeated on the 7th day, 14th,... on 7 multiple days.

Hence, Friday is on the 63rd day, as 63 is multiple of 7.

Now, the required day of the week on the 65th day is Sunday.

5M + 2B = 4(1M + 1B)

5M + 2B = 4M + 4B

1M = 2B

The required ratio of work done by a man and a boy = 2:1

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total. 

Work done by P and Q, working together in one day = 1/6 + 1/18  = 4/18 = 2/9 

They would take 9/2 days = 4 (1/2) days to complete the work working together.

Better is compared with worst and slower must be compared with quickest since fast is the extreme of it.

Hence, Better is to worst as slower is to quickest.

Here in the above given question, the mistake is "the is reapeated twice" before mistake.

Let the pipe B be closed after 'K' minutes.

30/16 - K/24 = 1 => K/24 = 30/16 - 1 = 14/16

=> K = 14/16 x 24 = 21 min.

If a number to be divisile by 88, it should be divisible by both "8" and "11"

Check for '8' :
For a number to be divisible by "8", the last 3-digit should be divisible by "8"
Here 72x23y --> last 3-digit is '23y'
So y=2 [ (i.e) 232 is absolutely divisible by "8"]

Chech for '11' :
For a number to be divisible by "11" , sum of odd digits - sum of even digits should be divisible by "11"
(7 + x + 3) - (2 + 2 + y)
(7 + x + 3) - (2 + 2 + 2)
(10 + x) - 6 should be divisible by "11"
for x = 7
=> 17 - 6 = 11 [ which is absolutely divisible by "11"]

So x = 7 , y= 2.