CAT Questions

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Let the present ages of the mother and daughter be 2x and x years respectively.
Then, (2x - 18) = 3(x - 18) => x = 36
Required sum = (2x + x) = 108 years.

Any two digit number can be written as (10P + Q), where P is the digit in the tens place and Q is the digit in the units place.
P + Q = 8 ----- (1)
(10Q + P) - (10P + Q) = 54
9(Q - P) = 54
(Q - P) = 6 ----- (2)
Solve (1) and (2) P = 1 and Q = 7
The required number is = 17

55 min. spaces are covered in 60 min.
60 min. spaces are covered in ((60/55) x 60) min = 720/11 min.
Loss in 65 min. = (720/11 - 65 ) = 5/11
Loss in 24 hrs = (5/11 x 1/65 x 24 x 60)min = 1440/143 min.

At 9 o’clock, the hour hand is at 9 and the minutes hand is at 12, i.e., the two hands are 15 min. spaces apart.
So, the minute hand should gain = (30 - 15) minutes = 15 minutes
55 minutes will be gained in 60 min.
15 minutes spaces will be gained in ((60/55) x 15) min. = 180/11 min.
The hands will be in the same straight line but not together i.e.,in 180 degrees at 180/11 min. past 9.

K's work in a day(1st day) = 1/30
L's work in a day(2nd day)= -1/60(demolishing)
hence in 2 days, combined work= 1/30 - 1/60 =1/60
since both works alternatively, K will work in odd days and L will work in even days.
1/60 unit work is done in 2 days
58/60 unit work will be done in 2 x 58 days = 116 days
Remaining work = 1-58/60 = 1/30
Next day, it will be K's turn and K will finish the remaining 1/30 work.
hence total days = 116 + 1 = 117.

In this problem , it has considered that 65 mins = 1hr
So mins has increased by 5 mins so multiply 5 x 24 = 120 mins extra ,
That is now per day it adds 2hr extra, so divide 1440/26 = 59.384 days =~ 60 days.