AIEEE Questions

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Number of boys in the row = number of boys uptil P + number of boys between P and Q + number of boys including Q and those behind Q = 14 + 4 + 7 = 25.

K = 100 L = 130
130------ 30
100------- ? => 23 1/13 %

Number of persons between Amrita and Mukul = 50 - (10 + 25) = 15. Since Mamta lies in middle of these 15 persons, so Mamta`s position is 8th from Amrita i.e. 18th from the front.

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

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.