Quantitative Aptitude - Arithmetic Ability Questions

P = 68000, R =503%  & T = 9 months (3/4 years)  

 S.I=p×r×t100=(6800×503×34)100=Rs.8500

We know that the day repeats every 7 days, 14 days, 21 days,...

So if today is Monday, after 7 days it is again Monday, after 14 days again it is Monday.

Hence, after 2 weeks i.e, 14 days the day repeats and is the same day. 

To get the least number of coconuts :
LCM = 30 => 30 + 1 = 31

Let the fixed expenditure of the hotel be Rs.x and the variable expenditure ( which is dependent on the guest ) is Rs.y , then 

          x + 10y = 600        ---------(1)

          x + 20y = 800       ----------(2)

From (1) & (2)

10y = 200 

=>  y = Rs. 20   and    x= 400

Hence the total expenditure when there are 40 guests = 400 + 40 x 20 = 1200

Therefore,  average expenditure = 1200/40 = Rs. 30

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.

log (0.0000134). Since there are four zeros between the decimal point and
the first significant digit, the characteristic is –5.

Given 11 questions of type True or False

Then, Each of these questions can be answered in 2 ways (True or false)

Therefore, no. of ways of answering 11 questions = 211 = 2048 ways.

Let C's age be x years.
Then, B's age = 2x years.
A's age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25 => x = 5
Hence, B's age = 2x = 10 years.