Quantitative Aptitude - Arithmetic Ability Questions

loss = C.P - S.P = 490 - 465.50 =24.50 

loss % = (loss x 100)/C.P = (24.50 x 100)/490 = 5%

(A+B) - (B-C) = 12 <=> A - C = 12.

For this we have to find the LCM of 24, 36 and 30

LCM of 24, 36 and 30 = 360 sec

360/60 min = 6 minutes.

In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

SP of first article = 1000
Profit = 30%
CP = (SP)x[100/(100+P)] = 10000/13

SP of second article = 1000
Loss = 20%
CP = (SP)x[100/(100-L)] = 5000/4 = 1250

Total SP = 2000

Total CP = 10000/13 + 1250 = 2019.23

CP is more than SP, he makes a loss.

Loss = CP-SP = 2019.23 - 2000 = 19.23
Loss Percent = [(19.23)/(2019.23)]x100 = 0.95 %.

Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year : 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017
Odd day : 1 2 1 1 1 2 1 1 1 2 1

Sum = 14 odd days => 0 odd days.
Calendar for the year 2018 will be the same as for the year 2007.

Let total amount = Rs. P

Total expenditure = (23 + 33 + 19 + 16)% of P = 91%ofP

Remaining money = (100 - 91)% of P = 9% of P

But given in the question that,

9% of P = 504

9p/100 = 504

=> P = 50400/9

=> P = Rs. 5600

Required money spent on Food & Insurance = 23 + 33 %

= 56% of 5600

= 56 x 5600/100 = 56 x 56 = Rs. 3136