Quantitative Aptitude - Arithmetic Ability Questions

we can treat every two consecutive terms as one.
So, we will have a total of 100 terms of the nature:
(2 + 5) + (7 + 6) + (12 + 7).... => 7, 13, 19,....

We know the sum of n terms  nn+12

Now, a= 7, d=6 and n=100
Hence the sum of the given series is
S= 100/2 x[2 x 7 + 99 x 6]
=> 50[608]
=> 30,400.

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 %.

3/15 + 4/16 + x/24 = 1

⇒x=1315

Given that,

 working days = 5

 working hours = 8

 A man get rupess per hour is Rs.2.40

 So in one day the man get total rupees is 2.40 x 8 = 19.2

 So in 5 days week the man get total rupees is 19.2 x 5 = 96

 So in 4 week the man get total rupees is 96 x 4 = 384

 So the man worked for = 160hours in 4 weeks

 But given that the man earned Rs.432

 Hence remaning money is (432-384 = 48)which is earn by doing overtime work

Overtime hours  = 48/3.20 = 15

So total worked hours is = 15 + 160 = 175.

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.

i=j/m

FV
= PV(1 + i)^n

Leap year means the year that can be divisible by 4. From the above options 2009 is not divisible by 4. So it is not a leap year.