Quantitative Aptitude - Arithmetic Ability Questions

Let each installment be Rs.x. Then, 

(P.W. of Rs.x due 1 year hence) + (P.W of Rs.x due 2 years hence) + (P.W of Rs. X due 3 years hence) = 7620.

x1+503x100+x1 +503x1002+x1 +503x1003 = 7620

 6x7+36x49+216x343=7620

 294x + 252x + 216x = 7620 x 343

 => x = 3430

 Amount of each installment = Rs.3430

Given the cost price of the articles = Rs. 450

To get overall 20% gain,

Total Selling Price = (20/100) x 450 = 540

One third of the CP = 1/3 x 450 = Rs. 150

But given 1/3 of articles are sold at 10% loss

S.P of 1/3 of articles = 90% of 150

= 90 x 150/100 = 135

Then, S.P on remaining 2/3 goods must be

= 450 - 135 = 405 ...........(1)

CP on remaining goods

= 2/3 x 450 = 300 ............(2)

Profit = SP - CP = 405 - 300 = 105

Profit %  = (105/300) x 100

= 35%.

Required probability:

 4C1*4C1*4C152C3=4*4*422100=165525

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

3 letters can be choosen out of 9 letters in 9C3 ways.

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in (4C2*5C1)+4C3 ways

Hence,required probability = 4C2*5C1+4C39C3 = 17/42

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Herein the given sequence, 4, 2, 1, 0.5, 0.25, 0.125,...

Common Ratio r = 2/4 = 1/2 = 0.5/1 = 0.25/0.5 = 0.125/0.25 == 0.5.