Quantitative Aptitude - Arithmetic Ability Questions

I. 2x2 + 19x + 42 = 0

2x2 + 12x + 7x + 42 = 0

2x (x+6) + 7(x +6) = 0

(2x+7) (x+6) = 0 x = -7/2, -6 I

I. 4y2 + 43y + 30 = 0

4y2 +40y + 3y + 30 = 0

4y(y + 10) + 3(y + 10) (4y + 3)(y + 10) = 0

y = -3/4, -10

So answer is no relation.

15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 = 4 odd days.

Jan. Feb. Mar.  Apr.  May.  Jun.  Jul.  Aug.

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 =  0 odd days.

Given day is Sunday

The given series follows the pattern that,

×5+5,   ×5+5,   ×5+5,   ×5+5,   ×5+5

38 x 5 + 5  = 195

195 x 5 + 5 = 980

980 x 5 + 5 = 4905

4905 x 5 + 5 = 24530

24530 x 5 + 5 = 122655

So the missing number is 980 .

Let the numbers be x, x + 2, x + 4, x + 6 and x + 8.

Then  [x + (x + 2) + (x + 4) + (x + 6) + (x + 8) ] / 5 = 61.

or 5x + 20 = 305 or x = 57.

So, required difference = (57 + 8) - 57  = 8

When R and W are the first and last letters of all the words then we can arrange them in 5!ways. Similarly When W and R are the first and last letters of the words then the remaining letters can be arrange in 5! ways.

Thus the total number of permutations = 2 x 5!  = 2 x 120 = 240

ANS:      log₅512 = log512/log5  =  log29log10/2  =9log2log10-log2 =9*0.30101-0.3010 =2.709/0.699 =2709/699 =3.876