Quantitative Aptitude - Arithmetic Ability Questions

Let the numbers be x and (100-x).

Then,x100-x=5*495

 =>  x2-100x+2475=0

 =>  (x-55) (x-45) = 0

 =>  x = 55 or x = 45

  The numbers are 45 and 55

Required difference = (55-45) = 10

Correct Sum = (36 * 50 + 48 - 23) = 1825.

Correct mean = = 1825/50 = 36.5

Middle numbers  =  [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5.

Let the total profit be Rs. z. Then,

B's share = Rs. 2z/3,  A's share = Rs. ( z - 2z/3 ) = Rs. z/3.

A : B = z/3 : 2z/3 = 1:2

Let the total capital be Rs, X and suppose B's money was used for x months. Then.

(1(x) / 4 * 15) / (3x) / 4 * y) = 1/2 <=> y = (15 * 2 / 3) = 10 .

 Thus, B's money was used for 10 months.

The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005.

So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday.

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

Given 56% of 225 + 20% of 150 = ? - 109

we know that a% of b = b% of a

Now,

225% of 56 + 20% of 150 = ? - 109

200% of 56 + 25% of 56 + 20% of 150 = ? - 109

112 + 14 + 30 = ? - 109

? = 112 + 14 + 30 + 109

? = 264

56% of 870 = 56x870/100 = 487.20

82% of 180 = 82x180/100 = 147.60

32% of 90 = 32x90/100

487.20 + 147.6 - 28.8  =  ?

? = 634.8  - 28.8

? = 606.