Quantitative Aptitude - Arithmetic Ability Questions

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).

 Since the month begins with a Sunday, so there will be five Sundays in the month,

Required average = (510 * 5 + 240 * 25) / 30 = 8550/30 = 285.

52000 ÷ 40 ÷ 65 × 30 = ? - √400

(52000 ÷ 40) ÷ 65 × 30 = ? - 20

(1300 ÷ 65) × 30 = ? - 20

20 × 30 = ? - 20

600 = ? - 20

? = 620

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 cost price of 1-liter pure milk be Re.1, then 

 6 liters (milk) → C.P = Rs. 62 liters (water) → C.P = Rs. 0→CP = Rs.6 only

8 litre mixture  =>

SP =>  8 x 2 = Rs. 16

Profit % = 16 - 66x100 = 10006 = 166.66%

Let the side of the square(ABCD) be x meters.

Then, AB + BC = 2x metres.

AC = 2x = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % =0.59x2x*100 = 30% (approx)

Let there be x voters and k votes goes to loser then 

      0.8x - 120 = k + (k + 200)

         k+200 = 0.41x

 => k = 1440

and  (k + 200) =1640

Therefore  1440/3200 x 100 = 45%