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).

Here the given number series is 2 4 27 256 3125 46656

It follows a pattern that,

$$\begin{gathered} 11 = 1 \\ 22 = 4 \\ 33 = 27 \\ 44 = 256 \\ 55 = 3125 \\ 66 = 46656 \end{gathered}$$

Hence the wrong number in the given number series is 2 instead of 1.

$$\begin{gathered} [(K-L)2-K+L2]4K = ab \\ K2+L2-2KL-K2+L2+2KL4K= ab \end{gathered}$$

=> - 4KL/4K = a/b

=> L = - a/b

=> bL = -a

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 original radius = R. 

New radius = 50100R= 50100R 

Original area =R2  and new area = πR2

3πR24*1πR2*100 

Decrease in area = πR22=πR24 = 75%