Latest Microsoft Aptitude Question SOLUTION: If you have two buckets, one with red paint and the other with blue paint, and you take one cup from the blue bucket and poor it into the red bucket. Then you take one cup from the

ANS :
Assuming both buckets have the same amount of respective paints;
Both the buckets will have the same ratio.
The red bucket will have ratio of red to blue paints as n:1;
And the blue bucket will have ratio of red to blue paints as 1:n.
[where n >=1]
If the Qn is to find which bucket has the highest ratio of red to blue paints,
then the answer is red bucket.
Reason :
Let the red bucket have n cups of red paint and the blue bucket also have n cups of blue paint.

Step 1: 1 cup of blue paint from blue bucket is poured into red bucket.
Now the red bucket has n cups of red and 1 cup of blue paints.
the paint is mixed thorougly and has uniform distribution.
And the blue bucket has n-1 cups of blue paint.
ratio of red to blue paints in the red bucket = n:1.

Step 2: 1 cup of paint from red bucket is poured into blue bucket.
Now the red bucket will still have the same ratio of paints in it b'coz if it has
the ratio of red to blue as n:1, then any portion we take will have the same
ratio.Therefore, ratio of red to blue paints in red bucket = n:1.
The cup of paint taken fron red bucket has n/(n+1) cup of red and 1/(n+1) cup of
blue paints. [B'coz after first step, the red bucket has n+1 cup of paints, out
of which n are red and 1 is blue.]
Now the blue bucket will have n/(n+1) cup of red and (n-1)+(1/(n+1)) cups of blue.
Ratio of red to blue paints in the blue bucket = [n/(n+1)]/[n-1+1/(n+1)]
= n/((n-1)*(n+1)+1) = n/(n^2-1+1) = n/(n^2) = 1/n = 1:n
Hence proved.

Assuming there are equal ratio of paint between the two buckets so, let's consider the two buckets 5 litres of paint each ,
lets assume 1 litre of paint taken from blue bucket,
I : Blue bucket left with 4L of Blue paint
II : Red bucket have 6L of paint(1L Blue,5L Red) with ratio 1:5 of Blue:Red , afterwards when we take 1L of paint from it , then 1L paint will have the same blue - red ratio i.e. 1:5 ,hence the quantity of Blue:Red in that will be :
Blue - (1/6) = 0.13 Red - (5/6) 0.87 . hence the quantity of blue paint remaining in red bucket = (1-0.13)=0.87 , quantity of red paint in red bucket =(5-0.87)=4.13 , hence the 1L paint consist of 0.13 Blue and 0.87 Red .
When we pour this 1L in Blue bucket , quantity of blue paint in blue bucket is =(4+0.13)=4.13 , quantity of red paint in blue bucket is = (0.87),
Therefore , in Blue Bucket - Blue:Red::4.13:0.87
in Red Bucket -
Quantity of red paint

not assuming anything..

red = p
blue = q

on first pouring, let x
red bucket = p+x
blue bucket = q-x

in red bucket , the ratio between red and blue = p/x
in blue bucket , the ratio between red and blue cant be measure as it only has blue paint;

on second pouring x;

in red bucket , the ratio between red and blue = p - (px/p+x) / x-(x'2/p+x)
in blue bucket , the ratio between red and blue = px / (p+x) * (q-x);

if p and q are equal in quantity, as we can measure only quantity in maths, even if we are measuring quality than also we have to define it in terms of quantity....

so if p>>1 and p>>x than p/x >> px / (p+x)(p-x);

hence proved