Sucheta

• Mar 29th, 2006

 

I solved this sum by trial and error method.

no of cabbage is directly proportional to the area.no of cabbage is C and side of square patch is X. Therefore,

?? C=kX^2

C has to be a perfect?square.

now the area is increased by sum amount say Y and no of cabbage is increased by 211

? C+211=K(X+Y)^2

C+211 also has to b a?perfect square

that means our job is to find the two perfect square whose difference is 211.

Consider the square root of first no as a+d and??square root of second no is a+e where a is the base part same for the both the nos and only e is greater than d.

now as the diff between their square is 211 therefore

(a+e)^2 - (a+d)^2 = 211

a^2+2ae+e^2-a^2-2ad-d^ 2=211

2a(e-d)+(e^2-d^2)=211

now i have used trail and error to split 211 as 200+11 where 11 is e^2-d^2 and 200 is 2a(e-d) and i got e=6, d=5, a=100 therefore apples in his patch are 106*106 and apples in her patch is 105*105

tweety

 Profile Answers by tweety Questions by tweety

• Mar 31st, 2006

 

thanks a lot sucheta....u have given me a clear solution for my question

Ashish Sharma

• Apr 6th, 2006

 

Here's the clearest solution i could think of :

With the simple matrix funda, the cabbages would have been arranged in a square matrix sort of fashion ( obviously because we have a square patch ). So let us assume that the matrix this year has 'n' rows and 'n' columns. Obviously the cabbages last year would form a matrix with say 'x' rows and 'x' columns where x

P.S : the answer i.e. 106*106 cabbages this year after 105*105 cabbages last year is also correct ( and can be achieved with the same simple equation )?but has an assumption that the two perfect squares should be consecutive which is not stated in the problem.