Quantitative aptitude


This query is : Resolved 

06 November 2011 divide 56 into two parts such that three times the first part exceeds one third of the second by 48. the parts are

06 November 2011 The parts are--

First part - 20
Second part - 36

Total (20+36)=56

Check--
(20*3)-(36/3)=60-12=48.


Method====

Let 1st part be X and 2nd part be Y...

So,

X+Y=56

or, Y=56-X

and

3X-(Y/3)=48

Put the value of Y in above equation and you will find the result.

Value of X will be =20
therefore Y =(56-20)=36


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