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
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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