divide 180 into two parts such that one-third of one part may exceed one-sixth of the other part by 6
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Answered by
20
let,two parts be X and 180-X
so,bye given condition,
x/3=(180-X)/ (6) +6
solve this....we will get
x=72
then,180-X=180-72
=108
So,ans is 108 and 72
so,bye given condition,
x/3=(180-X)/ (6) +6
solve this....we will get
x=72
then,180-X=180-72
=108
So,ans is 108 and 72
Answered by
6
Divide 120 into two parts such that one -tenth of one part may exceed one -fifth of the other part by 2
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