Find a number such that 7 more than one–fifthof a number is two –third of the number.
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Step-by-step explanation:
Let the number required is x
According to the question we can write as
(4/5)x−10=(2/3)x
On rearranging,](4/5)x−(2/3)x=10
Now taking the L.C.M of 3 and 5 is 15
(12x−10x)/15=10
On cross multiplication,
2x=150
x=150/2=75
So the required number is 75
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