please solve this statement question in maths?
1.divide 60 into two parts such that 3 times the smaller part may exceed 100 by as much as 9 times the bigger Falls short of 200?
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Let one part be x.
∴Second part = 60 - x.
Let (60 - x) be the smaller part.
According to the question,
3(60 - x) - 100 = 200 - 9(x)
⇒180 - 3x - 100 = 200 - 9x
⇒6x = 120
∴x = 20
Thus, 2 parts are 20 and (60 - 20) = 40
∴Second part = 60 - x.
Let (60 - x) be the smaller part.
According to the question,
3(60 - x) - 100 = 200 - 9(x)
⇒180 - 3x - 100 = 200 - 9x
⇒6x = 120
∴x = 20
Thus, 2 parts are 20 and (60 - 20) = 40
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