divide 184 into two parts such that one third of one part may exceed one seventh of another part by 8
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Let the two parts of 184 be "x" and "y" respectively.
So, we have: x + y = 184 → (1)
Also, 1/3rd of one part exceeds 1/7th of another part by 8.
i.e, 1/3x - 1/7y = 8
⇒7x - 3y = 168 [Multiplying throughout by 21] → (2)
Solving Eqs. (1) and (2), we get,
x=72 and y=112
Hence, the two parts of 184 are 72 and 112.
Hope this helped.
So, we have: x + y = 184 → (1)
Also, 1/3rd of one part exceeds 1/7th of another part by 8.
i.e, 1/3x - 1/7y = 8
⇒7x - 3y = 168 [Multiplying throughout by 21] → (2)
Solving Eqs. (1) and (2), we get,
x=72 and y=112
Hence, the two parts of 184 are 72 and 112.
Hope this helped.
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