Question

Divide 184 in two parts such that one-third of one part may exceed one-seventh of another part by 8​

Answer 1

Answer:

Let one half be x

∴ the opposite half is (184 – x)

One-third of x = (x/3)

One-seventh of alternative half  

=> $\frac{184-X}{7}$

Given that tierce of 1 half exceeds common fraction of alternative half by eight.

=>  $\frac{x}{3} =\frac{184-X}{7} +8$

=> $\frac{x}{3} =\frac{184-X+56}{7}$

⇒ 7x = 3(240 – x)

⇒ 7x = 720 – 3x

⇒ 10x = 720

∴ x = 72

The 2 components seventy two and 112.

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@GauravSaxena01