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A Managers paid a different amount to six Employees – A, B, C, D, P and R, not necessarily in the same order. The amount paid by him to each person is not necessarily a whole number. He paid them one after the other and paid no two persons at the same time.

Further, it is known that
1. P received 1200 which was the least amount paid by the Managers.
2. Except the first and second payments, the amount paid by the Managers was the arithmetic mean of the previous two amounts paid by him.
3. He paid at least two of the Employees an amount more than 2200.
4. R received his payment before B and D. Also, D received 150 more than R.
5. P and C, who did not receive the highest payment, received their payments consecutively.

Source : Me

Please tell the logic.

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Solution (in order)

$$A=3600, P=1200, C=2400, R=1800, B=2100, D=1950$$

Reasoning

Given constraint number 2, it must be the case that the highest and lowest paid employees are the first two. Since $P$ is the lowest paid employee, he must be in the first two. $C$ is not the highest paid but is consecutive with $P$ so must be the third person paid. $R$ is also not the highest paid but is paid before $B$ and $D$ so $R$ must be fourth. This leaves $A$ as being paid first i.e, the order is $$A, P, C, R, B/D, D/B$$ Now, by rule 2, we have $R = \frac{P+C}{2} = 600 + \frac{C}{2}$.
If $D$ is paid 5th, then we have $D = \frac{C+R}{2} = \frac{3R}{2} - 600$ which makes $R = 1500$ by rule 4. From here we could expand the whole solution set using rule 2 but only one employee, $A$, will have more than $2200$, contravening rule 3. So we must have that $B$ is 5th and $D$ is 6th.

This means that $B = \frac{3R}{2} - 600$ and subsequently, $D = \frac{B+R}{2} = \frac{5R}{4} - 300$ and from rule 4, we get $R=1800$.

Expanding this out using rule 2, we get $$A=3600, P=1200, C=2400, R=1800, B=2100, D=1950$$

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