1 Task : p + (h-1)q amount shall be charged ........... (i)

N tasks & H hrs. of service: Each task will be of (h/n) hrs. of service -> n(p+((h/n) - 1)q) amount shall be charged ..... (ii)

(ii) - (i) = pn+hq-nq - p - hq + q = pn-qn - (p-q) = (p-q)(n-1) dollars => amount shall increase => C ans.
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Amount billed initially with just one task in h hours : p + (h-1)q

If the task is divided in n tasks, assuming each task will take h/n hours.
So, for one task, amount billed : p+ (h/n -1)q
Total amount billed for n tasks together : np + nq(h/n-1)
= np +q(h-n)

Extra billing which has to be paid : np + q(h-n) - p - (h-1)q
= (n-1)p + q(h-n-h+1)
= (n-1)p -q(n-1)
= (n-1)(p-q)
Hence C.

Given: A legal service bills p dollars for the first hour of service on a task and q dollars for each additional hour, where p > q.

Asked: If the client requires a specific number of hours h of service, how does the amount billed to the client change if the work is broken up into n tasks rather than one task?

Bill amount for the whole work (A)= p + (h-1)q
Bill amount for the broken up work (B)= np + (h-n)q

Difference (B-A) = np + (h-n)q - p - (h-1)q = (n-1)p - (n-1)q = (n-1)(p-q)

IMO C
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We can let p = 5, q = 2, h = 7 and n = 3. Furthermore, let’s say that the 3 tasks last 2, 2, and 3 hours, respectively.

Therefore, as one task, the total amount billed would be:

5 + 2(7 - 1) = 5 + 12 = 17 dollars

However, as three tasks, the total amount billed would be:

[5 + 2(2 - 1)] + [5 + 2(2 - 1)] + [5 + 2(3 - 1)] = 7 + 7 + 9 = 23 dollars.

We see that the total amount billed would increase by 6 dollars (and we can reject choices D and E). Let’s look at the other 3 choices.

A. p - q = 5 - 2 = 3 → This is not 6.

B. h(p - q) = 7(5 - 2) = 21 → This is not 6.

C. (n - 1)(p - q) = (3 - 1)(5 - 2) = 6 → This is 6.

Answer: C
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