Question

(a) worked out incorrectly; or (b) worked out correctly but using a method not specified in the problem statement; or (c) not solved completely, but the answer is stated with inadequate or no work shown, EVEN IF THE FINAL ANSWER IS CORRECT.

Answer 1

Answer:- Darren's total pay for the entire 12 months is $ 10260 (Ans.) The detailed calculation is given below:- This is basically a problem based on Arithmetic Progression, where we have an initial term (the first month's salary) a common difference (fixed amount of increment each month) and we need to find the sum of the series (the total salary from the first till the last month). Formulae of arithmetic progression:- Formula for the nth term is, T) = a + (m - 1)d (1) Formula for the sum of n term is, Sn = (a + 1) .(2) where, a = the first term 1 = the last term d = common difference Tn the nth term Sn = sum of the n terms Now, according to the problem, The salary of the 3rd month= $ 820 The salary of the 12th month= $ 910 Now, from the 3rd month to the 12th month salary increment has happened 12 – 3 = 9 times. And, the total increment in this case = $ (910 – 820) = $ 90 90 Therefore, each month the increase in the salary = $ $ 10 9 Now, in this situation we understand that, for the given arithmetic progression, we first need to find out the first term a, which is the first month's salary. And we are given, Tz = 820 i.e, the third month's salary and the common difference d 10 i.e, the fixed increment of salary each month. = Therefore, from the equation (1) we get,

Tz = a + (3 – 1)d or, a = Tz - 2d or, a = 820 – 2 x 10 or, a = 820 – 20 or, a = 800 Therefore, we have the first month's salary, a = 800 And, the last month's (i.e. the 12th month) salary, 1 = 910 (given) 12 Therefore, total payment made in 12 months = = (800 + 910) [using equ". (2)] 2 = = 6 X 1710 = 10260 Hence, Darren's total pay for the entire 12 months is $ 10260 (Ans.)