Correct answer should be C

Let temp from 12 ...19 Jul be a,b,c....h and that from 13 July to 18 July be H

a +H +h/9 = H/6 + 5

2a+ 2h - H = 18*5


1. H/6 = h - 4 or H = 6h - 24 .....equation 1

Putting the in equation 1

2a + 2h - 6h - 24 = 18 * 5 We still have a two variable equation so we cannot solve this equation BCE remain cross out AD


2. a +H +h/9 = 84 Still useless.

Combine Put the value of H from equation 1 in 2. We will have a two variable equation in a and h

Now we have two equations in variables a and h , because we have replaced H with h, thus we can solve them to find a and h and hence the answer should be C

Let the daily max temp from July 12 to July 19 be T12, T13, T14 ... T19.

Then according to question stem - (T12+T13...+T19)/8 - (T13+T14+...T18)/6 = 5 ----------(1)

We need T12.

Statement 1
(T13+T14+...T18)/6 = T19-4 ---------------(2)

It is not possible to get T12 from 1 and 2.

Insufficient.

Statement 2
(T12+T13+...T19)/8 = 84 --------------------(3)

It is not possible to get T12 from (1) and (3).

Insufficient.

Together,
Substituting (3) and (2) in (1), we get -
84 - (T19-4) = 5
=> T19 = 83

So, (2) can be rewritten as
(T13+T14+...T18)/6 = 79
=> T13+T14+...T18 = 474 -----------(4)

(3) can be rewritten as
(T12+T13+...T18+83)/8 = 84
=> T12+T13+...T18 = 589 -------------(5)

Subtracting (4) from (5)
(T12+T13+...T18) - (T13+T14+...T18) = 589-474 = 115
or, T12 = 115.

Sufficient.

Thus the answer is C.