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avg daily max temperature

by neerajkumar1_1 » Sat Sep 04, 2010 9:42 pm
The avg daily max temperature at a certain weather stationfrom July 12 to July 19 was 5 degrees higher than the avg daily max temp from July 13 to July 18. What was the max temp on July 12???

1) the avg daily max temp from July 13 to July 18 was 4 degrees less than the max temp on July 19.
2) the avg daily max temp from July 12 to July 19 was 84 degrees.

Please get me the steps also...
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by mohit11 » Sat Sep 04, 2010 11:38 pm
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
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by niksworth » Sat Sep 04, 2010 11:49 pm
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.
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