During the month of April, the average temperatures on day n

I didnt get the solution to this problem in the first attempt. However, on my second attempt, the underlying concept which is tested in the question became crystal clear to me.

Evaluate TA and TB upto n=4, then draw out a number line- to visualize the concept of averages.

30---32.5---35

Remove 30

0-----2.5----5

Now the no line is divided into equal parts as we keep on taking the averages:


0---1.25---2.5---3.75--5

TA4= 3.75
TB4= 1.75

Hence TA4 will keep decreasing towards 2.5 and TB4 will keep increasing towards 2.5 . Therefore the first statement can be proved- TA29 <TA4 and TB29 >TB4.

Now lets directly take statement 3.

since both Ta and TB are moving towards 2.5 in equal amounts. On any day, the difference from the previous day for both A and B will be same ( Please refer the image) and statement 2 is a direct result of statement 3.

Kudos the post if you liked my solution.

Thanks

I solved it like this, still time consuming one

A:

day 1 temp: 60/2 , diff from prev: 0
day 2 temp: 70/2 = (140/4) , diff from prev: +10/(2)
day 3 temp: 130/4 = (260/8) , diff from prev: -10/ (2 ^ 2)
day 4 temp: 270/8 = (540/16) , diff from prev: +10/(2^3)
.
. and goes on

So the point is diff between (n th) and (n-1 th) day is -------------------------(1)
+10/(2^(n-1)) => if n is even
-10/(2^(n-1)) => if n is odd

Now B:
day 1 temp: 70/2 , diff from prev: 0
day 2 temp: 60/2 = (120/4) , diff from prev: -10/(2)
day 3 temp: 130/4 = (260/8) , diff from prev: +10/ (2 ^ 2)
day 4 temp: 250/8 = (500/16) , diff from prev: -10/(2^3)
day 5 temp: 510/16 , diff from prev: +10/(2^4)
.
. and goes on

So the point is diff between (n th) and (n-1 th) day is -------------------------(1)
-10/(2^(n-1)) => if n is even
+10/(2^(n-1)) => if n is odd

TA(4) = 30 + (10/2) - (10/2^2) + (10/2^3)
TA(29) = 30 + (10/2) - (10/2^2) + (10/2^3) - ...................................... -(10/2^28)

TB(4) = 35 - (10/2) + (10/2^2) - (10/2^3)
TB(29) = 35 - (10/2) + (10/2^2) - (10/2^3) - ...................................... +(10/2^28)

Let a = (10/2) - (10/2^2) + (10/2^3),
b = (10/2) - (10/2^2) + (10/2^3) - (10/2^4)...................................... -(10/2^28)

Also let us check if a > b,
(10/2) - (10/2^2) + (10/2^3) > (10/2) - (10/2^2) + (10/2^3) - (10/2^4)...................................... -(10/2^28)
=> 0 > - (10/2^4) + (10/2^5) - ....................... -(10/2^28) => this is definitely true => so a > b ----------------------------------- eqn (2)

so TA(4) = 30 + a, TA(29) = 30 + b
TB(4) = 35 - a , TB(29) = 35 - b

Now let us see the answer choices

I. TA(4) - TB(4) > TA(29) - TB(29)

from above, (30 + a) - (35 - a) > (30 + b) - (35 - b)
=> rearranging => 2a > 2b => a > b
from eqn -----------------(2) => a > b
So this choice is true

II. TA(4) + TB(4) = TA(29) + TB(29)
from above, (30 + a) + (35 - a) = (30 + b) + (35 - b) => rearranging 65 = 65 => which is true.
so this choice also true

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

From above eqn ------------------(1),
for all n for both the cities, the absolute of diff between two days , |10/2^(n-1)|.
So this choice also true

Hence E

Let us try to visualise this problem by assuming two variables in place of real values.

Let us consider TA(1)=a, TA(2)=b and therefore TB(1)=b, TB(2)=a.

Now look at the series:
TA(3) = (a/2)+(b/2)
TB(3) = (a/2)+(b/2)
TA(4) = (a/4)+(3b/4)
TB(4) = (3a/4)+(b/4)
TA(5) = (3a/8)+(5b/8)
TB(5) = (5a/8)+(3b/8)

If you look closely, coefficients of a and b are getting interchanged in both the expressions TA and TB. Also, the interchanged coefficients in TB are (1-coefficient of TA)

Therefore, if we assume TA(n) = La+Mb, then TB(n) = (1-L)a+(1-M)b; L and M are assumed coefficients of a and b respectively in nth expression.

Now, coming to options.

Option (2) = TA(n)+TB(n) = La+Mb+ (1-L)a+(1-M)b = a+b, which is constant and independent of n
Hence option(2) is correct

Option (3): |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

Let us consider TA(n+1) = La+Mb
TA(n) = Pa+Qb

Hence, TB(n+1) = (1-L)a+(1-M)b
TB(n) = (1-P)a+(1-Q)b

Put both the values in expression of Option 3:
L.H.S = (L-P)a+(M-Q)b
R.H.S = (P-L)a+(Q-M)b

Hence |L.H.S|= |R.H.S|

Option 1:TA(4) - TB(4) > TA(29) - TB(29)
Let us consider TA(n) = La+Mb
TB(n) = (1-L)a+(1-M)b

Therefore, TA(n) - TB(n) = 2(La+Mb)-(a+b)= 2TA(n) - (a+b)

Hence, TA(n) - TB(n) is proportional to TA(n)
Please note that TA(n) decreased as n increases, therefore TA(4) - TB(4) > TA(29) - TB(29).

TA(n) --> even > odd
so ; TA(4) > TA(29)
TB(n) --> odd > even
so ; TB(4) < TB(29)

I. TA(4) - TB(4) > TA(29) - TB(29)
--> TA(4) - TA(29) > TB(4) - TB(29)
--> (+) > (-) ; True

II. TA(4) + TB(4) = TA(29) + TB(29)
--> I try if n = 3, 4
--> TA(n) + TB(n) = 65 ; True

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3
--> ΔTA(n) = ΔTB(n) ; True

The quickest way to approach this would be graphical.

We can see the when n>3, the temperatures TA(n) and TB(n) progress symmetrically about y=32.5 and tends to flatten out

I. TA(4) - TB(4) > TA(29) - TB(29)

As for the first statement, the gap between TA(4) and TB(4) will obviously be much larger than that between TA(29) and TB(29)

So I is true

II. TA(4) + TB(4) = TA(29) + TB(29)

Since the projections are symmetrical about y=32.5, the average temperatures of TA(n) and TB(n) for any n>3 will be the same

So [TA(4)+TB(4)]/2 = [TA(29)+TB(29)]/2

=> TA(4)+TB(4) = TA(29)+TB(29)

So II is also true

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

For any n>3, the rise of TA(n) will be equal to fall of TB(n) and vice versa

So, III is also true

Answer is (E)

Posted from my mobile device

The Strategy to above question lies like this-
1.Don't solve it (Based on factors like if you are not doing very well )
2.Solve it - In that case, i think this approach is good -

Since the option includes I in C and E, try to eliminate by calculating I -
Following the explanation given by chetan2u follow till end of second point and rearrange the equation as -
TB(29) - TB(4) > TA (29) -TA (4).

So I is in the answer which eliminates A,B and D
Now just check the validity of III.

I checked till 4,5 and realized why the modulus of difference will always be the same.
So E it is.


Please press kudos, if I was able to help you. Thank you.

Maybe not the hardest question here, IMO, but definitely tough. I think the most exhausting rate/time issues are the hardest to solve reasonably fast when the answer is supposed to be a fraction, I mean when we have like four unknown variables.

Anyway, here I did like this:

Starting by writing down the first values in both sets I realised the following points along the way:

Serie A: 30-35-32,5-33,75-33,125
Serie B: 35-30-32,5-31,25-31,875

• That both sets will eventually approximate a value (n) with an inifinitely small change from the value before (n+1)
• That for a given value of (n) the absolute change between two terms will always be the same for both sets, but will have opposite signs

I. TA(4) - TB(4) > TA(29) - TB(29)

• That the difference between TA(n) and TB(n) will be less than between the previous terms (n-1) if (n) is odd and more if (n) is even.
• That the biggest difference then is for TA(4) and the smallest difference is for TA(5). So, if the statement would instead have been TA(5) - TB(5) > TA(29) - TB(29), then the statement would have been false.

II. TA(4) + TB(4) = TA(29) + TB(29)

• That the sum of TA(n) + TB(n) will never change. The values will always be the same distance from the respective infinite average. What we add to A, we subtract from B, and vice versa.

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

• This last statement was the easiest to "see". It's the same concept as for validating statement 2. The absolute value signs only confirmed what I already knew.

Look at the two sequences. The first two terms are the same - they are just interchanged.
TA - 30, 35, 32.5, ...
TB - 35, 30, 32.5,...
Why will the fourth term be different? Because it is the avg of 35 and 32.5 in case of A and that of 30 and 32.5 in case of B. Avg is the middle value. Let's put these on the number line.



Each smaller line is the avg of previous two lines. Avg of 30 and 35 is 32.5, the green line common to both A and B. Next day avg for B is the left black line while that for A is the right black line. Next day avg for B is the left yellow line while for A is the right yellow line etc.
The avgs for each day for A and B are equidistant from 32.5.


I. TA(4) - TB(4) > TA(29) - TB(29)
T(4) avgs are given by the black lines. Subsequent avgs are all closer to each other (corresponding) than the black lines i.e. the red lines, the yellow lines etc have less distance between them that the black lines.
Hence TA(4) - TB(4) is greater than TA(n) - TB(n) for all subsequent values of n.
Hence TA(4) - TB(4) is greater than TA(29) - TB(29). Must be true.

II. TA(4) + TB(4) = TA(29) + TB(29)
Each set of avgs is equidistant from 32.5. So the sum of each set of avgs will be 65.
TA(4) + TB(4) = TA(29) + TB(29) = 65
Must be true

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3
The distance between subsequent avgs on each side will be the same for all n > 3. The distance between black and yellow lines on the left is the same as the distance between black and yellow lines on the right. The distance between yellow and purple lines on the left is the same as the distance between yellow and purple lines on the right and so on...
Hence this must be true too.

Answer (E)

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