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

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During the month of April, the average temperatures on day n  [#permalink]

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New post 12 Feb 2016, 23:32
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During the month of April, the average temperatures on day n in two cities A and B are denoted by TA(n) and TB(n) respectively. The average temperature in city A on any day is the average of the average temperatures of the previous two days in city A. The average temperature in city B on any day is the average of the average temperatures of the previous two days in city B. If TA(1) = 30, TA(2) = 35, TB(1) = 35 and TB(2) = 30, which of the following statements must be true?

I. TA(4) - TB(4) > TA(29) - TB(29)
II. TA(4) + TB(4) = TA(29) + TB(29)
III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

A. II only
B. III only
C. I and II only
D. II and III only
E. I, II and III
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 13 Feb 2016, 03:58
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TeamGMATIFY wrote:
During the month of April, the average temperatures on day n in two cities A and B are denoted by TA(n) and TB(n) respectively. The average temperature in city A on any day is the average of the average temperatures of the previous two days in city A. The average temperature in city B on any day is the average of the average temperatures of the previous two days in city B. If TA(1) = 30, TA(2) = 35, TB(1) = 35 and TB(2) = 30, which of the following statements must be true?

I. TA(4) - TB(4) > TA(29) - TB(29)
II. TA(4) + TB(4) = TA(29) + TB(29)
III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

A. II only
B. III only
C. I and II only
D. II and III only
E. I, II and III


Hi,
Although its testing our visualization power to realize a certain activity will make what difference to an equation, it is a tough Q, which is not likely to be confronted, if you are not doing extremely -2 well

Now lets VISUALIZE what is happening in each case...
Important take aways, which mat be helpful in such Qs
Quote:

FIRST IMPORTANT POINT:-


The difference is 5 in first two terms and since the next terms are average of the pervious two terms, the difference in alternate terms will be based on 5..

SECOND:-


if the pervious two terms are big and small, the next will be bigger than previous..
and if the pervious two terms are small and big, the next will be smaller than previous..
so in A since first is 30 and next is 35, the next will be smaller, then bigger... etc
so even numbers will be larger than the previous odd..
in B it will be opposite...

THIRD..


in A, odd numbers will increase from 32.5 to nearly 33.25, and even numbers will decrease from 33.75 to 33.25..
in B, even numbers will increase from 31.25 to 31.75, and odd numbers will decrease from 32.5 to 31.75 ..

FOURTH


A will have no number<32.5 and B will have no number>32.5 after the two terms..

There will be many points which can clear off with the above observations too ..



but lets work out the nth term in each sequence to hit the answer straight



A.. 30, 30+5, 30+5-5/2, 30+5-5/2+5/4...
.. 30, 30+5, 30+5/2, 30+5/2+5/4,30+5/2+5/4-5/8...
.. 30, 30+5, 30+5/2, 30+15/4,30+25/8...
now lets remove 30 and work out.. 5,5/2,15/4,25/8..
5, 5*1/2, 5*3/4, 5* 5/8..
so 2nd term= 5..
3rd term= 5*1/2= 5* {2*3
-5}/2^(3-2)
4th term= 5*3/4= 5*{2*4[/b]-5}/2^(4-2)..
so nth term= 5*{2*n-5}/2^(n-2)..

so NTH term= 30 +5*{2n-5}/2^{(n-2)}

B... 35, 35-5, 35-5+5/2, 35-5+5/2-5/4..
here Nth term= 35 - 5*{2n-5}/2^{(n-2)}


let see the choices..


I. TA(4) - TB(4) > TA(29) - TB(29)
TA4= 30+5*25/4..
TB4= 35-5*25/4..
TA29= 30+5*53/2^27= nearly 30..
TB29= 35-5*53/2^27=nearly 35...

substitute values.. TRUE

II. TA(4) + TB(4) = TA(29) + TB(29)
TA4= 30+5*25/4..
TB4= 35-5*25/4..
TA29= 30+5*53/2^27= nearly 30..
TB29= 35-5*53/2^27=nearly 35...

substitute values..
30+35=30+35..
TRUE

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)|
it basically is asking difference in two consecutive terms
the values are equal for two sides but in opposite sign..
TRUE
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 18 Mar 2017, 10:05
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I'd suggust approaching it differently:
1) from the given we can calculate that the 1st is true
2) second, let's look at answer choices: only C and E left, we need to test the 3rd case
3) 66,25/2-67,5/2=-0,75
63,25-62,5=0,75
hence 3rd is correct
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During the month of April, the average temperatures on day n  [#permalink]

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New post 24 Mar 2017, 06:07
Hi Chetan2u,

Is the formula you mentioned is applicable for any difference in values between the terms mentioned.?
In this case the difference is 5 and hence the formula is = 5*(2n-5)/2^(n-2)
In case the difference is 3 can the formula be = 3*(2n-3)/2*(n-2)

TIA
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 25 Apr 2017, 11:02
1
1
chetan2u wrote:
TeamGMATIFY wrote:
During the month of April, the average temperatures on day n in two cities A and B are denoted by TA(n) and TB(n) respectively. The average temperature in city A on any day is the average of the average temperatures of the previous two days in city A. The average temperature in city B on any day is the average of the average temperatures of the previous two days in city B. If TA(1) = 30, TA(2) = 35, TB(1) = 35 and TB(2) = 30, which of the following statements must be true?

I. TA(4) - TB(4) > TA(29) - TB(29)
II. TA(4) + TB(4) = TA(29) + TB(29)
III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)| for n>3

A. II only
B. III only
C. I and II only
D. II and III only
E. I, II and III


Hi ,
Although its testing our visualization power to realize a certain activity will make what difference to an equation, it is a tough Q, which is not likely to be confronted, if you are not doing extremely -2 well

Now lets VISUALIZE what is happening in each case...
Important take aways, which mat be helpful in such Qs
Quote:

FIRST IMPORTANT POINT:-


The difference is 5 in first two terms and since the next terms are average of the pervious two terms, the difference in alternate terms will be based on 5..

SECOND:-


if the pervious two terms are big and small, the next will be bigger than previous..
and if the pervious two terms are small and big, the next will be smaller than previous..
so in A since first is 30 and next is 35, the next will be smaller, then bigger... etc
so even numbers will be larger than the previous odd..
in B it will be opposite...

THIRD..


in A, odd numbers will increase from 32.5 to nearly 33.25, and even numbers will decrease from 33.75 to 33.25..
in B, even numbers will increase from 31.25 to 31.75, and odd numbers will decrease from 32.5 to 31.75 ..

FOURTH


A will have no number<32.5 and B will have no number>32.5 after the two terms..

There will be many points which can clear off with the above observations too ..



but lets work out the nth term in each sequence to hit the answer straight



A.. 30, 30+5, 30+5-5/2, 30+5-5/2+5/4...
.. 30, 30+5, 30+5/2, 30+5/2+5/4,30+5/2+5/4-5/8...
.. 30, 30+5, 30+5/2, 30+15/4,30+25/8...
now lets remove 30 and work out.. 5,5/2,15/4,25/8..
5, 5*1/2, 5*3/4, 5* 5/8..
so 2nd term= 5..
3rd term= 5*1/2= 5* {2*3
-5}/2^(3-2)
4th term= 5*3/4= 5*{2*4[/b]-5}/2^(4-2)..
so nth term= 5*{2*n-5}/2^(n-2)..

so NTH term= 30 +5*{2n-5}/2^{(n-2)}

B... 35, 35-5, 35-5+5/2, 35-5+5/2-5/4..
here Nth term= 35 - 5*{2n-5}/2^{(n-2)}


let see the choices..


I. TA(4) - TB(4) > TA(29) - TB(29)
TA4= 30+5*25/4..
TB4= 35-5*25/4..
TA29= 30+5*53/2^27= nearly 30..
TB29= 35-5*53/2^27=nearly 35...

substitute values.. TRUE

II. TA(4) + TB(4) = TA(29) + TB(29)
TA4= 30+5*25/4..
TB4= 35-5*25/4..
TA29= 30+5*53/2^27= nearly 30..
TB29= 35-5*53/2^27=nearly 35...

substitute values..
30+35=30+35..
TRUE

III. |TA(n+1) - TA(n)| = |TB(n+1) - TB(n)|
it basically is asking difference in two consecutive terms
the values are equal for two sides but in opposite sign..
TRUE


Hi chetan2u ,

Please explain how did you think in this way. Thanks

30, 30+5, 30+5-5/2, 30+5-5/2+5/4...
.. 30, 30+5, 30+5/2, 30+5/2+5/4,30+5/2+5/4-5/8...
.. 30, 30+5, 30+5/2, 30+15/4,30+25/8...
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During the month of April, the average temperatures on day n  [#permalink]

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New post 04 Oct 2017, 23:40
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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
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 28 Jun 2018, 10:52
VeritasPrepKarishma is there an easy and quick approach to this question?
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 20 Jul 2018, 01:58
Is there any short approach for this question
KarishmaB Bunuel or chetan2u please explain.
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 31 Jul 2018, 22:39
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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).
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 03 Aug 2018, 09:38
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
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Re: During the month of April, the average temperatures on day n  [#permalink]

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New post 12 Aug 2018, 19:56
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
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Re: During the month of April, the average temperatures on day n &nbs [#permalink] 12 Aug 2018, 19:56
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