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