Given p=|q| and q=3^a-(27^40+27^8), which of the following values of '

Hi....
Since 1 is very very small as compared to 27^{32}, we can discard 1 as we are looking at an approximate value

Hi,

I think it could also be solved in the following way:
q=3^a−(27^40+27^8)=3^a- ((3^3)^40 + (3^3)^8

= 3^a - (3^120 + 3^24)

= 3^a - 3^120 - 3^24

if a = 120 then the result would be - 3^24. For all other values, it will be greater than this.

So, Answer is A.

Bunuel pushpitkc Bunuel niks18 gmatbusters KarishmaB

I have a small query regarding the penultimate step by chetan2u as below:


Here is the question stem again which asks for a definite value and not approximation.



I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.

What I did not get was: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which voilates Q stem.
Can you suggest why we did not take next positive value i.e. 140 as correct OA?

Hi

Query1:

I believe chetan2u took the highlighted text to infer that \(3^a\) must be a positive value.. it is not to be inferred from the question stem.

3^a is always positive- 3^a can never be negative and 3^a tends to zero only if a approaches negative infinity.

Concept: (positive)^x, is always positive, it can not be negative. (positive)^x tends to zero only if a approaches negative infinity.

Query2: if a is actually equal to 120 then the final value of q will be 0 and p will be 0 , which violates Q stem
I am afraid, I couldn't understand this point. How it violates the Question stem.- please explain the query further to let me expand.

In fact , p=|q|. Hence the minimum value of p can be equal to zero.( it cannot be negative). So , as explained in above posts, p approaches zero, when a is 120. (out of the given 5 options).

Hi adkikani,

As you know, |q|=q when q>0
|q|=-q when q<0
So, p=|q|=positive
Hence, determining the least value of 'p' is same as determining the least value(minimum) of q.

Given \(q=3^a-(27^{40}+27^8)=3^a-(3^{120}-3^{24})=3^a-3^{120}-3^{24}\)--------------(1)



when a=120, \(q=3^{120}-3^{120}+3^{24}\)=\(-{3^{24}}\neq0\)



Here a=120 is not approximated. When a=120, the given expression has the least value of q, i.e. \(-3^{24}\). So, \(p=|q|=-q=-(-3^{24})=3^{24}\)



Let's take a=140, so (1) becomes, \(q=3^{140}-3^{120}-3^{24}\). As \(3^{140}-3^{120}\)is positive, so \(|q|> 3^{24}\)
But we want the least value. Therefore, we have to discard a=140.



No, the highlighted portion signifies 'p' must be positive. (q<0 (say -5) implies p=|q|=-q=-(-5)=positive; q>0 implies p=|q|=positive)

P.S:- We can have value of the expression(|q|) less than \(3^{24}\) but we are told to determine the least value of 'p' from the given 5 answer options.(a=120, 140,180,185, and 190)

Most of the points have been made amply clear in above posts. However, the point on approximation..
What is the least value of p, it is 0 as p=|q|
And at a=120, it is not 0, it is 3^120-(3^120+3^24)=-3^24
So p =|-3^24|=3^24

So that is not equal to 0..
But are we looking for 'a' to get the least value of p...NO then a will be 120.xyz, something in decimals.
We are looking for the least value for the FOLLOWING values

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