Is 7^x greater than 2,500?

Question

(1) 7^(x+1) > 16,500
(2) 7^(x+2) = 7^x + 16464

Kudos for a correct  solution .

AbdurRakib · Answer

Here the question is very clear to me I directly go through the Statements .My first job is to transform the statements into 7^x to the left side.

Statement 1 :

7^ (x+1) >16500

=(7^x )(7^1) > 16500
=7^x> 16500/7 ,the amount is whatever but amounts are not specified.. Not Sufficient

Statement 2 :

7^(x+2)=7^x + 16464

or,(7^x)(7^2)=7^x+16464
or,49(7^x)-(7^x)=16464
or,48(7^x) =16464
or,7^x=16464/48
or,7^x=343 ,This answer confirms that 7^x=343, not greater than 2,500 , Sufficient

So,  Correct answer is B

I am worried about my approach.Please suggest me

dina98 · Answer

I pick B too. 7^x can be obtained with just the second option.

noTh1ng · Answer

I got lost into prime factorization with this problem..

Of course you could just say from statement 1 that:
 7^x > 16500/7 
 7^x > 2357,712

but: Is there a neater approach?

ENGRTOMBA2018 · Answer

The way you are doing is the most straightforward way but you can save yourself the time for calculating 16500/7 by noticing that 25*7=175.

Thus 7^x must be greater than a value less than 2500, say 2300 or 2400 (it doesn't matter what value you pick as you will get 2 answers with this statement making it not sufficient!)

For statement 2, Your method again is straightforward. Again you can save time with calculations by recognizing that 50*2500 =12500 and as 16k is > 12500 and as 48<50, a combination of these 2 facts will make 16k/48 >> 2500.

Hope this helps.

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

Is 7^x greater than 2,500?

(1) 7^(x+1) > 16,500
(2) 7^(x+2) = 7^x + 16464 
 
We need some preliminary knowledge to solve this problem. Remember that if the Que's scope includes the con scope, that con is sufficient. Essentially, the key is to approach the question using inclusion relation, instead of direct substitutions. 
 
In the original condition, we have 1 variable (x) and since we need to match the number of variables and equations, we need 1 equations and we have 1 each in 1), 2). The answer is likely D.

In case of 1), 7^(x+1)>16,500, 7^x*7>16,500, 7^x>16,500/7=2,357 and since 2,500>2,357, it is not sufficient. '
In case of 2), we can actually find the value of x. the x value is either smaller or bigger than 2500, therefore the answer is always yes or no, and thus it is sufficient. Therefore the answer is B.

itwarriorkarve · Answer

Guys, the answer should be D for this right?
We clearly have 7^x> 16500/7, which will clearly answer the question "is 7^x > 2500". Both of the statements alone are sufficient to answer the asked questions. Please highlight if I'm missing anything! :shock:

ENGRTOMBA2018 · Answer

You are not completely correct.

Yes,  7^x  > 16500/7 ----> you dont need to solve for 16500/7 but realise that 25*7=175 and thus 2500*7=17500. This means that 16500/7 will be a bit less than 2500.

With actual numbers, you see that 16500/7 = 2357.1 --->  7^x  > 2357

Now consider 2 cases:

If  7^x  = 2400 , this will give a "no" for "is 7^x > 2500" but

If  7^x  = 2700 , this will give a "yes" for "is 7^x > 2500"

both these values of 2400 and 2700 are >2357. As you get 2 different answers for statement 1, this statement is NOT sufficient.

itwarriorkarve · Answer

Amazing!! How'd I miss that? Thank you so much for your answer!!

anudeep133 · Answer

Solution :

Statement1 : 7^x > 16500/7 ==> 7^x > 2357.
We know that 2500 = 50(50) which is closer to (49)(49).
So lets check 7^4 which is 2401 which is less 2500 but greater than 2357 and 7^5 will obviously be greater than 2500.
So, Insufficient.

Statement2 : 49(7^x) = 7^x + 16464 ==> 48(7^x) = 16464 ==> 7^x = 343.
So, NO. Sufficient

Option B

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Is 7^x greater than 2,500?

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