I guess statement (1) should be: \(z<25\) and \(w=7^x\).

"x is an integer and x raised to any odd integer is greater than zero" means \(x=integer>0\).
Q: is \(w-z>5(7^{x-1}-5^x)\)?

(1) \(z<25\) and \(w=7^x\) --> is \(7^x-z>5(7^{x-1}-5^x)\)? --> is \(7^x-z>5*7^{x-1}-5^{x+1}\)? --> is \(7^x-5*7^{x-1}+5^{x+1}>z\)? --> as the lowest value of \(x\) is 1, then the lowest value of LHS is when \(x=1\): \(LHS_{min}=7^x-5*7^{x-1}+5^{x+1}=7-5+25=27\) --> so the lowest value of LHS is 27, which is more than \(z\) as \(z<25\) --> hence \(7^x-5*7^{x-1}+5^{x+1}>z\) is true. Sufficient.

(2) \(x = 4\). No info about \(w\) and \(z\). Not sufficient.

Answer: A.
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Could you please reply with the detailed explanation for the above question.
i am confused as to how just Statement I is sufficient to answer the question(A), without knowing the value of X?

Hello, let me try and explain.

paraphrasing the questions gives us, is w-z > 5.7^(x-1) - 5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0.

taking statement 1, you have the question as 7^x-z > 5.7^(x-1) - 5^(x+1)

Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x-1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1.

So, 7^x > 5.7^x-1
and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get:

7^x - z > 5.7^x-1 - 5^(x+1),

Hence answer A.

the question is not simple but if you think a bit abstract and test number you can arrive to the solution.

The FIRST important thing, breaking the problem, is to understand tha our X is positive. Infact, if a number raised to power of 3 is >0 that means X itself is positive because the odd powr maintain the original sign of the number, so X must be positive.

At this point using the exponent rules we have w - z > 5*7^x-1 - 5^x+1 ----> 7^x - 24 (from stem x<25) > 5*7^x-1 - 5^x+1

testing number 1 ------> 7 -24 > 1 - 25 ---> - 17 > -24 this is TRUE. also if you test some value < 25 positive but also negative the result is the same.

1) sufficient

x = 4 -------> w - z > 5*7^3 - 5^5..........at this point you know nothing about w and z.

2) insufficient

A is the answer.
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why can't x be zero apart from even ? any odd number raised to the power zero is 1 which is an interger greater than zero. I think x is given as an interger and not only positive integer. if you work with x = 0 , statement 1 is not sufficient ?

We are told that integer x raised to to any odd integer is greater than zero: \(x^{odd}>0\). Now, if \(x=0\) then \(x^{odd}=0^{odd}=0\) (for odd>0), which violates given condition, so \(x\) cannot be zero.

Hope it's clear.
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Thanks, I misunderstood the question. I thought it was odd int raised to the power x. It is clear now.

I don't understand the function of statement z<25. If lowest value of LHS is 27 which is more than z, then it doesn't fulfill the condition given?

Please explain why the lowest value of X is 1.

See the highlighted part: x is an integer and x raised to any odd integer is greater than zero means that x is a positive integer, thus its lowest possible value is 1.

Hope it's clear.
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Responding to a pm:

What is the meaning of: x is an integer and x raised to any odd integer is greater than zero?
It is just a convoluted way of saying 'x is a positive integer'. Since x raised to any odd power is positive, x must be positive. x can take values 1, 2, 3, 4 and so on.

Question: Is \(w - z > 5*7^{x-1}-5^{x+1}\)

Using statement 1: Is \(7^x - 25 > 5*7^{x-1} - 5^{x+1}\)

Even though we are given that z is less than 25, let's assume it to be 25. In case we can prove that left hand side is greater than right hand side for z = 25, we can prove that left hand side will remain greater than right hand side for any other value of z. If z has a smaller value, say 9, left had side will become even larger (since a smaller number will be subtracted) than the right hand side.

Now note that \(7^x\) will be greater than \(5*7^{x - 1}\) since \(7^x\) can be written as \(7*7^{x - 1}\) (x is a positive integer).
Also note that minimum value of x is 1 so \(5^{x + 1}\) will always be equal to or larger than 25 (x is a positive integer).

So comparing the left and right hand side expressions, \(7^x\) is greater than \(5*7^{x - 1}\) and 25 is less than or equal to \(5^{x+1}\). So left hand side must be greater than the right hand side in all cases when x is a positive integer.

Answer (A)
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paraphrasing the questions gives us, is w-z > 5.7^(x-1) - 5^(x+1). Further, x is a positive integer as an odd power of x is greater than 0.

taking statement 1, you have the question as 7^x-z > 5.7^(x-1) - 5^(x+1)

Now, for all cases where x is a positive integer, 7^x would be greater than 5.7^(x-1). Next, 5^(x+1) would always be >= 25 as the least value of x can be 1.

So, 7^x > 5.7^x-1
and z < 5^(x+1). Reversing this inequality and adding both the inequalities, we get:

7^x - z > 5.7^x-1 - 5^(x+1),

Hence answer A.
tnx for this
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x raised to any odd integer is greater than 0 --> x = integer >0 --> x>=1

(1) --> w-z > 7^x - 5^2 = 7.7^(x-1) - 5^2 > 5.7^(x-1) - 5^2

in order to w-z> 5.7^(x-1) - 5^(x+1), x+1 >=2 --> x >=1 --> sufficient.

(2) insufficient

=> A.

Given: x is an integer and x raised to any odd integer is greater than zero
KEY CONCEPT: A number raised to an ODD power retains its sign.
That is, POSITIVE^ODD = POSITIVE and NEGATIVE^ODD = NEGATIVE
The given info tells us that x^ ODD = POSITIVE
So, it must be the case that x is a POSITIVE integer

Target question: Is w - z > 5[7^(x-1) - 5^x?]
Let's tidy the target question by expanding the right side of the inequality to get....
REPHRASED target question: Is w - z > 5[7^(x-1)] - 5^(x+1)?

Statement 1: z < 25 and w = 7^x
Take the REPHRASED target question and replace w with 7^x to get:
Is 7^x - z > 5[7^(x-1)] - 5^(x+1)?

IMPORTANT: Notice that, on the left side of the inequality, we're subtracting z
We're told that z < 25
So, we can help MINIMIZE the value of the left side by making z as big as possible
So, let's make z = 25 (aside: we could make z equal something really close to 25, like 24.99999999999, but let's make things easy on ourselves and make z EQUAL 25.
We get: Is 7^x - 25 > 5[7^(x-1)] - 5^(x+1)?

Now let's get like terms on the same side of the inequality.
First subtract 5[7^(x-1)] from both sides to get: Is 7^x - 5[7^(x-1)] - 25 > -5^(x+1)?
Then add 25 to both sides to get: Is 7^x - 5[7^(x-1)] > 25 - 5^(x+1)?
Factor 7^(x-1) from left side to get: Is 7^(x-1)[7 - 5] > 25 - 5^(x+1)?
Simplify left side to get: Is 7^(x-1)[2] > 25 - 5^(x+1)?

Now recognize that 7^(x-1) will be POSITIVE for all values of x.
So, 7^(x-1)[2] must be POSITIVE

Now recognize that since x is a POSITIVE integer, 5^(x+1) must be greater than or equal to 25 (since x = 1, is the smallest possible value of x)
So, 25 - 5^(x+1) must be less than or equal to zero.

So, our question becomes Is some POSITIVE number > some number that's less than or equal to zero?
The answer to this REPHRASED target question is a definitive YES
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: x = 4
Since we have no information about z or w, we cannot answer the REPHRASED target question with certainty.
So, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent
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Hi Karishma,

For this problem, you essentially tried to find if the LHS is ALWAYS greater than the RHS (if you could say Yes, with the information given in S1, you are THEN ONLY saying S1 is sufficient)

For that you minimized the LHS as much as possible which is why x = 1 and z = 25

I understood that part

HOWEVER

My question is

Should you not also check the opposite, i.e. ?

You need to also check if the LHS is ALWAYS lower than the RHS (if you could say Yes, with the information given in S1, you could THEN ONLY also S1 is sufficient as well)

For this new scenario, one would have to maximize the LHS...

Why isn't this scenario part of your calculations

Thank you !

The simple trick is to realize that if x is a positive integer, then (7^x-1 - 5^x) is always negative -> RHS is always negative

Please reattempt this question after knowing the above fact... you will be able to solve it quickly without the complete solution.

Complete Solution

W-Z > 5*(7^x-1 - 5^x)
LHS = W-Z
RHS = 5*(7^x-1 - 5^x) (Always negative)

A) now notice that W-Z is always positive if x > 1
for eg : X= 2 , z= 25
means, 49 - 25 > 0 (always posiiive)

so LHS > RHS for x > 1

now lets see X = 1

We want to see the smallest value of LHS (w-z)

Possible when
W is lowest = 7^1 = 7
Z is highest = 25

W-Z = 7-25 = -18
RHS = -25
so LHS > RHS
If the smallest value of LHS (w-z) is bigger than any value of RHS then it’s always true

SUFFICIENT


B) We do not know the value of W and Z

hi Bunuel - 5 * 5^x = 5^x+1?

Please use proper formatting when posting.

\(5*5^x = 5^{x+1}\)

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An odd number can be negative also. In that case, won't the answer be E.