Is x 0? (1) x^6 x^7 (2) x^7 x^8

Question

Is x > 0?

(1) x^6 > x^7
(2) x^7 > x^8

nt2010 · Accepted Answer

Stmt 1 is Not Sufficient

x^6 > x^7 means x < 1, so x > 0 when 0< x < 1 and when x < 0 so x < 0

Stmt 2 is Sufficient

x^7 > x ^8 means 0<x<1 so x > 0

A variable with smaller power is more than a variable with larger power iff the variable is in the range 0 < x < 1

Answer is B

//kudos please, if the above explanation is good.

Zarrolou · Answer

Any number with an even exponent is >0 , but I wanna give you an alternative solution: 1. x^6>x^7 so 0>x^7-x^6, 0>x^6(x-1) at this point we can divide by x^6 because we know that does not equal 0(otherwise x^6=x^7 and not >) 0>x-1 and finally 1>x . Is x positive? it could be(0,5) or not (-1) 2. x^7>x^8 becomes 0>x^6(x^2-x) divide 0>x^2-x so x(x-1)<0 and 0

Bunuel · Answer

Is x > 0?

(1) x^6 > x^7. This implies that \(x\neq{0}\), thus \(x^6>0\). Divide both part by x^6: \(1>x\) (\(x\neq{0}\)). Not sufficient.

(2) x^7 > x^8. The same here. Divide both part by x^6: \(x>x^2\) --> \(0<x<1\). Sufficient.

Answer: B.

Solving inequalities:
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data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

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Hope it helps.

aceacharya · Answer

Felt the following would be a simpler explanation

The two option would go as follows

1. x^6 > x^7
This option is only possible either when 1>x>0 or x<0
Like if x is <1 but >0 then the values will keep on decreasing with every increasing exponent
and if x<0 then the even powers will be positive and odd ones would be -ve therefore irrespective of the value of |x| the even exponents will always be greater

2. x^7 > x^8
This is only possible when 1>x>0, as the situation would be as in case 1 but when X<0 the even exponents will always be greater

Only 2 is sufficient but 1 is not B

summer101 · Answer

Sorry this is probably as silly as it gets but i need help in understanding the below

In x^7 > x^8, say we divide by x^6,
x> x^2
0>x(x-1)
Now 0>x and 0>(x-1)
for 0>x-1 ==> x <1, till here i am fine
but how do we get x>0 to make 0<x<1 ??

DEEPAKCHANDRA · Answer

x*(x-1)<0 implies either x<0 and x-1 > 0 or x>0 and x-1<0 first is impossible and hence from second we have 0

Zarrolou · Answer

Till here you are fine x> x^2 so x^2-x<0 we have to solve this, and to solve let me use an old trick. Lets solve x^2-x=0,x(x-1)=0 so x=1 or x=0. Now because the sign of x^2 is + and the operator is < we take the INTERNAL values: 0 with = ) then, once you have the results take a look at the sign of x^2 an the operator. (<,-) or (>,+) take ESTERNAL values. (if they are the "same" ) (>,-) or (<,+) take INTERNAL values(like this case). Let me know if it's clear now

unverifiedvoracity · Answer

1. x^6 > x^7 , either 0 < x< 1 or x < -1 (the result of even power of a negative number is positive). Not Sufficient. 2. x^7 > x^8 , 0 < x< 1. x cannot be negative, If x were negative, x^7 will be negative and x^8 will be positive and this statement won't hold true. Sufficient. Answer is B.

adiagr · Answer

x^6 > x^7 x^6 - x^7 > 0 x^6 *(1 - x) > 0 x^6 is always a positive value. so (1-x)> 0 or x < 1 x can be (1/2) a positive value or x can be -2 , a negative value, so not sufficient. Statement (2) x^7 > x^8 x^7 - x^8 > 0 x^7 *(1 - x) > 0 x^6.x.(1-x) > 0 Now x^6 is always a positive value. so x(1-x)> 0 or x(x-1) < 0 0< x <1 Thus x is always positive. B is sufficient.

shreyagupta1401 · Answer

Simply assume values. Let's say x = 0.5 or -0.5

From statement 1, X^6>X^7
Case 1, if X is 0.5, X^6 > X^7 is true
Case 2, if X is -0.5, X^>X^7 is true
We cannot prove is X is >0. Insufficient.

From Statement 2, X^7>X^8
Case 1, if X is 0.5, X^7>X^8 is true
Case 2, if X is -0.5, X^7>X^8 is false
Therefore, sufficient. X>0

Answer is B.

priyaarorastep · Answer

Official Explanation: 
Statement (1) tells us that xˆ6>xˆ7
: this is true of any fraction between 0 and 1 and any negative integer; INSUFFICIENT. 
Statement (2) tells us that xˆ7>xˆ8
; this is only true of fractions between 0 and 1. Since this guarantees that the lower limit for xis above 0, it definitively answers the question "yes"; SUFFICIENT. 
If you’re wondering why statement (2) worked when statement (1) didn’t, note that in the second statement the first exponent is odd and the second is even. The even one must be 0 or positive, so if the odd one is greater, it must be greater than 0.

sood1596 · Answer

In statement 2, why are we diving by x^6.

If we take x^7 common. 1>x is what we get which is similar to A.

Can someone help me explain?

Thanks

Basshead · Answer

(1) x^6 - x^7 > 0 x^6 (1 - x) > 0 x^6 is always positive, therefore 1 > x . x < 0 or 0 < x < 1 . INSUFFICIENT. (2) x^7 - x^8 > 0 x^7 (1 - x) > 0 x^6*x (1 - x) > 0 x > 0 and 1 > x OR x < 0 and 1 < x ; the second range is not possible. Therefore 1 > x > 0 . SUFFICIENT. Answer is B.

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Is x > 0? (1) x^6 > x^7 (2) x^7 > x^8

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