If 0

Question

If 0<x<1, which of the following must be true?

1. x^5< X^3
2. X^4+ X^5 < X^3+ X ^ 2
3. X^4 - X^5 < X^3 - X ^ 2

Can some one help me with the working for option 3?

gk3.14 · Answer

1. x^5< X^3 
2. X^4+ X^5 < X^3+ X ^ 2 
3. X^4 - X^5 < X^3 - X ^ 2 

1. True : x is a positive fraction 
2. True : x^2(x^3 + x^2) < x^3 + x^2
3. True : -x^2(x^3 -x^2) < x^3 - x^2 : -x^2 < 1 (x is positive)


Can you confirm the answer?

2times · Answer

gk3.14

Yup you got it correct. Also I really like you solution.

Thanks

gmatornot · Answer

Only 1 and 2 must be true. For 0x^2>x^3 .. and so on. Hence statements 1 and 2 are always true. For Statement III .. X^4 - X^5 < X^3 - X ^ 2 X^4+X^2 < X^3 + X^5 X^4 + X ^ 2 < X(X^2 + x^4) which implies that x>1 hence 3 is not possible.

2times · Answer

gmatornot,

I used the same logic but OA is all three are correct.

Professor · Answer

i donot think 3 is correct. lets use plug-in with x = 1/2
 
LHS = X^4 - X^5 
= (1/2)^4 - (1/2)^5
= (1/2)^4 (1 - 1/2) 
= (1/2)^4 (1/2) 
= (1/2)^5

RHS = X^3 - X^2
= (1/2)^3 - (1/2)^2
= (1/2)^2 (1/2 - 1)
= (1/2)^2 (-1/2)
= - (1/2)^3

therefore (X^4 - X^5) > (X^3 - X^2)

you cannot solve the inequality as it is. you are solving the inequality as if the given inequality is correct. also when you divide the inequality by -ves, the inequality should be fliped.

2times · Answer

check this...

heman · Answer

1. x^5-x^3 < 0 x^3(x^2-1) < 0 since 0 x^3 cannot be -ve Hence (x^2-1) < 0 --> 0 x^2 cannot be -ve Hence (x^2-1) < 0 --> 0 x^2 cannot be -ve (x^2 +1) < 0 x^ 2 < -1 which is not possible Hence Only 1 & 2 are True. I dont buy the OA. Heman

Professor · Answer

lets not go after who said what and what is OA given? do you agree that (X^4 - X^5) is +ve and (X^3 - X^2) is -ve since x is a fraction. you tell me, which is greater? i am 100% sure that statememnt 3 is not correct as given by you. note: make sure you posted question correctly. :wink:

heman · Answer

2 times I checked the attached Q
Statement 3 should be

X^4 - X^5 < x^2 - x^3
In that case
 x^4(1-x) < x^2(1-x)
 x^4-x^2<0
 x^2( x^2-1) < 0
 x^2 cannot be -ve
Hence x^2-1 < 0
which implies 0<x<1 Hence True
Hence (3) is true.
(1) & (2) are true as shown be4

Heman

2times · Answer

no wonder I get so many quant questions incorrect due to silly mistakes

Thanks guys!

gk3.14 · Answer

Ah.. Thanks Prof.. I was treating these inequalites as if they were true..

GMATT73 · Answer

As originally written, only satements I and II are true. 

After being edited, all three statements stand.

Just pick 2 for x if you don`t know the rule of fractional exponents.

Sometimes Kaplan strategies work (Kan Always Pick Little Abstract Numbers)

yessuresh · Answer

Heman, I guess you went wrong in one place.
when x^4(1-x) < x^2(1-x)
You divided by (1-x) on both sides, to get x^4-x^2<0.
But you cant do this in inequalities, since you dont know what the value of 1-x is. If the value is negative the inequality sign needs to be reversed.
Hence (3) is false.

heman · Answer

yessuresh

U can divide in this case since Qstem states that x is +ve 0<x<1

Heman

If 0<x<1, which of the following must be true? 1.

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