If the integer y 6, is x^3y 56? (1) 4 x^3

Question

If the integer  y > 6 , is  x^3y > 56 ?

(1)  4 < x^2 ≤ 9

(2)  x^3 + x > x^3

chetan2u · Accepted Answer

Meaning of the question Three cases... Case I:- If x>2.....answer is a DEFINITE YES as Least value of y is 7 . Case I:- But if 02 ?.....Definite YES (2) Is x\leq 0 ?....Definite NO Now the Choices (1) 4 < x^2 ≤ 9 so x^2>4.......x<-2..and...x>2 and x^2\leq 9.......x\geq -3..and...x\leq 3 Common range we get is -3\leq x<-2 and 2 x^3 x>0... If x>2, surely YES, otherwise cannot say Combined.. x>0 and 2

Staphyk · Answer

If the integer y>6 , is x^3y >56? Yes
Or x^3y<=56 No
So we know y= 7,8,9 etc.

(1) 4 < x^2 <=9
2^2 < x^2 <= 3^2
|2| < |x| <= |3|
Case 1: -2 < x <= -3
Case 2: 2< x<= 3
Now case 1: x= -ve , (-ve)^3•y<56 so No
Case 2: x= 2.01,2.02 etc. (2.01)^3•y>56 yes
(Not sufficient)

(2) x^3 + x > x^3
(1)^3 +1 >(1)^3
Now x^3•y< 56 No
(3)^3 +3 > (3)^3
Now x^3•y >56 Yes

(1+2) x can’t be negative as x^3+x > x^3 so case 1 in (1) is ruled out
Now case 2: 2< x <=3
and in (2) x^3+x > x^3 so x must be 2<x<=3
so x^3•y >56 (Sufficient)

Answer is C like Corona-virus

Posted from my mobile device

preetamsaha · Answer

ST:1

x^2 <= 9
i.e. -3<=x<=3
also x^2 >4
i.e. x>2 and x<-2
combining both , x belongs to [-3,-2) U (2,3]
now , y>6
if x=2.5 , y=7 , x^3y >56 [ any one can take 2.01 , 2.001 , 2.0001 and so on the value will be always >56 since if we consider x=2 and y =7 (which is an integer, >6) the value will be 56 . so anything greater than 2 of x will give us x^3y>56 ]
but if x=-2.5 , y=7 , x^3y <56
so ST 1 is not sufficient

ST:2
x^3+x >x^3
i.e. x>0
if x =1 , y =7 , x^3y <56
but if x= 2.5 ,y=7 , x^3y >56
so ST 2 is not sufficient .

combining ST 1 & 2
x belongs to (2,3]
so from the above example we can conclude that x^3y> 56

correct answer is C

GMATWhizTeam · Answer

Solution Step 1: Analyse Question Stem • y > 6 where y is an integer o So minimum value of y = 7 • We need to find if x^3y > 56 • Now, x^3y > 56 only if o x > 0 and x^3* 7 > 56 ⟹x^3 > 8 ⟹ x > 2 So, we need to find whether x >2 or not. Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE Statement 1: 4x^3 • According to this statement : x^3 -x^3 + x >0 ⟹ x > 0 • However we don’t know if x > 2 or not. For example: o Case 1 : x can be 0.5 or 1 etc.  in such cases x < 2 o Case 2: x can be 3, 4, 4.5 etc.  in such cases x > 2 So, from this statement we cannot decidedly say if x > 2. Hence, statement 2 is also NOT sufficient and we can eliminate answer Option B. Step 3: Analyse Statements by combining. • From statement 1: o 2 < x ≤ 3 or o -3 ≤ x < -2 • From statement 2: x > 0 • On combining both we got 2 < x ≤ 3 ⟹ 2 < x Thus, the correct answer is Option C.

CrackverbalGMAT · Answer

From the question data, we know that the smallest possible value for y is 7 since y is an INTEGER greater than 6. If we take this as the boundary condition and plug in y=7 in the inequality \(x^3\)*y>56, the question becomes is \(x^3\)>8 which is equivalent to finding out if x>2.

As far as possible, when you have an equation/ inequality given in the question stem, we try to break it down to a stage where we know what to look for in the statements.

From statement I alone, 4<\(x^2\)≤9. This means that 2<x≤3 OR -3≤x<2. This is insufficient to answer the question asked since x may be greater than 2 or lesser than -2.

For example, if x = 3 and y = 7, \(x^3\)*y = 27*7 = 189 which is definitely greater than 56. But, if x=-3 and y = 7, \(x^3\)*y = -189 which is lesser than 56.
Statement I alone is insufficient. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, \(x^3\)+x>\(x^3\) which can be simplified to conclude x>0. Knowing that x is positive is insufficient to establish if \(x^3\)*y>56. The best value to prove statement II insufficient is x=1. If x=1, \(x^3\)*y will not be more than 56 till y reaches 56; post this stage, the expression will be more than 56.
Statement II alone is insufficient. Answer option B can be eliminated. The possible answer options are C or E.

Combining statements I and II, we have the following:
From statement II, we know that x has to be positive; from statement I alone, we know 2<x≤3 OR -3≤x<2. Clearly, when we combine both conditions, the range of x that satisfies both is 2<x≤3. This helps us uniquely answer the question and say x>2 OR \(x^3\)*y>56.
The combination of statements is sufficient. Answer option E can be eliminated.

The correct answer option is C.

Hope that helps!

chetan2u · Answer

The colored portion is wrong..
 x can be 1 and y = 57, then also answer is YES.. So we look for x<0 or x>2

chetan2u · Answer

Same as the member above

The colored portion is wrong..
 x can be 1 and y = 57, then also answer is YES.. So we look for x<0 or x>2

Kinshook · Answer

Asked: If the integer y > 6 , is x^3y > 56 ? (1) 4 < x^2 ≤ 9 If x>0 2 x^3 x>0 For all x>0 x^3y is NOT NECESSARILTY > 56 if y>6 NOT SUFFICIENT (1) + (2) (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 or x>0 256 MUST BE TRUE SUFFICIENT IMO C

CrackverbalGMAT · Answer

Hello Chetan,

Thanks for bringing this up. I agree with you. However, I have used that portion of my reply to highlight the fact that breaking down the question stem is a good thing to do on Algebra questions.

Notice that I have taken the value of x as 1 when dealing with the 2nd statement, which is essentially the point you are making. I admit that I could have paraphrased that paragraph better to include x=1 so that this ambiguity could have been avoided. Thanks for the feedback Chetan.

chetan2u · Answer

My point was that we are surely looking for x>2, but also if x<0, then the statement would be sufficient as it would give us a NO as the answer.

Bunuel · Answer

This question is a part of  Are You Up For the Challenge: 700 Level Questions  collection.

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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3

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