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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 18 Mar 2020, 10:41
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C
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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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C
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 00:23
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Bunuel
If the integer \(y > 6\), is \(x^3y > 56\)?


(1) \(4 < x^2 ≤ 9\)

(2) \(x^3 + x > x^3\)
Show more

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 \(0<x\leq 2\), the answer will depend on the value of y. For example if x=1 and y=57, answer is yes but if x=2, and y=7, answer is no.
Case I:- If \(x\leq 0\), answer is definite NO.

Now we can see the choices do not talk of value of y, so we look at only two cases in our choice.
(1) Is \(x>2\) ?.....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 \(2<x\leq 3\)
so both yes and no

(2) \(x^3 + x > x^3\)
x>0...
If x>2, surely YES, otherwise cannot say

Combined..
x>0 and \(2<x\leq 3\)
suff

C
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 18 Mar 2020, 14:19
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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

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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 18 Mar 2020, 14:55
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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
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 18 Mar 2020, 22:54
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Bunuel
If the integer \(y > 6\), is \(x^3y > 56\)?


(1) \(4 < x^2 ≤ 9\)

(2) \(x^3 + x > x^3\)
Show more

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: \(4<x^2≤9\)
    • According to this statement, there can be two cases:
      o Case 1: \(2^2 < x^2 ≤ 3^2 ⟹ 2 < x ≤ 3 ⟹ 2 < x\)
      o Case 2: \((-2)^2 < x^2 ≤ (-3)^2 ⟹ -3 ≤ x < -2 ⟹ x < 2\)
We can see that the result of the above two cases are contradictory.
Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.
Statement 2: \(x^3+x>x^3\)
    • 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.
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 00:04
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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!
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 00:25
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Bunuel
If the integer \(y > 6\), is \(x^3y > 56\)?


(1) \(4 < x^2 ≤ 9\)

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

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.


Show more

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
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 00:26
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ArvindCrackVerbal
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.


The correct answer option is C.

Hope that helps!
Show more

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
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 00:47
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Bunuel
If the integer \(y > 6\), is \(x^3y > 56\)?


(1) \(4 < x^2 ≤ 9\)

(2) \(x^3 + x > x^3\)
Show more

Asked: If the integer \(y > 6\), is \(x^3y > 56\)?


(1) \(4 < x^2 ≤ 9\)
If x>0
2<x<=3
8<x^3<=27
56<x^3y;
For x<0
x^3y<0<56
NOT SUFFICIENT

(2) \(x^3 + x > x^3\)
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
2<x<=3
8<x^3<=27
56<x^3y; x^3y>56 MUST BE TRUE
SUFFICIENT

IMO C
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 06:51
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chetan2u
ArvindCrackVerbal
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.


The correct answer option is C.

Hope that helps!

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
Show more

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.
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 19 Mar 2020, 08:58
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ArvindCrackVerbal
chetan2u
ArvindCrackVerbal
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.


The correct answer option is C.

Hope that helps!

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

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.
Show more

:thumbup: :)
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. :please :please
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If the integer y > 6, is x^3y > 56? (1) 4 < x^2 ≤ 9 (2) x^3 + x > x^3 03 Nov 2021, 07:11
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If the integer \(y > 6\), is \(x^3y > 56\)?


(1) \(4 < x^2 ≤ 9\)

(2) \(x^3 + x > x^3\)
Show more



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 31 May 2024, 12:17
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