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

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

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.

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!

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

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

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

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.

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.

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

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