For integers x and y, if 91x = 8y, which of the following must be true

Question

For integers x and y, if 91x = 8y, which of the following must be true?

I. y > x
II. y/7 is an integer
III. The cube root of x is an integer

A) I only
B) II only
C) III only
D) I and II
E) II and III

kunal555 · Accepted Answer

Statement 1: y>x

When y=x=0, equation holds but y is not greater than x 
When x=-8 and y=-91, equation again holds but x>y
NOT TRUE

Statement 2: y/7 is an integer

Since x and y are integers, 91x and 8y must also be integers.
It is given that 91x=8y
or 13*7*x = 8 y
or 13x = 8y/7
To balance the equation, y/7 must be an integer
TRUE

Statement 3: The cube root of x is an integer

x can be equal to 2*2*2*3 and for this value of x,y will be 13*7*3
So, x may or may not be a cube root.
NOT TRUE

Answer:-B

AkshdeepS · Answer

Given 91*x=8*y
 
It can be equal when x = 8 or -8
and y = 91 or -91

So Statement (I)  y < x if we take x = -8 and y =-91
but y > x if we take x = 8 and y = -91, so (I) is not true always

Statement (2) in any case

91/7 or -91/7 is an integer that is 13 or -13. So it is true

Statement (3)  cube root of 8 or -8 is 2 and -2 respectively. So it is true always.

So IMO answer is E

ggurface · Answer

1) 
91 * 8 = 8 * 91, y > x
91 * -8 = 8 * -91, y < x
False

2) 
The valid numbers for x 0, and plus/minus multiples of 8. Therefore, the valid numbers for y are 0 and multiples of +-91, which are divisible by 3.
True

3)
Consider the case where x = 8, y = 81. The root of x is not an integer, and rooting it again does not make it an integer.
False

Answer B

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

Solution: B.

Statement 1 is not necessarily true. x could equal -8 and y could equal -91, for example, in which case the equation holds but x > y” title=”x > y”/>.

Statement 2 is true: for <img src= to equal 91x, then the prime factorization: 2*2*2*y = 13*7*x. y must then be able to account for the prime factor of 7 on the other side of the equation.

And statement 3 is not necessarily true. While x must account for the factors 2*2*2, it could also include a non-cubed factor as well. For example, x could be 2*2*2*5 and y could be 13*7*5. The equation would hold because the extra 5 is accounted for on both sides. x MUST account for 2*2*2, but need not be limited to just that, as x and y could have duplicate factors on either side. Beware the statements that look very likely to be true when you face these “must be true” problems – the GMAT is a master of misdirection.

bhartiyaayus · Answer

Hi Bunuel,

I have one doubt in this question. Why can't y be a factor of 13. If Y is a factor of 13, it will still solve the equation. Can you please explain why y/7 will be an integer.

Thanks.

Bunuel · Answer

91x = 8y

\frac{x}{y} = \frac{8}{91} = \frac{2^3}{7*13} .

x is a multiply of 2^3 and y is a multiple of 7*13.

ScottTargetTestPrep · Answer

Without actually solving the equation algebraically, we can see that x could be 8 and y could be 91, since 91(8) = 8(91). However, this is not the only possibility. We see that x could be 0 and y could be 0, since 91(0) = 8(0), or x could be -8 and y could be -91, since 91(-8) = 8(-91). In any event, we see that x is a multiple of 8 (including 0 and the negative multiples) and y is a multiple of 91 (including 0 and the negative multiples).

Let’s analyze each Roman numeral:

I. y > x

Since both x and y could be 0, y is not necessarily greater than x.

Roman numeral I is not true.

II. y/7 is an integer

Since y is a multiple of 91, y/7 is an integer.

Roman numeral II must be true.

III. The cube root of x is an integer

We’ve mentioned that x is a multiple of 8. If x = 8, then the cube root of x is an integer. However, if x = 16, then the cube root of x is not an integer.

Roman numeral III is not true.

Answer: B

dasoisheretorule · Answer

y could very easily be written as (91 * x)/8 , and y/7 then becomes (91 * x)/(8 * 7) which can then be shortened to (13 * x)/8.
It can be very easily be proven that this expression is not an integer.

Kinshook · Answer

Asked: For integers x and y, if 91x = 8y, which of the following must be true? y = 91x/8; x is a multiple of 8 x= 8y/91 I. y > x If x>0; y>x But if x<0;y

ProfChaos · Answer

(III) The cube root of x is an integer 
This cannot always be true

We have been given 91x=8y
Since there is no common factor between 8 and 91 the first pair of numbers for x and y that will satisfy this equation is 8 and 91 (-8 and -91 too)
In both these cases cube root of x is an integer (2 and -2 respectively)

But what if we keep x=8*2 and y=91*2
Equation is still satisfied but the values of x and y have changed
Now the cube root of x=16 is not an integer

Thus Statement III cannot be always true.

Fdambro294 · Answer

X, Y = Integers (can be (-)Neg. , (+)Pos., or 0 Integers)

91*X = 8*Y

X = 8 * Y / 91

Analysis: Since 8 is NOT Divisible by 91 and X must = an Integer ---- we know that Y must be a MULTIPLE of 91. From the Factor Foundation Rule, if Y is a Multiple of 91, then Y is a Multiple of any Factors of 91 (7 and 13)

II. must be True

I. Y > X

we know that Y must be an Integer Divisible by 91, but it could be a (-)Negative Integer

If y = (-)91

then X = (-)8

y < X

I. does NOT have to be True

III. Just because X must be Divisible by 8 (which is a Perfect Cube) does NOT mean a Multiple of 8 must be a Perfect Cube.

for instance, X = 16 and Y = 182 ---- the Constraints are satisfied and X = 16 is NOT a Perfect Cube

Only II must be True

-B-

Basshead · Answer

(1) y could be -91 and x could be -8. Doesn't have to be true.

(2) Lets break the equation into its prime factorization form:

7 * 13 * x = 2 * 2 * 2 * y

Since each side must be equal, y must contain a 7 in its prime factorization. Therefore y is a multiple of 7.

(3) If x = 8, then the cube root is an integer. However, we can't say for certain that x = 8. x could be 8 * 3. Doesn't have to be true.

Answer is B.

rvgmat12 · Answer

OE:
Statement 1 is not necessarily true. x could equal -8 and y could equal -91, for example, in which case the equation holds but x > y.

Statement 2 is true: for 8y to equal 91x, then the prime factorization: 2 * 2 * 2 * y = 13 * 7x. y must, then, be able to account for the prime factor of 7 on the other side of the equation.

And statement 3 is not necessarily true. While x must account for the factors 2 * 2 * 2, it could also include a non-cubed factor as well. For example, x could be 2 * 2 * 2 * 5 and y could be 13 * 7 * 5. The equation would hold because the extra 5 is accounted for on both sides. x MUST account for 2 * 2 * 2, but need not be limited to just that, as x and y could have duplicate factors on either side.

aditijain1507 · Answer

ScottTargetTestPrep  in second eqn, I went to y/7=13/8*x -- this does not give me a clear idea that y is divisible by 7 as both sides can have decimal values in this case. Please tell where am I going wrong. what is a better approach for such questions?  gmatwhiz   Bunuel

Wadree · Answer

91x = 8y ...so it seems if x = 8 and y = 91 ...then equation will match...so y > x ...but what if x = - 8 and y = - 91 
Then equation will also match...but then .... x > y ..... so ...y > x ....is not a must.......

Then we got.....y / 7 ..... now.....91x = 8y.....in other words 13 × 7x = 8y .....
So....left side got 13 and 7 as prime factor....so right side must also have 13 and 7 as factor...so those 13 and 7 must be inside y.....in other words those 13 and 7 must be factor of y.....so..... y / 7 ... must be integer......

Lastly.....91x = 8y .... so it might seem x = 8 ... so cube root of x is 2 ...an integer... but wut if....
y = 13 × 7 × 3 ..... cuz then... x =   / 91 = 24 ...... so then cube root of x = 24 wont be an integer......
so cube root of x is an integer....aint a must......

so .... only || .....is must......

! nah id win!

For integers x and y, if 91x = 8y, which of the following must be true

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For integers x and y, if 91x = 8y, which of the following must be true

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