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arjtryarjtry
) If x, y and z are integers greater than 1, what is the value of x + y + z?
1) xyz = 70
2) x/yz = 7/10


1) xyz = 70 (suffcient)
= 2 * 5 * 7 ( is only combination which leads to 70)

2) x/yz = 7/10 (not suffcient)

x/yz = 7/10 ( x=7 y = 5 z=2)
x/yz = (7/10)*(k/k) also true for ( x=14 y = 10 z=2)


A is the answer
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arjtryarjtry
) If x, y and z are integers greater than 1, what is the value of x + y + z?
1) xyz = 70
2) x/yz = 7/10

Given : x,y,z are integers >1
(1)xyz=70 => xyz=5*7*2 => say 5,2,7 are values of x,y,z in any order
hence x+y+z is answered =>SUFFI
(2)x/yz =7/10 this can be any values for x,y,z say,
x=7,y=5,z=2 or x=14,y=5,z=4 hence x+y+z will change INSUFFI

IMO A
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If integers x, y and z are greater than 1 what is the value of x + y + z ?

A. xyz = 70
B. x/yz = 7/10

I got this right. but with 2nd statement. I went to only 10x = 7yz
is that the end of 2nd statement or we can drive something with that?

x,y,z >1

(1) xyz = 70 = 2*5*7 so we can find x+y+z

(2) \(\frac{x}{yz} = \frac{7}{10}\)

this just means x is a multiple of 7 and the product yz is a multiple of 10. We have no info to determine values of x,y and z.

A
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xyz = 70
so the values could be 1 7 10
x+y+z = 18
if the values are 2 5 7
x+y+z =14
so a not sufficient
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anilnandyala
xyz = 70
so the values could be 1 7 10
x+y+z = 18

if the values are 2 5 7
x+y+z =14
so a not sufficient

"If x, y and z are integers greater than 1 ..."

Correct answer is A.
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Bunuel, if the question states that x, y and z are greater then 1... then is ok to say that one of thoes is 1? as I read it , none of them can be one " GREATER THEN 1"???
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marijose
Bunuel, if the question states that x, y and z are greater then 1... then is ok to say that one of thoes is 1? as I read it , none of them can be one " GREATER THEN 1"???

Again: "If x, y and z are integers greater than 1 ..." So none of them is one.
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Bunuel,

The problem doesnt specifically state that the integers are unique or different. Is it possible for A to be insufficient as xyz to be 7*7*5 with a sum of 19 or the answer provided above with 2*5*7 and a sum of 14?

I'm assuming the variables need to be unique integer values, but this threw me off at first.
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Bunuel,

The problem doesnt specifically state that the integers are unique or different. Is it possible for A to be insufficient as xyz to be 7*7*5 with a sum of 19 or the answer provided above with 2*5*7 and a sum of 14?

I'm assuming the variables need to be unique integer values, but this threw me off at first.

Notice that 7*7*5 = 245 not 70.
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Bunuel,
Yes, good correction. Obviously. For some reason I was thinking 7+7 = 14 and 14*5 = 70. Dumb mistake by me. Thank you!
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arjtryarjtry
If x, y and z are integers greater than 1, what is the value of x + y + z?

(1) xyz = 70
(2) x/yz = 7/10

We know x,y,z > 1

Read St.A --> xyz = 70 = 2*5*7
as x,y and z are greater than 1 we cant have (10,7,1), (70,1,1) or other combinations
Hence x+y+z = 2+5+7 = 14
A is sufficient

St. B --> x/yz = 7/10
now x is 7k, y is 2m and z is 5n
Hence x+y+z will be variable
Hence B is useless.

Only A is sufficient
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\(X , Y , Z integers > 1\)

\(x + y + z = ?\)

Statement 1

\(xyz = 70\)

Prime factorizing 70 we get 2, 5 and 7 ( we can not take any value to be 1 because all integers are greater than 1)

so \(x + y + z = 14\)

No matter what value we take for x, y, z , sum will always be 14

Hence statement 1 is Sufficient

Statement 2

\(\frac{x}{yz} = \frac{7}{10}\)

here if \(x = 7\) then y be 2 and z be 5 then \(x + y + z = 14\)

x can also be 14 , then y be 2 and z be 10 then \(x + y + z = 26\)

so clearly statement 2 is not sufficient.

Hence Answer is A
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A video explanation can be found here:
https://www.youtube.com/watch?v=AKkm1HfkPOM&t=9s

Practically every GMAT will ask you a question that requires you to think about prime factorization, so always have than in the back of your mind.

70 = 2*5*7, so the sum of primes is 14. Statement (1) is sufficient

In statement (2) x, y and z COULD be 7, 2 and 5, but could also be a multiple of x and a multiple of y or z, such that the expression simplifies to 7/10 (for e.g. x = 14, y = 4, and z = 5), so (2) not sufficient
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Bunuel
anilnandyala
xyz = 70
so the values could be 1 7 10
x+y+z = 18

if the values are 2 5 7
x+y+z =14
so a not sufficient

"If x, y and z are integers greater than 1 ..."

Correct answer is A.

Bunuel
I factored 105 to get --> 7*3*5
But then I was thinking that technically you can have 21*1*5=105
or 105*1*1=105 etc. To clarify, this problem basically restricts those possibilities simply by 7+3+5 being the only answer that appears... the other options I mentioned are much bigger than the answer choices provided.
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Bunuel
anilnandyala
xyz = 70
so the values could be 1 7 10
x+y+z = 18

if the values are 2 5 7
x+y+z =14
so a not sufficient

"If x, y and z are integers greater than 1 ..."

Correct answer is A.

Bunuel
I factored 105 to get --> 7*3*5
But then I was thinking that technically you can have 21*1*5=105
or 105*1*1=105 etc. To clarify, this problem basically restricts those possibilities simply by 7+3+5 being the only answer that appears... the other options I mentioned are much bigger than the answer choices provided.

First of all, were did you get 105? We have xyz = 70, not xyz = 105.

Next, you are quoting the reply that answers your query. We are told that x, y and z are integers greater than 1, so none of them can be 1.
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