If integers x, y and z are greater than 1 what is the value

statement 1 : 70 = 2*5*7
x+y+z = 2+5+7 = 14 suff

statement 2 : x/yz = 7/10 = 14/20 = 21/30 ... not suff

option A

Hi All,

DS questions test a variety of skills, including the thoroughness of your thinking. In other words, do you notice when there's more than one possibility? Also, do you realize when enough "restrictions" exist to limit the number of possible answers to just one possibility? Those ideas are at the heart of this DS question.

Right from the beginning, we're given restrictions to work with:
X, Y and Z are INTEGERS
X, Y and Z are all > 1

We're asked for the value of X + Y + Z.

Fact 1: XYZ = 70

At first glance, many Test Takers would think that this was insufficient information since there are "so many" ways to get a product of 70. HOWEVER, we have to work with those initial restrictions (X, Y and Z are INTEGERS and they're ALL > 1). How many different sets of numbers will get us a product of 70 now?

With a bit of pattern-matching, you should notice that since 70 "ends in 0", one of the numbers MUST be a multiple of 5 and at least one of the numbers MUST be even. Prime-factorization can be used to quickly break the 70 "into pieces":

70 = (2)(5)(7)

A 2, a 5 and a 7 represent the ONLY possible numbers. Since the question asks us for the SUM of the 3 numbers, it doesn't matter which variable is which number. The final answer is 2 + 5 + 7 = 14.
Fact 1 is SUFFICIENT

Fact 2: X/YZ = 7/10

With the prior work form Fact 1, the Fact seems to have an obvious solution:
X = 7
Y and Z = 2 and 5
So X + Y + Z = 14

The "catch" here is that fractions can be reduced, so while X/YZ = 7/10 gave us this first outcome, X/YZ = 14/20 = 21/30 = 28/40 = etc. are ALSO possibilities that would lead to different answers.

For example:
X/YZ = 14/20
X = 14
Y = 2
Z = 10
So X + Y + Z now = 26 and we have a different answer.
Fact 2 is INSUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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

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

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

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"???

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.

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!

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

\(X , Y , Z integers > 1\)

\(x + y + z = ?\)

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

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

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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