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

My answer is d. however the correct answer is a.

I can't figure why you cannot solve it using 2.

since x/yz = 7/10 implies that

10 x = 7 yz, or

2 * 5 * x = 7 * y * z

so x + y + z = 2 + 5 + 7

Second condition is not sufficient . consider below values for x, y and z
x=14. y=2, z=10
Still we will get x/yz = 14/2*10 = 7/10

A.

2nd statement cannot be solved - the question by the person posting the question assumes that xyz = 70.

2*5*x = 7* y* z

x can be 80 and y*z can be 10 * 8 or x can be 100 and y* z can be 10*10

hence they yield different solns for x+y+z

1. XYZ = 70

X>1 ; Y>1 Z>1

consider XYZ as one number i.e. 70

prime factors of 70 = 2 * 5 * 7

Other Factors of 70 can be formed using several combinations above 3 prime numbers,

For example, 14 * 5 * 1 ----> X = 14 , Y = 5 and Z = 1 but Z > 1 so these values of X, Y n Z are not valid.

2nd example, X*Y*Z = 10 * 1 * 7 = 70 ,however, Y = 1 invalidates condition Y > 1.

The only possible combination that satisfies the given condition i.e. X > 1; Y>1 and Z >1

is X*Y*Z = 2 * 5 * 7

giving X + Y + Z = 14.

Hence statement 1 alone is sufficient.


- Rahul

Your way ----->

My answer is d. however the correct answer is a.

I can't figure why you cannot solve it using 2.

since x/yz = 7/10 implies that

10 x = 7 yz, or

2 * 5 * x = 7 * y * z

so x + y + z = 2 + 5 + 7

---------------------------------------------

2 * 5 * X = 7 * Y * Z

X could be anything = 14, 28, 49

giving rt. side = 140, 280, 49

left side

Y = 2 z = 10 giving 140 (X + Y + Z = 14 + 2 + 10 = 26)

or

Y = 4 Z = 10 giving 280 (X + Y + Z = 28 + 4 + 10 = 42)

or

Y = 7 z = 10 giving 490 (X + Y + Z = 49 + 7 + 10 = 66)

So 2 alone is not sufficient.

But 1 alone is sufficient.

ah, yeah, ok, i see what you're doing.

here's the deal: if x, y, and z had to be PRIMES, then this solution would be entirely 100% correct. specifically, x would have to be 7 because there's a 7 on the right-hand side of the equation, and y and z (in some order) would have to be 2 and 5 because those two primes are on the left side of the equation.

HOWEVER, unfortunately for your "solution", the 3 variables don't have to stand for primes, and so you can add in extra factors to create additional possibilities.
for instance, you COULD have x = 7, y = 2, z = 5 (the "primes" solution).
but you could also have x = 70, y = 2, and z = 50 (multiply one number on each side by 10, so that the products remain the same). or you could have one of an infinity of other possibilities, all along the same lines.
therefore, insufficient.

--

moral of the story:
don't assume that the restrictions that apply to primes can apply just as well to integers in general. primes are very special animals.