Problem Solving

shankar.ashwin wrote:

• If x < y < z but x^2 > y^2 > z^2 > 0, which of the following must be positive?
  
  (A) (x^3) * (y^4) * (z^5)
  (B) (x^3) * (y^5) * (z^4)
  (C) (x^4) * (y^3) * (z^5)
  (D) (x^4) * (y^5) * (z^3)
  (E) (x^5) * (y^4) * (z^3)

Plug in combinations that satisfy the conditions in the question stem yet at the same time prove that the answers DON'T have to be positive.

Let x=-3, y=-2, z=1.
-3 < -2 < 1, satisfying the condition that x<y<z.
(-3)² > (-2)² > 1², satisfying the condition that x² > y² > z² > 0.

When we plug into the answers, our only concern is the sign of the resulting product.

(A) (x^3) * (y^4) * (z^5) = (negative)(positive)(positive) = negative. Eliminate A.

(B) (x^3) * (y^5) * (z^4) = (negative)(negative)(positive) = positive. Hold onto B.

(C) (x^4) * (y^3) * (z^5) = (positive)(negative)(positive) = negative. Eliminate C.

(D) (x^4) * (y^5) * (z^3) = (positive)(negative)(positive) = negative. Eliminate D.

(E) (x^5) * (y^4) * (z^3) = (negative)(positive)(positive) = negative. Eliminate E.

The correct answer is B.

If x < y < z
Possibilities
x = -ve, y = -ve z = -ve - (x=-5, y=-4, z=-3,- Satisfies the condition x^2 > y^2 > z^2 (25>16>9))
x = +ve, y = +ve z = +ve - (x=3, y=4, z=5,- Doesn't satisfy the equation x^2 > y^2 > z^2(9<16<25))
x = -ve, y = -ve z = +ve - (x=-5, y=-4, z=3 - Satisfies the condition x^2 > y^2 > z^2 (25>16>9))
x = -ve, y = +ve z = +ve - (x=-A, y=4, z=5 - Doesn't satisfy the equation y^2 > z^2(16<25))

Here we have two possibilities, let us test them against the options

which of the following must be positive?

If x = -ve, y = -ve z = -ve

(A) (x^3) * (y^4) * (z^5) - +ve
(B) (x^3) * (y^5) * (z^4) - +ve
(C) (x^4) * (y^3) * (z^5) - +ve
(D) (x^4) * (y^5) * (z^3) - +ve
(E) (x^5) * (y^4) * (z^3) - +ve

If x = -ve, y = -ve z = +ve

(A) (x^3) * (y^4) * (z^5) - -ve
(B) (x^3) * (y^5) * (z^4) - +ve
(C) (x^4) * (y^3) * (z^5) - -ve
(D) (x^4) * (y^5) * (z^3) - -ve
(E) (x^5) * (y^4) * (z^3) - -ve

Since the question reads, MUST be positive, the answer is B

IMO, answer is B)

I am sorry for such an inappropriate way of solving this question, but it is good for a lazy one (like me hehe)
I just take a quick look to the answer choices,and see that 4 out of 5 choices are identical.I mean:

(A) (x^3) * (y^4) * (z^5) and
(E) (x^5) * (y^4) * (z^3)

(C) (x^4) * (y^3) * (z^5) and
(D) (x^4) * (y^5) * (z^3)
are identical (in A and E x and z have odd exponents, and Ys have the same even exponent; in C and D Xes have the same even exponent 4, and y and z have odd exponents)

so, eliminating these choices we get that only (B) (x^3) * (y^5) * (z^4) is the answer.

0 ---- z^2 ---- y^2 ---- x^2 ----
we know that prime roots are +ve but need to find non-prime roots too (i.e. negative roots)

let us simplify, 2 ways are possible: either x>0 or x<0
a) x>0, all numbers are +ve ignore as we have more than one ans choice correct
b) x<0
~ y can be -ve or +ve
~ z can be -ve or +ve
Consider x=-3 always and y=(-2,2), z=(-1,1)??? and x<y<z Correctly noting z cannot be -ve/+ve, we state z is +ve only and if x=-ve (some -3), y=-ve (some -2) or +ve (some 2)??? but x<y<z, z=+ve (some 1), then y can be only -ve
now testing ans.choices: x=-ve,y=-ve,z=+ve
(A) (x^3) * (y^4) * (z^5) -> -*+*+=-
(B) (x^3) * (y^5) * (z^4) -> -*-*+=+ good choice
(C) (x^4) * (y^3) * (z^5) -> +*-*+=-
(D) (x^4) * (y^5) * (z^3) -> +*-*+=-
(E) (x^5) * (y^4) * (z^3) -> -*+*+=-

b

shankar.ashwin wrote:

• If x < y < z but x^2 > y^2 > z^2 > 0, which of the following must be positive?
  
  (A) (x^3) * (y^4) * (z^5)
  (B) (x^3) * (y^5) * (z^4)
  (C) (x^4) * (y^3) * (z^5)
  (D) (x^4) * (y^5) * (z^3)
  (E) (x^5) * (y^4) * (z^3)