Examkin

Question

Difficulty: Very hardAlgebraic Equations and Systems in Data Sufficiency

If mm and nn are non-zero real numbers, what is the value of m2+n2mn\frac{m^2 + n^2}{mn}?

(1) m2n+mn2=6(m+n)m^2 n + m n^2 = 6(m + n)
(2) m2−n2=3(m−n)m^2 - n^2 = 3(m - n)

  1. A
    Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
  2. B
    Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
  3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
  4. D
    EACH statement ALONE is sufficient.
  5. E
    Statements (1) and (2) TOGETHER are NOT sufficient.

Answer

Both statements together are sufficient, but neither statement alone is sufficient.
Evaluating each statement alone yields multiple possible numerical outcomes because factoring produces independent cases (m+n=0m+n=0 vs mn=6mn=6 for the first statement, and m−n=0m-n=0 vs m+n=3m+n=3 for the second). However, when combining both statements, three of the four sub-cases fail: two violate non-zero or consistency conditions, and the sub-case where mn=6mn=6 and m+n=3m+n=3 has a negative discriminant (32−4(1)(6)=−15<03^2 - 4(1)(6) = -15 < 0), yielding no real solutions. This leaves only the sub-case mn=6mn=6 and m−n=0m-n=0, which uniquely determines the value of the target expression to be 2. Therefore, both statements together are sufficient.

Step-by-Step Solution

1
Rephrase the target expression.
The target expression m2+n2mn\frac{m^2 + n^2}{mn} can be rewritten as mn+nm\frac{m}{n} + \frac{n}{m}. We need a unique numerical value for this expression.
Simplifying the target helps evaluate what parameters or relationships are required.
2
Evaluate Statement (1) independently.
Rearrange Statement (1): mn(m+n)−6(m+n)=0  ⟹  (mn−6)(m+n)=0mn(m + n) - 6(m + n) = 0 \implies (mn - 6)(m + n) = 0. This gives two cases: Case 1: m+n=0  ⟹  n=−mm + n = 0 \implies n = -m. Since m,n≠0m, n \neq 0, m2+(−m)2m(−m)=2m2−m2=−2\frac{m^2 + (-m)^2}{m(-m)} = \frac{2m^2}{-m^2} = -2. Case 2: mn=6mn = 6. For instance, if m=2,n=3m = 2, n = 3, then 4+96=136\frac{4+9}{6} = \frac{13}{6}. Multiple values exist (−2-2 and 136\frac{13}{6}), so Statement (1) alone is NOT sufficient.
Factoring instead of dividing by (m+n)(m+n) preserves the root m+n=0m+n=0.
3
Evaluate Statement (2) independently.
Rearrange Statement (2): (m−n)(m+n)−3(m−n)=0  ⟹  (m−n)(m+n−3)=0(m - n)(m + n) - 3(m - n) = 0 \implies (m - n)(m + n - 3) = 0. This gives two cases: Case 1: m−n=0  ⟹  m=nm - n = 0 \implies m = n. Since m,n≠0m, n \neq 0, m2+m2m2=2\frac{m^2 + m^2}{m^2} = 2. Case 2: m+n=3m + n = 3. For instance, if m=1,n=2m = 1, n = 2, then 1+42=52\frac{1+4}{2} = \frac{5}{2}. Multiple values exist (22 and 52\frac{5}{2}), so Statement (2) alone is NOT sufficient.
Factoring preserves the root m−n=0m-n=0.
4
Combine Statement (1) and Statement (2).
We test the four combinations of cases:
- Case A (m+n=0m + n = 0) & Case X (m−n=0m - n = 0): System gives m=0,n=0m = 0, n = 0. Contradicts the condition that m,nm, n are non-zero.
- Case A (m+n=0m + n = 0) & Case Y (m+n=3m + n = 3): 0=30 = 3, impossible.
- Case B (mn=6mn = 6) & Case X (m−n=0m - n = 0): m=n  ⟹  m2=6  ⟹  m=n=±6m = n \implies m^2 = 6 \implies m = n = \pm\sqrt{6}. Here m2+n2mn=6+66=2\frac{m^2+n^2}{mn} = \frac{6+6}{6} = 2.
- Case B (mn=6mn = 6) & Case Y (m+n=3m + n = 3): n=3−m  ⟹  m(3−m)=6  ⟹  m2−3m+6=0n = 3 - m \implies m(3-m) = 6 \implies m^2 - 3m + 6 = 0. The discriminant is b2−4ac=9−24=−15<0b^2 - 4ac = 9 - 24 = -15 < 0, which yields no real solutions for mm and nn.

Only Case B & Case X yields valid real non-zero solutions, providing a unique value of 2.
Checking real constraints eliminates non-real systems and isolates a single valid numerical outcome.

Key Concept

Factoring non-linear algebraic systems without illegal variable division, and verifying real-number constraints using discriminants in Data Sufficiency.
Estimated Time:2m 30s
Rate this question

Topics

Algebraic Equations and Systems in Data Sufficiency

More questions from this topic