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enjoylife1788
- Senior | Next Rank: 100 Posts
- Posts: 38
- Joined: Sat Aug 15, 2009 10:31 pm
- Location: India
Looking at the manhattan's challenge problem for this week, I found it too easy to be a challenge problem. So it made me think, is it actually so easy or am I wrong somewhere?
If abc ≠0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =
(A) (1 + bc)/(b + c)
(B) (1 - bc)/(b + c)
(C) (1 + b + c)/(bc)
(D) (1 - b - c)/(bc)
(E) (1 - b - c)/(b + c)
Here is how i solved:
1/a + 1/b + 1/c = 1/abc
(bc+ac+ab)/abc = 1/abc
Multiplying both sides by abc (since abc≠0)
bc+ac+ab=1
ac+ab = 1-bc
a(b+c)=1-bc
a=(1-bc)/(b+c)
Answer is B
[/i]
If abc ≠0 and the sum of the reciprocals of a, b, and c equals the reciprocal of the product of a, b, and c, then a =
(A) (1 + bc)/(b + c)
(B) (1 - bc)/(b + c)
(C) (1 + b + c)/(bc)
(D) (1 - b - c)/(bc)
(E) (1 - b - c)/(b + c)
Here is how i solved:
1/a + 1/b + 1/c = 1/abc
(bc+ac+ab)/abc = 1/abc
Multiplying both sides by abc (since abc≠0)
bc+ac+ab=1
ac+ab = 1-bc
a(b+c)=1-bc
a=(1-bc)/(b+c)
Answer is B
[/i]
