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13 Oct 2020, 00:15
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D
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Difficulty:
65%
(hard)
Question Stats:
50% (01:45) correct
50%
(01:57)
wrong
based on 212
sessions
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If a, b, and c are positive, is a > b?
(1) \(\frac{a}{(b+c)}>\frac{b}{(a+c)}\)
(2) \(b+c < a\)
(1) \(\frac{a}{(b+c)}>\frac{b}{(a+c)}\)
(2) \(b+c < a\)
13 Oct 2020, 03:22
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Quote:
Step 1: Understanding statement 1 alone
(1)\( \frac{a}{(b+c)}>\frac{b}{(a+c)}\)
As a, b and c are positive, a(a+c) > b(b+c)
\(a^2 + ac > b^2 + bc\)
\(a^2 - b^2\) + ac - bc > 0
(a + b) (a - b) + c (a - b) > 0
(a - b) ( a + b + c) > 0
As ( a + b + c) is positive, (a - b) > 0; a > b
Sufficient
Step 2: Understanding statement 2 alone
(2) b+c<a
As a, b and c are positive and b+c is less than a, hence b < a
Sufficient
IMO D
13 Oct 2020, 03:46
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since, a,b,c are positive we can cross multiply.
1) a^2=ac>b^2+bc
=> a^2-b^2+ac-bc >0
=>(a+b)(a-b) +b(a-b)>0
=> (a+b+c)(a-b) >0
since a,b,c are positive a+b+c will always be positive
=> a-b>0 , Sufficient
2) a>b+c
since, a,b,c are positive numbers and since a is greater than the sum of b and c then a will definitely greater than b, Sufficient.
D.
1) a^2=ac>b^2+bc
=> a^2-b^2+ac-bc >0
=>(a+b)(a-b) +b(a-b)>0
=> (a+b+c)(a-b) >0
since a,b,c are positive a+b+c will always be positive
=> a-b>0 , Sufficient
2) a>b+c
since, a,b,c are positive numbers and since a is greater than the sum of b and c then a will definitely greater than b, Sufficient.
D.
