Latest Elitmus Aptitude Question SOLUTION: . If 1/a + 1/b + 1/c = 1/(a+b+c) where a + b + c != 0 , abc != 0, what is the value of (a+b)(b+c)(c+a)? A. Equals 0 B. Greater than 0 C. Less than 0 D. Cannot be determined

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(#M40037399) ELITMUS Numerical Ability question prediction Keep an EYE

. If 1/a + 1/b + 1/c = 1/(a+b+c) where a + b + c != 0 , abc != 0, what is the value of (a+b)(b+c)(c+a)?

A. Equals 0

B. Greater than 0

C. Less than 0

D. Cannot be determined

Asked In Elitmus likhith vallabhaneni (10 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

Indira P Nayak  (10 years ago)

first solve the eq-
1/a+1/b+1/c=1/(a+b+c)
=>(bc+ac+ab)/abc=1/(a+b+c)
=>(a+b+c)(bc+ac+ab)=abc
=>abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2+abc=abc
=>2abc+a^2c+a^2b+b^2c+ab^2+bc^2+ac^2=0 ----eq (1)

Now solve the eq-
(a+b)(b+c)(c+a)
=>(ab+ac+b^2+bc)(c+a)
=>abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2+abc
=>2abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2 ---eq(2)
Now compare eq(1) and eq(2)
Since both are equal ao we conclude that
(a+b)(b+c)(c+a)=0
So the ans is optn (a)

Is this solution Helpfull? Yes (30) | No (1)  

Akash Agrawal  (10 years ago)

put a=1,b=-1,c=1 then 1/a+1/b+1/c=1 and 1/(a+b+c)=1
so (a+b)(b+c)(c+a)=0 ans

Is this solution Helpfull? Yes (7) | No (8)  

Bala Krishna  (10 years ago)

Since abc != 0 and a+b+c !=0 ; when we take modify the term 1/a + 1/b + 1/c = 1/(a+b+c) , we can simply multiply (a+b+c) and abc on numberator. That will yield us, (a+b+c)(bc+ca+ab) = abc or 2abc + a^2(b+c) + b^2(a+c) + c^2(a+b) = 0;

when we simplify (a+b)(b+c)(c+a), we get 2abc + a^2(b+c) + b^2(c+a) +c^2(a+b) . So, the answer is simply 0.

Is this solution Helpfull? Yes (2) | No (2)  

deepu kumar singh  (10 years ago)

equals to zero.


first solve the eq-
1/a+1/b+1/c=1/(a+b+c)
=>(bc+ac+ab)/abc=1/(a+b+c)
=>(a+b+c)(bc+ac+ab)=abc
=>abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2+abc=abc
=>2abc+a^2c+a^2b+b^2c+ab^2+bc^2+ac^2=0 ----eq (1)

Now solve the eq-
(a+b)(b+c)(c+a)
=>(ab+ac+b^2+bc)(c+a)
=>abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2+abc
=>2abc+a^b+ac^2+a^2c+b^2c+b^2a+bc^2 ---eq(2)
Now compare eq(1) and eq(2)
Since both are equal ao we conclude that
(a+b)(b+c)(c+a)=0
So the ans is optn (a)

Is this solution Helpfull? Yes (2) | No   

Akash Kar  (10 years ago)

let S=(a+b)(b+c)(c+a)

putting different values of a,b,c in 'S' we will get different values of it.
like putting a=3,b=-1,c=-1 we get -ve value of S.

if we put a=1,b=-1,c=1 we get S=0

we will get different values of 'S' changing the values of a,b,c satisfying the equations a+b+c!=0 and abc!=0

so answer may be option D.

Is this solution Helpfull? Yes (1) | No (2)  

Vikash Kumar  (10 years ago)

Ans will be B

Is this solution Helpfull? Yes | No (1)  

shah uchit a  (10 years ago)

a equals 0

Is this solution Helpfull? Yes | No (1)  

Shashank Sharma  (10 years ago)

b) greater then 0

Is this solution Helpfull? Yes | No (1)  

RAHUL KUMAR  (10 years ago)

c less than 0

Is this solution Helpfull? Yes | No (1)  

RAHUL KUMAR  (10 years ago)

A.equals 0

Is this solution Helpfull? Yes | No (2)  

Gowthu Reddy  (9 years ago)

equal to 0...assume a=-1,b=1,c=1

Is this solution Helpfull? Yes | No   

kritika gupta  (8 years ago)

1/a + 1/b + 1/c =1/ (a+b+c)
=> 1/a + 1/b =1/ (a+b+c) - 1/c
=> (b+a)/ab = (c - (a+b+c)) / c(a+b+c)

Cross multiplying (Since, a,b,c !=0 and a+b+c != 0)

=> c(a+b+c)(b+a) = - ab(a+b)
=> (a+b)( ca +cb +cc + ab ) = 0
=> (a+b)( ca + cc + cb + ab) = 0
=> (a+b)(b+c)(c+a)=0

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