Latest Elitmus Aptitude Question SOLUTION: If 1/a + 1/b + 1/c = 1/(a+b+c)where a+b+c is not equal to 0, what is the value of (a+b)(b+c)(c+a)... (a) 0 (b) >0 (c) 1

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(#M40021862) ELITMUS Verbal Ability question Aptitude Keep an EYE

If 1/a + 1/b + 1/c = 1/(a+b+c)where a+b+c is not equal to 0, what is the value of (a+b)(b+c)(c+a)...
(a) 0
(b) >0
(c) 1

Asked In Elitmus prayas goel (11 years ago)

Solved Rashmi bharti Read Solution (10)

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Read Solution (10) :

BAPI MODAK  (11 years ago)

(a+b)(b+c)(c+a)=a2b+a2c+b2a+b2c+c2a+c2b+2abc.........(1)

now, solve
wll get..
a2b+a2c+b2a+b2c+c2a+c2b+2abc=0...(2)

from (1) & (2)
(a+b)(b+c)(c+a)=0

Is this solution Helpfull? Yes (28) | No (2)  

Amit kumar  (11 years ago)

let a= 1, b= -1 and c= 2
verifing a+b+c is not equals to zero. and 1/a + 1/b + 1/c = 1/(a+b+c)
So, (a+b)(b+c)(c+a)= 0

Is this solution Helpfull? Yes (23) | No (1)  

Rashmi bharti  (11 years ago)

Ans is o i.e, (a)
As 1/a+1/b+1/c=1/(a+b+c),
=>(a+b+c)(ab+bc+ac)=abc
=>a^2(b+c)+b^2(a+c)+c^2(a+b)-2abc=0.
Similarly on solving (a+b)(b+c)(c+a)=a^2(b+c)+b^2(a+c)+c^2(a+b)=0

Is this solution Helpfull? Yes (11) | No   

Sam Sunder Golla  (10 years ago)

Answer is 0.
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

Is this solution Helpfull? Yes (4) | No   

siva  (10 years ago)

it's simple ANS is :0
now assume a=-1,b=1,c=1
then If 1/a + 1/b + 1/c = 1/(a+b+c) is:1/-1 +1/1 +1/1=1/(-1+1+1)
:1=1
condition 1: a+b+c!=0 ,-1+1+1=1!=0 satisfied
(a+b)(b+c)(c+a)=(-1+1)(1+1)(1-1)=0
so ANS is 0

so

Is this solution Helpfull? Yes (2) | No   

prayas goel  (11 years ago)

It must be 0 na???
Plz let me know if u have different options ...
(c)

Is this solution Helpfull? Yes (1) | No   

Amitabha Ghosh  (11 years ago)

ans is 0

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

Ashwani Kumar  (10 years ago)

(a) 1/a+1/b+1/c=1/(a+b+c) =>(bc+ac+ab)/abc=1/(a+b+c)
=>0

Is this solution Helpfull? Yes (1) | No   

sravan  (11 years ago)

Ans :0
by simplifying The Equation we can Get the Result

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shruti srivastava  (10 years ago)

c) 1
since (a+b)(b+c)(c+a)=(1/a+ 1/b + 1/c)(a+b+c)
u can check it.

Is this solution Helpfull? Yes | No (1)  

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