Latest Elitmus Aptitude Question SOLUTION: If 1/a+1/b+1/c= 1/(a+b+c) where a+b+c is not equal to zero,what is the value of (a+b)(b+c)(C+a) A)equals zero B) Greater than zero C) less than zero D) can not be determined

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(#M40018467) ELITMUS question Quadratic Equation Keep an EYE

If 1/a+1/b+1/c= 1/(a+b+c) where a+b+c is not equal to zero,what is the value of (a+b)(b+c)(C+a)
A)equals zero
B) Greater than zero
C) less than zero
D) can not be determined

Asked In Elitmus Sumit Rawal (11 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

aruna  (11 years ago)

i think ....Answer is A
let a=1,b=-1,c=1
then the above equation 1/a + 1/b+ 1/c=1/(a+b+c) satisfies
then (a+b)(b+c)(c+a)=0

Is this solution Helpfull? Yes (8) | No (3)  

Chandni Maurya  (10 years ago)

ans=(a) equals zero
=>1/a+1/b+1/c= 1/(a+b+c)
=> (a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1
subtracting 3 both side,
=> [(a+b+c)/a]-1+[(a+b+c)/b]-1+[(a+b+c)/c]-1=1-3
=> (b+c)/a+(a+c)/b+(a+b)/c = -2
adding 2 both side,
=> [(b+c)/a+(a+c)/b+(a+b)/c]+2 = -2+2
=> [bc(b+c)+ ac(a+c)+ab(a+b)]/abc + 2 = 0
=> [bc(b+c)+ ac(a+c)+ab(a+b)]+2abc = 0*abc
//[bc(b+c)+ ac(a+c)+ab(a+b)]+2abc=(b^2*c+c^2*b)+(a^2*c+c^2*a)+(a^2*b+b^2*a)+2abc
= (a+b)(b+c)(c+a)//
=> (a+b)(b+c)(c+a)= 0
=> equal to zero

Is this solution Helpfull? Yes (4) | No   

Md Intshamuddin  (10 years ago)

@diksha u can call on 9916020682,will explain on phone.

Is this solution Helpfull? Yes (3) | No   

Md Intshamuddin  (11 years ago)

(b)
1/a+1/b+1/c= 1/(a+b+c)
=> (a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1
=> [(a+b+c)/a]-1+[(a+b+c)/b]-1+[(a+b+c)/c]-1=1+3
=> (b+c)/a+(a+c)/b+(a+b)/c=4
=> [(b+c)/a+(a+c)/b+(a+b)/c]+2=4+2
=> [(a+b)(b+c)(C+a)]/abc=6
=> (a+b)(b+c)(C+a)]=6abc
which is always greater than zero

Is this solution Helpfull? Yes (1) | No (7)  

Saiyam Behl  (10 years ago)

put a=-b
then try

Is this solution Helpfull? Yes (1) | No   

SAMBIT KUMAR GHOSH  (10 years ago)

equal to 0

Is this solution Helpfull? Yes (1) | No   

DIKSHA RANGNANI  (10 years ago)

couldn't understand from step 5 to 6

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Krishanu chatterjee  (10 years ago)

will be less than 0

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VIKASH KUMAR SINGH  (10 years ago)

A)equals zero
1/a + 1/b + 1/c= 1/(a+b+c)
divide both side by 'abc'. You will get,
(bc+ca+ab)(a+b+c) = abc
multiply the lef side braced elements... you will get...
a^2c + ac^2 + a^2b + ab^2 + b^2c + bc^2 + 2abc = 0 ------ > 1
now, multiply the braced elements for which value is being asked... you will get
the left side elements of the eqution "1" that will give you a value= 0.

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manish  (7 years ago)

expand (a+b)(b+C)(c+a)
we get
2abc+c^2(a+b)+b^2(a+c)+a^2(b+c)
now solve the 1/a+1/b+1/c= 1/(a+b+c)
you will get
2abc=--c^2(a+b)-b^2(a+c)-a^2(b+c)
on putting this value we get ans=0

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