Elitmus
Exam
Numerical Ability
if 1/a + 1/b + 1/c = 1/(a+b+c) where a+b+c !=(not equals) 0, abc != 0 , then what is the value of (a+b) (b+c) (c+a)?
a)equal to 0
b) greater than 0
c) less than 0
d) can't determine
Read Solution (Total 8)
-
- 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 - 9 years agoHelpfull: Yes(4) No(3)
- Taking 11 common from LHS separating z and breaking 151001 as a factor of 11 with some remainder in RHS
11(113v+112w+11x+y) + z = 11x13727 + 4 [hence; z = 4]
Equating z =4 and cancelling 11 from each side we get the following simplified equation and again repeating the above procedure;
11(112v + 11w +x) + y =11x1247 + 10 [y = 10]
11(11v + w ) + x = 11x113 + 4 [x = 4]
11v + w = 11x10 + 3 [v=10 , w=3]
So we get the unique solution for the above equaton. - 9 years agoHelpfull: Yes(2) No(1)
- Sorry this solution for above question.
- 9 years agoHelpfull: Yes(2) No(0)
- ans is equal to 0
let a=-1,b=1,c=2
1/-1+1/1+1/2=1/(-1+1+2)
we get 1/2=1/2 condition satifies - 9 years agoHelpfull: Yes(2) No(4)
- ab+bc+ba/abc=1/(a+b+c)
=> (ab+bc+ca)(a+b+c)=abc
=> a^2b+ab^2+abc+abc+b^2a+bc^2+abc+a^2c+ac^2=abc
=> a^2b+ab^2+bc^2+b^2a+a^2c+ac^2+2abc=0
=>(a+b)(b+c)(c+a)=0 - 8 years agoHelpfull: Yes(1) No(0)
- a) equals to 0
equates both equations...we get 0
- 9 years agoHelpfull: Yes(0) No(0)
- I got answer as (-a(c^2))
- 9 years agoHelpfull: Yes(0) No(0)
- equal to 0
- 8 years agoHelpfull: Yes(0) No(0)
Elitmus Other Question
Choice (a) if the question can be answered by using one of the statements alone, but cannot be answered by using the other statement alone.
Choice (b) if the question can be answered by using either statement alone.
Choice (c) if the question can be answered by using both the statements together, but cannot beanswered by using either statement alone.
Choice (d) if the question cannot be answered even by using both the statements together.
Is x+y> √2 ? (The square root of 2)
1)x^2 + y^2 =1
2)x= log a base b and y =log b base a ,where a,b,x,y are positive real number and a & b is not equal is 1
If v,w,x,y, and z are non negative integers, each less than 11, then how many distinct combinations of (v,w,x,y,z) satisfy v(11^4) + w(11^3) +x(11^2) + y(11) + z =151001 ?
a)0
b)1
c)2
d)3