Elitmus
Exam
Numerical Ability
Quadratic Equations
Q. If x,y,z are non negative integers and x=6 then find the number of solutions of 1/x +1/y = 1/z
Read Solution (Total 12)
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- 1/x+1/y=1/z
1/6+1/y=1/z
(y+6)/6y=1/z
z=6y/(6+y)
z=6(y+6-6)/(6+y)
z=6{1-6/(6+y)}
z=6-36/(y+6)
for z to be integer
y=3 so z=2
y=6 so z=3
y=12 so z=4
y=30 so z=5
hence,no. of nonnegative integral solutions=4
(x,y,z)=(6,3,2),(6,6,3),(6,12,4),(6,30,5)
ans 4 - 11 years agoHelpfull: Yes(61) No(5)
- Rakesh ji your answer is correct but we should also include zero in the solution(i.e y=0 z=0 x=6) because it is given that x,y,z are non negative.
so there will be 5 solutions. - 10 years agoHelpfull: Yes(4) No(7)
- 1/6+1/y=1/z
(y+6)/(6y)=1/z
z=(6y)/(6+y)
partition fraction
z=6-36/(y+6)
now put the value of y=3,6,12,30
and get the value of z..... - 11 years agoHelpfull: Yes(3) No(0)
- solving the equation you will get 6-z=6*z/y
means the range of z will be from 1 to 6
from eq y=6*z/6-z calculate y for every z if it is integer than it will be true..
Hence there will 4 sol that will satisfy the y equal to integer for z = 2,3,4,5
4 solutions - 7 years agoHelpfull: Yes(2) No(1)
- no of solution are 4
{(x,y,z)= (6,3,2),(6,6,3),(6,12,4),(6,30,5)} - 11 years agoHelpfull: Yes(1) No(8)
- it is a condition in which there is 1 eqtn and two variable , hence infinite solutions
- 10 years agoHelpfull: Yes(1) No(3)
- @chaurasia jee when you take y=0,x=0,x=6 then it will not satisfied the eqn:
1/x=(1/z-1/y).so i think we cant take that set.Is it clear DEEPAK BABU.. :) - 10 years agoHelpfull: Yes(1) No(0)
- it is infinite solutions.
- 9 years agoHelpfull: Yes(1) No(0)
- @Naveen Pandey...frnd how did you come up with this solution ?Is there any standard approach you used ?
- 11 years agoHelpfull: Yes(0) No(1)
- what r the options.
- 11 years agoHelpfull: Yes(0) No(0)
- 3,2
6,3
12,4
30,6 - 11 years agoHelpfull: Yes(0) No(1)
- How it comes z=6-36/(y+6)...cn u explain.........
- 10 years agoHelpfull: Yes(0) No(0)
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