Elitmus
Exam
Numerical Ability
Quadratic Equations
Q. If x and y are two integers then number of possible solutions for 5|x|+2|y|=100 ?
Read Solution (Total 8)
-
- 5|x|+2|y|=100
=> 5|x|=100-2|y|
(100-2|y|) is even so 5|x| should even
for
x=0,y=50
x=2,y=45
x=4,y=40
x=6,y=35
....
x=18,y=5
x=20,y=0
for x=0,y=50,(x,y)=(0,50)&(0,-50) => 2 soln
for x=20,y=0,(x,y)=(20,0)&(-20,0) => 2 soln
for remaining 9 soln, (x,y)=(2,4,6..,18, 45,40,35,..5)
4 soln for each pair
for x=2,y=45, (x,y)=(2,45)(-2,45)(-2,-45)(2,-45)
similarly 4 soln for remaining 8 pair
total no. of soln= 2+2+9*4= 40
- 10 years agoHelpfull: Yes(19) No(1)
- 5|x|+2|y|=100
solution for integer solution x is always even
possible value of x=(-20,-18,-16.........0,2,4,6.....20){total 21 valus
for every value of x y have 2 solution + or - remaining at -20 and 20 (y=0)
so total solution =21*2-2=40
so ans is 40 - 11 years agoHelpfull: Yes(4) No(3)
- @Rakesh kindly solve all ques ur solution seems logical.....
- 10 years agoHelpfull: Yes(4) No(0)
- Ans:- 40
I checked by programming.. they given mod it means we need to take -ve and +ve numbers. - 11 years agoHelpfull: Yes(2) No(0)
- 100/lcm of 5&2
100/10=10 - 11 years agoHelpfull: Yes(1) No(0)
- 2*anything is an even number.
hence 2|y|=100-5|x| has to be an even number.
which can be possible if and only if 5|x| is an even number. hence x can be of range 2,4,6,8,10,12,14,16,18,20 and its negative representations along with a 0(zero). hence total possible solutions is 21 - 11 years agoHelpfull: Yes(1) No(0)
- @Arun,but 21 wasn't there among options..
- 11 years agoHelpfull: Yes(0) No(0)
- good job rakesh
- 10 years agoHelpfull: Yes(0) No(0)
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