if |x -4 | + |y - 4| =4 then how many integers value the sets (x,y) have
a infinte
b 5
c 16
d 25

• mod of x-4 should be 0 to 4 and also mod of y-4 should be 0 to 4.So first think all the possible value of x so that x-4=0 and think all the possible value of so that y-4=4.So to make |x-4|=0 and |y-4|=4 value of x and y is 1.(4,0),2.(4,8).simillarly to make |y-4|=0 and |x-4|=4 value of (x,y) is 3.(0,4),4.(0,8).to make |x-4|=1 and |y-4|=3, values are 5.(3,1),6.(3,7),7.(5,1),8.(5,7) and |x-4|=3,|y-4|=1 values are9.(1,3),10.(7,3),11.(1,5),12.(7,5).for |x-2|=2 and |y-2|=2 values are 13.(2,2),14.(2,6),15.(6,6),16.(6,2). so total 16.
• 9 years agoHelpfull: Yes(13) No(1)
• c) 16
  as |x-4| and |y-4| both will be positive so, total no of possible values both set of mod is (0+4)=4, (1+4)=4, (2+2)=4, (3+1)=4, (4+0)=4.
  so when (0+4) or (4+0) only 2 set of values will satisfy that counts 2+2=total 4 set of values,
  
  now for (1+3), (2+2), (3+1) each case will produce 4 possible set of x,y as we have 2 mods. so total counts 4+4+4=12.
  
  so final total x,y solution counts to 4+12=16.
• 9 years agoHelpfull: Yes(7) No(4)
• a) infite..................sets include both positive and negative integers
• 9 years agoHelpfull: Yes(2) No(7)
• considering |x-4| be positive and |y-4| be negative
  therefore expansion will be
  x-4-y+4=4
  x-y=4
  there are infinite set of x and y for this case.
  thus ans is (A)
• 9 years agoHelpfull: Yes(2) No(0)
• b) 5
  as there is mode so both |x-4| and |y-4| will always be positive
  So only possible combinations for
  |x-4|+|y-4| are 0,4 (x=4, y=0)
  4,0 (x=0, y=4)
  1,3 (x=5, y=7)
  3,1(x=7, y=5)
  2,2(x=6, y=6)
• 9 years agoHelpfull: Yes(1) No(7)
• 10
  
  (x,y)=(0,4) (4,0) (1,3) (2,2) (3,1) (4,8) (5,7) (6,6) (7,5) (8,4)
• 9 years agoHelpfull: Yes(1) No(3)
• Sry, My Fault My Above Solution Is Wrong
• 9 years agoHelpfull: Yes(1) No(0)
• Ans:--10
  
  (x,y)=(0,4) (4,0) (1,3) (2,2) (3,1) (4,8) (5,7) (6,6) (7,5) (8,4)
  No other combination is possible.....All the option is wrong
• 9 years agoHelpfull: Yes(0) No(7)
• 16
  as put x=0 to 8 as after putting x>8 or x
• 9 years agoHelpfull: Yes(0) No(0)
• All The Answers Above Are Incorrect.
  Because , | X | >=0 Always
  So If We Take Any Negative Values For X or Y Can Lead To Sum>4.
  Hence, Values Should Be Positive For X And Y.
  So. Possible Combinations Are (4,0),(3,1),(2,2),(1,3)&(0,4)
  Hence , Answer Is 5
• 9 years agoHelpfull: Yes(0) No(3)
• this mode question |x-4|+|y-4| it always produce positive values
  values of |x-4| or |y-4| not greater than 4
  so possible combination is :-
  (0,4),(1,3),(2,2),(3,1),(4,0),(4,8),(5,7),(6,6),(7,5),(8,4)
• 8 years agoHelpfull: Yes(0) No(0)
• option a ..it is infinite...u can draw graph for it
• 8 years agoHelpfull: Yes(0) No(0)