X and Y are the integers. Possible different solutions of 2|x| + 3|y|=100 ...
(a) 16 (b) 33 (c) 66 (d)132

• The solution would be 66. Let's first consider the first and fourth quadrant only
  
  x= 2,5,8,......,50
  
  Now corresponding to these values we have:-
  
  y=32,30,.......,0
  and
  y=-32,-30, ...... 0
  
  and now values of x in 2nd and 3rd quadrant
  
  x=-2,-5,-8,.......,-50
  
  and
  
  y=32,30,.......,0
  and
  y=-32,-30, ...... 0 again
  
  So total number of soltions to it are
  
  (17*4) - 2=66
  
  Some of the solutions are
  (2,32) , (-2,32) , (-2,-32) , (2,-32)
• 8 years agoHelpfull: Yes(11) No(0)
• Ans- 33
  |y|=(100-2|x|)/3
  at y=0,x=50 or -50
  there are only 16 values till x=50 for which y will be an integeri.e (2,32),(5,30)(8,42),(11,26),(14,24),(17,22)(20,20),(23,18),(26,16),(29,14),(32,12),(35,10),(38,8),(41,6),(44,4),(47,2)
  similarly for -ve 16 values, x and y will be integer
  one another value at x=0,y=0
  so there are total of 33 possible values.
  
• 8 years agoHelpfull: Yes(8) No(0)
• correction to my prev answer... Ans: (b) 33
  b'coz ans will b 16 only when we want positive integer solutions..
  The integer values of x that would satisfy the equation will be in an Arithmetic Progression where the common difference is the co-efficient of y and vice-versa.
  
  For example, 2x + 3y = 100
  One of the solutions to the equation is x = 50 and y = 0
  Now, the other integer values of x would be 53, 56, 59,.. and also in the other direction like 47, 44, 41, 38.... and so on.
  Very similarly, values of y would be 2, 4, 6, 8... and also in the other direction like -2, -4, -6, -8...
• 8 years agoHelpfull: Yes(4) No(4)
• there are 4 case:
  1st : when both x and y are negative
  2nd: both +ve
  3rd : x= - ve , y= + ve
  4th: x= + ve , y=- ve
• 8 years agoHelpfull: Yes(2) No(3)
• lets look at the equation,
  
  2|X|+3|Y| = 100
  or, |X| = 50 -3/2|Y|
  look at the R.H.S which can't be negative because because |X| acquire its value {0,+ve} .
  now, equating R.H.S to zero.
  50 -3/2|Y|= 0
  i.e |Y|= 33.33 but we want only integer as said in question
  so, |Y|= 33
  By eliminating mod we will get range of y . i.e +33 to -33 .
  
  ANS = 66
• 8 years agoHelpfull: Yes(2) No(1)
• Ans c ) 66
  
  Y X
  
  0 50,-50
  2,-2 47,-47
  4,-4 44,-44
  .
  .
  .
  .
  .
  .
  
  till
  
  32,-32 2,-2
  
  And therefore all 17 sol except 1st has 4 combinations and the 1st one has only 2 combinations...i.e
  total =(16 *4)+(1*2)=66
• 8 years agoHelpfull: Yes(1) No(0)
• Ans : (a) 16
• 8 years agoHelpfull: Yes(0) No(3)
• Sir Please Explain How 16
• 8 years agoHelpfull: Yes(0) No(1)
• Answer is 66
  
  16 Possible combinations are:-
  
  (x,y) = (50,0), (47,2), (44,4), (41,6), (38,8), (35,10), (32,12), (29,14), (26,16), (23,18), (20,20), (17,22), (14,24), (11,26), (8,28), (5,30), (2,32)
  
  From the above combination, 16 combinations can have the negative values also (like (-47,-2), (-44,-4)...and so on)... hence the mod(x) and mod(y) will make the values always be +ve. so there will be 16 *4 combinations = 64 combinations possible as considering the negative and positive symbols for x and y (x,y) = (+,+), (+,-), (-,+),(-,-)
  
  For (x,y) = (50,0) there can be only 2 combinations possible
  
  Hence the answer will be 64 + 2 combinations = 66 combinations...
• 3 years agoHelpfull: Yes(0) No(0)