how many ways a batsman can score 200 with only 4's and 6's?

• The solution is very simple.
  just form the equation as said.
  i.e. 4x+6y=200
  or
  2x+3y=100
  or 2x=100-3y
  100-3y has to be an even number.
  y can be 0, 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32
  which makes total of 17 cases,
  hence 17 is the answer.
• 9 years agoHelpfull: Yes(11) No(0)
• First Case:
  LCM of 4 and 6 is 12.
  So, total runs made by 4's and 6's =200/12= 16.
  Second Case:
  Only 4's= 50*4.
  
  Total ways =16+1= 17.
• 9 years agoHelpfull: Yes(7) No(0)
• batsman can score 200 in 16 ways....
  
  1) 32*6+2*4=200
  2) 30*6+5*4=200
  3) 28*6+8*4=200
  4) 26*6+11*4=200
  5) 24*6+14*4=200
  6) 22*6+17*4=200
  7) 20*6+20*4=200
  8)18*6+23*4=200
  9) 16*6+26*4=200
  10) 14*6+29*4=200
  11) 12*6+32*4=200
  12) 10*6+35*4=200
  13) 8*6+38*4=200
  14) 6*6+41*4=200
  15) 4*6+44*4=200
  16) 2*6+47*4=200
  
  As maximum no. of 6's possible within 200 is....33 that is 33*6=198...but for this case 200-198=2 ,which is not possible.....so max no of 6's can be 33-1=32.....that is 6*32=192 for this min no. of 4's required is 2 as.....200-196=8,that is 2*4.
  
  and again we know, 6*2=4*3 =12 .
  
  so ,the max no. 6's will be 32 and it will be decreased by two that is (30,32,30......till it comes 2) and accordingly 4 will increase by 3 that is(2,5,8,11,......till no of sixes comes 2 ).
  
  as, the batsman have to score 200 with 4's and 6's .....we can't take ....no. of 4's or 6's as 0.
  
• 9 years agoHelpfull: Yes(3) No(2)
• 17
  ananya u missed 50*4+6*0=200
• 9 years agoHelpfull: Yes(1) No(0)
• batsman can score 200 in 3 ways
  1)50*4=200
  2)4*20+6*20=200
  3)4*5+6*30=200
• 9 years agoHelpfull: Yes(0) No(6)
• let x=no.of 4,y=no. of 6
  4x+6y= 200
  2x+3y= 100
  2x= (100-3y)/2
  x= 50 - 3y/2
  here x and should be whole number
  therefore y will be multiple of 2
  therefore maximum value of y will be 32
  and minimum value of x will be 2
  but all y is multiple of 2 or 0
  therefore total number of ways=32/2+1 =17
  
• 9 years agoHelpfull: Yes(0) No(1)
• Ananya...i like your solution. but you did not consider the case of 50*4=200. then the no. of ways become 17. should we consider this case because then we are taking only 4's?
• 9 years agoHelpfull: Yes(0) No(0)
• 200 score can be achieved by hitting 4's only can be done in 1 way that is
  50*4 +0*6 =200
  now 3 fours = 2 sixes
  so we can take 3 fours from that 50 and convert into 2 sixes
  so the combinations will be like this
  (50 - 3)*4 +(0+2)*6=200
  (50-3-3) * 4 +(0+2+2)*6=200
  so there are 50/3= 16 ways of scoring 200 with sixes and 1 without any 6 (i.e the 50*4 +0*6=200)
  so there are 17 combinations
• 9 years agoHelpfull: Yes(0) No(0)
• For this one we can write 4X+6y=200
  
  If we put 2 for x and 32 for y then it will satisfy.
  again for x=5 and y=30 it will be satisfy,
  if we continue this series we get that value of x increases 3 whereas value of y increases only 2.
  so possible cases are (2,32)(5,30)(8,28)(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)
• 9 years agoHelpfull: Yes(0) No(0)
• there are lot of way to score 200 with only 4's & 6's,which are following
  1) 50*4=200
  2) 10*6+35*4=200
  3) 20*6+20*4=200
  4) 30*6+5*4=200
  5) 32*6+2*4=200
  .
  .
  .
  .
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  so on ......
  
• 9 years agoHelpfull: Yes(0) No(0)
• 17 ways.
  50*4+0*6
  47*4+2*6
  44*4+4*6
  .
  .
  .
  .
  .
  2*4+32*6
  for every 2 sixes he hits, he has to hit 3 less fours.
  min. no of fours needed is 2, 200 is not divisible by 6,
• 9 years agoHelpfull: Yes(0) No(0)
• In such Cases, Take the higher Number and find out the it's maximum Multiple , Here it is 32
  Check with 6*1,6*2, 6*3 and so on, We see that 1,3,5,7... do not fit in .
  So Divide 32 by 2 and add 1 (1 is added for the special Case i.e. 4*50) = 17
• 9 years agoHelpfull: Yes(0) No(0)
• For this one we can write 4X+6y=200
  
  If we put 2 for x and 32 for y then it will satisfy.
  again for x=5 and y=30 it will be satisfy,
  if we continue this series we get that value of x increases 3 whereas value of y increases only 2.
  so possible cases are (2,32)(5,30)(8,28)(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)and (25,0)
  
• 9 years agoHelpfull: Yes(0) No(0)