anand packs 304 marbles int packets of 9 or 11 so that no marble is left. anand wants to maximize the number of bags with 9 marbles . how many bags does he need if there should be atleast one bag with 11 marbles.

• let x is number of bag contains 11 marbles, and y is number of bag contains 9 marbles.
  we find that (x+y) value ,, maximize the number of bags with 9 marbles . it means y is maximize..
  
  11x+9y=304
  
  y=(304-11x)/9
  
  y is maximum .. when x is less value and also (304-11x ) is divisible by 9
  
  so we choose minimum x value is 8. ( here we choose 8 becoz when i put x value is 1, 2,3,4,5,6,7 that case (304 - 11*x) is not divisible by 9.. )
  when y= (304 - 8 *11)/9
  y=24
  so total bag is x+y =8+24 =32 ans...
  
• 8 years agoHelpfull: Yes(40) No(0)
• 304/9 we get remainder as 7. ( we have to packs into packets of 9, 11. so we remove a no multiple of 11 and after dividing the result by 9 ,the remainder becomes zero)
  .".( 304 - ( 81+7))/9 = 24
  so he needs 24 "9 marbles packets" & 8 "11 marbles pockets"
  total bags = 32
• 8 years agoHelpfull: Yes(5) No(0)
• 304/9 gives reminder 7.. i.e 7 marbles are remaining
  so to make the remaining marbles multiple of 11
  check the remainder of smallest multiple of 11 which gives remainder 7 when divided by 9
  88/9 gives reminder 7
  
  so 304-88=216
  216/9=24
  24 bags is the maximum.
• 8 years agoHelpfull: Yes(3) No(3)
• 304=9^33+7
  304=9^32+16
  304=9^31+25
  .......................
  ......................
  
  
  304=9^24+88=9^24+11x8 which is true so 24 is the answer
• 8 years agoHelpfull: Yes(1) No(0)
• 304/9 gives reminder 7.. i.e 7 marbles are remaining
  so to make the remaining marbles multiple of 11
  check the remainder of smallest multiple of 11 which gives remainder 7 when divided by 9
  88/9 gives reminder 7
  
  so 304-88=216
  216/9=24
  24 bags and 8 bags==>32 bags
• 8 years agoHelpfull: Yes(1) No(0)
• 304/9=33 and remaining marbles are 7.
  then less the packet of 9 marbles one by one and increase marbles up to marbles are not multiple of 11
• 8 years agoHelpfull: Yes(0) No(0)
• 9*35=315>304
  9*34=306>304(there is no atleast one bag with 11 marbles)
  9*33=297+11*1=308>304
  9*32+11*1=288+11*1=299
• 8 years agoHelpfull: Yes(0) No(0)
• the sum of 3 from the 4 numbers a,b,c,d are 4024,4087,4524,4573.what is the largest of the numbers a,b,c,d?
• 8 years agoHelpfull: Yes(0) No(0)
• given 9x + 11y = 304.
  x = 304−11y9 = 33+7−2y9−y
  So y = - 1 satisfies. Now x = 35. But y cannot be negative.
  Now other solutions of this equation will be like this. Increase or decrease x by 11, decrease or increase y by 9. So we have to maximise x. next solution is x = 24 and y = 8. So bags required are 32.
  
• 8 years agoHelpfull: Yes(0) No(1)
• Given 9x + 11y = 304.
  x = 304−11y9 = 33+7−2y9−y
  So y = - 1 satisfies. Now x = 35. But y cannot be negative.
  Now other solutions of this equation will be like this. Increase or decrease x by 11, decrease or increase y by 9. So we have to maximise x. next solution is x = 24 and y = 8. So bags required are 32.
• 7 years agoHelpfull: Yes(0) No(1)