Oranges can be packed in sets of 10 oranges in box type a or 25 oranges in box type b. A carton comprising of 1000 oranges of type a and b is packed. How many different combinations are possible in the number of type A and type B boxes while organizing the oranges?
option
a. 21
b. 20
c. 19
d. 18

• answer is 21..
  
  we can write problem in form of eqtn..
  
  25*b+10*a=1000
  now we have to find all posiible value of (a,b)..
  so,
  for satisfying eqtn "b" must be even & the range of "b" will varry from
  0 to 40 so the total number is.... 21
• 10 years agoHelpfull: Yes(35) No(28)
• c.19
  25 oranges boxes can be packed in this fashion
  (2,4,6,8,.......38) (since 0 and 40 cant be possible because in dat case type a become 0 or in later one type b become 0)
  so 2 to 38 will comprises of 19 box
  
  
• 10 years agoHelpfull: Yes(31) No(21)
• ans is 21
  25x+10y=1000
  from x=0 to x=40 with an increment of 2 and y=100 to y=0 with a decrement of 5 satisfies the equation in 21 ways.
• 10 years agoHelpfull: Yes(14) No(8)
• Box A can contain 10 oranges and Box B can contain 25 oranges so the equation is XA + YB=1000, we know the values for A&B so Equ becomes 10X + 25B=1000 => 2X + 5B=200, so fro the equation it is clear that the basic two values of X,Y are (0,40) & (100,0). Clearly if X increases by 5 then Y should decrease by 2 in order to maintain the constant value i.e 1000 oranges. so the different combinations will be (50,20) (55,18) (60,16) (65,14) (70,12) (75,10) (80,8) (85,6) (90,4) (95,2) (100,0) (45,22) (40,24) (35,26) (30,28) (25,30) (20,32) (15,34) (10,36) (5,38) (0,40 ) so there are 21 different combination by which we can arrange the oranges in the carton.
• 8 years agoHelpfull: Yes(13) No(3)
• Here we have 10a +25b=1000
  => a=(200-5b)/2
  Now a and b both should be integer so we can take all the even value for b lying between 0 and 40(excluding 0 and 40).
  Thus getting total of 19 possible combinations.
• 10 years agoHelpfull: Yes(9) No(8)
• answer is c.(19)
  check for different combinations of box type a and b like
  95 box of type a and 2 of type b
  90 box of type a and 4 of type b
  ............
  5 box of type a and 38 of type b
  avoid boundary conditions(100 box of type a and 0 of type b && 0 box of type a and 40 of type b)
• 10 years agoHelpfull: Yes(4) No(3)
• for 100 oranges
  take the boxes like as follows
  A:10 and B:25 (capacity)
  now find possible comibanations
  A=0 B=25*4
  A=10*5 B=25*2
  A=10*10 B=25*0
  
  add possiblilities
  (0+4)+(5+2)+(10)=21 boxes
• 6 years agoHelpfull: Yes(3) No(0)
• the expression will be 10x+25y=1000
  the above exp will be satisfied only with multiples of 2 up to 38.
  so no. of combinations are 19.
  if oranges are packed in type A boxes only= 1 way
  if oranges are packed in type B boxes only= 1 way
  so total = 19+1+1=21
• 7 years agoHelpfull: Yes(2) No(0)
• The answer is 19.
• 10 years agoHelpfull: Yes(1) No(12)
• Answer: a
  
  Given that 10a + 25b = 1000 ⇒ 2a + 5b = 200
  One general solution for the above equation is when a = 0, then b = 40.
  To get other solutions, a is increased by 5 and b is decreased by 2. So next solution is (5, 38)
  Now b is reduced upto 0.
  So total solutions = 40−02+1 = 21
• 8 years agoHelpfull: Yes(1) No(2)
• c.)19
  Soln.
  1000=25b+10a
  The pairs possible
  (2,95)
  (4,90)
  (6,whatever)
  .
  .
  .
  .
  (38,whatever)
  as odd values of box b results in no proper box of a
  so count no of terms in series 2,4,6,8,10...........upto 38
  hence no of terms=19
• 8 years agoHelpfull: Yes(0) No(3)
• please say the correct answer pls....
• 8 years agoHelpfull: Yes(0) No(0)
• ans is 19
  
  10a+25b=1000
  2a+5b=200
  b=40-2/5*a
  so we get when a=0 b=40
  a=5 b=38
  a=10 b=36
  .............
  a=100 b=0
  here we exclude a=0 and a=100
  so last term,Tn=95=5+(n-1)5
  so,n=19
• 7 years agoHelpfull: Yes(0) No(1)
• Given that 10a + 25b = 1000 ⇒ 2a + 5b = 200
  One general solution for the above equation is when a = 0, then b = 40.
  To get other solutions,
  a
  is increased by 5 and
  b
  is decreased by 2. So next solution is (5, 38)
  Now
  b
  is reduced upto 0.
  So total solutions =
  40
  −
  0
  2
  +
  1
  = 21
• 5 years agoHelpfull: Yes(0) No(1)