There are $50$ notes with value $\large\bf{₹}$ $100$ but no $\large\bf{₹}$ $2$ note in $50$ notes. How many notes can be there? If there are $50$ notes whose total value is $\large\bf{₹}$$100$ but $\large\bf{₹}$$2$ note should not be there in the count of those $50$ notes. How many notes can be there?
Available Denomination are

*

*$\large\bf{₹}1$

*$\large\bf{₹}2$ (cannot be used)

*$\large\bf{₹}5$

*$\large\bf{₹}10$

*$\large\bf{₹}20$

*$\large\bf{₹}50$

*$\large\bf{₹}100$  (Obvious that not required)

*$\large\bf{₹}200$  (Obvious that not required)

*$\large\bf{₹}500$    (Obvious that not required)

*$\large\bf{₹}2000$    (Obvious that not required)

Putting it other way
Let us assume $a,b,c,d$ and $e$ as variables satisfying following conditions

*

*$a+b+c+d+e = 50$

*$a + 5b + 10c + 20d + 50e = 100$
Then what can be values of $a,b,c,d$ & $e$?
(Given that they must be non-negative integers!)
 A: Since you cannot use 2 Rupee notes, you must use lower values to counter the higher values. Since there is only one lower value (1 Rupee) and 4 higher values (5, 10, 20, 50), you need to balance how many 1 Rupee notes can be balanced out by which higher values to equal the same value for the same number of 2 Rupee notes. And how many of the higher values can be swapped to substitute how many of the smaller notes.
For example, replacing a 1 with a 5 saves you 4 notes, 5 with a 10 will allow you to save on 5 additional 1 Rupee notes, replacing a 5 with a 20 yields 15, etc.
If you then target a factor of 50 total notes, then it becomes a matter of simply repeating this value once you hit it.
The second highest factor of 50 is 25, which yields 50 Rupees in 2 Rupee notes. You'd need 50 1 Rupee notes to match that. You can replace 4 at a time with 5 Rupee notes to get 45 + 1*5 (46), 40 + 2*5 (42), 35 + 3*5 (38), 30 + 4*5 (34), 25 + 5*5 (30). At this point, you have 30 notes, which is only 5 away from the target of 25 notes. Since replacing a 5 Rupee note with a 10 Rupee note saves us exactly 5 notes, then we can get 25 notes totalling 50 Rupees with the combination of 20 + 4*5 + 1*10.
Since we started with the second highest factor, we only need to double the amount. 40 notes worth 1 Rupee, 8 notes worth 5 Rupees, and 2 notes worth 10 Rupees.
A: One way to think about this is that the average note has value 2 rupees and only one note is any lower.  So each large note has to be paired with enough 1 notes to bring the average down to 2, and the group can be viewed as a "note".  A 5 rupee note has to be paired with 3 1 rupee notes making a package that is worth 8 rupees and has 4 notes.  Similarly a 10 rupee note must be paired with 8 1 rupee notes to make a package worth 18.
Your new notes have value 8, 18, 38, 98 and you need to make 100 (the count will take care of itself).  You can't use 98 at all.  Working mod 8 can make it easy.  If you use two 38's, you need 24 more, which can only be done with 3 8's.  So this gives 2*20+3*5+45*1 (in the original notes).  If you use one 38, that is 6 mod 8, so to get to 4 mod 8 we need 3 18's and 1 8, giving 20+3*10+5+45*1.  Finally, if we use no 38's, we need 2 18's and 8 8's, which leads to 2*10+8*5+40*1.
A: 1  Rupee  40 Notes
5  Rupees 08 notes
10 Rupees 02 Notes
1*  40=40
   5*  08=40
   10* 02=20

Total  100.00 Rupees
A: $a = 5, b = 1, c = 2, d = 1, e = 1$ hence $a + 5b + 10c + 20d + 50e$ will be $5 + (5\times 1) + (10\times 2) + (20\times 1) + (50\times 1) = 5 + 5 + 20 + 20 + 50 = 100$.
