During a competition, numismatist Aditi has …

for Rs.1 aditi needs Rs.1 in one pouch.
 for Rs.2 aditi needs Rs.2 in second pouch.
now,

• from pouch 1 and 2 aditi can easily count = 1 + 2 = Rs.3 .

so,
for Rs.4 now, aditi needs Rs.4 in 3rd pouch.
then,

• from all three pouches she can count = 4 + 1 = 5, 4 + 2 = 6 and 4 + 3 = 7 also.
• or, 4 + 3 = upto 7 .

next,
for Rs.8 now, aditi needs Rs.8 in 4th pouch.
then,

• from all four pouches she can count upto = 7 + 8 = Rs.15

next,
for Rs.16 now, aditi needs Rs.16 in 5th pouch.
then,

• from all five pouches she can count upto = 15 + 16 = Rs.31 .

next,
for Rs.32 now, aditi needs Rs.32 in 6th pouch.
then,

• from all six pouches she can count upto = 31 + 32 = Rs.63

next,
for Rs.64 now, aditi needs Rs.64 in 7th pouch.
then,

• from all seven pouches she can count upto = 63 + 64 = Rs.127

next,
for Rs.128 now, aditi needs Rs.128 in 8th pouch.
then,

• from all eigh pouches she can count upto = 127 + 128 = Rs.255.

as aditi needs to allocate upto Rs.200 only.
therefore, we can conclude that, the minimum number of bags required to keep the total money is 8.