Question

During a competition, numismatist Paheli has been given Rs. 200 in Rs.1 denominations. The judge asks Pheli to allocate the Rs. 1 denominations into a number of pouches such that any amount required between Rs.1 & Rs. 200 can be given by giving out a certain number of pouches without opening them. Pheli thinks and asks the judge to give her 'x' number of pouches to keep the money, where 'x' is the minimum number of bags required to keep the total money. Can you guess the value of 'x'? answer in integer .
please give correct answer​

Answer 1

Solution :-

→ 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 .

Learn more :-

in a bag , number one rupee coin is three time then two rupees coin if there are only Rs 150 in this bag , then find out...

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