Question

A man goes on dividing a certain sum of money into half. He does this 9 times. He gives the
amount in round 1 and round 4 and round 7 to A. To B he gives the amount in rounds 2, 5 and
8. To C he gives the amount in Rounds 3, 6 and 9. Assume that the balance sum at the end of
Round 9 is Rs. 195.31.
How much does A get in all (nearest 100)?
A] Rs. 45100
B] Rs. 48100
C] Rs. 57000
D] Rs. 62000​

Answer 1

Answer:

$ 57,000

Step-by-step explanation:

Answer 2

10 x 10⁴ is the total does A receive when the Rs. 195.31 will be the remaining amount after Round 9.

Given that,

A man continues to split a certain amount of money in two. He repeats it nine times. He awards A with the money from rounds 1, 4, and 7. He transfers the money to B in phases 2, 5, and 8. He transfers the sum to C in Rounds 3, 6, and 9. Assume that Rs. 195.31 will be the remaining amount after Round 9.

We have to find how much total does A receive.

We know that,

Let us take the original amount is x at the end of round 9,

The amount left is $\frac{x}{2^9}$  = $\frac{x}{512}$

Since the amount is Rs. 195.31 the original sum,

x = 195.31 x 512

Approximately 200 x 500 = 10 x 10⁴.

Therefore, 10 x 10⁴ is the total does A receive.

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