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a large tank filled with water is to be emptied by removing half of the water present in it every day, after how many days will there be closest to $10$ percent water left in the tank?

okay what I understood here: suppose the tank is filled $100%$ percent, then after one day it is left with $50$ percent then number of days left with $10$ percent water will be $1/2\times 10=5$ days?

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Hint:

After 1 day the tank will be $\frac{1}{2}$ or 50% full.

After 2 days the tank will be $\frac{1}{4}$ or 25% full

Can you see a pattern.

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HINT:

After $n$th day, the remaining water will be $$\left(1-\sum_{1\le r\le n}\frac1{2^r}\right)\cdot100\%=\frac1{2^n}\cdot100\%$$

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In this case, by far the easiest approach is simply to calculate the amount of water left after 1,2,3,4,... days and see which of the numbers is closest to 10%. (After 4 days there will be less than 10% left because $2^4>10$, so the answer cannot be more than 4).

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