Tricky Word Problem: Solve It with These Tips | Help with Calculations

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The discussion focuses on solving a tricky word problem involving a stone block that loses 0.5% of its mass with each chisel strike. To determine how many strikes are needed to reduce the block to less than half its original size, the formula 0.995^n < 0.5 is used, leading to the conclusion that at least 139 strikes are required. Participants clarify the mathematical approach and confirm the calculations involved in reaching this conclusion. The conversation also humorously suggests that the problem might be better suited for a K-12 homework section. Overall, the key takeaway is the mathematical solution to the word problem involving exponential decay.
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Tricky word problem.

Hi I'm having trouble figuring out how to approach this word problem and wonder if anyone had tips.

Problem:
Just suppose, for the sake of argument, that you have a large block of stone. While staring at this block of stone, it dawns on you that perhaps you should carve it into something. Luckily, you have a hammer and chisel handy.

But it's no ordinary hammer and chisel. Somehow, no matter how hard you hit the chisel, exactly one-half of one percent of the block chips off.

How many times do you have to hit the chisel before the block is less than half of its original size?
 
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:smile: Basically, since 0.5% is chipped off each successive strike, you thus retain 99.5% of the block on each successive strike.
*You need to find out how many strikes are needed--->how many 0.5% chips are needed to reduce the block to half of its mass.
*Therefore, letting n be the number of strikes thus far, you will retain \left( {99.5\% } \right)^n of the mass after n strikes.
---To find out the minimal n strikes needed to reduce to less than half the mass will require the inequality:

0.995^n &lt; \frac{1}{2} \Rightarrow n \geqslant \left\lceil {\frac{{\log 0.5}}{{\log 0.995}}} \right\rceil \Rightarrow n \geqslant 139

*Thus, the minimum quantity of strikes needed is 139.

(since n \in \mathbb{N} \cup \left\{ 0 \right\}) :biggrin:
(...tho shouldn't this be in the Homework K-12 section? :redface:)
 
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After one blow, there is 1* .995 of the block left. After 2 blows there is 2 * .995 of the block left. So after 'n' blows there is .995^n left. So you need to solve: 0.5=0.995^n
 
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