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This is from a regional math olympiad paper in Bangladesh (in case you didn't know—taka is the Bangladeshi currency, the short form being $\text{tk}$.):

Tamim works $4$ hours a day. Every hour for each of his life’s completed years, he earns $250\space \text{tk}$. Once he worked $92$ days in $9$ months, for which he earned $2200000\space \text{tk}$. What was his age on the last day of these $92$ days?

I don't understand the question clearly. It's a bit confusing.

Does he earn 250 tk. for all of the years he lived combined? If he's $n$ years old, does he get $\frac{250}{n}$ for each year of his life?

Or he gets 250 tk for each year he's old?

The wording is bound to sound unnatural, as the problem-makers aren't native speakers. It's a bit wrong at times too. They also make a lot of typos: honestly, a lot. So the intended meaning of this problem is probably different than what it seems at first glance. I tried to adjust some of the phrasings as per the Bengali version so as to retain the original meaning. Still, there could be something wrong in it.

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    $\begingroup$ We could probably have guessed that $\text{tk}$ was a currency of some sort, but cool that you specified it. The less we have to guess, the better. And the way I read it, they mean that as long as he is 30 years old, he makes $30\cdot 250\text{tk} = 7\,500\text{tk}$ each hour, but as soon as he turns $31$, he starts making $7\,7500\text{tk}$ each hour instead. $\endgroup$ – Arthur Jan 24 '18 at 8:31
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Let $x_{dh}$ be Tamim's age at work day $d$ and hour $h$.

\begin{aligned} \text{day}_1 &: 250 \cdot (x_{11} + x_{12} + x_{13} + x_{14}) \\ \text{day}_2 &: 250 \cdot (x_{21} + x_{22} + x_{23} + x_{24}) \\ \ldots & \\ \text{day}_n &: 250 \cdot (x_{n1} + x_{n2} + x_{n3} + x_{n4}) \\ \ldots & \\ \text{day}_{92} &: 250 \cdot (x_{92_1} + x_{92_2} + x_{92_3} + x_{92_4}) & \\ \end{aligned}

The total he earned is \begin{aligned} 250 \cdot \sum^ {d = 92} _{d = 1} \left( \sum^ {h = 4} _{h = 1} x_{dh} \right) = 2200000 & \rightarrow \\ \sum^ {d = 92} _{d = 1} \left( \sum^ {h = 4} _{h = 1} x_{dh} \right) = \frac{2200000}{250} = 8800 & \rightarrow \\ \text{average } x_{dh} = \frac{8800}{92 \cdot 4} = 23.91 \ldots \end{aligned}

and since he worked for $9 < 12$ months he could have just one birhtday in between the first and the last day of work; thus its age at the end of the $92$th day is at most $24 \ldots$ and at least $22 \rightarrow$ it's 24.

PS: Indeed we can work out the exact point of his birthday. Letting $x$ be the days since he started working, we have $$23 \cdot 250 \cdot 4x + 24 \cdot 250 \cdot (92 \cdot 4 - (4x)) = 2200000$$ and solving for $x$ we find that $x = 8$: he was 23 until his 8th work day.

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  • $\begingroup$ Thanks. Could you check out this other question math.stackexchange.com/questions/2618874/…? $\endgroup$ – Soha Farhin Pine Jan 24 '18 at 9:38
  • $\begingroup$ Why do you add his age at each hour of his 4-hour working period? $\endgroup$ – Soha Farhin Pine Jan 24 '18 at 9:41
  • $\begingroup$ @SohaFarhinPine the 4h profit is (by definition) $250 \cdot x_{11} + 250 \cdot x_{12} + 250 \cdot x_{13} + 250 \cdot x_{14} = 250 \cdot (x_{11} + x_{12} + x_{13} + x_{14})$ $\endgroup$ – sirfoga Jan 24 '18 at 9:47

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