A question of arithmetic regarding erection cost of a structure. [closed]

The current erection cost of a structure is Rs. $$13,200$$. If the labour wages per day increase by $$\frac 1 5$$ of the current wages and the working hours decrease by $$\frac 1 {24}$$ of the current period, then the new cost of erection in Rs. is

$$(A)\ 16,500$$.

$$(B)\ 15,180$$.

$$(C)\ 11,000$$.

$$(D)\ 10,120$$.

I have got an answer different from the answer which is approximately equal to $$16528$$. Which is very close to option $$(A)$$. Is it correct? Please help me in this regard.

Thank you very much.

Attempt:

Labour wages increment is proportional to erection cost and working hours decrement is reverse proportional to the erection cost.

So the required erection cost is $$13200×\frac{\frac65}{\frac{23}{24}}$$ which simplifies to $$16528$$ (approx.)

closed as off-topic by Xander Henderson, Cesareo, Key Flex, Shailesh, Chinnapparaj ROct 21 '18 at 9:43

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• Welcome to Math StackExchange! Please tell us how you got the answer of $16528$. Explaining what you did will help us understand where you went wrong and allow us to give you more assistance. – Noble Mushtak Oct 20 '18 at 19:47
• What's that? Just at the very beginning $3$ downvotes. Who are bloodies? – math maniac. Oct 20 '18 at 19:47
• Unfortunately, the community can be kind of harsh against questions that don't follow our Guidelines. Your question would be higher-quality if you showed your work and explained how you got to your answer. I apologize for the swift downvotes from others, but I would suggest adding your work to your question in order to improve it. – Noble Mushtak Oct 20 '18 at 19:49
• Labour wages increment is proportional to erection cost and working hours decrement is reverse proportional to the erection cost. – math maniac. Oct 20 '18 at 19:49
• So the required erection cost is $\frac {13200 \times \frac 6 5} {\frac {23} {24}}$ which simplifies to $16528$ (approx.). – math maniac. Oct 20 '18 at 19:52

Suppose we assuming that the total number of hours needed remain the same, then indeed your solution is correct, that is the answer is

$$13200 \times \frac65 \times \frac{24}{23} \approx 16528$$

which unfortunately not one of the option.

However, suppose for some reason, their efficiency improves and the days needed remains the same, then the answer is $$13200 \times \frac65 \times \frac{23}{24}=15180.$$

This is a badly framed question where the setting is not clear.

Just FYI, $$13200 \times \frac65 \times \frac{25}{24}$$ gives you the first option but it is not correct.

I am aware of the background of the question which is from GATE, the answer key is $$B$$, hence they have assumed that the number of days required remain the same.

• Yeah this is what I have said. The efficiency of the workers cannot be improved all of a sudden. So the second case that you and Noble Mushtak have mentioned is somewhat absurd unless some extra conditions were given in the question. Isn't it so @Siong Thye Goh. – math maniac. Oct 22 '18 at 8:43
• Seems like a question understanding problem now, rather than a maths problem. – Siong Thye Goh Oct 22 '18 at 8:56
• Yeah! 😕😕 You are right! – math maniac. Oct 22 '18 at 14:05

As you said, the increase in erection cost is proportional to the increase in wages, so the new cost after wages increase is:

$$13,200\cdot \left(1+\frac{1}{5}\right)=15,840$$

Now, the decrease in working hours is actually proportional to the decrease in wages because if the working hours decrease, then since the workers are on an hourly wages, the amount you need to pay the workers also decreases, so the erection cost decreases. Thus, the new cost after wages increase and hours decrease is:

$$13,200\cdot \left(1+\frac{1}{5}\right)\cdot (1-\frac{1}{24})=13,200\cdot\frac{6}{5}\cdot\frac{23}{24} =15,180$$