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A shopkeeper sells a file and a notebook for Rs. 27 to the first customer, a notebook and a pen for Rs. 31 to a second customer, and a pen and a file for Rs. 29 to a third customer. The prices of the items are rounded in rupees.

Which of the following inference is correct?

  1. The pen is the costliest of the three.
  2. The file is the costliest of the three.
  3. The notebook is the costliest of the three.
  4. The shopkeeper sold the items to different customers at different rates.

My problem : question is correct and I think that if I fix any price for some item then that will not satisfy the equations so I think 4th option is correct.

I don't know which tag will be appropriate.

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4 Answers 4

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If the question is saying that each item costs a whole number of rupees (is this what "rounded in rupees" means?) then you are right, it is not possible for the shopkeeper to have charged the same prices to each customer (so the fourth option is correct).

This is because in total he has sold two of each item, but he has charged a total of Rs $87$. If he had charged $a$ for each file, $b$ for each notebook and $c$ for each pen then we would have $87=2(a+b+c)$, but $87$ is odd so this is impossible.

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  • $\begingroup$ I can't upvote or thanks then what should I do in these posts to show my gratitude ?... But still thanks ur ans was not long and was enough for me to understand. $\endgroup$ Jul 11, 2017 at 13:35
  • $\begingroup$ You're welcome! You should be able to accept one answer (not necessarily this one; just pick whichever you found most helpful). $\endgroup$ Jul 11, 2017 at 13:52
  • $\begingroup$ Will I be able to accept only one answer? $\endgroup$ Jul 11, 2017 at 15:15
  • $\begingroup$ Yes, I'm afraid so. You can unaccept and then accept a different answer, if you change your mind, but you can't have more than one accepted answer. $\endgroup$ Jul 11, 2017 at 15:23
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We have

$$F+N=27\\ F+P=29\\ N+P=31$$

From the first two equations, we know that $P=N+2$, and likewise from the second and third equations, we know that $N=F+2$. Adding the first two equations and subtracting the third from that sum, we get a value for $F$, and from there the rest is straightforward.

It is required that values be whole numbers of Rupees? If so, then there is a problem, but if half-Rupees are allowed, then there is a solution.

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  • $\begingroup$ There is problem means fourth is right ..right? $\endgroup$ Jul 11, 2017 at 13:28
  • $\begingroup$ Yes, I would agree $\endgroup$ Jul 11, 2017 at 13:29
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Pretty straight forward. You have three equations in three unknowns so should be able to solve for the cost of each item separately then answer the question.

Call the price of the pen "p", the cost of the screwdriver "s", and the cost of the notebook "n".

"A shopkeeper sells a file and a notebook for Rs. 27 to the first customer" so f+ n= 27.

"A notebook and a pen for Rs. 31 to second customer" so n+ p= 31.

"and a pen and a file for Rs. 29 to third customer" so p+ f= 29.

Subtract "f+ n= 27" from "n+ p= 31" to get "p- f= 2".

Add "p+ f= 29" to "p- f= 2" to get "2p= 31" so p= 31/2 R. Then p- f= 31/2- f= 2 so f= 31/2- 2= 27/2 R. Finally, n+ p= n+ 31/2= 31 so n= 31- 31/2= 31/2 R.

The pen cost 31/2= 15.5 R, the file cost 27/2= 13.5 R, and the notebook cost 31/2= 15.5 R.

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  • $\begingroup$ But it can't be called rounded I think. very confusing this "rounded" word. $\endgroup$ Jul 11, 2017 at 13:36
  • $\begingroup$ You have an arithmetic error here: 31-27=4, not 2 $\endgroup$ Jul 11, 2017 at 14:47
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Take the price of a note, a pen, a file to be N, P, F respectively. Then using the given data, you can formulate a system of 3 simultaneous equations given below, $$P+N=31$$ $$ F+N=27$$ $$P+F=29$$. Use this system to find the signs of $P- N$, $P-F$, $N-F.$ You'll get which one is costlier.

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