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A girl went to market to buy brinjal,onion and coconut.........she gives


Rs 2 and buys 40 brianjals
Rs 1 and buys 01 onion
Rs 5 and buys 01 coconut..........
....................BUT................... The total amount of rupees spent and the number of vegetables bought should be 100....

Question: How many of each onion, brinjal, coconut does the girl buy with the above prices so that the total rupees spent and the net quantity bought is exactly 100??

Example:-
Rs Quantity (indicative)
2 $\quad$ 40 Brinjals
1 $\quad$ 1 Onion
5 $\quad$ 1 Coconut
--- ----
100 100 ----> Expected sum(Once the quantities are provided by the solution)

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    $\begingroup$ What’s unclear about this? What’s desired is obviously a solution in non-negative integers to the system $$\left\{\begin{align*}&x+y+z=100\\&2x+y+5z=100\;.\end{align*}\right.$$ And the unique solution is clearly $x=z=0,y=100$: she buys $100$ onions, no brianjals, and no coconuts. $\endgroup$
    – Brian M. Scott
    Jul 28, 2013 at 22:23
  • $\begingroup$ @BrianM.Scott why are the equations like this, for 2 rs she buys 40 brinjals not 1. $\endgroup$
    – quid ♦
    Jul 29, 2013 at 0:18
  • $\begingroup$ @quid: No, she doesn’t. That was merely an example illustrating what quantities are supposed to be equal; as noted in the parenthetical remark, it does not have the right quantities. $\endgroup$
    – Brian M. Scott
    Jul 29, 2013 at 0:20
  • $\begingroup$ @BrianM.Scott: what is in your opinion the price of 1 brinjal? And, why? In my opinion it is 0.05 rs as for 2rs she buys 40 (see at the start). $\endgroup$
    – quid ♦
    Jul 29, 2013 at 0:28
  • $\begingroup$ Correction to my original comment: I misread one line of the post. The second equation should be $0.05x+y+5z=100$, and there are two solutions, $x=80,y=1,z=19$ and $x=0,y=100,z=0$. The only thing requiring clarification is whether we’re to assume that the girl actually did buy at least one of each item. $\endgroup$
    – Brian M. Scott
    Jul 29, 2013 at 5:59

1 Answer 1

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Let $x,y$, and $z$ be respectively the numbers of brinjals, onions, and coconuts bought. The requirement that $100$ items be bought means that $x+y+z=100$. If $40$ brinjals cost $2$ rupees, the cost of one brinjal is $\frac2{40}=0.05$ rupees. Each onion costs one rupee, and each coconut costs $5$ rupees, so the total cost is $0.05x+y+5z$. Thus, we want a solution in non-negative integers to the system

$$\left\{\begin{align*} &x+y+z=100\\ &0.05x+y+5z=100\;. \end{align*}\right.\tag{1}$$

Multiply the second equation by $20$ to get $x+20z+100z=2000$ and subtract the first to get $19y+99z=1900$. Solving for $z$ in terms of $y$, we find that $$z=\frac{1900-19y}{99}=\frac{19}{99}(100-y)\;.$$

This is non-negative if and only if $y\le 100$, and since we also want $y\ge 0$, the only possible values of $y$ are the integers from $0$ through $100$. In order for $z$ to be an integer, $100-y$ must be a multiple of $99$, so either $y=100$, or $y=1$.

  • If $y=100$, then $z=0$, and from $(1)$ we see that $x=0$ as well. One solution, then, is to buy $100$ onions and nothing else.

  • If $y=1$, then $z=19$, and from $(1)$ we find that $x=80$: she can buy $80$ brinjals (for $4$ rupees), one onion (for $1$ rupee), and $19$ coconuts (for $95$ rupees).

If she is required to buy a non-zero amount of each of the three items, the only solution is the second one.

(For those who are unfamiliar with the term, brindjal, like aubergine, is another name for eggplant.)

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  • $\begingroup$ Leave those rs and quantity .. but try to use the numbers by usin ratio and proportional $\endgroup$
    – chethana
    Jul 31, 2013 at 13:09
  • $\begingroup$ Guys even i am confused abt the question ill try to make it clear for you........ leave the rupees and the quantity ........ . use the numbers and their total should be 100... By using Ratio And Propotional $\endgroup$
    – chethana
    Jul 31, 2013 at 13:18
  • $\begingroup$ @chethana: I don’t see any way to solve it by applying ratios and proportionality to the original $40,1,5$ numbers; there is no solution that is a multiple of that set. $\endgroup$
    – Brian M. Scott
    Jul 31, 2013 at 21:03

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