How can a Shopkeeper Break Even on a Business

[KaizerSoze]

A shopkeeper buys 100 apples at $10 per apple from a farmer. The shopkeeper is selling apples at $15 retail price.

If the apples do not sell in 7 days, the Farmer will purchase the remaining apples at a 10% discount to the price at which the farmer sold to the shopkeeper.

Q: How many apples should the SHOPKEEPER sell at $15/apple to breakeven (no loss or profit) before 7days.

 

[Dr.Peterson]

A shopkeeper buys 100 apples at $10 per apple from a farmer. The shopkeeper is selling apples at $15 retail price.

If the apples do not sell in 7 days, the Farmer will purchase the remaining apples at a 10% discount to the price at which the farmer sold to the shopkeeper.

Q: How many apples should the SHOPKEEPER sell at $15/apple to breakeven (no loss or profit) before 7days.

This is most easily solved using algebra; does the fact that you put this under arithmetic indicate that you don't know any algebra?

In any case, in order to know what sort of help you need, we have to see what you have tried, or at least what you have learned. We don't just give answers with no visible effort on your part:

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If you do know basic algebra, I'd suggest defining a variable representing the number of apples sold, and writing an equation that sets the profit equal to zero.

 

[The Highlander]

A shopkeeper buys 100 apples at $10 per apple from a farmer. The shopkeeper is selling apples at $15 retail price.

If the apples do not sell in 7 days, the Farmer will purchase the remaining apples at a 10% discount to the price at which the farmer sold to the shopkeeper.

Q: How many apples should the SHOPKEEPER sell at $15/apple to breakeven (no loss or profit) before 7days.

I get the answer to be fractional (XX.666...)! Another 'defective question?

I do wish people would take more care when constructing questions like this (to ensure that the numbers 'work out' evenly).

Clearly, under normal circumstances, only whole apples can be sold, so, if the shopkeeper sells one fewer apple than my answer, s/he makes a $4 loss on the operation while if s/he sells one apple more s/he makes a $2 profit. The only way to break even (ie: make "no loss or profit") is to sell someone ⅔ of an apple for $10!

 

[KaizerSoze]

Thank you for the response. I did not know that my effort counted on forums, so did not bother putting that in. But here is how I tried to solve it, But I don't know where I am going wrong in my formulas. Here, let me try solving this again and may be it will help, where I am going wrong OR stuck in my logical deduction.


1. Cost Price (cp) for shopkeeper per apple = total cost of apples / cp per apple = 100/10 = $10 per apple
2. Profit per apple: Selling price (sp) - cost price (cp) = 15-10 = $5/apple
3. Total profit for 100 apples = 100* 5 = $500
4. The total risk (TR) for the shopkeeper if no apples are sold in 7 days is to sell all apples to the farmer at a 10% discount, which is $1 per apple
a total of $100 risk on all apples

How many apples he needs to sell at $5 profit to cover for $100 initial risk = total risk / profit per apple = 100/5 = 20 apples.

Firstly, I do not know if my answer is correct - algebraic expression is TR/(sp-cp).

But here is how I know the answer is wrong: If he sells ONLY 20 apples by the end of 7 days, then the remaining apples (80) will be sold back to the shopkeeper at $9 per apple = 9*80 = $720. Shopkeeper's cost of 80 apples is $800 and $800-$720 = $80.

Gain on 20 apples is $100 and loss on 80 apples is $80. So selling 20 apples is an excessive and I am stuck here trying to find the correct approach and answer.

 

[Dr.Peterson]

1. Cost Price (cp) for shopkeeper per apple = total cost of apples / cp per apple = 100/10 = $10 per apple
2. Profit per apple: Selling price (sp) - cost price (cp) = 15-10 = $5/apple
3. Total profit for 100 apples = 100* 5 = $500
4. The total risk (TR) for the shopkeeper if no apples are sold in 7 days is to sell all apples to the farmer at a 10% discount, which is $1 per apple
a total of $100 risk on all apples

How many apples he needs to sell at $5 profit to cover for $100 initial risk = total risk / profit per apple = 100/5 = 20 apples.

Firstly, I do not know if my answer is correct - algebraic expression is TR/(sp-cp).

But here is how I know the answer is wrong: If he sells ONLY 20 apples by the end of 7 days, then the remaining apples (80) will be sold back to the shopkeeper at $9 per apple = 9*80 = $720. Shopkeeper's cost of 80 apples is $800 and $800-$720 = $80.

Gain on 20 apples is $100 and loss on 80 apples is $80. So selling 20 apples is an excessive and I am stuck here trying to find the correct approach and answer.

I'm not familiar with the concept of total risk as you are using it; this is not how I would approach the problem, so I can't be sure where you are making an error in your method (e.g. if your formula ignores the repurchase). Perhaps someone here is familiar with how it is done in business courses.

The equation I wrote just sets the net gain (revenue minus cost plus gain from the discounted repurchase) to zero. I did not calculate the profit per apple, in part because there are two groups of apples, sold and unsold, which would confuse me a bit. I did get an answer smaller than 20.

In fact, one way to get an equation is to do what you did as a check (a very wise thing to have done) as a model for the equation, just replacing 20 with a variable.

I get the answer to be fractional (XX.666...)! Another 'defective question?

I do wish people would take more care when constructing questions like this (to ensure that the numbers 'work out' evenly).

Clearly, under normal circumstances, only whole apples can be sold, so, if the shopkeeper sells one fewer apple than my answer, s/he makes a $4 loss on the operation while if s/he sells one apple more s/he makes a $2 profit. The only way to break even (ie: make "no loss or profit") is to sell someone ⅔ of an apple for $10!

As I see it, it is common in the real world to have the break-even amount not be a whole number, so I think, if anything, this is a better question than what you envision, because the real world is not as kind as you. (Of course, in the real world, apples don't cost $10 each (not the ones we buy).

What I would actually do is to write an inequality, which says that we have at least broken even (that is, did not lose money), and round the solution in the appropriate direction to get a whole number that should be sold. But I don't think it's wrong to say "We will break even if we sell xx.66 apples, so we want to sell at least xy whole apples to make a profit."

 

[KaizerSoze]

I get the answer to be fractional (XX.666...)! Another 'defective question?

I do wish people would take more care when constructing questions like this (to ensure that the numbers 'work out' evenly).

Clearly, under normal circumstances, only whole apples can be sold, so, if the shopkeeper sells one fewer apple than my answer, s/he makes a $4 loss on the operation while if s/he sells one apple more s/he makes a $2 profit. The only way to break even (ie: make "no loss or profit") is to sell someone ⅔ of an apple for $10!

Sorry. I should have framed the question as " How many apples should the SHOPKEEPER sell at $15/apple to breakeven or just better (no loss) before 7days." So that rounding off to the next whole number would be understood.

 

[KaizerSoze]

I think I found the solution after digging my heels into it for quite some time. Thank you for the help

 

[The Highlander]

I'm not familiar with the concept of total risk as you are using it; this is not how I would approach the problem, so I can't be sure where you are making an error in your method (e.g. if your formula ignores the repurchase). Perhaps someone here is familiar with how it is done in business courses.

I did an undergraduate business course and we were always taught to deal with break even analyses as graphs, eg:-

​

but such analyses were always aimed at entire business operations involving both fixed and variable costs, sales & turnover whereas this question essentially deals with a single 'deal' within the shopkeeper's overall operations.

I've never come across this use of the term "risk" in the sense it appears to be used by the OP; the closest I would think of it would be profitability or just plain profit/loss.

As I see it, it is common in the real world to have the break-even amount not be a whole number, so I think, if anything, this is a better question than what you envision, because the real world is not as kind as you. (Of course, in the real world, apples don't cost $10 each (not the ones we buy).

What I would actually do is to write an inequality, which says that we have at least broken even (that is, did not lose money), and round the solution in the appropriate direction to get a whole number that should be sold. But I don't think it's wrong to say "We will break even if we sell xx.66 apples, so we want to sell at least xy whole apples to make a profit."

Indeed, in the real world things very rarely work out to be nice round numbers and the suggestion that the answer might be best expressed as an inequality is a very good one.

My point was that when a problem is constructed like this (producing a fractional answer when discrete data are involved) it may shake the confidence of the student.

If they get an answer like X⅔ then they may think: "Well, that can't be right! You can't sell ⅔ of an apple!".

However, it doesn't affect the ability of the problem to be solved algebraically, it just makes the answer a bit incongruous.

Thank you for the response. I did not know that my effort counted on forums, so did not bother putting that in. But here is how I tried to solve it, But I don't know where I am going wrong in my formulas. Here, let me try solving this again and may be it will help, where I am going wrong OR stuck in my logical deduction.


1. Cost Price (cp) for shopkeeper per apple = total cost of apples / cp per apple = 100/10 = $10 per apple
2. Profit per apple: Selling price (sp) - cost price (cp) = 15-10 = $5/apple
3. Total profit for 100 apples = 100* 5 = $500
4. The total risk (TR) for the shopkeeper if no apples are sold in 7 days is to sell all apples to the farmer at a 10% discount, which is $1 per apple
a total of $100 risk on all apples

How many apples he needs to sell at $5 profit to cover for $100 initial risk = total risk / profit per apple = 100/5 = 20 apples.

Firstly, I do not know if my answer is correct - algebraic expression is TR/(sp-cp).

But here is how I know the answer is wrong: If he sells ONLY 20 apples by the end of 7 days, then the remaining apples (80) will be sold back to the shopkeeper at $9 per apple = 9*80 = $720. Shopkeeper's cost of 80 apples is $800 and $800-$720 = $80.

Gain on 20 apples is $100 and loss on 80 apples is $80. So selling 20 apples is an excessive and I am stuck here trying to find the correct approach and answer.

I think you are making a stick for your own back in adopting the approach you took. The simplest and (IMNSHO ) best approach is to ignore altogether the individual profit (or loss) on the sale of the apples.

This is just a deal where the shopkeeper has outgoings (what s/he pays to buy the apples, ie: 100 @ $10) and income (what s/he gets when s/he 'disposes' of them, (whether it is to a retail customer, for $15 each or back to the farmer for $9 each).

If the total income exceeds the total outgoings then s/he makes a profit or if the reverse is true then a loss is made so, basically, you are tasked with finding the number apples sold (@ $15) such that the income exactly equals the initial outlay.

The shopkeeper starts the deal by buying 100 apples @ $10 so the total outgoings are (as I'm sure you can see): $1,000. (That doesn't change and remains constant throughout!)

If no apples are sold to customers then all 100 go back to the farmer @ $9 ⇒ total income of $900 (and, therefore, a loss of $100).

(Income - Outgoings = $900 - $1,000 = -$100)​

If one apple is sold for $15 then 99 go back to the farmer @ $9 ⇒ total income of $890 + $15 = $905 (and, therefore, a, slightly smaller, loss of $95).

If two apples are sold for $15 then 98 go back to the farmer @ $9 ⇒ total income of $880 + $30 = $910 (and, therefore, an even smaller loss of $90).

If ten apples are sold for $15 then 90 go back to the farmer @ $9 ⇒ total income of $810 + $150 = $960 (and, therefore, a much smaller loss of just $40).

As you can see, the more s/he sells to customers (@ $15) the smaller the loss becomes until, eventually, the total income (number sold @ $15 plus number returned @ $9) will be greater than the intial ($1,000) outlay.

Thinking of the situation in those terms, can you now figure out where the break even point will lie?
(Formulating an equation, where an expression of income - initial outlay = 0. would bethe algebraic approach to solving this. You could start by letting x equal the number of apples sold @ $15; how many are then returned to the farmer @ $9? )

Hope that helps.

 

[KaizerSoze]

I did an undergraduate business course and we were always taught to deal with break even analyses as graphs, eg:-

View attachment 37979​

but such analyses were always aimed at entire business operations involving both fixed and variable costs, sales & turnover whereas this question essentially deals with a single 'deal' within the shopkeeper's overall operations.

I've never come across this use of the term "risk" in the sense it appears to be used by the OP; the closest I would think of it would be profitability or just plain profit/loss.


Indeed, in the real world things very rarely work out to be nice round numbers and the suggestion that the answer might be best expressed as an inequality is a very good one.

My point was that when a problem is constructed like this (producing a fractional answer when discrete data are involved) it may shake the confidence of the student.

If they get an answer like X⅔ then they may think: "Well, that can't be right! You can't sell ⅔ of an apple!".

However, it doesn't affect the ability of the problem to be solved algebraically, it just makes the answer a bit incongruous.


I think you are making a stick for your own back in adopting the approach you took. The simplest and (IMNSHO ) best approach is to ignore altogether the individual profit (or loss) on the sale of the apples.

This is just a deal where the shopkeeper has outgoings (what s/he pays to buy the apples, ie: 100 @ $10) and income (what s/he gets when s/he 'disposes' of them, (whether it is to a retail customer, for $15 each or back to the farmer for $9 each).

If the total income exceeds the total outgoings then s/he makes a profit or if the reverse is true then a loss is made so, basically, you are tasked with finding the number apples sold (@ $15) such that the income exactly equals the initial outlay.

The shopkeeper starts the deal by buying 100 apples @ $10 so the total outgoings are (as I'm sure you can see): $1,000. (That doesn't change and remains constant throughout!)

If no apples are sold to customers then all 100 go back to the farmer @ $9 ⇒ total income of $900 (and, therefore, a loss of $100).

(Income - Outgoings = $900 - $1,000 = -$100)​

If one apple is sold for $15 then 99 go back to the farmer @ $9 ⇒ total income of $890 + $15 = $905 (and, therefore, a, slightly smaller, loss of $95).

If two apples are sold for $15 then 98 go back to the farmer @ $9 ⇒ total income of $880 + $30 = $910 (and, therefore, an even smaller loss of $90).

If ten apples are sold for $15 then 90 go back to the farmer @ $9 ⇒ total income of $810 + $150 = $960 (and, therefore, a much smaller loss of just $40).

As you can see, the more s/he sells to customers (@ $15) the smaller the loss becomes until, eventually, the total income (number sold @ $15 plus number returned @ $9) will be greater than the intial ($1,000) outlay.

Thinking of the situation in those terms, can you now figure out where the break even point will lie?
(Formulating an equation, where an expression of income - initial outlay = 0. would bethe algebraic approach to solving this. You could start by letting x equal the number of apples sold @ $15; how many are then returned to the farmer @ $9? )

Hope that helps.

This approach requires over shooting and under shooting the exact BEP, then plot it --> You are going nuclear when a simple bow and arrow would suffice. Below is the step by step approach, that can be put together on a excel sheet and does not require a graph.

Initial Cost Calculation​

• Total cost of 100 apples at $10 each:
  100 apples×10 dollars/apple=1,000 dollars

Revenue and Loss Scenarios​

1. Selling Apples at Retail Price ($15/apple)
   • Revenue per apple sold: $15
2. Buyback Price by Farmer ($9/apple)
   • Revenue per apple if not sold in 7 days: $9

Break-Even Analysis​

To break even, the shopkeeper's total revenue should equal the initial cost of $1,000.

Let x be the number of apples sold at $15 each.​

Remaining apples that the farmer will buy back: 100−x​

• Revenue from selling x apples at $15 each:
  15x
• Revenue from the remaining 100−x apples bought back at $9 each:
  9(100−x)

Total revenue to break even:​

15x+9(100−x)=1,000

Simplifying the Equation​

1. Expand and combine like terms:
   15x+900−9x=1,000
   6x+900=1,000
2. Solve for xx:
   6x=1,000−900
   6x=100
   x=100
   x=16.67

Since the shopkeeper cannot sell a fraction of an apple, rounding up to ensure no loss:

x=17

 

[KaizerSoze]

from a formula stand point the number of apples x that must be sold at $15 each to BE is given by

x= (Purchase price total - total loss total)/ (Per unit selling price - per unit loss price) = (1000-900)/(15-9) = 100/6 = 16.66 ~ 17 (Rounded up)

 

[Dr.Peterson]

Let x be the number of apples sold at $15 each.​

Remaining apples that the farmer will buy back: 100−x​

• Revenue from selling x apples at $15 each:
  15x
• Revenue from the remaining 100−x apples bought back at $9 each:
  9(100−x)

Total revenue to break even:​

15x+9(100−x)=1,000

Yes, this is exactly the equation I suggested. Good.