# How many of the mixed sweets do you get for 10N?

#### chijioke

##### Full Member
Equal quantity of sweets at two for 10 naira and six for 10 naira are mixed together. How many of the mixed sweets do you get for N10?

This is what I did,
$\text{ for the sweets sold for two for 10N, one cost}~ \frac{10}{2}=N5$$\text{ for the sweets sold for six for 10N, one cost}~ \frac{10}{6}=\frac{5}{3}$Am I doing it right? How should I go about it?

Equal quantity of sweets at two for 10 naira and six for 10 naira are mixed together. How many of the mixed sweets do you get for N10?

This is what I did,
$\text{ for the sweets sold for two for 10N, one cost}~ \frac{10}{2}=N5$$\text{ for the sweets sold for six for 10N, one cost}~ \frac{10}{6}=\frac{5}{3}$Am I doing it right? How should I go about it?
That's a valid start (though not how I did it). What do you plan to do next? What are your ideas?

You might, for example, now ask, how much will it cost for one of each? Then what is the unit cost of the mix?

My own method avoided fractions, by asking how much 6 of each cost, rather than one of each.

My own method avoided fractions, by asking how much 6 of each cost, rather than one of each
suppose I decide to borrow your method, six of the sweet sold for two 10N cost
$\frac{10}{2} \times 6 = N 30$We have already being told that the second type of sweet is sold for six at N10.
So total cost of sweets sold for six at N10 = N10 + N30 = N40
I am getting right? If I am getting right, what else do I do?

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suppose I decide to borrow your method, six of the sweet sold for two 10N cost
$\frac{10}{2} \times 6 = N 30$We have already being told that the second type of sweet is sold for six at N10.
So total cost of sweets sold for six at N10 = N10 + N30 = N40
I am getting right? If I am getting right, what else do I do?
That's right, except that you haven't said that this is a total of 12 sweets, 6 of each type. Now what?

Just to make sure we're clear, I'll rewrite what you said here a little more carefully:

Six of the sweets sold at two for N10 cost [imath]\frac{10}{2} \times 6 = N 30[/imath].​
We have already been told that the second type of sweet is sold at six for N10.​
So total cost of 12 of the mixed sweets is N10 + N30 = N40.​

So how many will you get for N10?

So total cost of 12 of the mixed sweets is N10 + N30 = N40.
So how many will you get for N10?
As 12 sweets cost N40, number of sweets that can be obtained for N10 is $\frac{40}{10}=4$Thank you I am cleared now.
I still have questions on this kind of concept. But is a worked example. What I just need is an explanation on it. So I will be creating a new thread for it shortly. Thank you sir.

As 12 sweets cost N40, number of sweets that can be obtained for N10 is $\frac{40}{10}=4$Thank you I am cleared now.
Not quite right. Give that another try! Maybe write a proportion. (You didn't use the 12.)

Not quite right. Give that another try! Maybe write a proportion. (You didn't use the 12.)
How I am supposed to use the 12? I am beginning to get confused.
six sweets cost N30 and another six cost N10

As you rightly pointed out total cost of 12 sweet is N40.
You now asked me, if the cost of sweets as we now have it is N40, how many sweets can be bought for N10? Which is now 10 +10 +10 +10 = 4 sweets
since one corresponds to N10.
Or maybe perhaps, 12 sweet cost N40 and we are interested in knowing the number of sweets that can be bought for 10N then the number of sweets will decrease in the ratio 10 : 40 =1 :4
That is the number of sweets that can be bought for N10 = $\frac{10}{40}=3$Is this what you want me to do?

How I am supposed to use the 12? I am beginning to get confused.
six sweets cost N30 and another six cost N10

As you rightly pointed out total cost of 12 sweet is N40.
You now asked me, if the cost of sweets as we now have it is N40, how many sweets can be bought for N10? Which is now 10 +10 +10 +10 = 4 sweets
since one corresponds to N10.
No, 1 sweet does not correspond to N10, and 10+10+10+10 is not 4.

Or maybe perhaps, 12 sweet cost N40 and we are interested in knowing the number of sweets that can be bought for 10N then the number of sweets will decrease in the ratio 10 : 40 =1 :4
Yes, this is one way to do it. Since N10 is 1/4 of N40, the number of sweets for N10 is 1/4 of the number you get for N40.

That is the number of sweets that can be bought for N10 = $\frac{10}{40}=3$Is this what you want me to do?
But 10/40 is not 3! I suppose you meant [imath]\frac{10}{40}\times12=3[/imath], which is the correct answer.

As I suggested before, another approach is to write a proportion: $\frac{x\text{ sweets}}{N10}=\frac{12\text{ sweets}}{N40}$
Solve that by cross-multiplying, or any way you prefer.

But 10/40 is not 3! I suppose you meant 1040×12=3\frac{10}{40}\times12=34010×12=3, which is the correct answer.
Yes. But how do I check to see that my answer is correct?

There is something very odd going on here.
Is it a rogue seller or just a very poorly thought up question?
The logic is impeccable to arrive at 3 for N10 (from the mixed batch of sweets).
But why would you pay N10 for only 3 sweets when you could buy 1 (of the 2 for N10) for N5, plus 3 (of the 6 for N10) for N5 and, thereby, get 4 sweets for the same money (N10)?
Furthermore, if you paid N10 for 3 sweets (from the mixed batch) what's to prevent you getting all 3 sweets of the 6 for N10 type? (Which means they're only worth N5!)
I'd stay well away from that shop!

Yes. But how do I check to see that my answer is correct?
Draw a picture?

If A is a sweet that costs N10 for 2, and B is a sweet that costs N10 for 6, then I can't make a picture of 3 sweets with equal numbers of each type, but I can draw a jar of 60, with 30 of each type:

AAAAA AAAAA​
AAAAA AAAAA​
AAAAA AAAAA =30 @ N10 for 2 = 15*N10 = N150​
BBBBB BBBBB​
BBBBB BBBBB​
BBBBB BBBBB = 30 @ N10 for 6 = 5*N10 = N50​

The total cost is N200 for 60 sweets; for N10, we'd get 1/20 of these, which is 3.

But @The Highlander's comment is very significant. No set of 3 that you get will really be worth N10:
• AAA would be worth N15,
• AAB would be worth N11 2/3,
• ABB would be worth N8 1/3, and
• BBB would be worth N5.
Clearly, they wouldn't let you serve yourself (which would be like selling mixed nuts and letting you pick out the best), but would sell randomly chosen bags, so it would be like a lottery. And I suppose some people like taking a chance. But they certainly wouldn't sell a bag of 3 for N10.

But on average, the value of 3 chosen randomly would be (1/8)N15 + (3/8)N11 2/3 + (3/8)N8 1/3 + (1/8)N5 = 15/8 + 35/8 + 25/8 + 5/8 = 80/8 = N10.

(Unrealistic word problems are a longstanding tradition among math teachers; look at some of the ancient Egyptian problems, and they're just as silly. They aren't meant to be real, but to challenge your thinking.)

Yes. But how do I check to see that my answer is correct?

The explanation (below) of what is going on here may help you to understand it better.

Draw a picture?

But on average, the value of 3 chosen randomly would be.... (see final paragraphs)
What @Dr.Peterson writes is both insightful and useful (as I trust you will see).

Let's assume that the question describes a real-life situation where a shopkeeper has these these sweets to sell and the ones that are 2 for N10 are green whilst the ones that are 6 for N10 are blue.

He decides that instead of (or perhaps as well as) selling them separately he will mix them together in equal quantities, so how much will he charge for sweets sold out of the mixed batch?

Well, lets say he fills a bowl with 60 of the green sweets and another bowl with 60 of the blue sweets and then empties both of those bowls into a larger bowl that will now contain 120 of the sweets all thoroughly mixed together.

First, let us (again) consider how much each sweet costs (is worth) as you did in your initial post (remember?).

The green sweets are 2 for N10 so each one is worth N5
$$\displaystyle \left(\frac{10}{2}\right)$$. Let's call that g, so g = N5.

and the blue sweets are 6 for N10 so each one is worth N
$$\displaystyle \frac{5}{3}\left(\frac{10}{6}\right)$$ which we could also write as: N1⅔ and we will call that b, so b = N1⅔.

and those are exactly the same numbers that you came up with yourself, yes?

So, now consider the value of each of the bowls I described above...

The first bowl has 60 green sweets in it each costing N5 so its value is 60 × N5 = N300

and the second bowl has 60 blue sweets in it each costing N1⅔ so its value is 60 × N1⅔ = N100...

$$\displaystyle \left(60\times \text{N}1⅔=60\times\frac{\text{N}5}{3}=\frac{\text{N}300}{3}=\text{N}100\right)$$

and so, the value of the bowl of mixed sweets is N400 (N300 + N100), so, what we have now (drawing a picture ) is this...

So, now he has to decide how much he will sell the sweets in his mixed bowl for...

Well, there are 120 sweets in that bowl and its total value is N400 so he figures that he should sell them for N3⅓ each. Why?

Well, $$\displaystyle \frac{\text{N}400}{120}=\frac{\text{N}40}{12}=\text{N}3\frac{4}{12}=\text{N}3\frac{1}{3}=\text{N}3⅓$$.

So, if you now buy 3 sweets they will cost you 3 × N3⅓ (= N10) which is why the answer to the question is that you get 3 sweets for N10.

Of course, he is now charging more for the blue ones than they are worth while, at the same time, he is now charging less for the green ones than they are worth!

But let's consider for a moment the average (mean) price of the sweets...

The average cost of 1 green & 1 blue sweet would be:-

(g + b) ÷ 2 = $$\displaystyle (\text{N}5 +\text{N}1⅔)\div 2=\left(\text{N}\frac{15}{3}+\text{N}\frac{5}{3}\right)\div 2=\text{N}\frac{20}{3}\div 2=\text{N}\frac{10}{3}=\text{N}3\frac{1}{3}=\text{N}3⅓$$

So what he is charging is actually an average price for the sweets and he could argue that is fair because, if he sells his whole bowl of mixed sweets, he will get N400 for it (120 × N3⅓ = N400) and he is, therefore (ostensibly) not overcharging for the sweets (because, although he's charging more for each blue one, he's charging less for each green one).

However, what he's doing (as @Dr.Peterson says) is creating what might be described as a lottery!

If you buy 3 sweets (from his mixed bowl) for N10 and...

they are all green then they are worth N15 so you have done very well

and, if two of them are green, then the 3 sweets are worth N11⅔ so you have still done quite well

but, if you get only 1 green sweet then it, plus the 2 blue ones, are only worth N8⅓ so you've overpaid

and, if all three are blue, then they're only worth N5 so you've paid double what they're really worth!

Hope that helps.

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Draw a picture?

If A is a sweet that costs N10 for 2, and B is a sweet that costs N10 for 6, then I can't make a picture of 3 sweets with equal numbers of each type, but I can draw a jar of 60, with 30 of each type:

AAAAA AAAAA​
AAAAA AAAAA​
AAAAA AAAAA =30 @ N10 for 2 = 15*N10 = N150​
BBBBB BBBBB​
BBBBB BBBBB​
BBBBB BBBBB = 30 @ N10 for 6 = 5*N10 = N50​

The total cost is N200 for 60 sweets; for N10, we'd get 1/20 of these, which is 3.

But @The Highlander's comment is very significant. No set of 3 that you get will really be worth N10:
• AAA would be worth N15,
• AAB would be worth N11 2/3,
• ABB would be worth N8 1/3, and
• BBB would be worth N5.
Clearly, they wouldn't let you serve yourself (which would be like selling mixed nuts and letting you pick out the best), but would sell randomly chosen bags, so it would be like a lottery. And I suppose some people like taking a chance. But they certainly wouldn't sell a bag of 3 for N10.
I understand everything you wrote until I came to this part below
But on average, the value of 3 chosen randomly would be (1/8)N15 + (3/8)N11 2/3 + (3/8)N8 1/3 + (1/8)N5 = 15/8 + 35/8 + 25/8 + 5/8 = 80/8 = N10.
How did you arrive at this average? I thought you obtained this average by dividing the cost of 3 type of sweets chosen at random by 8.
If what I am thinking is true, then
$\frac{15}{8}+\frac{11 \frac{2}{3} }{8} +\frac{8 \frac{1}{3}}{8} +\frac{5}{8}$$=\frac{15}{8}+\frac{5}{8}+\frac{35}{24}+\frac{25}{24}$$=\frac{45+15+35+25}{24}=\frac{120}{24}=N5$Could you please explain?

I understand everything you wrote until I came to this part below

How did you arrive at this average? I thought you obtained this average by dividing the cost of 3 type of sweets chosen at random by 8.
If what I am thinking is true, then
$\frac{15}{8}+\frac{11 \frac{2}{3} }{8} +\frac{8 \frac{1}{3}}{8} +\frac{5}{8}$$=\frac{15}{8}+\frac{5}{8}+\frac{35}{24}+\frac{25}{24}$$=\frac{45+15+35+25}{24}=\frac{120}{24}=N5$Could you please explain?
I didn't explain that in detail, because it involves the mean of a distribution, and I didn't want to take a lot of time to explain. Here is what I said:
No set of 3 that you get will really be worth N10:
• AAA would be worth N15,
• AAB would be worth N11 2/3,
• ABB would be worth N8 1/3, and
• BBB would be worth N5.
Clearly, they wouldn't let you serve yourself (which would be like selling mixed nuts and letting you pick out the best), but would sell randomly chosen bags, so it would be like a lottery. And I suppose some people like taking a chance. But they certainly wouldn't sell a bag of 3 for N10.

But on average, the value of 3 chosen randomly would be (1/8)N15 + (3/8)N11 2/3 + (3/8)N8 1/3 + (1/8)N5 = 15/8 + 35/8 + 25/8 + 5/8 = 80/8 = N10.

(Unrealistic word problems are a longstanding tradition among math teachers; look at some of the ancient Egyptian problems, and they're just as silly. They aren't meant to be real, but to challenge your thinking.)
When we select 3 sweets, in order, we might get any of 8 equally likely results: AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB.
• Of these, 1 is all A's; so the probability of that is 1/8, and we multiply that by the value, N15. That is, 1/8 of the time we get a value of N15.
• 3 are 2 A's and 1 B; so the probability of that is 3/8, and we multiply that by the value, N11 2/3. That is, 3/8 of the time we get a value of N11 2/3.
• 3 are 1 A and 2 B's; so the probability of that is 3/8, and we multiply that by the value, N8 1/3. That is, 3/8 of the time we get a value of N8 1/3.
• 1 is all B's; so the probability of that is 1/8, and we multiply that by the value, N5. That is, 1/8 of the time we get a value of N5.
Add those products and you have the mean of the probability distribution.

You didn't take into account the different probabilities of getting the different values.