There was a chocolate bar which weighed 0.65kg. If Daniel and Jacob split it 1:3 and Jacob and Antonia split it 1:2, how much chocolate did Antonia get.
A- 150g
B- 565g
C- 433g
D- 400g
E- 390g
Anyone have a clue as to how E is the answer? Been trying to figure this out for the last hour.
Added pic:
That is how i feel about this question :)
Yes: F = 325g
…now I'll read on and see where I went wrong
I got F = 325g as well.
Now i'm intrigued.
Maybe this is why our education system is screwed.
That's what I got.
The error is the word "split". Split implies that the bar is broken at 1:3 and then Jacob splits his 3/4ths further at 1:2, i.e. Antonia gets 2/4ths (1/2 the bar).
The question SHOULD have been phrased as follows:
There was a chocolate bar which weighed 0.65kg. If Daniel's share and Jacob's share has a ratio of 1:3; and Jacob's share and Antonia's share has a ratio of 1:2, how much chocolate did Antonia get?
Then it's correctly E.
Half the bar is still 325g though, not E(390g).
@YadaYadaYada: Antonia = x
jacob = x/2
daniel=jacob/3=x/2/3=x/6
0.65=x+x/2+x/6=6x/6+3x/6+x/6=10x/6
x=0.65*6/10=0.390
@Bren20: As it is a primary school test, assume no need for algebra, J:A = 1:2 = 3:6 so D:J:A = 1:3:6, so Antonia = 6 parts, 1 part is 65 g so 6 parts 390g.
@YadaYadaYada: That's why I was saying split is the wrong word, if you use the phrasing as I suggested the answer is below
So the real ratio is 1:3:6
The original question's wording is so atrocious
Selective schools test?
Yep
Time to find one with a proper maths department
Its called Eddie Woo
I can't tell if I am daft or the question is. inb4 mexican girl
Why not- oh.
First split is 1 part to Daniel, 3 to Jacob (3/4 of whole).
Jacob then splits remaining 1 part to 2 for Antonia (2/3 of 3/4 is 6/12 or 1/2).
Answer is 0.325kg.
That’s what I got too, 325g
This is also the answer I arrived at
Ah I get it now. I read it as Daniel and Jacob sharing ⅓ the chocolate bar.
I also get 325g.
Yeah same here. Misprints or miscalculations strike again.
Yep, worked it out before I scrolled down and got 325g. Thinking none of the answers were correct.
But then again, I’ve spent 20 years trying to work out how the Monty Hall problem works and just can’t, no matter how many times it is explained to me.
The crux of the Classic Game "Let's Make a Deal" is that the presenter/host (the famous Mr Monty Hall), he actually knows which doors have which prizes. And he is bound by the rules of the game to ALWAYS (keyword) to open a second door that does NOT have a prize.
That's not made clear in the original wording.
Imagine there are 1000 doors instead of 3. You choose a door and then Monty opens 998 of them. Should you switch?
This boils down to, "what are the chances I picked right on the first go". With a thousand doors, incredibly unlikely.
The prize has to be behind one of the doors, so the probability is split between the two, if the chances of your door being correct are incredibly unlikely, that makes the other door very very likely indeed.
Same for 3 doors only not such an extreme difference. You only had 1/3 of a chance of getting it right first go, and the prize has to be behind one of the doors, so when Monty opens one the other door has to be 2/3 since 1/3 and 2/3 add to 1.
Hope that helps
@SlickMick: That's not how it works. Chance isn't changed by what you know - if your eyes were closed the whole time you still have a 1/1000 chance your door is correct, same as if you have your eyes open, see what happened and decide to stick with your original door. Your first choice is still 1/1000. Every single door is individually 1/1000. When the doors without goats are opened, all of those 1/1000 chances coalesce into the single remaining door you can switch to.
Another way of explaining it is to simply list out all the possibilities. Say you open each door twice to see what Monty does. Goats are behind doors 1 & 2 and the prize is behind door 3.
You choose the first door: Monty must open door 2 (as it is the other goat)
You choose the first door: Monty still must open door 2.
You choose the second door: Monty must open door 1 (as it is the other goat).
You choose the second door again: Monty still must open door 1.
You choose door 3: Monty has a choice as there are 2 doors with goats, he chooses door 1.
You choose door 3 again: Monty still has a choice, he chooses door 2.
In 4 out of the 6 scenarios, you are better off switching. This is because anytime you chose a goat to start with the other door is a prize. And you had a 2/3 chance of choosing a goat to start with, as there are two goats and one prize.
Saying that the choice is 50/50 because there are two options would be the same as saying you have a 50% chance of winning the lottery because you either win or you don't.
Maybe it would be easier with lottery tickets? Imagine everyone in Australia chose some lottery numbers and you did too, and one person in the whole country has the winning numbers. Someone comes along and burns 25 million non winning lottery papers (but never yours no matter if winning or not) leaving just you and one other person's ticket. Should you switch tickets with them? Do you still think the chances of you having picked the right numbers is 50/50 because someone burned all the non winning numbers except for yours and one other person's?
@Quantumcat: Okay, I clearly shouldn't have stuck my beak in where I clearly have no idea what we're talking about. How did goats come into the equation??
I feel that you might have left some relevant information out of your previous post?? :)
@SlickMick: In the original Monty Hall problem two doors contain goats and one contains the big prize. It is assumed you want the prize and not a goat.
@Quantumcat: I kinda get the situation, but in your last lottery ticket scenario, say you were left with just two people with tickets - you and Bob. It appears that both you and Bob feel you would be better off switching tickets?
@pangwen: The difference is that your ticket never gets burned no matter whether it is winning or not, (you are still 1 in 25 million chance of having been correct) whereas Bob is only there because all the other non winning tickets got burned. Bob could have been Alice or Sophie or David, whoever happened to have the winning ticket
@Quantumcat: @Quantumcat yes, but isn’t it the same from Bob’s perspective? They’re sitting there watching everyone else’s tickets get burned, and now they also think there’s a better chance of winning the prize if they switch.
@pangwen: If you replay the event 25 million times, your ticket is safe all 25 million times and Bob's is burned every time or maybe saved once or twice, same as each of the others. If Bob realises that your ticket is always safe no matter what then he should realise his is very likely to be the winner, the only way switching would be right for him would be if you picked the winning numbers outright, whereas he didn't - his ticket is safe because it is the winner not because the rules of the game are his is safe no matter what.
If the game is played 25 million times with you as the winner, anybody else could have been the random one left as the person in charge of this weird game has to pick 24,999,999 people at random to burn the tickets of. If the game is played 25 million times with Bob as the winner, his ticket gets left every time. If you were the winner, there is only the tiniest of chances that Bob happened to be the one left standing as it could have been anybody. If Bob is the winner, he had to be the one left standing every time.
I came across the problem from the card game bridge, there's a situation where you have AKT in one hand and small cards in the other, and you are missing the Q, J, and two small cards. You play the ace first, and you see either the queen or jack get played from the hand after the AKT. The question is, do you now play the king (hoping the hand that played the Q or J also has the other one with no small cards to play) or play small from the other hand towards the KT, playing the T, hoping the hand before the KT has the other one (then the ten will win and you will also win the king)? It looks like it is a 50/50 guess, but actually you are more likely to win by expecting the person who played the Q or J to have only that card and no other. Why? Because if they had both, they could have played either one, but if they only had one, they had to play that one. It is called "the principle of restricted choice" if you want to look it up. Kind of similar to the lottery situation, if Bob is left standing, either he is the winner or he isn't (if you picked the numbers right). If he is the winner then he had to be the one left standing, if he wasn't, he only had a 1/24,999,999 chance of being left.
@SlickMick: I know it's hard to get your head around, but it actually isn't a gamblers fallacy ("26 reds in a row, the next has to be black." is a gamblers fallacy) and it certainly isnt 50:50.
The way I was shown was with a table that made it a bit easier to visualise how it works and what is happening.
In this example, we will pick Door 1. There is a 1/3 (33%) chance we picked the car. There is a 2/3 (66%) chance we picked nothing.
| Door 1 | Door 2 | Door 3 | Staying with Door #1 | Switching door |
|---|---|---|---|---|
| Nothing | Open | Car | Nothing | Wins car |
| Nothing | Car | Open | Nothing | Wins car |
| Car | Open | Nothing | Wins car | Nothing |
The show host is never going to pick to open the door with the prize behind it and wont open the door we picked. So, when they open the only door they can pick, it will always be the empty one. This reduces the open door's probability down to 0%, but our initial choice still only has a 1/3 chance of being right.
So, you can now see how out of the 3 possible scenarios, 2 out of 3 win you the car if you swap, only 1 scenario wins if you stay.
While you wont always win by swapping, you have a much higher chance of winning if you swap doors. 1/3 vs. 2/3.
The only time it would be 50:50 is if you walked in, door 2 was already open and you only had door 1 or 3 to pick from.
And if you get bored and want to play it yourself and/or run a simulation, you can play the Monty Hall game yourself :)
@Eeples: I'm not sure what gender has to do with it (though a couple of those respondents didn't seem to like that a female was smarter than them).
I think she explained it really well. After reading her response, @Quantumcat's post makes sense. :)
the key is what the host chooses to reveal.. if he says "I'll now show you what's behind a random door" x998.. the chances just reduce to 50/50 on the last go and it's no longer in accordance with the original Monty Hall problem.
If he says strictly says "I'll now reveal a door WHICH IT IS NOT BEHIND AND IS NOT THE DOOR YOU PICKED" x998… only THEN should you definitely switch as the chances that you picked the right one is still 1/999 whereas the remaining door is 998/999..
There's a hidden inference.. as it would be a less entertaining game show with scenario A which would likely end up opening either the door containing the car or your door at go 499.. A game show would leave the excitement till last likely be scenario B, even though it won't be strictly worded that way
Yeah the Monty Hall problem took a long time to click with me. It's very psychologically challenging I think.
I think the only time it clicked for me was instead of explaining it, someone drew me a picture of how it works. Once it was visualised, I found it easier to get my head around, but reading or listening to someone explain it, it's just going over my head.
The Monte hall problem comes down to the fact that your selection is random, but the hosts selection is made with perfect knowledge. The odds change after his selection because you know that he selects a dud door with certainty.
Agree also. Someone’s made a boo-boo with their answer selections, I think.
Exactly. First part is 1:3. exactly 4 parts
Second part is 1:2. exactly 3 parts.
But what is the chance of any of the kids sharing chocolate. Zero.
Why is Jacob touching Daniel's chocolate?
.325 None of the above lol
It's ALL in the wording. You can easily misunderstand it as "one-third" instead of "1-to-3", or as "one-part-out-of-four". OP maybe take a photo of the sheet so we can see it, since a single word or character can make a difference in the meaning.
Example/
650g x 1/3 = 217g each for Jacob and Daniel and a third-party
217g x 1/2 = 108g each to Jacob and Antonia
Example 2:
650g x 1/3 = 216.7 Daniel's Portion
650g x 2/3 = 433.3 Jacob Initial Portion
433.3 x 1/2 = 216.7 Jacob Last Portion
433.3 x 1/2 = 216.7 Antonia Last Portion
Example 3=
650g 1/4 (Daniel 1) = 162.5
650g 3/4 (Jacob 1) = 487.5
487.5 1/3 (Jacob 2) = 162.5
487.5 2/3 (Antonia) = 325
You can't misunderstand it as 1 part to 3 parts is what 1:3 means.
See synergy's comment below. This maths problem is worded poorly by the teachers because they want the students to suffer instead of learning.
@Kangal: Definitely worded badly but that is in relation to the order things are happening in rather than to whether it is a ratio or a fraction
@Kangal: The real redpill is learning that this sort of poor wording of problems continues right up through high school, and then into university too, and then (entertainingly enough) well into academic/professional settings. Often you're 99% of the way to solving a problem if you can just figure out what the heck it is that you're even being asked.
@Kangal: These sorts of questions are examples of maths teachers demonstrating their total lack of English composition skills.
@ganchan: Those that can't do, teach.
And clearly the person who wrote this can't do anything properly, English nor maths.
you're not helping ;)
325 still isn't one of he options though. I think that's the real issue.
650/4 = 162.5g.
From the first split, Daniel gets 162.5g, Jacob gets 487.5g (162.5 x 3)
For the second split, 487.5g/3 = 162.5g
Jacob gets 162.5g and Antonia gets 325g.
No idea how they're getting 390g?
Let's work it backwards…
390 x 2 = 780grams
…is it possible in any way that OP read "0.78kg" instead as "0.65kg" ??!
I think the question is faulty. Maybe it's not split twice, but they are simple ratios of the total.
edit: see comment below by synergy, he's solved it. Very bad wording on teachers part.
Daniel O
Jacob OOO
Antonia OOOOOO
10 pieces of chocolate -> "O", Antonia has 6 pieces.
650 / 10 * 6 = 390g
This is a 'solution' but the phrasing of the question is awful if this is the intended answer.
If they said the ratio of each of the shares of the chocolate between the 3 people is so, then this is the right answer
Even then, the phrasing is wrong. I think the OP has possibly typed this out inaccurately.
No, the selective school test is like this.
Daniel and Jacob split it 1:3 and Jacob and Antonia split it
It says "and" not "then" on purpose, meaning both statements are for the whole block - not two sequential events.
@supersour: Think about it from the perspective of the three individuals who have the bar in front of them. How does Daniel and Jacob know in advance how much the second split was going to be? How does it work if this was a real life situation? Also, the fact that Jacob is part of two splits and there is only one bar, physically requires it be a sequence. You can't simply apply pure maths interpretation here.
There's a third interpretation that the wording invites: the definition of 'it'. They say split 'it' as if the same original bar is being split separately twice.
It's not stated that Daniel or Jacob took their split. Or Jacob/Antonia. So you could say Daniel took 1/4 of the bar, Jacob took 3/4, and Jacob took another 1/3, and Antonia took 2/3.
It doesn't make sense because the question is ambiguous.
@ozbargainsam: Yeah the order of operations is not intuitive.
My initial calculation came out same as ozbargainsam's - 325g for Antonia, 162.5g each for Daniel & Jacob. Calculation being:
Danial/Jacob = 1:3 split
Daniel = 1/4 = 162.5g
Jacob = 3/4 = 487.5g
Jacob/Antonia = 1:2 split
Jacob = 1/3 of his 3/4 = 1/4 = 162.5g
Antonia = 2/3 of Jacob's 3/4 = 2/4 = 1/2 = 325g
But if we work backwards from Antonia, we get:
Jacob/Antonia = 1:2 split
Jacob = 1
Antonia = 2
Daniel/Jacob = 1:3 split
Daniel = 1
Jacob = 3
If we take Jacob's 3 back up to Antonia's split, we get:
Jacob/Antonia = 1:2 split
Jacob = 3
Antonia = 6
So in total:
Daniel = 1/10 = 65g
Jacob = 3/10 = 195g
Antonia = 6/10 = 390g
@ozbargainsam: It’s not ambiguous. Split it a verb, not a noun. The answer to this question is 325g. The teacher thought they were talking about “shares”, but the teacher doesn’t understand words.
@supersour: No, the question is ambiguous- at best. “And” has a sequential sense as well. Or are you saying that “I arrived home and had a bath” means the same as “I had a bath and arrived home”?
@AddNinja: English is a pretty ambiguous language.
If only there was a symbolic language that tried to get rid of as much ambiguity as possible by using strict definitions and clear logical steps of deduction. Oh well, a man can dream.
@whatwasherproblem: I know - but even if someone managed to invent that, some dope of an examiner would write the test questions in English anyway :)
@supersour: As a conjunction "and" can mean one thing follows another. If they're trying to claim than AND does not indicate sequential events, they are wrong and need to read a dictionary.
https://dictionary.cambridge.org/dictionary/english/and
"used to join two parts of a sentence, one part happening after the other part:"
• I got dressed and had my breakfast.
https://www.collinsdictionary.com/dictionary/english/and
"You use and to link two statements about events when one of the events follows the other."
• I waved goodbye and went down the stone harbour steps.
• I pulled the door closed and did up my boots on the landing.
https://www.macmillandictionary.com/dictionary/british/and
"used for showing that one thing happens after another"
• He switched off the television and went to bed.
Did you make up your own question to suit the answer, or did I miss something.
That looks like " If Daniel and Jacob split it 1:9 then Jacob and Antonia split it 3:6, how much chocolate did Antonia get.".
I totally acknowledge I switched "then" for "and" for clarity. What's the alternative? Are they in parallel universes simultaneously sharing the same chocolate?
This how I did it too to reach 390g
1:3
Then in the next one 1:2 - so the 1 for Jacob = 3
Then you see that Antonia is double that of Jacob = 6
And finally the 1 goes to Daniel
Need to divide it in to 10 parts (1 to Daniel, 3 to Jacob, 6 to Antonia), so 65g each. Antonia gets 6x65g = 390g
Correct. Poorly worded question as some are understandably implying the 1:3 happens first, and then the 1:2. It is talking about relative ratios.
We know Jacob has 3x Daniel’s amount, and Antonia has 2x Jacob’s amount. Therefore Antonia has 6 times Daniel’s amount (2x3).
Daniel has 1 part
Jacob has 3 parts (1:3 vs. Daniel)
Antonia has something over Daniel and Jacob so get 6 parts (1:2 vs. Jacob)
10 parts in total (1+3+6)
1 part = 65g (650g /10)
Antonia has 6*65g = 390g
ahhh… is it trying to say Jacob gets 3x Daniel and 1/2 Antonia? Because that is not what it says.
I would be telling the student to challenge.
This is a reflection no the quality of teacher's college more so than year 6.
If that is the answer they are aiming for then they have worded the question poorly.
Daniel got ¼ (162.5g) and Jacob got ¾ (487.5g). If Jacob wanted to share his portion with his girlfriend Antonia then that's on him. He doesn't get to steal some of Daniel's pitiful portion of chocolate to feed her.
If that is the answer they are aiming for then they have worded the question poorly.
Yeah agreed it is sh*t wording. Could have just gone with Jacob got 3x as much as Daniel, and Antonia got 2x as much as Jacob, total is 650g.
Yep, that's the way I thought about it. If they said they split it in a ratio of 1:3:6 maybe that would make the question more obvious. It's just promoting you to get those 3 numbers. Once you have then it should be straightforward.
It is poorly worded in that you could read it as the splits happening consecutively, rather than simultaneously.
Ohhhh so it's not two splits
So the way I think I would do it.
Part A (Daniel and Jacob) 1:3
total = 0.65
0.65/4 = 0.1625 (Daniels part)
Jacobs part therefore = 0.4875
Part B (Jacob and and Antonia) 1:2
total = 0.4875
0.4875/3 = 0.1625(jacobs part)
Antonia's part therefore 0.3250
I hate these questions. Fortunately you can derive the answer indisputably using algebra.
Let d, j, and a be the amount in grams that each person got
We know d + j + a = 650
by the ratios, 3d = j (as d/j = 1/3)
and 2j = a (as j/a = 1/2)
it's then just a matter of solving these equations
d + j + a = 650
j/3 + j + a = 650
(a/2)/3 + j + a = 650
a/6 + j + a = 650
and
a/6 + a/2 + a + 650
(1/6 + 1/2 + 1)a = 650
a = 390g
Arrgghhh, most of us misunderstood the question.
I assumed Daniel and Jacob split it 1:3 and THEN Jacob split it with Antonia 1:2
BUT
Daniels and Jacob's split is 1:3 AND Jacob and Antonia's split is 1:2
It was not split again.
So basically 1:3:6 Daniel/Jacob/Antonia
Or in other words 650g/10=65g per part so Antonia gets 6X65g=390g
Not that hard and most year 6s could do that EXCEPT for how badly the question was worded 🤔
Yeh, shit question
A glaring example of how some maths teachers are terrible at retiring intelligible sentences.
Yep. If a primary school teacher could set a bad question what ratio of the students would understand what the hell they were on about?
Yep same here. I think the wording was fairly unambigious that it was split twice, not COULD BE split one of two ways. What a BS question
I even started working it out exactly like synergy and decided I was overthinking it, and after rereading the question ended up with 325g
Sadly this is commonplace in high school level tests. Some of the teachers aren't talented enough to write genuinely challenging questions which will provide the bell-curve result they're aiming for, or, they're constrained by a syllabus which is too 'easy'. So, they resort to petty tactics, writing trick/ambiguous questions, which are low-effort and reliably provide the desired result, without opening the door for complaints that they were outside of the scope.
Half of doing well on final high school tests is understanding what tricks to look for and avoiding them.
F?