Poll of the Day > Can you solve this ASIAN Style MATH Problem for 6 Year Olds? What's the answer?

(+) Yeah! (+)

These new Asian style math problems are designed to make it simpler for 6 year old kids to learn Math as it focuses on deepening understanding by using physical objects and pictures to explore traditional math symbols and spending more lesson time developing reasoning and problem-solving skills..

So can you solve this "simple" Math problem? let's see if people will get it right

http://i.dailymail.co.uk/i/pix/2017/08/21/12/436F58E200000578-4809178-image-a-76_1503315116146.jpg
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first equation = 40
second equation = 40
multiply by commen factor
cross out a variable

..
..
I legit had real trouble with this when i was 18 because high school made me stupid
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It's simple algebra, though that is tough for a six year old...


2nd jar has x bugs
1st jar has x+7 bugs
3rd jar has x-3 bugs
All three added together totals 40, so...

x+7 +x +x-3 = 40
3x = 36
x = 12
x + 7 = 19

Darn it, that took me way longer than it should have because I kept thinking 27 after forgetting two whole steps, and I knew that wasn't right.

Here's the way I did it.

0 in third jar, 3 in second jar, and 10 in first jar for 13 total.
Subtract the 13 from the 40 total to know that there is 27 other bugs.
Divide those other bugs by 3 and for 9 add that many to each jar. (Though, the only one you care about is the first.)
10+9=19.
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streamofthesky posted...

It's simple algebra, though that is tough for a six year old...


2nd jar has x bugs
1st jar has x+7 bugs
3rd jar has x-3 bugs
All three added together totals 40, so...

x+7 +x +x-3 = 40
3x = 36
x = 12
x + 7 = 19

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Yeah fuck this one

16 bugs in the first jar if you have eyes
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Whoops, I got 37 because I messed up my algebra and swapped the order of things then I redid it to get the answer.
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that seems super hard for a 6 year old

the jars are an example of the right differences, but don't add up to 40. 13 + 6 + 3 = 22. (oops, never counted the first jar, it actually has 16, but 13 + 6 + 3 still has the right differences.) so we need 40 - 22 = 18 more bugs. distributed evenly would be 18 / 3 = 6 more bugs for each jar, which makes it 19 + 12 + 9 = 40.
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I got my answer by telling my calculator to do rref on a 3-by-4 matrix. I really liked linear algebra.
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streamofthesky posted...

It's simple algebra, though that is tough for a six year old...

Pretty much this. I can't imagine anyone in my first grade class having understood this. I might have understood if my teacher tried to explain it, since I was really good at math, but this still might have been more advanced than I could've handled at the time.
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The question and picture don't match up. You basically just have to plug and play numbers. Since the first jar has 7 more than the second and the third jar has 3 less than then second, this answer works perfectly:
first:19
second:12
third:9
19+12=31
31+9=40

However, if you base it on the jars, none of the answers are correct and the verbal expressions aren't accurate. There are 16 bugs in the first jar, 6 in the second, and 3 in the third. That does NOT equal 40, it equals 25. If you take 25 from 40, that leaves 15 left not accounted 4, which would logically require giving them amount to equal the 40 while following the verbal expressions. This adds 4 to third for 19, 6 to second for 12, and 6 to first for 9, which is 6+6=12+3=15

In other words, there are multiple ways to do it, but plugging in numbers is the easiest way, once you know the final result, but only after making the expressions stated by the verbal expressions..

3. there are clearly 3 in the jar.
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also its 19. Surprisingly didn't take me as long as i thought it would at 6am.

just pick a low number like 15. 7 less than that? 9, 3 less than that? 6.

15. 9. 6.

Just increase each one by 1 until you reach a total of 40.
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Ragnawind posted...

There are 16 bugs in the first jar,

hah, I didn't count the first jar and assumed it was 13. so the picture doesn't even have the right differences and is just completely misleading.
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Ragnawind posted...

The question and picture don't match up. You basically just have to plug and play numbers. Since the first jar has 7 more than the second and the third jar has 3 less than then second, this answer works perfectly:
first:19
second:12
third:9
19+12=31
31+9=40

However, if you base it on the jars, none of the answers are correct and the verbal expressions aren't accurate. There are 16 bugs in the first jar, 6 in the second, and 3 in the third. That does NOT equal 40, it equals 25. If you take 25 from 40, that leaves 15 left not accounted 4, which would logically require giving them amount to equal the 40 while following the verbal expressions. This adds 4 to third for 19, 6 to second for 12, and 6 to first for 9, which is 6+6=12+3=15

In other words, there are multiple ways to do it, but plugging in numbers is the easiest way, once you know the final result, but only after making the expressions stated by the verbal expressions..

How old or dumb are you that you never learned algebra or systems of equations?
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streamofthesky posted...

It's simple algebra, though that is tough for a six year old...

SmokeMassTree posted...

Yeah fuck this one

16 bugs in the first jar if you have eyes

Not gonna lie at first I read this as there are 7 more bugs in the second than the first and I was very confused.
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You guys think this would be hard for a 6-year-old? In addition to what the problem states, they'd only need to understand that if you add anything to 1 bottle, the same needs to be added to each of the other 2, or simply that adding to 1 bottle is adding 3 times as much to the total (yes, 6-year-olds can learn multiplication). Going that route, the picture provided makes it even easier to solve.
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SmokeMassTree posted...

Yeah fuck this one

16 bugs in the first jar if you have eyes

My brotha knows what's up
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AbsoluteDenial posted...

You guys think this would be hard for a 6-year-old? In addition to what the problem states, they'd only need to understand that if you add anything to 1 bottle, the same needs to be added to each of the other 2, or simply that adding to 1 bottle is adding 3 times as much to the total (yes, 6-year-olds can learn multiplication). Going that route, the picture provided makes it even easier to solve.

I don't know many or any 6 year olds that know multiplication, much less within an algebra problem. My 10 year old niece that just graduated elementary would be hopelessly confused. Just explaining that paragraph you wrote to her would make her head spin.

This isn't even an algebra problem. You guys did all this math to solve a problem that didn't need it.
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They messed up, there are 16 bugs in the 1st jar which is 10 more than the 2nd. The total number of bugs is 25, not 40.
It's more of a "don't blindly accept what you're told" question.
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SKARDAVNELNATE posted...

They messed up, there are 16 bugs in the 1st jar which is 10 more than the 2nd. The total number of bugs is 21, not 40.
It's more of a "don't blindly accept what you're told" question.

Ignore the diagram, that threw me off as well. All it does is show proportions, not how to actually solve the problem.
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Dikitain posted...

All it does is show proportions

The proportions it shows are different from those in the problem. How is that suppose to deepen understanding?
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This actually is an easy way for younger students to solve such problems, without having to use algebra, and it's a way that is used early on for many other math problems (Addition/subtraction etc).

In fact we often have counters they can use, along with boxes on their work sheets etc, so they can start with say 4 counters, then take 2 away, and tell how many counters they have left.

In this case you would have a first box with 7 more counters then the second box, which has 3 more counters then the third. Any child who can count would be able to eventually find the answer there. No higher math is needed, just being able to count counters.
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AbsoluteDenial posted...

You guys think this would be hard for a 6-year-old? In addition to what the problem states, they'd only need to understand that if you add anything to 1 bottle, the same needs to be added to each of the other 2, or simply that adding to 1 bottle is adding 3 times as much to the total (yes, 6-year-olds can learn multiplication). Going that route, the picture provided makes it even easier to solve.

6-year-olds are just learning how to do scary word problems and 2-digit addition, and somehow you think they can solve not just algebra, but an algebra word problem?
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Golden Road posted...

AbsoluteDenial posted...

You guys think this would be hard for a 6-year-old? In addition to what the problem states, they'd only need to understand that if you add anything to 1 bottle, the same needs to be added to each of the other 2, or simply that adding to 1 bottle is adding 3 times as much to the total (yes, 6-year-olds can learn multiplication). Going that route, the picture provided makes it even easier to solve.

6-year-olds are just learning how to do scary word problems and 2-digit addition, and somehow you think they can solve not just algebra, but an algebra word problem?

Yes. Kids are quick to learn. If you build up properly and don't limit the difficulty of the problems you'd be surprised at how advanced the math is that they can understand.
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This is why it's stupid to underestimate children, they are just as capable of learning as we are. This is only surprisingly difficult for the age range because we weren't taught as early on about this and instead were left with finger paints.
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so what's the correct answer then, tc?
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second jar = x
first jar = x + 7
third jar = x - 3


if there was 12 bugs in the second jar. then the first jar had 19. third jar would have 9.

Done!
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Kyuubi4269 posted...

This is why it's stupid to underestimate children, they are just as capable of learning as we are. This is only surprisingly difficult for the age range because we weren't taught as early on about this and instead were left with finger paints.

It's kinda' funny to see people who have absolutely no memory from when they were 6. There's a huge middle ground between algebra word problems and nothing but finger painting.

It's not underestimating kids to realize that kids who have difficulty solving "Ann had 8 apples. Ben had 12 apples. How many more apples did Ben have than Ann?" are going to have far more trouble with the word problem in the original post.
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Kyuubi4269 posted...

they are just as capable of learning as we are

They're actually better at learning than we are.

Duck-I-Says posted...

Kyuubi4269 posted...

they are just as capable of learning as we are

They're actually better at learning than we are.

We just have hang ups and conflicting information, when learning from scratch we learn just as well, and on established subjects with a little reassessment of thinking we can match as well.
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Did it all in my head:

3x + 4 = 40
3x = 36
x = 12

So x is the 2nd jar. To get the first jar, add 7 to it. 19.
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thecolorgreen posted...

Did it all in my head:

3x + 4 = 40
3x = 36
x = 12

So x is the 2nd jar. To get the first jar, add 7 to it. 19.

Most 6 year olds have not learned multiplication or division yet:)
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Golden Road posted...

Kyuubi4269 posted...

This is why it's stupid to underestimate children, they are just as capable of learning as we are. This is only surprisingly difficult for the age range because we weren't taught as early on about this and instead were left with finger paints.

It's kinda' funny to see people who have absolutely no memory from when they were 6. There's a huge middle ground between algebra word problems and nothing but finger painting.
It's not underestimating kids to realize that kids who have difficulty solving "Ann had 8 apples. Ben had 12 apples. How many more apples did Ben have than Ann?" are going to have far more trouble with the word problem in the original post.

In the United States, most children aren't taught any math at all until they are 5 or 6, so of course they'd have difficulty when first starting out. The kids who are solving this problem, however, likely started learning arithmetic much earlier.

I myself learned how to solve single digit multiplication when I was 6, then exponents and the concept of negative integers when I was 7, all just from watching my older sister do her homework. The ability of children to learn new things is incredible, but parents and the education system let that potential go to waste to instead do finger paints and such like Kyuubi said.
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Dikitain posted...

SKARDAVNELNATE posted...

They messed up, there are 16 bugs in the 1st jar which is 10 more than the 2nd. The total number of bugs is 21, not 40.
It's more of a "don't blindly accept what you're told" question.

Ignore the diagram, that threw me off as well. All it does is show proportions, not how to actually solve the problem.

Same here. The diagram probably help imagine the scenario but I had to force myself to not rely on counting the actual insect in the diagram and instead rely on the text.

AbsoluteDenial posted...

In the United States, most children aren't taught any math at all until they are 5 or 6, so of course they'd have difficulty when first starting out. The kids who are solving this problem, however, likely started learning arithmetic much earlier.

Funny enough, my friend who immigrated from China was accused of cheating when he easily breezed through a test because he was taught the material at an earlier age in China compared to his fellow students in the United States.
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wolfy42 posted...

thecolorgreen posted...

Did it all in my head:

3x + 4 = 40
3x = 36
x = 12

So x is the 2nd jar. To get the first jar, add 7 to it. 19.

Most 6 year olds have not learned multiplication or division yet:)

Yes but it's not that you can't teach them it. When you do teach children more advanced math at a younger age they're eager to learn and take longer to be discouraged. We should be teaching these concepts at a younger age.
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I don't recall learning anything like this at 6 years old, but I'm glad they're trying to push the kids harder:

1st Jar
X+7
2nd Jar
X
3rd Jar
X-3

(X+7) + (X) + (X-3) = 40
3X + 4 = 40
3X = 36
X = 12

1st Jar
12 + 7
2nd Jar
12
3rd Jar
12 - 3

1st Jar
19
2nd Jar
12
3rd Jar
9

Why are people bringing variables into this? Just "brute force" it.
Jar 3 has the least number of bugs, start at 1
Jar 2 has 3 more, 4.
Jar 1 has yet 7 more, 11.
11-4-1 shows proper proportion to each other, but not the total sum of bugs.
Sum 16, target 40, difference 24.
Divide the difference by the number of jars, 24 / 3 = 8
Add 8 to each jar, 19-12-9.
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Been ages since I've actually done math problems, so I don't really remember how to actually show the work, plus I'm used to doing basic math in my head anyway, so I just plugged in numbers until I figured out what worked. Got 17, 10, 13, then realized I misread the question and got 19, 12, 9.
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codman4 posted...

I don't recall learning anything like this at 6 years old, but I'm glad they're trying to push the kids harder:

In the West, most aren't taught at an early age. Asia has a high education level mostly due to learning from such a young age in comparison.

Ragnawind posted...

codman4 posted...

I don't recall learning anything like this at 6 years old, but I'm glad they're trying to push the kids harder:

In the West, most aren't taught at an early age. Asia has a high education level mostly due to learning from such a young age in comparison.

They need to make things more difficult for kids in the U.S. I think people have been worrying too much about the kids that fail, and not so much on the kids that don't get enough education.

streamofthesky posted...

It's simple algebra, though that is tough for a six year old...


2nd jar has x bugs
1st jar has x+7 bugs
3rd jar has x-3 bugs
All three added together totals 40, so...

x+7 +x +x-3 = 40
3x = 36
x = 12
x + 7 = 19

This is the best and easiest way to calculate it. But if you don't know how to do that, since its such small numbers, you can just start with a guess and use trial and error and get it in like 2 minutes.