# 2nd grade math problem

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**1**DVD Talk Legend

Thread Starter

**2nd grade math problem**

Without using division how would you come up with the answer?

The perimeter of a rectangle is 48cm. One side is 14cm. How long are the other sides?

In my mind you subtract 48 by 28 and then take half of 20. However if you don't know what half is or how to divide be 2 how would you find the length of the other 2 sides?

The perimeter of a rectangle is 48cm. One side is 14cm. How long are the other sides?

In my mind you subtract 48 by 28 and then take half of 20. However if you don't know what half is or how to divide be 2 how would you find the length of the other 2 sides?

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**3**DVD Talk Legend

**Re: 2nd grade math problem**

2X + 2Y = 48

X = 14

2 x 14 +2Y = 48

28 +2Y = 48

2Y = 48-28

2Y = 20

Y = 20/2

Y = 10.

I had no idea they were teaching algebra to second graders. I wish they had done that back in ye olden tymes when I were a wee tot.

X = 14

2 x 14 +2Y = 48

28 +2Y = 48

2Y = 48-28

2Y = 20

Y = 20/2

Y = 10.

I had no idea they were teaching algebra to second graders. I wish they had done that back in ye olden tymes when I were a wee tot.

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**5**DVD Talk Legend

Thread Starter

**Re: 2nd grade math problem**

There is still a division in there. If they haven’t learned fractions or division I’m not sure how they’d get the answer.

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**6****Re: 2nd grade math problem**

The perimeter of a rectangle is 48cm. One side is 14cm. How long are the other sides?

In my mind you subtract 48 by 28 and then take half of 20. However if you don't know what half is or how to divide be 2 how would you find the length of the other 2 sides?

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**8**DVD Talk Legend

**Re: 2nd grade math problem**

Yeah, it's very simple division but you still need it to divide the remaining 20 cm by two. If there's a way to figure this out without using any division at all I can't think of it.

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**Re: 2nd grade math problem**

4th grade teacher here. Obviously, I can't speak for all teachers, especially 2nd grade teachers, but it's my understanding the intent of this type of question for a 2nd grader is to figure out the lengths of the other two sides by using division but not in the rote sense that an adult might go to like the algebraic equations Jason used. On the last step, the student might ask "What's half of 20?". While this is technically division, it's the concept of fractions' connection to division that's probably the intent of the problem. I'm pretty sure many if not most 2nd graders could figure out half of 20 - at least in my teaching experience.

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**11**DVD Talk Limited Edition

**Re: 2nd grade math problem**

______

14|_____|14

14+14=28

48-28=20

So you have 20 and both sides need to be equal.

1+19=20

2+18=20

3+17=20

. . .

10+10=20 ---here's the answer, because both sides have to be the same since rectangles have opposite sides of the same length.

9 +11

8+12 etc

___10___

14|________|14

10

That's how I'd do it.

14|_____|14

14+14=28

48-28=20

So you have 20 and both sides need to be equal.

1+19=20

2+18=20

3+17=20

. . .

10+10=20 ---here's the answer, because both sides have to be the same since rectangles have opposite sides of the same length.

9 +11

8+12 etc

___10___

14|________|14

10

That's how I'd do it.

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**13****Re: 2nd grade math problem**

wow.. The Flynn effect must be alive and thriving, or all the kids (yes, 7 year-old KIDS) in that class must have IQs in the "very superior" range.

That sounds like a 5th or 6th grade problem to me. I don't think I even learned about perimeter until at least 3rd or possibly even 4th grade, and this was in the "gifted" class in math. Let alone how to compute what amounts to an algebraic problem using unknowns and "division" like this one!

So, they're really pushing things these days, it seems.

Or, teaching by some alien form (core?) which I guess tries to expose kids to as much as possible as early as possible in putting the whole picture of math together.

That sounds like a 5th or 6th grade problem to me. I don't think I even learned about perimeter until at least 3rd or possibly even 4th grade, and this was in the "gifted" class in math. Let alone how to compute what amounts to an algebraic problem using unknowns and "division" like this one!

So, they're really pushing things these days, it seems.

Or, teaching by some alien form (core?) which I guess tries to expose kids to as much as possible as early as possible in putting the whole picture of math together.

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**14**DVD Talk Platinum Edition

**Re: 2nd grade math problem**

Sum of the other two sides = 20.

We know two 14cm sides give us 28, so the unknown side must be less than or equal to 13.

So now just go through 13, 12, 11,...till you get a number which when doubled gives 20.

We know two 14cm sides give us 28, so the unknown side must be less than or equal to 13.

So now just go through 13, 12, 11,...till you get a number which when doubled gives 20.

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**15****Re: 2nd grade math problem**

Simplest way without dividing is to straight up make the rectangle yourself:

You could use graph paper (with two sides known, brief trial and error for the other two) and count out the perimeter.

If you want to explain it with blocks for a more hands on approach: You can lay down 14 and 14 for the top and bottom of the rectangle, and have 28. Then you put one block for each side and count out in pairs (29,30; 31,32) until you get to 48 and both sides “connect”, forming the rectangle. Then you can count how many blocks you used to go down each side. You can do a final check and count all the way around to 48 again.

You could use graph paper (with two sides known, brief trial and error for the other two) and count out the perimeter.

If you want to explain it with blocks for a more hands on approach: You can lay down 14 and 14 for the top and bottom of the rectangle, and have 28. Then you put one block for each side and count out in pairs (29,30; 31,32) until you get to 48 and both sides “connect”, forming the rectangle. Then you can count how many blocks you used to go down each side. You can do a final check and count all the way around to 48 again.

*Last edited by bluetoast; 04-16-18 at 11:30 PM.*

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**16**DVD Talk Hero

**Re: 2nd grade math problem**

The perimeter of a rectangle is 48cm. One side is 14cm. How long are the other sides?

In my mind you subtract 48 by 28 and then take half of 20. However if you don't know what half is or how to divide be 2 how would you find the length of the other 2 sides?

With more folds, works great for division by 4, 8, and so on.

Or by trial and error. You need two equal numbers that add to 20. Guess (wrong), like 8, subtract from 20 to get 12, not quite equal. But you know the number is bigger than 8, smaller than 12, Repeat. You quickly get to 10 and 10. (For more complicated problems, this becomes Newton's method, but that requires calculus and derivatives, and they can't even divide yet)

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**17**DVD Talk Hero

**Re: 2nd grade math problem**

Without any division?

Though I would consider knowing that half of twenty is to ten to barely qualify as division, you could probably physically construct a rectangle out of blocks or tiles, and then count the units out.

ETA: bluetoast beat me to the punch.

Though I would consider knowing that half of twenty is to ten to barely qualify as division, you could probably physically construct a rectangle out of blocks or tiles, and then count the units out.

ETA: bluetoast beat me to the punch.

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**18**DVD Talk Hero

**Re: 2nd grade math problem**

14|_____|14

14+14=28

48-28=20

So you have 20 and both sides need to be equal.

1+19=20

2+18=20

3+17=20

. . .

10+10=20 ---here's the answer, because both sides have to be the same since rectangles have opposite sides of the same length.

9 +11

8+12 etc

___10___

14|________|14

10

That's how I'd do it.

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**19**DVD Talk Legend

Thread Starter

**Re: 2nd grade math problem**

edit: But...you can "show your work" without dividing by 2.

*Last edited by Timber; 04-17-18 at 06:29 AM.*

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**20**DVD Talk Legend

**Re: 2nd grade math problem**

WTF they trying to teach 2nd graders this shit for? For a kid 7 years old, seems like they're trying too hard for college prep.

In my mind, if this is a 48cm rectangle and one side is 14cm, the opposite side has to be 14cm as well. 14+14=28. 48-28=20. The other 2 sides share that 20, so 10 each. Without simple division, and without some complicated fucking algebra (for a 7 year old), I'd have a harder time solving the problem.

In my mind, if this is a 48cm rectangle and one side is 14cm, the opposite side has to be 14cm as well. 14+14=28. 48-28=20. The other 2 sides share that 20, so 10 each. Without simple division, and without some complicated fucking algebra (for a 7 year old), I'd have a harder time solving the problem.

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**21**DVD Talk Godfather

**Re: 2nd grade math problem**

I was thinking blocks like mentioned, or just simply tally marks.sketch out the rectangle then put one tally on one side, then one on the other side. Alternate until you get to twenty

Gotta show your work.

Gotta show your work.

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**22**DVD Talk Special Edition

**Re: 2nd grade math problem**

My son is in the 2nd grade, and with some prodding, I got him to work the problem.

Two of the sides are 14.

He didn't know 14+14, but he does know 15+15 = 30. So 14+14 = 30-2 = 28.

Then 48-28 = 20

And 10+10 = 20, so the other two sides are each 10.

Two of the sides are 14.

He didn't know 14+14, but he does know 15+15 = 30. So 14+14 = 30-2 = 28.

Then 48-28 = 20

And 10+10 = 20, so the other two sides are each 10.

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**25****Re: 2nd grade math problem**

Easy.

Instead of dividing by 2, multiply by 0.5

"2Y = 20"

"(2Y x 0.5) = (20 x 0.5)"

"Y = 10"

You might think this is cheating, but no part of the rules said you can't multiply by half.

Instead of dividing by 2, multiply by 0.5

"2Y = 20"

"(2Y x 0.5) = (20 x 0.5)"

"Y = 10"

You might think this is cheating, but no part of the rules said you can't multiply by half.