Liberty's Flame

#1: Roscoe To: no gnu taxes (#0)

It's prbolematic.

Roscoe posted on 2016-05-24 16:52:23 Reply Private Reply

#2: no gnu taxes To: Roscoe (#1)

I wish the edit function worked on Titles.

no gnu taxes posted on 2016-05-24 16:52:53 Reply Private Reply

#3: ConservingFreedom To: no gnu taxes (#0)

It'd take a sharp 4th grader.

Why do you ask?

ConservingFreedom posted on 2016-05-24 16:53:27 Reply Private Reply

#4: no gnu taxes To: ConservingFreedom (#3)

This and other similar problems are on my daughter's 4th grade end of year math test.

I doubt many of the parents could solve this problem.

BTW, this is a public school. Ironically, the private schools here are much easier.

no gnu taxes posted on 2016-05-24 16:58:03 Reply Private Reply

#5: rlk To: no gnu taxes (#0)

John saves $1 on day 1, $3 on day 2, $6 dollars on day 3, 10 dollars on day 4 and so on.
Given this pattern, how long will it take John to save $165?

It depends on the definition of "and so on".

rlk posted on 2016-05-24 17:16:00 Reply Private Reply

#6: TheFireBert To: no gnu taxes (#0)

John saves $1 on day 1, $3 on day 2, $6 dollars on day 3, 10 dollars on day 4 and so on.

1,3,6,10?

What is the pattern? Add $3 to the amount saved each day? 6+3 does not equal 10.

TheFireBert posted on 2016-05-24 17:16:54 Reply Private Reply

#7: ConservingFreedom To: TheFireBert (#6)

1,3,6,10?
What is the pattern?

1 + 2 = 3; 3 + 3 = 6; 6 + 4 = 10.

10 + 5 = 15; 15 + 6 = 21; ...

ConservingFreedom posted on 2016-05-24 17:25:15 Reply Private Reply

#8: no gnu taxes To: ConservingFreedom (#7)

Yeah, comes out to 9 days.

The public schools here are hard because they expect parents to carry half the water for them.

My daughter has at least an hour of homework every night on issues they expect us to teach her.

no gnu taxes posted on 2016-05-24 17:30:48 Reply Private Reply

#9: misterwhite To: TheFireBert (#6)

First you have to solve the pattern:

Day: ..1 .... 2 ... 3 ..... 4 ....... 5 ...
Amt: $1 .. $3 .. $6 .. $10 ... ($15)

The pattern is the previous day's amount plus the day #. Day 6 would be $15 + 6 = $21.

I'd say tough for a 4th grader.

misterwhite posted on 2016-05-24 17:32:01 Reply Private Reply

#10: Vicomte13 To: TheFireBert (#6)

What is the pattern?

The pattern is "Add the number of the day to the total from the day before".

It would be easier to find the answer by mechanical computation than to derive an algebraic expression and solve it.

Day 1 = $1 2 = $3 3 = $6 4 = $10 5 = $15 6 = $21 7 = $28 8 = $36 9 = $45 10 = $55 11 = $66 12 = $78 13 = $91 14 = $105 15 = $120 16 = $136 17 = $153 18 = $171

So, it would take 18 days to save $165 (or $166, or $169, all the way up to $171).

This is a pretty hard problem. You either "see" the pattern outright (which I did) or you spend endless time trying to find some sort of progression.

Worse, the end-point doesn't land on $165, so you have to have the broader sense to recognize that $165 is within the set of solutions for Day 18.

If the number were set, instead at $16,304, one would have to derive a formula. There is a formula for this sort of repeating sequence it resembles an interest problem), so once again one will either remember that formula, or one will have to derive it. It would take me a long time to do that.

This is a bit much for a 9 year old.

Vicomte13 posted on 2016-05-24 17:36:44 Reply Private Reply

#11: no gnu taxes To: Vicomte13 (#10)

It actually comes out to 9 days.

1+3+6+10+15+21+28+36+45=165

no gnu taxes posted on 2016-05-24 17:39:27 Reply Private Reply

#12: BobCeleste To: no gnu taxes (#0)

With the dept of education it is a 7th grade problem, get rid of the Dept of education and it will be a second grade problem.

BobCeleste posted on 2016-05-24 19:40:16 Reply Private Reply

#13: paraclete To: no gnu taxes (#0)

That depends on the child and good teachers apparently some adults here didn't quite get the question which might mean comprehension isn't being taught.

paraclete posted on 2016-05-24 20:28:34 Reply Private Reply

#14: SOSO To: no gnu taxes (#0)

Do you think this is a reasonable math prbolem for a 4th grader?

No, it is basically an algebra problem though it can be solved by brute force addition if the student recognizes the pattern and can add.

SOSO posted on 2016-05-24 21:59:02 Reply Private Reply

#15: Fred Mertz To: SOSO (#14)

Add what?

Fred Mertz posted on 2016-05-24 22:12:44 Reply Private Reply

#16: SOSO To: Fred Mertz (#15)

The sum of numbers starting with 1, 1+2=3, 3+3=6, 6+4=10, 10+5=15, etc. etc. That is an inelegant way for solve the problem that was proposed using just arithmetic. There is a more elegant algebraic approach using mathematics. Both are well above 4th grade level expectations.

SOSO posted on 2016-05-24 23:26:30 Reply Private Reply

#17: SOSO To: Fred Mertz (#16)

The sum of numbers starting with 1, 1+2=3, 3+3=6, 6+4=10, 10+5=15, etc. etc. That is an inelegant way for solve the problem that was proposed using just arithmetic. There is a more elegant algebraic approach using mathematics. Both are well above 4th grade level expectations.

Before you ask the answer is n(n+1)/2.

SOSO posted on 2016-05-24 23:30:25 Reply Private Reply

#18: nolu chan To: SOSO (#17)

Before you ask the answer is n(n+1)/2.

That gives a digit, for n=9 it would yield 45 (9*10/2), correct for the ninth digit in the sequence, but not the sum of the digits.

The formula for the sum is:

f(n) = n(n+1)(n+2)/6

For n=9:

f(9) = 9(10)(11)/6

f(9) = 990/6 = 165

nolu chan posted on 2016-05-25 01:29:30 Reply Private Reply

#19: Vicomte13 To: no gnu taxes (#11)

It actually comes out to 9 days.
1+3+6+10+15+21+28+36+45=165

You're right. I was ignoring the cash already accumulated and just looking at the day's number.

Vicomte13 posted on 2016-05-25 09:35:50 Reply Private Reply

#20: Vicomte13 To: nolu chan (#18)

The formula for the sum is:
f(n) = n(n+1)(n+2)/6

Why divided by 6? That is not intuitive.

Vicomte13 posted on 2016-05-25 09:52:40 Reply Private Reply

#21: no gnu taxes To: Vicomte13 (#20)

I'm certainly not going to through the algebraic manipulations it would take to form an equation. My guess is you would have to solve it like forming the equation for figuring each period payment to pay off an interest loan. You form a general equation, multiply by a factor, and then subtract the original equation. You eliminate all the middle terms, allowing you come up with an equation based on the number of terms.

That is at the very least high school level math.

no gnu taxes posted on 2016-05-25 10:18:05 Reply Private Reply

#22: SOSO To: nolu chan (#18)

That gives a digit, for n=9 it would yield 45 (9*10/2),

Correct.

SOSO posted on 2016-05-25 20:16:53 Reply Private Reply

#23: nolu chan To: Vicomte13 (#20)

Why divided by 6? That is not intuitive.

1*2*3... Not easy to explain.

The sum of the first positive integers 1+2+3 ... (n-2)+(n-1)+(n)

for n=100

1+100
2+99
3+98
[...]
98+3
99+2
100+1

50x101 = 5050

50 × 101 = (100/2) × 101 = (100/2) × (100 + 1)

Substituting n for 100, we get (n/2) × (n + 1) for the SUM of the first n integers.

Or n(n+1)/2

Now let us look at the 4th grade problem (not):

(1)   1     1            1
(2)   3     4           12
(3)   6    10          123
(4)  10    20         1234
(5)  15    35        12345
(6)  21    56       123456
(7)  28    84      1234567
(8)  36   120     12345678
(9)  45   165    123456789

For n=9, there is 9*1, 8*2, 7*3, 6*4, 5*5, 4*5, 3*6, 2*7, and 1*9. A fourth grader might just add them up.

Each of the numbers 1 thru 9 of the sequence is made by the sum of the first n integers, or n(n+1)/2.

For n=4, the sequence number is 10 (1+2+3+4).

For n=4, the sum of n=1 thru n=4 is (1) + (1+2) + (1+2+3) + (1+2+3+4) or 1+3+6+10 = 20

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45

(1)  9*0 + 1 = 1       1
(2)  9*0 + 3 = 3       4
(3)  9*0 + 6 = 6      10
(4)  9*1 + 1 = 10     20
(5)  9*1 + 6 = 15     35
(6)  9*2 + 3 = 21     56
(7)  9*3 + 1 = 28     84
(8)  9*4 + 0 = 36    120
(9)  9*5 + 0 = 45    165

9*16=144 plus 1+3+6 + 1+3+6 + 1 = 21. Total 165.

To get the sequence number, multiply n*n+1 (two factors) and divide by 2 (1*2).

For n=4, multiply 4 x 5 = 20, divide by 2, and the number is 10.

For the sum of all sequence numbers to n, multiply n*n+1*n+2 (three factors) and divide by 6 (1*2*3).

For sum n=4, multiply 4 x 5 x 6 = 120, divide by 6, and the sum is 20.

g(n) = n(n+1)/2

f(n) is the sum of the numbers 1 through n.

So:

f(n) = g(1) + g(2) + ... + g(n)

And:

f(n) = n(n+1)(n+2)/6

nolu chan posted on 2016-05-25 20:30:39 Reply Private Reply

#24: goldilucky To: nolu chan (#23)

goldilucky posted on 2016-05-27 14:58:17 Reply Private Reply

#25: goldilucky To: nolu chan (#23)

This is calculus and is advanced math more suitable for college students. In order for one to understand calculus you have to understand the basic concept of algebra and the FOIL method before you can integrate this into calculus. Therefore, it is not suitable for a fourth grader.

goldilucky posted on 2016-05-27 15:01:43 Reply Private Reply

#26: goldilucky To: nolu chan (#23)

goldilucky posted on 2016-05-27 15:02:35 Reply Private Reply

#27: nolu chan To: goldilucky (#25)

it is not suitable for a fourth grader.

I agree and did sort of signify that I do not see it as a fourth-grade question. It is nothing I ever saw in fourth grade.

Now let us look at the 4th grade problem (not):

nolu chan posted on 2016-05-27 18:01:26 Reply Private Reply

#28: goldilucky To: nolu chan (#27)

I missed that one. Thanks.

goldilucky posted on 2016-05-28 12:49:04 Reply Private Reply

#29: TheFireBert To: misterwhite (#9)

I'd say tough for a 4th grader.

Exactly!

Thinking like a 4th grader I would wonder what the question was getting at. They have their head filled with so much other crap that extrapolating a pattern out of a seemingly random (at first) set of numbers... seems like insanity. And with no algebraic background to back up the higher level of thinking, I would assume they have little basis to solve the problem with.

No wonder they go on to political science or psychology; at least then they can make stuff up and no one will call them out on it!

TheFireBert posted on 2016-05-31 21:28:41 Reply Private Reply

#30: misterwhite To: TheFireBert (#29)

This is a common question on the SATs -- determining the pattern of numbers. But if the kids were being taught patterns in class, this problem isn't that bad.

But throwing this problem out of the blue at a ten-year-old would stump him.

misterwhite posted on 2016-06-01 09:00:33 Reply Private Reply

Source:	None
Published:	May 24, 2016
Author:	me
Post Date:	2016-05-24 16:48:35 by no gnu taxes
Views:	3367