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front end estimation 3rd grade math
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Question:Please explain how to estimate by using front end estimation
546234
Do I need to round off to the nearest....? Please help and thanks
Answers:I would loosely say 550  250 = 300; .... or 540  240 = 300 whereas actual answer = 312 ... so semiclose and in the ballpark. if you could get closer, you would probably be able to do the actual problem in your head.
Answers:I would loosely say 550  250 = 300; .... or 540  240 = 300 whereas actual answer = 312 ... so semiclose and in the ballpark. if you could get closer, you would probably be able to do the actual problem in your head.
Question:My daughter has been working in math on estimation and rounding. (3rd grade) She brought home an assignment on Front end estimation.
I understand the concept ex: 789 + 545 = 700 + 500
But why would you use it?
It seems so much more natural to use regular estimation/rounding and it seems like regular estimation and round would give you a more correct estimate.
When would you use this as opposed to regular estimation?
Answers:I have no idea. One of the teachers I worked with even admitted that some of the things that they teach won't be used that much if at all in the real world, and that only certain occupations will use certain things. Call her up and ask her.
Answers:I have no idea. One of the teachers I worked with even admitted that some of the things that they teach won't be used that much if at all in the real world, and that only certain occupations will use certain things. Call her up and ask her.
Question:Hi! My 9 years old son had this problem to solve yesterday as homework:
abcd+bcd+cd+d=1986; find out the number "abcd".
While I quickly found out the number (by trial and error), which is 1459 (1459+459+59+9=1986), I don't have the patience (nor the brains) to easily find the rule which applies in this case. So please someone enlighten me. Thanks! Edit:
ok... everybody says to drop the first digit but this is not the... problem; the problem is how to find the number 1459 in the first place knowing that abcd+bcd+cd+d=1986; so 1459 is not known; it is the answer to the problem; but how do you find it? Edit 2: HOW DO YOU FIND "abcd"? @to denise: I'm sure there must be some sort of formula or trick to solve this. Edit 3:
So... an attempt to formulate a general rule should be that the fourth digit should be divided by four, the third by three, the second by two and the one by one. But of course you should take into consideration the fact that the sum of the numbers may be higher than ten (10). For the number in the problem (1986) the following rule works:
1986:
36/4=9
15/3=5
8/2=4
1/1=1
(three[3] taken from eight and added in front of six works in this case; for some three digit numbers it should be two[2]).
However, you must admit that a 9 year old should probably be close to a genius to formulate a rule and its exceptions for a problem like this. And guessing is not what math should be about. School sucks... Edit 4:
Correction: I meant "the first (digit should be divided) by one". Well... it took me a while but I checked Matt's solution and it is indeed correct. Thanks Matt!
You're the winner! I think the next step will be to ask my son to write this solution down and show it to his teacher (hihihihihihihi  evil, high pitched laughter).
Answers:Well a only shows up once...so let a=1. Or in this case, 1000. Also, d shows in every term, so 4*d has the last digit = 6. Out of 110, only 4 and 9 have the last number equal to 6 when multiplied by 4, being 16 and 46 respectively. First scenario: d=9Second: d=4 1bcd + bcd +cd +9=1986bcd+bcd+cd+4=1986 I remove 1000 from the first term and subtract the remainder on the end bcd+bcd+cd=977bcd+bcd+cd=982 2*(bcd)+cd=9772*(bcd)+cd=982 2*(bc9)+c9=9772*(bc4)+c4=982 3*9=27. 97727=9503*4=12 98212=970 2*(bc0)+c0=9502*(bc0)+c0=970 Divide both sides by 10 to cancel the 0 2*(bc)+c=952*(bc)+c=97 2*b0+3*c=952*b0+3*c=97 2*10*b+3c=952*10*b+3c=97 20*b+3c=9520*b+3c=97 Again, 3*c must have the last number = to 5, or 7 It now branches again. Numbers 1:10 that have a remainder of 5 or a 7 are: 5 for the first, and 9 for the second, so it turns out there's no need to keep branching First scenario againSecond scenario Let c=5c=9 20*b+3*5=9520*b+3*9=97 20*b=8020*b=70 b=4b=70/20. Since b can't equal a number with a decimal place, The first solution must be the correct choice. To be honest, I'm a third year Chemical Engineering student staying up late studying for his transient chemical/thermodynamic process final in 5 hours, and it took me 20+ minutes to think this through, which seems a little ridiculous. I don't think your son's teacher was thinking this particular assignment through too carefully when he/she assigned it. Thanks for the brain exercise! :)
Answers:Well a only shows up once...so let a=1. Or in this case, 1000. Also, d shows in every term, so 4*d has the last digit = 6. Out of 110, only 4 and 9 have the last number equal to 6 when multiplied by 4, being 16 and 46 respectively. First scenario: d=9Second: d=4 1bcd + bcd +cd +9=1986bcd+bcd+cd+4=1986 I remove 1000 from the first term and subtract the remainder on the end bcd+bcd+cd=977bcd+bcd+cd=982 2*(bcd)+cd=9772*(bcd)+cd=982 2*(bc9)+c9=9772*(bc4)+c4=982 3*9=27. 97727=9503*4=12 98212=970 2*(bc0)+c0=9502*(bc0)+c0=970 Divide both sides by 10 to cancel the 0 2*(bc)+c=952*(bc)+c=97 2*b0+3*c=952*b0+3*c=97 2*10*b+3c=952*10*b+3c=97 20*b+3c=9520*b+3c=97 Again, 3*c must have the last number = to 5, or 7 It now branches again. Numbers 1:10 that have a remainder of 5 or a 7 are: 5 for the first, and 9 for the second, so it turns out there's no need to keep branching First scenario againSecond scenario Let c=5c=9 20*b+3*5=9520*b+3*9=97 20*b=8020*b=70 b=4b=70/20. Since b can't equal a number with a decimal place, The first solution must be the correct choice. To be honest, I'm a third year Chemical Engineering student staying up late studying for his transient chemical/thermodynamic process final in 5 hours, and it took me 20+ minutes to think this through, which seems a little ridiculous. I don't think your son's teacher was thinking this particular assignment through too carefully when he/she assigned it. Thanks for the brain exercise! :)
Question:Its for my 6th grade math class. I looked every were but i cant find a definition.
Answers:Front end estimation mostly produces a closer estimate of sums or differences than the answer produced by adding or subtracting rounded numbers. How to estimate a sum by front end estimation: Add the digits of the two highest place values Insert zeros for the other place values Example 1: 4496 + 3745 is estimated to be 8100 by front end estimation (i.e. 4400 + 3700). Example 2: 4496 + 745 is estimated to be 5100 by front end estimation (i.e. 4400 + 700).  Using the leading, or leftmost, digits to make an estimate quickly and easily. Example:Using frontend estimation to estimate the sum of 594, 32, and 221, an initial estimate would be 5 + 0 + 2 hundreds or 700.
Answers:Front end estimation mostly produces a closer estimate of sums or differences than the answer produced by adding or subtracting rounded numbers. How to estimate a sum by front end estimation: Add the digits of the two highest place values Insert zeros for the other place values Example 1: 4496 + 3745 is estimated to be 8100 by front end estimation (i.e. 4400 + 3700). Example 2: 4496 + 745 is estimated to be 5100 by front end estimation (i.e. 4400 + 700).  Using the leading, or leftmost, digits to make an estimate quickly and easily. Example:Using frontend estimation to estimate the sum of 594, 32, and 221, an initial estimate would be 5 + 0 + 2 hundreds or 700.