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hi, regardless of adding, subtraction, division, multiply. i can't hardly do able to do it when come to 3 digit and above.
1 to 2 digit is ok for me.
Is there a effective way to train up my mental calculation????

2007-02-27 19:23:50 · 4 answers · asked by noti69000 3 in Science & Mathematics Mathematics

4 answers

There are tons of tecniques. And then you need to practice them.

For addition and subraction i go left to right.

192 + 234
start left. 1 + 2 is 3. So you have 3.
Then 3 + 9 is 12. You know to quickly just add 1 to your 3 to make it 4. So 4 2 is in your head.
finally 2+4 = 6
so 4 2 6 = 426

Works for subtraction

932 - 770
so start with 2 in your head. THen 3 becomes 13
13 - 7 = 6 and you just automatically subtract 1 from your first number.

1 and 6 are the numbers you have now.
then 2-0 =
162


Multiplication and division....works easier for smaller numbers. but take something you can figure out with your 12 tables.

18 * 8. In this case, take the number 18 and divide it by 2. YOu get 9

All you have to figure out is 9*8...which should be commited to memory by now to be 72. Then mulitply that number by 2. Again left to right.
7*2 = 14. 2*2 = 4
144.

Can pretend the number came out to be 78
Same thing. 7*2 = 14. 8*2 =16. so you get 146 +10 is basically what happens in your head really fast.

I got this information math tape back in teh 90's.....about an hour long.....and after some practice, i'm a lot faster doign it this way.

I'm an engineer and found out most of my friends also go left to right insstead of right to left.

2007-02-27 20:29:38 · answer #1 · answered by My name is not bruce 7 · 0 0

If you are interested in fast mental calculation, try Vedic Mathematics, it could be useful also to develop intuitive thinking

2007-02-27 19:48:46 · answer #2 · answered by 11:11 3 · 1 0

the more you practice using your calculator,
the more you perform operations mentally.

2007-02-27 19:40:49 · answer #3 · answered by iyiogrenci 6 · 0 1

practice more and don't rush up....

2007-02-27 20:01:35 · answer #4 · answered by krypton9 1 · 0 0

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