How to solve this type of prob in easy way? (43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷47

Ganesh Ujwal

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How to solve this type of prob in easy way? (43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷47

(43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷ 47 = ?

Right answer is 660 approx.

I should not use calculators etc, I should do this sum in few seconds. what is the trick behind it?
 
(43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷ 47 = ?

Right answer is 660 approx.

I should not use calculators etc, I should do this sum in few seconds. what is the trick behind it?
The trick is to approximate.

I will work this out for you this time.

I would first factor out 57 as 60 (since 57~ 60).

This gives us 60[(43) ÷ (7 X 13) + (18 X 29) ÷ 47]

Now 43 ÷ (7 X 13) = 43 ÷ 91 ~ 45 ÷ 90 = 1/2

(18 X 29) ÷ 47 ~ (15 X 30) ÷ 45 = 10

Now(43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷ 47 ~ 60[(43) ÷ (7 X 13) + (18 X 29) ÷ 47] ~ 60[1/2 + 10]~ 60[11] = 660
 
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The trick is to approximate.

I will work this out for you this time.

I would first factor out 57 as 60 (since 57~ 60).

This gives us 60[(43) ÷ (7 X 13) + (18 X 29) ÷ 47]

Now 43 ÷ (7 X 13) = 43 ÷ 91 ~ 45 ÷ 90 = 1/2

(18 X 29) ÷ 47 ~ (15 X 30) ÷ 45 = 10

Now(43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷ 47 ~ 60[(43) ÷ (7 X 13) + (18 X 29) ÷ 47] ~ 60[1/2 + 10]~ 60[11] = 660

This is the sort of thing I first thought; but nothing is said about how accurate the approximation has to be (or how certain we need to be that it is). It happens that 660 is a very close approximation in this case (659.9978957); is the answer meant to be accurate to the nearest integer, or hundredth, or ten? The kind of approximation you did (and that I would have done) does not guarantee anything better than, say, the nearest ten or hundred, as numbers are being rounded to only one significant digit. (The last step, approximating 1/2 as 1, in effect gave a possible error of 30!) The answer appears to be a matter of luck.

The real question is, what are the real requirements? There are mental math tricks stronger than this sort of approximation, but they take a lot of learning, and probably still take more than a few seconds to do. Doing the calculations by hand to the nearest integer, it probably took me no more than a minute or two. There is no real sense in demanding speed and accuracy at the level apparently expected, even if there are no calculators existing in the country where this is done. And if that is really demanded, then they would be teaching the appropriate methods.
 
This is the sort of thing I first thought; but nothing is said about how accurate the approximation has to be (or how certain we need to be that it is). It happens that 660 is a very close approximation in this case (659.9978957); is the answer meant to be accurate to the nearest integer, or hundredth, or ten? The kind of approximation you did (and that I would have done) does not guarantee anything better than, say, the nearest ten or hundred, as numbers are being rounded to only one significant digit. (The last step, approximating 1/2 as 1, in effect gave a possible error of 30!) The answer appears to be a matter of luck.

The real question is, what are the real requirements? There are mental math tricks stronger than this sort of approximation, but they take a lot of learning, and probably still take more than a few seconds to do. Doing the calculations by hand to the nearest integer, it probably took me no more than a minute or two. There is no real sense in demanding speed and accuracy at the level apparently expected, even if there are no calculators existing in the country where this is done. And if that is really demanded, then they would be teaching the appropriate methods.
Yes, the answer appears to be a matter of luck and a bit of trying to get the answer to be 660. If I had not been given the desired result I would certainly have multiplied the 60 by 10 rather than 11. If speed was really the requirement (which I doubt), then the method I used (and you thought of as your first attempt at solving it) is the best for speed but certainly not for accuracy. I am shocked at how close I got to the exact result with my willy nilly approximations.
 
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Yes, the answer appears to be a matter of luck and a bit of trying to get the answer to be 660. If I had not been given the desired result I would certainly have multiplied the 60 by 10 rather than 11. If speed was really the requirement (which I doubt), then the method I used (and you thought of as your first attempt at solving it) is the best for speed but certainly not for accuracy. I am socked at how close I got to the exact result with my willy nilly approximations.
I suspect the example was contrived in an effort to "motivate" approximation.
 
I suspect the example was contrived in an effort to "motivate" approximation.
I guess, but this was a terrible example to approximate with. I rounded 5 times! One can round off 4*48 to 4*50. I am ok with that, though I prefer students to really use the distribute law and compute, in their heads, 4*(50-2) = 200 - 8 = 192.
 
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I should not use calculators etc, I should do this sum in few seconds. what is the trick behind it?

Ganesh, can you tell us something about the context? Is this for some test, or a contest, or a course, or what? Is the example real, such as from a sample test, or did you make it up? What is said about use of calculators, how long you are allowed, and so on? What reason is given for testing this ability? Is anything taught in your context about methods for rapid calculation or estimation?

Any information like that might suggest what you really need to learn, and what "tricks" we could propose.
 
(43 X 57) ÷ (7 X 13) + (18 X 29 X 57) ÷ 47 = ?



I should do this sum in few seconds.

Let's see:

One one thousand

Two one thousand

Three one thousand

No, after seeing Jomo's suggested solution, which would include the time to organize the steps,
almost as lemmas in one's head and/or on paper, I would state it would take well more than
"a few seconds."

If a mathematical savant were asked for the sum in a "few seconds," then that is a different story.
 
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