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Digital Riddle

Date: 10/04/2016 at 12:39:11
From: amey
Subject: doubt for word problem expert

The digit in the units place of a number is equal to the digit in the tens
place of half of that number, and the digit in the tens place of that
number is less than the digit in the units place of half of the number by
1. If the sum of the digits of the number is 7, then what is the number?

In my book, I have the solution, but it does not show what math
corresponds to which English word.

Please explain the meaning of the first sentence. How do you create an
equation from that? And which phrase contains the operation?

In general, how do you translate English like this into math?



Date: 10/05/2016 at 10:58:39
From: Doctor Peterson
Subject: Re: doubt for word problem expert

Hi, Amey.

Wow, that's a challenging problem. It's definitely complicated! 

I see one place where my first reading of it was wrong, so I want to make
it clearer.

The phrase "less than ___ by 1" is awkward, with closely related parts
distantly separated. Let me change that to "1 less than ___":

   The digit in the units place of a number is equal to the digit in
   the tens place of half of that number, and the digit in the tens
   place of that number is 1 less than the digit in the units place
   of half of the number. If the sum of the digits of the number is 7,
   then what is the number?

Now we have two different numbers: "a number" and "half of that number."
The latter phrase interferes with the flow of the sentence. It will be
easier to read if we just call the two numbers X and Y. (I won't be using
those as the variables I'm actually solving for, just as temporary
placeholders for readability.)

   The digit in the units place of X is equal to the digit in the
   tens place of Y (where Y = X/2), and the digit in the tens place of
   X is 1 less than the digit in the units place of Y. If the sum of
   the digits of X is 7, then what is X?

Now, the problem is not really about X and Y as a whole, but about
individual digits; and the standard way to do that is to write the number
X as 10a + b, where

    the digit in the tens place of X = a
   the digit in the units place of X = b

We also have a second number, and information about its component digits.
We can call that Y = 10c + d:

    the digit in the tens place of Y = c
   the digit in the units place of Y = d

The fact that Y = X/2 (or X = 2Y) can now be written as

          10a + b = 2(10c + d)         [1]

Now let's replace those phrases in the problem with the variables:

   "b" is equal to "c", and "a" is 1 less than "d." If the sum of
   "a" and "b" is 7, then what is 10a + b?

This is getting a lot clearer now, isn't it? 

We can write some equations in our four variables:

                b = c                  [2]
                a = d - 1              [3]
            a + b = 7                  [4]

We now have four equations in four variables, so we are ready to start
solving.

I would first simplify the big equation:

          10a + b = 20c + 2d           [1']

Next, I would replace c with b from equation [2], and d with a + 1 from
equation [3], to eliminate those variables and keep only new versions of
equations [1] and [4]:

          10a + b = 20b + 2(a + 1)     [1"]
            a + b = 7                  [4]

Simplifying [1"],

         8a - 19b = 2

Replacing "b" with 7 - a from [4],

   8a - 19(7 - a) = 2
              27a = 135
                a = 5
                
That means that b = 7 - 5, hence b = 2.

So the solution is

                X = 10a + b 
                  = 52

Half of this number is 26; and, checking, we see that

   (1) The digit in the units place of 52 is 2; this is equal to the
       digit in the tens place of 26.
   (2) The digit in the tens place of 52 is 5; this is 1 less than the 
       digit in the units place of 26, 6.

So we have the solution.

Notice that the process of understanding the equation was most of the
work, and was not accomplished by just matching up each phrase in the
problem statement with a symbol. That is not how translation between
languages works, and it is not how we translate word problems to
equations. We just focus on understanding the whole thing, step by step,
restating and clarifying until we know exactly what it is saying because
we have written it symbolically.

I would not normally rewrite the whole problem three times as I did here
(though it did help a lot to do so, and may be necessary in complex
problems like this one). I might mark up the paragraph, underlining parts
and writing in the variables I want to use for them, and then writing the
equations carefully.

I should also mention that there are probably many ways to solve this. For
example, I could have used only two variables rather than four. Even the
way I solved it on paper was different from what I did when I typed it up.

And the only way you will learn to be confident in solving these problems
is to keep trying, so that you gain experience. There are no magic methods
that will let you automatically translate a problem into equations without
thinking. So in the future, when you have a question like this, please
show us your own attempt, even if you are sure it is wrong. You will
benefit much more from doing that, because you will be developing your own
ways of thinking rather than trying to accumulate "rules" that are not
really yours.

- Doctor Peterson, The Math Forum at NCTM
  



Date: 10/05/2016 at 11:38:24
From: amey
Subject: Thank you (doubt for word problem expert)

Thank you, sir, for explaining this.
Associated Topics:
Middle School Equations
Middle School Word Problems

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