Lesson Miscellaneous evaluating problems

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Miscellaneous evaluating problems


Problem 1

If  a-b = 4  and  ab+c^2+4 = 0,  find a+b.

Solution 

From a-b = 4  you have  a = b+4.   


Sunstitute it into the second equation, replacing "a". You will have


(b+4)*b + c^2 + 4 = 0,   or


b^2 + 4b + c^2 = 0,


(b^2 + 4b + 4) + c^2 = 0


(b+2)^2 + c^2 = 0.


The sum of two squares of real numbers can be equal to zero if and only iff both addends are zeros.


Thus (b+2)^2 = 0,  which implies  b= -2.


Then from very first equation,  a = 4 + b = 4 + (-2) = 2.


Then  a + b = 2 + (-2) = 0,  and the solution is completed.

Problem 2

If   (2006-a)(2004-a)=2005,  find (2006-a)^2 + (2004-a)^2.

Solution 

From  (2006-a)(2004-a)=2005  you have


    2006*2004 - 2006a - 2004a + a^2 = 2005,

    2006*2004 - 2*2005*a + a^2 = 2005,

    -2*2005*a + a^2 = 2005 - 2006*2004.     (1)


Next,  

     (2006-a)^2 + (2004-a)^2 = 2006^2 -2*2006*a + a^2 + 2004^2 -2*2004*a + a^2 = 

   = 2006^2 + 2004^2 -4*2005a + 2a^2 = 2006^2 + 2004^2 + (2*(-2*2005a  + a^2)) = 


        Now replace the expression  (-2*2005a  + a^2)  by 2005 - 2006*2007, based on (1), and then you get


   = 2006^2 + 2004^2 + 2*(2005 - 2006*2004) = (2006-2004)^2 + 2*2005 = 2^2 + 4010 = 4014.


Answer.  4014.

Problem 3

If  a-b = -6  and  b+c = 9,   find the value of   3a%5E2-b%5E2-2c%5E2.

Solution 

If a-b = -6  and  b+c = 9,  then a = -6 + b,  c = 9 - b,


      3a^2 - b^2 - 2c^2 = 3*(-6+b)^2 - b^2 - 2*(9-b)^2 = 

    = 3*(36 - 12b + b^2) - b^2 - 2*(81 - 18b + b^2) = 108 - 36b + 3b^2 - b^2 - 162 + 36b - 2b^2 =

    = combine like terms = -54. 


ANSWER.  If  a-b = -6  and  b+c = 9,  then the value of  3a%5E2-b%5E2-2c%5E2  is  -54.

Problem 4

Evaluate the following sums

    a)   1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203

    b)   300 - 299 + 298 - 297 + ... + 100 - 99

Solution

a)   1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203 

Group the numbers in pairs, using parentheses:


    1 - 3 + 5 - 7 + 9 -11 + ... + 201 - 203 = 


  = (1-3) + (5-7) + (9-11) + . . + (201-203) = 

  =  -2   +  (-2)   + (-2) + . . . + (-2) = 51 times (-2) = 51*(-2) = -102.


b)   300 - 299 + 298 - 297 + ... + 100 - 99 

The same idea works:  Group the numbers in pairs, using parentheses


    300 - 299 + 298 - 297 + . . . +  100 - 99  = 


  = (300-299) + (298-297) + . . .  + (100-99) =


  =    1     +     1     + . . . +     1     =  101 times 1 = 101*1 = 101.

Problem 5

Express the following product as reduced fraction   %281-1%2F2%29%281-1%2F3%29%281-1%2F4%29%2Aellipsis%2A%281-1%2F2008%29.

Solution 

%281-1%2F2%29%281-1%2F3%29%281-1%2F4%29......%281-1%2F2008%29 = %281%2F2%29%2A%282%2F3%29%2A%283%2F4%29%2Aellipsis%2A%282007%2F2008%29 = 


    Notice that the denominator of the first fraction is the numerator of the second fraction.

    Next, the denominator of the second fraction is the numerator of the third fraction.

    This pattern continues: the denominator of the third fraction is the numerator of the fourth fraction.

    And so on . . . 


Cancel all common factors in numerators and denominators to get final answer = 1%2F2008.

Problem 6

Find the value of the product   %281-1%2F2%5E2%29%281-1%2F3%5E2%29%281-1%2F4%5E2%29%2Aellipsis%2A%281-1%2F999%5E2%29.

Solution 

First, it is clear that


    P = %281-1%2F2%5E2%29%281-1%2F3%5E2%29%281-1%2F4%5E2%29%2Aellipsis%2A%281-1%2F999%5E2%29 = 

         %281-1%2F2%29%2A%281-1%2F3%29%2A%281-1%2F4%29%2Aellipsis%2A%281-1%2F999%29*(%281%2B1%2F2%29%2A%281%2B1%2F3%29%2A%281%2B1%2F4%29%2Aellipsis%2A%281%2B1%2F999%29



Now, consider  Q = %281-1%2F2%29%2A%281-1%2F3%29%2A%281-1%2F4%29%2Aellipsis%2A%281-1%2F999%29.


    Notice that the denominator of the first fraction is the numerator of the second fraction.

    Next, the denominator of the second fraction is the numerator of the third fraction.

    This pattern continues: the denominator of the third fraction is the numerator of the fourth fraction.

    And so on . . . 


    Cancel all common factors in numerators and denominators to get the value Q = 1%2F999.



Next, consider  R = %281%2B1%2F2%29%2A%281%2B1%2F3%29%2A%281%2B1%2F4%29%2Aellipsis%2A%281%2B1%2F999%29.


    Notice that the numerator of the first fraction is the denominator of the second fraction.

    Next, the numerator of the second fraction is the denominator of the third fraction.

    This pattern continues: the numerator of the third fraction is the denominator of the fourth fraction.

    And so on . . . 


    Cancel all common factors in numerators and denominators to get the value R = %281%2F2%29%2A1000 = 500.



So, the final answer is  P = Q*R = %281%2F999%29%2A500 = 500%2F999.    ANSWER.

Problem 7

Find the sum of the sequences  8,  88,  888,  . . .  up to  n  terms.

Solution 

So, we want to find the sum


    R = 8 + 88 + 888 + . . . + 888...8,     (1)


where the last number is written using "n" digits of 8.


Consider another, closely related sum


    T = %289%2F8%29%2AR = 9 + 99 + 999 + . . . + 999...9,    (2)


where the last number is written using "n" digits of 9.


Add n to T. It is the same as to add "1" to each term of the sum (2). Therefore, 


    T + n = 10 + 100 + 1000 + . . . + 10^n.      (3)


The sum (3) is the sum of  n  first terms of the geometric progression with the first term 10 and the common ratio of 10.  So, the sum (3) is equal to


    T + n = 10%2A%28%2810%5En-1%29%2F%2810-1%29%29 = %2810%2F9%29%2A%2810%5En-1%29.


Thus you have 


    T = %2810%2F9%29%2A%2810%5En-1%29+-n,   or


    %289%2F8%29%2AR = %2810%2F9%29%2A%2810%5En-1%29+-n,   which implies


    R = %2880%2F81%29%2A%2810%5En-1%29+-+%288%2F9%29%2An.


It is the final formula and the ANSWER :  R = 8 + 88 + 888 + . . . + 888...8 = %2880%2F81%29%2A%2810%5En-1%29+-+%288%2F9%29%2An.


Check.  Let us take n = 2.  Then from one side,  R = 8 + 88 = 96.

                            From the other side,  R = %2880%2F81%29%2A%28100-1%29+-+%288%2F9%29%2A2 = %2880%2F81%29%2A99+-+%288%2F9%29%2A2 = %2880%2A11%29%2F9+-+16%2F9 = %2880%2A11-16%29%2F9 = 864%2F9 = 88.   ! Correct !



        Let us take n = 3.  Then from one side,  R = 8 + 88 + 888 = 984.

                            From the other side,  R = %2880%2F81%29%2A%281000-1%29+-+%288%2F9%29%2A3 = %2880%2F81%29%2A999+-+%288%2F9%29%2A3 = %2880%2A111%29%2F9+-+24%2F9 = %2880%2A111-24%29%2F9 = 864%2F9 = 984.   ! Correct !

Problem 8

Find the sum of all possible  4-digit numbers that can be formed using the digits  2,  4,  5,  6,  7,  and  8,  with no repeated digits.

Solution 

In forming one of those 4-digit numbers, there are 6 choices for the first digit, then 5 for the second, 
4 for the third, and 3 for the fourth.  So, there are 6*5*4*3 = 360  four-digit numbers that can be formed in this way.


Imagine the list of those 360 numbers, ready to be added.  


Consider, for example the digit 4 in the "hundreds" position.
There are  5*4*3 = 60 numbers among all 360 numbers that contain the digit "4" in this position.
    (The factor 5 corresponds to any of 5 remaining digits in the "thousands" position; 
         factor 4 corresponds to any of 4 remaining digits in the "tens" position;
         factor 3 corresponds to any of 3 remaining digits in the "ones" position).


It is true for any of 6 digits in any of four positions: Each of the 6 digits is used 60 times in each of four positions.
It means that in each column of the list of the 360 numbers, each of the given digits is used 60 times.


The sum of the given digits is 2+4+5+6+7+8 = 32.  So the sum of the digits in each column is 60*32 = 1920.


And then the sum of the 360 numbers is


    1920*(1000+100+10+1) = 2133120.    


Answer.  The sum is   2133120.

Problem 9

If   x + y + z = 20  and   x^2 + y^2 + z^2 = 100,  find the value of   xy + xz + yz.

Solution 

Use an identity


    %28x%2By%2Bz%29%5E2 = x%5E2+%2B+y%5E2+%2B+z%5E2+%2B+2xy+%2B+2xz+%2B+2yz.


Substitute the given data. You will get


    20%5E2 = 100 + 2*(xy + xz + yz),


which implies


    xy + xz + yz = %28400-100%29%2F2 = 300%2F2 = 150.     ANSWER


Answer.  Then the value of  xy + xz + yz  is equal to  150.

Problem 10

If   F(x) = ax^2 +bx+c   and   F(x+5) = x^2+9x-7,   find the sum of   a+b+c.

Solution 

You have two sources of information:


    1)  F(x) = ax^2 + bx + c   (a general definition of the polynomial F(x)).

        From it,  a + b + c = F(1).



    2)  You also are given the second piece of information:  F(x+5) = x^2 + 9x - 7.

        To get F(1), you need to take x = -4 in the last identity, and then you will get


        a + b + c = F(1) = F(-4 + 5) = (-4)^2 + 9*(-4) - 7 = 16 - 36 - 7 = -27.



ANSWER.  a + b + c = -27.

Problem 11

For positive numbers  a,  b,  and c,  if 2ab = 1,  3bc = 2,  and 4ca = 3,  find the value of  a + b + c.

Solution 

From the given info, we have these equations

    ab = 1%2F2     (1)

    bc = 2%2F3     (2)

    ac = 3%2F4     (3)


Multiply all these three equations (their left sides and their right sides separately).  You will get

    a^2*b^2*c^2 = %281%2F2%29%2A%282%2F3%29%2A%283%2F4%29 = 1%2F4,

or

    (a*b*c)^2 = 1%2F4.


Take the square root of both sides.  Since "a", "b", and "c" are positive, it gives you

    a*b*c = 1%2F2.   (4)


At this point, you may find "a", "b" and "c" separately.

    To find "a", divide (4) by (2);

    To find "b", divide (4) by (3);

    To find "c", divide (4) by (1).


You will get  a = %28%281%2F2%29%29%2F%28%282%2F3%29%29 = 3%2F4;  b = %28%281%2F2%29%29%2F%28%283%2F4%29%29 = 2%2F3;  c = %28%281%2F2%29%29%2F%28%281%2F2%29%29 = 1.

Now a + b + c = 3%2F4 + 2%2F3 + 1 = 9%2F12 + 8%2F12 + 1 = 17%2F12 + 1 = 29%2F12.    ANSWER

Problem 12

All students in  Ms.  Fay's  Spanish class are either going on the  Spain trip,  the  Mexico trip,  or both.
1/4 of the students going to  Spain are also going to  Mexico,
and  2/7  of the students going on the  Mexico trip are also going to  Spain.
Which of the following could be the total number of students in  Ms.  Fay's  Spanish class?
    (A)   26;     (B)   27;     (C)   28;     (D)   29;     (E)   30.

Solution 

Let x be the number of students going to the Spain trip;
    y                           going to the Mexico trip;
    z                           going to both trips.


We are given

    %281%2F4%29x = z,

    %282%2F7%29y = z.


or, equivalently,

     x = 4z,
    2y = 7z.


The total is  x + y - z = 4z + %287%2F2%29z - z = 3z + %287%2F2%29z.

So we should check the appropriate values of z, if they can produce the values of the option's list.


For short, let's make a Table

    T   A   B   L   E
  -----------------------------

    z		3z + (7/2)z
		
    3		19.5
    4		26
    5		32.5
    6		39
    7		45.5
    8		52


Looking in this Table, we see only ONE opportunity for possible solution: 26 students.


ANSWER.  Under given conditions, the only possible solution is 26 students in the class.

Problem 13

If   2%5Ex = 4%5Ey = 8%5Ez   and   1%2F2x + 1%2F4y + 1%2F8z = 22%2F7,   show that   x = 7%2F16,  y = 7%2F32  and  z = 7%2F48.

Solution 

From the given part  2%5Ex = 4%5Ey = 8%5Ez,  we have

    2%5Ex = 2%5E%282y%29 = 2%5E%283z%29.


It gives us  x = 2y = 3z,  or  y = x%2F2,  z = x%2F3.


Now substitute it into equation

    1%2F%282x%29 + 1%2F%284y%29 + 1%2F%288z%29 = 22%2F7.


You will get

    1%2F%282x%29 + 1%2F%284%2A%28x%2F2%29%29 + 1%2F%288%2A%28x%2F3%29%29 = 22%2F7,

or

    1%2F%282x%29 + 2%2F%284x%29 + 3%2F%288x%29 = 22%2F7,

    4%2F%288x%29 + 4%2F%288x%29 + 3%2F%288x%29 = 22%2F7,

    %284%2B4%2B3%29%2F%288x%29 = 22%2F7,

    11%2F%288x%29 = 22%2F7,

    x = %2811%2A7%29%2F%2822%2A8%29 = 7%2F16.


Then   y = x%2F2 = 7%2F32,  z = x%2F3 = 7%2F48,   QED.


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