Lesson The tricks to solve some word problems with three and more unknowns using mental math

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The tricks to solve some word problems with three and more unknowns using mental math


In this lesson you will learn the couple of the tricks that allow to solve some word problems with three and more unknowns using mental math.
These problems are formally on systems of linear equations with three and more unknowns,  but the trick allows to solve them without using the general theory.

Problem 1

Alan,  Bob and Cleve weight together  255 pounds.  Bob,  Cleve and Dan weight together  267 pounds.
Cleve,  Dan and Alan weight together  262 pounds.  Dan,  Alan and Bob weight together  257 pounds.
What is the weight of each of the boys?


Solution 1   (easy solution explained "in fingers")

We can easily calculate the total weight of all four boys.  For it,  sum up the given weights of each triple of the boys:

255 lbs + 267 lbs + 262 lbs + 257 lbs = 1041 lbs.

It is three times the total weight of all four boys.  Hence,  the total weight of all four boys is one third of it,  i.e.  1041%2F3 = 347 lbs.

Now,  the Alan's weight is equal to the weight of all four boys  (347 lbs)  minus the weight of the triple which does not include Alan,  i.e.       347 - 267 = 80 lbs.
Similarly,  the Bob's weight is equal to the weight of all four boys  (347 lbs)  minus the weight of the triple which does not include Bob,  i.e.  347 - 262 = 85 lbs.
The Cleve's weight is equal to the weight of all four boys  (347 lbs)  minus the weight of the triple which does not include Cleve,  i.e.             347 - 257 = 90 lbs.
The Dan's weight is equal to the weight of all four boys  (347 lbs)  minus the weight of the triple which does not include Dan,  i.e.                  347 - 255 = 92 lbs.

Answer.  The Alan's weight is  80 lbs,  the Bob's weight is  85 lbs,  the Cleve's weight is  90 lbs, and  the Dan's weight is  92 lbs.


Solution 2   (a bit more formal,  but still easy to understand)

Let  x  be the Alan's weight  (in pounds),  y  be the Bob's weight,  z  be the Cleve's weight and  u  be the Dan's weight.
So,  you have four equations for four unknowns

x + y + z       = 255,                      (1)
      y + z + u = 267,                      (2)
x       + z + u = 262,                      (3)
x + y       + u = 257.                      (4)

Let us sum up left sides and right sides of these equations.  You will get

3(x + y + z + u) = 1041.

Hence
x + y + z + u = 1041%2F3 = 347.            (5).

Now,  subtract the equation  (2)  from the equation  (5).  You will get
x = 80.
Thus the Alan's weight is  80 lbs.  Next,  subtract the equation  (3)  from the equation  (5).  You will get
y = 85.
Thus the Bob's weight is  85 lbs.  Next,  subtract the equation  (4)  from the equation  (5).  You will get
z = 90.
Thus the Cleve's weight is  90 lbs.  As a last step,  subtract the equation  (1)  from the equation  (5).  You will get
u = 92.
Hence,  the Dan's weight is  90 lbs.  The problem is solved,  and you got the same answer as in the  Solution 1.


Problem 2

Jonny has a collection of coins in nickels  (5 cents),  dimes  (10 cents) and the quarters  (25 cents).
The amount of money in nickels and dimes in the collection is of  130  cents.
The amount of money in nickels and quarters is of  285  cents.  The amount of money in dimes and quarters is of  295  cents.
How many coins of each nomination are there in the collection?

Solution 1   (easy solution explained "in fingers")

We can easily calculate the total amount of money in the collection.  For it,  sum up the three given amounts of money:

130 c + 285 c + 295 c = 710 c.

It is doubled the total amount of the money in the collection.  Hence,  the entire collection has the amount of  710%2F2 = 355 c.

Now,  the amount in quarters is equal to the total amount  (355 c)  minus the amount in nickels and dimes,  i.e.  355 - 130 = 225 c.
It means that the number of quarters is  225%2F25 = 9.

Similarly,  the amount in dimes is equal to the total amount  (355 c)  minus the amount in nickels and quarters,  i.e.  355 - 285 = 70 c.
It means that the number of dimes is  70%2F10 = 7.

The amount in nickels is equal to the total amount  (355 c)  minus the amount in dimes and quarters,  i.e.  355 - 295 = 60 c.
It means that the number of nickels is  60%2F5 = 12.

Answer.  There are  12  nickels,  7  dimes and  9  quarters in the collection.


Solution 2   (a bit more formal,  but still easy to understand)

Let  n  be the number of nickels,  d  be the number of dimes,  and  q  be the number of quarters in the collection,  where  n,  d  and  q  are non-negative integers.
So,  you have three equations for three unknowns

5c + 10d           = 130,                      (1)
5c +           25q = 285,                     (2)
        10d + 25q = 295.                     (3)

Let us sum up left sides and right sides of these equations.  You will get

2(5n + 10d + 25q) = 710.

Hence
5n + 10d + 25q = 710%2F2 = 355.            (4).

Now,  subtract the equation  (1)  from the equation  (4).  You will get
25q = 355 - 130 = 225 cents.   Hence,  q = 225%2F25 = 9.   Thus there are  9 quarters in the collection.

Next,  subtract the equation  (2)  from the equation  (4).  You will get
10d = 355 - 285 = 70 cents.   Hence,  d = 70%2F10 = 7.   Thus there are  9 dimes in the collection.

As a last step,  subtract the equation  (3)  from the equation  (4).  You will get
5n = 355 - 295 = 60 cents.   Hence,  n = 60%2F5 = 12.   Thus there are  12 nickels in the collection.

The problem is solved.  You got the same answer as in the  Solution 1.

Problem 3

Alice and Barbara weight together in  50  pounds more than Christi weights.
Alice and Christi weight together in  52  pounds more than Barbara weights.
Barbara and Christi weight together in  54  pounds more than Alice weights.

Find the weight of each of the girls.

Solution

Let  x  be the Alice's weight  (in pounds),  y  be the Barbara's weight,  z  be the Christi's weight.
So,  you have three equations for three unknowns

  x + y - z = 50,                      (1)
  x - y + z = 52,                      (2)
-x + y + z = 54,                      (3)

Let us sum up left sides and right sides of these equations.  You will get

2x + 2y + 2z - x - y - z = 50 + 52 + 54,   or
x + y + z = 156.                      (4)

Now,  distract the equation  (1)  from the equation  (4).  You will get

2z = 156 - 50 = 106.

Hence,  z = 106%2F2 = 53 pounds.  This is the Christi's weight.

Next,  distract the equation  (2)  from the equation  (4).  You will get

2y = 156 - 52 = 104.

Hence,  y = 104%2F2 = 52 pounds.  This is the Barbara's weight.

As a last step,  distract the equation  (3)  from the equation  (4).  You will get

2x = 156 - 54 = 102.

Hence,  x = 102%2F2 = 51 pounds.  This is the Alice's weight.

Answer.  The Alice's weight is  51 lbs,  the Barbara's weight is  52 lbs,  and  the Christi's weight is  53 lbs.

Problem 4

Five baskets contain coconuts.  The first and second baskets together have a total of  52 coconuts.
The second and third baskets have  43 coconuts.  The third and fourth baskets have  34 coconuts.
The fourth and fifth baskets have  30 coconuts,  and the first and fifth baskets have  47 coconuts.
How many coconuts are in each basket?

Solution

Let  x%5Bi%5D  be the number of coconuts in the i-th basket,  i = 1, 2, 3, 4, 5.
Then we have this system of  5 equations in  5 unknowns:
x%5B1%5D + x%5B2%5D = 52,       (1)
x%5B2%5D + x%5B3%5D = 43,       (2)
x%5B3%5D + x%5B4%5D = 34,       (3)
x%5B4%5D + x%5B5%5D = 30,       (4)
x%5B1%5D + x%5B5%5D = 47.       (5)

Add all  5  equations  (1)  to  (5)  (both sides).  You will get
2%2A%28x%5B1%5D+%2B+x%5B2%5D+%2B+x%5B3%5D+%2B+x%5B4%5D+%2B+x%5B5%5D%29 = 206,   or, dividing both sides by 2
x%5B1%5D+%2B+x%5B2%5D+%2B+x%5B3%5D+%2B+x%5B4%5D+%2B+x%5B5%5D = 103.       (6).

Now, add equations  (1)  and  (4) and distract this sum from  (6).  You will get
x%5B3%5D = 103 - 52 - 30 = 21.
So,  there are  21  coconuts in basket  #3.

Next,  from equation  (2)  you can easily get get
x%5B2%5D = 43 - x%5B3%5D = 43 - 21 = 22.
So,  there are  22  coconuts in basket  #2.

Then from equation  (1)  you can easily get
x%5B1%5D = 52 - x%5B2%5D = 52 - 22 = 30.
So,  there are  30  coconuts in basket  #1.

Similarly,  from equation  (3)  you can easily get
x%5B4%5D = 34 - x%5B3%5D = 34 - 21 = 13.
So,  there are  13  coconuts in basket  #4.

Then from  (4)  you have  x%5B5%5D = 30 - x%5B4%5D = 30 - 13 = 17 coconuts in basket #5.

Answer.  30,  22,  21,  13  and  17  coconuts in baskets  1  to  5  respectively.


Solve yourself the following problems.

Problem 5

In a triangle,  the sum of the measures of its first and the second side is in  2 cm  greater than the measure of the third side.
The sum of the measures of its first and the third side is in  4 cm  greater than the measure of the second side.
The sum of the measures of its second and the third side is in  6 cm  greater than the measure of the first side.
Find the perimeter and the measure of each side of the triangle.

Problem 6

In a quadrilateral,  the sum of the measures of its first,  the second and the third side is  12 cm.
The sum of the measures of its first,  the third side and the fourth side is  14 cm.
The sum of the measures of its first,  the third and the fourth side is  16 cm.
The sum of the measures of its second,  the third and the fourth side is  18 cm.
Find the perimeter and the measure of each side of the quadrilateral.


My lessons in this site on determinants of  3x3-matrices and the Cramer's rule for solving systems of linear equations in three unknowns are
    - Determinant of a 3x3 matrix
    - Co-factoring the determinant of a 3x3 matrix
    - HOW TO solve system of linear equations in three unknowns using determinant (Cramer's rule)
    - Solving systems of linear equations in three unknowns using determinant (Cramer's rule)
    - Solving word problems by reducing to systems of linear equations in three unknowns
    - Solving systems of non-linear equations in three unknowns using Cramer's rule
    - Sometime two equations are enough to find three unknowns by an UNIQUE way
    - Two very different approaches to one word problem
    - Solving word problems in three unknowns by the backward method
    - Solving system of linear equation in 17 unknowns
    - Solving system of linear equation in 19 unknowns
    - OVERVIEW of LESSONS on determinants of 3x3-matrices and Cramer's rule for systems in 3 unknowns
under the current topic  Matrices, determinant, Cramer rule  of the section  Algebra-II.

My other lessons in this site on solving systems of linear equations in three unknowns are
    - Solving systems of linear equations in 3 unknowns by the Substitution method,
    - BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method,
    - Solving systems of linear equations in 3 unknowns by the Elimination method  and
    - BRIEFLY on solving systems of linear equations in 3 unknowns by the Elimination method
under the current topic  Matrices, determinant, Cramer rule  of the section  Algebra-II.


Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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