WORD PROBLEMS INVOLVING LINEAR EQUATIONS IN THREE VARIABLES

Word Problems Involving Linear Equations in Three Variables :

Here we are going to see some practice questions on linear equations in three variables.

Word Problems Involving Linear Equations in Three Variables - Questions

Question 1 :

Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now ?

Solution :

Let "x", "y" and "z" be the ages of Vani, her father, her grand father respectively.

Average age of them  =  53

(x + y + z)/3  =  53

  x + y + z  =  159 -----(1)

One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. 

(1/2)z + (1/3)y + (1/4)x  =  65

(x/4) + (y/3) + (z/2)  =  65

Taking L.C.M

3x + 4y + 6z  =  65(12)

3x + 4y + 6z  =  780 -----(2)

Four years ago if Vani’s grandfather was four times as old as Vani.

z - 4  =  4(x - 4)

  z - 4  =  4x - 16

4x - z  =  16 - 4

4x - z  =  12  -----(3)

1(6) - (2)

 6x + 6y + 6z  =  954 

 3x + 4y + 6z  =  780

  (-)   (-)   (-)    (-)

----------------------

  3x + 2y  =  174  ---(4)


(1) + (3)

   x + y + z  =  159

  4x + 0y - z  = 12

  -------------------

   5x + y  =  171  ---(5)

(4) - 2(5)

  3x + 2y  =  174

 10x + 2y  = 342

  (-)    (-)    (-)

 ------------------ 

  -7x  =  -168

  x  =  24

By applying the value of x in (5), we get

 5(24) + y  =  171  

 y = 171 - 120

 y = 51

By applying the values of x and y in (1), we get

x + y + z = 159

24 + 51 + z  =  159

z = 159 - 75

z = 84

Hence, Vani's age  =  24, his father's age = 51 and his grandfather's age = 84.

Let us look into the next example on "Word Problems Involving Linear Equations in Three Variables".

Question 2 :

The sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?

Solution :

Let "xyz" be the the required three digit number

x + y + z  =  11  ------(1)

If the digits are reversed, the new number is 46 more than five times the former number.

zyx  =  5(xyz) + 46

100z + 10y + 1x  =  5(100x + 10y + z) + 46

x - 500x + 10y - 50y + 100z - 5z  =  46

-499x - 40y + 95z  =  46 ------(2)

x + 2y  =  z 

x + 2y - z  = 0 ----(3)

40 (1) + (2)

       40x + 40y + 40z  =  440

     -499x - 40y + 95z  =  46 

     ----------------------------

      -459x + 135z  =  486  ----(4)

2 (1) - (3)

      2x + 2y + 2z  =  22

         x + 2y - z   =  0

       (-)   (-)   (+)

     ----------------------  

     x + 3z  =  22  -----(5)

(4) - 45(5)

              -459x + 135z  =  486 

45(5)==>   45 x + 135z  =  990

                (-)    (-)     (-)    (-)

             -----------------------

             -504x = -504  ==> x = 1

By applying x = 1 in (5), we get

1 + 3z  =  22

3z  =  21

z  =  7

By applying x = 1, and z = 7 in (1), we get

x + y + z  =  11

1 + y + 7  =  11

y  =  11 - 8 

y = 3

Hence the required number is 137.

Let us look into the next example on "Word Problems Involving Linear Equations in Three Variables".

Question 3 :

There are 12 pieces of five, ten and twenty rupee currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its value will be increased by ₹20. Find the number of currencies in each sort

Solution :

Let "x", "y" and "z" be the number of pieces of five, ten, and twenty rupee currencies.

x + y + z  =  12  ----(1)

5x + 10y + 20z  =  105

Divide it by 5  ==>  x + 2y + 4z  =  21 ----(2)

10x + 5y + 20z  =  105 + 20

10x + 5y + 20z  =  125

Divide it by 5  ==>  2x + y + 4z  =  25 ----(3)

(1) - (2)

  x + y + z  =  12

x + 2y + 4z  =  21

(-)   (-)  (-)    (-)

-------------------

   - y - 3z  =  -9 ---(4)

2(1) - (3)

  2x + 2y + 2z  =  24

  2x + y + 4z  =  25

  (-)   (-)   (-)    (-)

 --------------------

  y - 2z  =  -1  ----(5)

(4) + (5)

 -y - 3z  =  -9

  y - 2z  =  -1

  -------------

     -5z  =  -10  ==>  z  =  2

By applying z = 2 in (5)

 y - 2(2)  =  -1

  y = -1 + 4  ==> y = 3

By applying the value of y and z in (1)

x + 3 + 2  =  12

x  =  12 - 5

x  =  7

Hence the number of 5 rupee note  = 7, number of ten rupee note  =  3 and number of twenty rupee note = 2.

After having gone through the stuff given above, we hope that the students would have understood, "Word Problems Involving Linear Equations in Three Variables". 

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