SOLVING WORD PROBLEMS WITH THREE VARIABLES

Problem 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.

Given : Average age of them is 53.

Then, we have

(x + y + z) / 3  =  53

Multiply each side by 3. 

  x + y + z  =  159 -----(1)

Given : 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. 

Then, we have

(1/2)z + (1/3)y + (1/4)x  =  65

(x/4) + (y/3) + (z/2)  =  65

L.C.M of (2, 3, 2) is 12. So, multiply each side by 12.

3x + 4y + 6z  =  65(12)

3x + 4y + 6z  =  780 -----(2)

Given : Four years ago, Vani’s grandfather was four times as old as Vani.

Then, we have

z - 4  =  4(x - 4)

  z - 4  =  4x - 16

4x - z  =  16 - 4

4x - z  =  12 -----(3)

In order to eliminate y, subtract (2) from 4 times of (1). 

4(1) - (2) :

- x + 2z  =  - 144 -----(4)

Add 2 times of (3) and (4).

2(3) + (4) : 

7x  =  168

Divide each side by 7. 

x  =  24

Substitute 24 for x in (3). 

(3)-----> 4(24) - z  =  12

96 - z  =  12

Subtract 96 from each side. 

- z  =  - 84

Multiply each side by (-1). 

z  =  84

Substitute 24 for x and 84 for y in (1). 

(1)----->  x + y + z  =  159

24 + y + 84  =  159

y + 108  =  159

Subtract 108 from each side. 

y  =  51

The values of x, y and z are 

x  =  24

y  =  51

z  =  84

So, Vani is 24 years old, his father is 51 years old and and his grandfather is 84 years old.

Problem 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 x, y and z be the digits in hundreds place, tens place and one place in the three digit number. 

Then, the required three digit number is 

xyz

Given : The sum of the digits 11.

Then, we have

x + y + z  =  11  -----(1)

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

Then, we have

zyx  =  5(xyz) + 46

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

Simplify. 

100z + 10y + 1x  =  500x + 50y + 5z + 46

-499x - 40y + 95z  =  46 -----(2)

Given : If the hundreds digit plus twice the tens digit is equal to the units digit.

Then, we have

x + 2y  =  z 

x + 2y - z  = 0 -----(3)

In order to eliminate y in (1) and (2), subtract (2) from 40 times (1).

40(1) - (2) : 

- 459x + 135z  =  486 -----(4)

In order to eliminate y in (1) and (3), subtract (3) from 2 times (1).

2(1) - (3) : 

x + 3z  =  22 -----(5)

In order to eliminate z in (4) and (5), subtract 45 times of (5) from (4).

(4) - 45(5) : 

- 504x  =  - 504

Divide each side by -504.

x  =  1

Substitute 1 for x in (5). 

(5)-----> 1 + 3z  =  22

Subtract 1 from each side.

3z  =  21

Divide each side by 3.

z  =  7

Substitute 1 for x and 7 for z in (1). 

(1)-----> 1 + y + 7  =  11

y + 8  =  11

Subtract 8 from each side. 

y  =  3

The values of x, y and z are 

x  =  1

y  =  3

z  =  7

xyz  =  137

So, the required three digit number is 137.

Problem 3 :

There are 12 pieces of five, ten and twenty dollar 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)

Given : The total value of the currencies is $108.

Then, we have

5x + 10y + 20z  =  105

Divide each side by 5. 

x + 2y + 4z  =  21 -----(2)

Given : When first 2 sorts are interchanged in their numbers, its value will be increased by $20.

Then, we have

10x + 5y + 20z  =  105 + 20

Simplify. 

10x + 5y + 20z  =  125

Divide each side by 5. 

2x + y + 4z  =  25 -----(3)

In order to eliminate z in (1) and (2), subtract (2) from 4 times of (1).

4(1) - (2) : 

3x + 2y  =  27 -----(4)

In order to eliminate z in (2) and (3), subtract (3) from (2).

(2) - (3) : 

- x + y  =  - 4 -----(5)

In order to eliminate x in (4) and (5), add (4) and 3 times of (5). 

(4) + 3(5) : 

5y  =  15

Divide each side by 5. 

y  =  3

Substitute 3 for y in (4).

(4)-----> 3x + 2(3)  =  27

3x + 6  =  27

Subtract 6 from each side. 

3x  =  21

Divide each side by 3.

x  =  7

Substitute 7 for x and 3 for y in (1). 

(1)-----> 7 + 3 + z  =  12

10 + z  =  12

Subtract 10 from each side. 

z  =  2

The values of x, y and z are 

x  =  7

y  =  3

z  =  2

So, the number of 5 rupee note is 7, number of ten rupee note is 3 and number of twenty rupee note is 2.

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