# CLASS-8WORD PROBLEM OF SIMULTANEOUS LINEAR EQUATION

Word Problem

Many word problems involving two unknown quantities can be solved by framing simultaneous equations for the unknown quantities.  The method is to represent each unknown quantities by a variable and then to write two linear equations using the conditions mentioned in the problem. Solving the simultaneous equations thus framed gives the required values of the quantities.

There are some examples are given below for your better understanding

Example.1)  The sum of two numbers is 12. Thrice the first number plus four times the second number equals 30. Find the numbers.

Ans.) Let the numbers are x & y.

As per the given condition,  x + y = 12 ………………………………..(1)

and,  3x + 4y = 30…………………………………(2)

now, multiplying the first equation with 3 we find, 3x + 3y = 36 ………………(3)

now we will subtract the 2nd equation from the 3rd equation, we find –

3x + 3y =  36

3x + 4y =  30

-    -      -

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

- y =  6    or      y = - 6

Now we should put the value of y  in the equations  (1)

So,   x + y =  12

Or,   x + (- 6) = 12

Or,   x  =  12 – (- 6) =  12 + 6 = 18

Hence the numbers are – 6 & 18.               (Ans.)

Example.2)  In a two-digit number, the digit in the unit's place is two more than thrice the digit in the tens place. When the digits are reversed, the number obtained is 90 more than the original number. Find the original number.

Ans.) Let the digit at the tens place = x and the digit at the units place = y

Then the number = 10x + y

Given, y = 3x + 2

Or,  y – 3x =  2 ……………………………(1)

The number obtained after reversing the digits = 10y + x

As per the given condition, 10y + x = 10x + y + 90

Or,  10y – y = 10x – x + 90

Or,     9y  =  9x + 90

Or,    y = x + 10          [Divide the equation by 9]

Or,    y – x  =  10 ………………………….(2)

Now, we have to subtract equation (1) from equation (2) and we find –

y – x  =  10

y – 3x =   2

-    +      -

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

2x =  8   or   x = 4

Substituting the value of x in equation number (2) and we find –

y  -  x  =   10

or,  y – 4 =  10

or,  y  =  10 + 4  =  14

so, we find the value x = 4 & y = 14

so, as per the given condition the original number is = 10x + y =  (10 X 4) + 14  =  40 + 14 =  54   (Ans.)

Example.3)  If 10 is added to the numerator of a fraction and 5 is subtracted from its denominator, the fraction equals 2. If 4 is subtracted from the numerator and 4 is added to the denominator, the fraction becomes 2/5. Find the fraction.

Ans.)  Let the fraction is  x/y

x + 10

As per the given condition,   -------------- = 2

y – 5

by cross multiplication  x + 10 =  2 (y - 5)

or,     x + 10 = 2y – 10

or,    x – 2y = - 20 ……………………………………..(1)

x – 4               2

also,   ------------ =  ---------

y + 4               5

by, cross multiplication –

or,    5 (x – 4)  =  2 (y + 4)

Or,    5x – 20 =  2y + 8

Or,    5x – 2y  =  28 …………………………..(2)

Now we multiply equation (1) by 5 and we get

5x – 10y = - 100 ……………………….(3)

Now we will subtract equation (2) from equation (3) and we get –

5x – 10y =  - 100

5x –  2y  =     28

-     +         -

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

- 8y  =  - 128     or    y = 16

Now we will substitute the value of y in the equation (1) and we find –

x – 2y = - 20

or,         x – (2 X 16) = - 20

or,         x -  32 = - 20

or,         x  = 32 – 20  =  12

so, the value of x & y is  12 & 16 respectively

12

so, the desired fraction is =  --------- =  3 / 4           (Ans.)

16

Example.4) The cost of 4 kg of rice and 6 kg of sugar is \$ 240 whereas the cost of 5kg of rice and 10 kg of sugar is \$ 450. Find the cost of rice & sugar per kg.

Ans.)   Let the cost of 1 kg Rice is \$ x and the cost of 1kg sugar is \$ y.

So the equation are,   4x + 6y = 240 ……………………………(1)

5x + 10y =  450 ……………………………(2)

Now we will multiply by 5 with equation (1) and we will multiply by 4 with equation (2) and we find –

5 (4x + 6y) = 240 X 5

Or,      20x + 30y = 1200………………………..(3)

And,         4 (5x + 10y) = 450 X 4

Or,       20x + 40y = 1800…………………………..(4)

Now, we will subtract equation (3) from equation (4) and we get –

20x + 40y =  1800

20x + 30y =  1200

-     -        -

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

10y =  600   or   y  =  60

Now we would like to put the value of y in equation (1) and we find -

4x + 6y =  240

Or,       4x  + (6 X 60) = 240

Or,       4x  +  360  =  240

Or,       4x =  - 360 + 240

Or,       4x  =  - 120

Or,       x  =  - 30  or  30

(because value of any product can’t be negative)

So the cost of rice is \$ 30 / Kg and the cost of sugar is \$ 60/Kg  (Ans.)

Example.5) 24 tickets were sold for a total of \$ 300 at a fair. If an adult ticket cost \$ 10 and a child’s ticket cost \$ 5, how many of each kind of ticket were sold ?

And.)  Let, the number of adult’s ticket = x and the number of child’s ticket is = y

As per the given condition,  x + y = 24 ……………………………..(1)

Since an adult ticket cost \$15 and a child’s ticket cost \$ 5,

So,        15x + 5y = 300

Or,         3x + y = 60………………………….(2)

Now we will subtract equation (1) from equation (2) and we get –

3x + y  =  60

x + y  =  24

-      -     -

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

2x       =  36   or    x = 18

Now we would like to substitute the value of x in equation (1) and we get –

x + y = 24

or,          18 + y = 24

or,            y = 24 – 18 =  6

so, here we can conclude that 18 adult tickets and 6 child tickets were sold.             (Ans.)

Example.6) A collection has 36 one-dollar and two-dollar coins. How many coins of each kind are there in the collection if the total value of coins is \$ 47 ?

Ans.) Let the number of one-dollar coins = x and the number of two-dollar coins = y

The value of one-dollar coins = \$ x

And the value of the two-dollar coins = \$ y

As per the given condition,  x + y = 36 …………………………………(1)

And,                   x + 2y = 47 ……………………………………..(2)

Now we would like to subtract equation (1) from equation (2).

So,                              x  +  2y =  47

x  +   y  =  36

-      -        -

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

y  =  11

now we will substitute the value of y in equation (1) and we get -

x + y = 36

or,             x + 11 =  36

or,             x  =  36 – 11  =  25

so, the number of one-dollar coin is = 25 and the number of two-dollar coin is = 11     (Ans.)

Example.7)  A plane flying with the wind takes 2 hours to make a 900 km trip from one city to another. On the return trip against the wind, it takes two hours and 15 minutes. Find the speed of the plane in still air and the speed of the wind ?

Ans.) Let the speed of the plane in still air = x km/h and the speed of the wind = y km/h

Then the speed of the plane with the wind = (x + y) km/h

And the speed of the plane against the wind = (x – y) km/h

distance

Speed =   --------------

Time

So, as per the given condition –

900km

(x + y) km/h =  --------------  =  450 km

2 hours

So,  (x + y)  =  450 ………………………………….(1)

900 km                               1

And,  (x – y) km/h = ----------------- =  900 km ÷ 2 ------- hours

2 hours 15 minutes                       4

9                        4

=   900 ÷  --------  =   900 X -------- =  400 km

4                        9

So,  (x – y) =  400 ………………………………….(2)

Now we will subtract equation (2) from (1) and we get –

x  +  y  =  450

x  -  y  =  400

-     +       -

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

2y  =  50   or  y  =  25 km/h

Now, we will substitute the value of y in equation (1) and we find the speed of plane -

So,     x + y  =   450

Or,     x + 25  =  450

Or,     x  =  450 – 25  = 425 km/h

So, the speed of the plane in the air and the speed of the wind is 425 km/h and  25 km/h respectively.         (Ans.)