# SUBSTITUTION METHOD WORD PROBLEMS AND ANSWERS

Substitution Method Word Problems and Answers :

In this section, we will see, how to solve word problems using the method substitution method.

## System of Equations Word Problems Examples

Example 1 :

The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later,she buys 3 bats and 5 balls for Rs.1750. Find the cost if each bat and each ball.

Solution :

Let "x" be the cost of each bat

Let "y" be the cost of  each ball

7 x + 6 y  =  3800 ----- (1)

3 x + 5 y  =  1750 ----- (2)

6y  =  3800 - 7x

y  =  (3800 - 7x)/6

Apply  y  =  (3800 - 7 x)/6 in the second equation

3x + 5(3800 - 7x)/6  =  1750

[18x + 5(3800 - 7x)]/6  =  1750

(18x + 19000 - 35x)/6  =  1750

-17x + 19000  =  1750(6)

-17x + 19000  =  10500

-17x  =  10500 - 19000

-17x  =  -8500

x  =  8500/17

x  =  500

Now, we have to apply the value of x in the equation

y  =  (3800 - 7 x)/6

y  =  [3800 - 7(500)]/6

y  =  (3800 - 3500)/6

y  =  300/6

y  =  50

So, the cost of each bat  =  Rs.500

Cost of each ball  =  Rs.50

Example 2 :

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km,the charge paid is \$105 and for a journey of 15 km, the charge paid is \$155. What are the fixed charge and charge per km ? How much does a person have to pay for traveling a distance of 25 km?

Solution :

Let "x" be the fixed charge

Let "y" be the charge  per km for the distance covered

x + 10 y = 105   ------ (1)

x + 15 y = 155   ------ (2)

x = 105 - 10 y

Substitute x = 105 - 10 y in the second equation

105 - 10 y + 15y  =  155

105 + 5y  =  155

5y  =  155 - 105

5y  =  50

y  =  50/5

y  =  10

x  =  105 -10 (10)

x  =  105 - 100

x  =  5

Therefore the fixed charge is \$5

the charge  per km for the distance covered = \$10

After having gone through the stuff given above, we hope that the students would have understood, substitution method word problems and answers.

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