SUBSTITUTION METHOD WORD PROBLEMS AND ANSWERS

Problem 1 :

The coach of a cricket team buys 7 bats and 6 balls for $3800. Later, she buys 3 bats and 5 balls for $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.

Then, 

7x + 6y  =  3800 -----(1)

3x + 5y  =  1750 -----(2)

Solve (1) for y.

6y  =  3800 - 7x

y  =  (3800 - 7x)/6 -----(3)

Substitute y  =  (3800 - 7 x)/6 in (2)

(2)-----> 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

Substitute x  =  500 in (3)

(3)-----> y  =  [3800 - 7(500)] / 6

y  =  (3800 - 3500) / 6

y  =  300/6

y  =  50

So, the cost of each bat is $500 and each ball is $50.

Problem 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 + 10y  =  105 ------(1)

 x + 15y  =  155 ------(2)

Solving (1) for x.

x  =  105 - 10y -----(3)

Substitute x  =  105 - 10y in (2).

(2)-----> 105 - 10y + 15y  =  155

105 + 5y  =  155

5y  =  50

y  =  10

Substitute y  =  10 (3).

(30-----> x  =  105 -10(10)

x  =  105 - 100

x  =  5

Therefore, the fixed charge is $5 and charge per km for the distance covered is $10.

Amount has to be paid for a travel of 25 km is 

=  5 + 25(10)

=  5 + 250

=  $255

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