Question

A team bought 20 basketballs for a total of Nu 8800. Practice balls cost Nu 400 and official balls cost Nu 600. Use a system of equations to determine how many of each type of ball they bought.​

Answer 1

Given : A team bought 20 basketballs for a total of Nu 8800. Practice balls cost Nu 400 and official balls cost Nu 600.

To Find : a system of equations to determine how many of each type of ball they bought.​

Solution:

Practice balls  bought = x

official balls bought = y

x + y = 20

Practice balls cost  = 400x

official balls cost = 600y

400x + 600y  = 8800

=> 4x + 6y  = 880

=> 2x + 3y = 44

 x + y = 20

2x + 3y = 44

=> y = 4

   x = 16

Practice balls  bought = 16

official balls bought = 4

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Answer 2

SOLUTION

GIVEN

• A team bought 20 basketballs for a total of Nu 8800.

• Practice balls cost Nu 400 and official balls cost Nu 600.

TO DETERMINE

Use a system of equations to determine the number of each type of ball they bought.

EVALUATION

Let the team bought m number of practice balls and n numbers of official balls

Now it is given that the team bought 20 basketballs

So by the given condition

$\sf{m + n = 20 \: \: \: - - - - (1)}$

Again it is also stated that the team bought 20 basketballs for a total of Nu 8800. Practice balls cost Nu 400 and official balls cost Nu 600

So by the given condition

$\sf{400m + 600n =8800 }$

$\sf{ \implies \: 4m + 6n =88 }$

$\sf{ \implies \: 2m + 3n =44 \: \: - - - - (2) }$

Hence the required system of equations are

$\sf{ m + n =20 \: \: \: - - - - (1) }$

$\sf{ 2m + 3n =44 \: \: \: - - - - (2) }$

We now solve for the system of equations

From Equation 1 we get

m = 20 - n

Equation 2 gives

$\sf{ 2(20 - n) + 3n =4 4\: \: \: }$

$\sf{ \implies \: 40 - 2n+ 3n =4 4\: \: \: }$

$\sf{ \implies \: 40 + n =4 4\: \: \: }$

$\sf{ \implies \: n =4 \: \: }$

$\sf{ \implies \: m =20 - 4 = 16 \: \: }$

FINAL ANSWER

Hence the team bought 16 number of practice balls and 4 number of official balls

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