Question

A sports teacher has to spend rupees 1300 in purchases 10 footballs and 20 hockey sticks from my shop however if he purchases 15 football and 15 hockey sticks the amounts spent in Rupees 1350 the cost of one foot ball is

Answer 1

Answer:

Cost of 1 football: Rs.50

Step-by-step explanation:

make 2 eqn

10f+20h=1300

15f+15h=1350

By solving , you will get f=50,h=40

Answer 2

Final solution:

Footballs price = ₹ 50

Footballs price = ₹ 50 Hockey Stick price = ₹ 40

Step-by-step explanation:

According to given question:

We have to find cost of one football.

Step-1:

Given ,

A sport teacher has ₹1300 for 10 footballs and 20 hockey sticks .

But he purchase 15 footballs and 15 hockey sticks in ₹1350.

Step-2:

Let x for football and y for hockey sticks .

According to the question:

Ist statement:

$10x+20y=₹ 1300 \: \: \: \: \: ....(1)$

2nd statement:

$15 x+15 y = ₹1350 \: \: \: \: \: ...(2)$

Step -3:

Multiply 1st statement by 3 and 2nd statement by 2 ,

The equation becomes

$30 x+60 y = ₹3900 \: \: \: \: \: ....(3)$

$30 x+30 y = ₹2700 \: \: \: \: \: ...(4)$

Now ,

Subtract equation (4) from (3) ,we get

$30 x+60 y - (30x + 30y) = ₹(3900 - 2700)$

Sign also change ,

$30x+60y-30x-30y = \\ ₹(3900-2700)$

$30y = 1200$

Shift the 30 in division toward right hand side ,

$y = \frac{1200}{30}$

Divide the terms ,

$y = 40$

Step-4:

Now substitute the value of y in equation form (1)

$10x+20y=₹ 1300$

$10x+20 \times 40=₹ 1300$

$10x + 800 = 1300$

Shift the term 800 toward right hand side ,

$10x = 1300 - 800$

$10x = 500$

$x = \frac{500}{10}$

$x = 50$

Therefore ,

$x= ₹50 \: and \: y = ₹ 40$

Footballs price = ₹ 50

Hockey Stick price = ₹ 40

Project code #SPJ2

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