Question

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 of each bat and each ball.

Answer 1

Given :-

The coach of cricket teams buys 7bats and 6 balls for 3800.

Later she buys 3 bats and 5 balls for rs 1750.

To Find :-

Cost of each ball and bat

Solution :-

Let the cost of each bat be = Rs x

Let the cost of each ball be = Rs y

According to the Question,

⇒ 7x + 6y = 3800

⇒ 6y = 3800 - 7x

Dividing by 6, we get

⇒ y = (3800 - 7x)/6 … (i)

⇒ 3x + 5y = 1750

Putting the value of y

⇒ 3x + 5 ((3800 - 7x)/6) = 1750

Multiplying by 6, we get

⇒ 18x + 19000 - 35x = 10500

⇒ -17x =10500 - 19000

⇒ -17x = - 8500

⇒ x = - 8500/-17

⇒ x = 500

Putting this value in equation (i) we get

⇒ y = ( 3800 - 7 × 500)/6

⇒ y = 300/6

⇒ y = 50

Hence the cost of each bat = Rs 500 and the cost of each balls = Rs 50.

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pranav90

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let's bat is x and ball is y.

so,

7x+6y=3800 eq.1

3x+5y=1750 eq.2

by substitute method.

x=3800-6y/7 eq.3

putting x=3800-6y/7 into eq.2

then,3(3800-6y/7)+5y=1750

11400-18y+35y=12250

17y=850

y=850/17

y=50

putting y =50 in eq.1

then,7x+6×50=3800

7x+300=3800

7x=3800-300

7x=3500

x=3500/7

x=500

hope it helps,mark as brainliest

Answer 2

Answer:

Step-by-step explanation:

Let cost of each bat = Rs x and let cost of each ball = Rs y

According to given conditions, we have

7x + 6y = 3800 … (1)

And, 3x + 5y = 1750 … (2)

Using equation (1), we can say that

7x = 3800 − 6y ⇒ x =  $\frac{3800-6y}{7}$

Putting this in equation (2), we get

3 $(\frac{3800-6y}{7})$ + 5y = 1750

⇒ $(\frac{11400-18y}{7}$ + 5y = 1750

⇒  $\frac{5y}{1} - \frac{18y}{7} = \frac{1750}{1} -\frac{11400}{7}$

⇒  $\frac{35y-18y}{7} = \frac{12250-11400}{7}$

⇒ 17y = 850 ⇒ y = 50

Putting value of y in (2), we get

3x + 250 = 1750

⇒ 3x = 1500 ⇒ x = 500  

Therefore, cost of each bat = Rs 500 and cost of each ball = Rs 50