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. Then the cost of each bat and each ball​

Answer 1

let the bat and the ball be x & y

7x +6y = 3800.....1

3x + 5y =1750......2

YOU CAN APPLY ANY METHOD TO SOLVE THIS LIKE ELIMINATION METHOD, SUBSTITUTE METHOD ETC.

Answer 2

Solution :

Let the cost price of each bat be Rs.r

Let the cost price of each ball be Rs.m

A/q

$\longrightarrow\sf{(Bat's)+(Balls)=3800}\\\\\longrightarrow\sf{7r+6m=3800....................(1)}$

&

$\longrightarrow\sf{(Bat's)+(Balls)=1750}\\\\\longrightarrow\sf{3r+5m=1750...................(2)}$

Using Substitution method :

From equation (1),we get;

$\longrightarrow\sf{7r+6m=3800}\\\\\longrightarrow\sf{7r=3800-6m}\\\\\longrightarrow\sf{r=3800-6m/7......................(3)}$

∴Putting the value of r in equation (2),we get;

$\longrightarrow\sf{3\bigg(\dfrac{3800-6m}{7} \bigg)+5m=1750}\\\\\\\longrightarrow\sf{\dfrac{11400-18m}{7} +5m=1750}\\\\\longrightarrow\sf{11400-18m+35m=12250}\\\\\longrightarrow\sf{11400+17m=12250}\\\\\longrightarrow\sf{17m=12250-11400}\\\\\longrightarrow\sf{17m=850}\\\\\longrightarrow\sf{m=\cancel{850/17}}\\\\\longrightarrow\bf{m=Rs.50}$

∴Putting the value of m in equation (3),we get;

$\longrightarrow\sf{r=\dfrac{3800-6(50)}{7} }\\\\\\\longrightarrow\sf{r=\dfrac{3800-300}{7}} \\\\\\\longrightarrow\sf{r=\cancel{3500/7}}\\\\\longrightarrow\bf{r=Rs.500}$

Thus;

$\boxed{\sf{The\:cost\:price\:of\:each\:bat\:will\:be\:r=\boxed{\bf{Rs.500}}}}}\\\boxed{\sf{The\:cost\:price\:of\:each\:ball\:will\:be\:m=\boxed{\bf{Rs.50}}}}}$