# Solve the following situations mathematically:(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.

To do:

We have to solve the given situations mathematically.

Solution:

(i) John and Jivani together have $45$ marbles.

Both of them lost $5$ marbles each, and the product of the number of marbles they now have is $128$.

Let the number of marbles with John be $x$.

This implies,

Number of marbles with Jivani $=45-x$

Number of marbles John had after losing 5 marbles $= x - 5$

Number of marbles Jivani had after losing 5 marbles $= (45 - x) - 5 = 40 - x$

The product of the marbles they now have $=128$.

Therefore,

$(x - 5)(40 - x) = 128$

$40x-x^2-200+5x = 128$

$x^2 - 45x + 128 + 200 = 0$

$x^2 - 45x + 328 = 0$

$x^2-9x-36x+328=0$

$x(x-9)-36(x-9)=0$

$(x-9)(x-36)=0$

$x-9=0$ or $x-36=0$

$x=9$ or $x=36$

(ii) A cottage industry produces a certain number of toys in a day.

The cost of production of each toy (in rupees) was found to be $55$ minus the number of articles produced in a day.

On a particular day, the total cost of production was Rs. $750$.

Let the number of toys produced in a day be $x$.

This implies,

The cost of production of each toy $= 55 - x$

The total cost of production is the product of the number of toys produced in a day and the cost of production of each toy $=x (55 - x)$

Therefore,

$x(55-x) = 750$

$55x-x^2 = 750$

$x^2-55x+750 = 0$

$x^2-25x-30x+750=0$

$x(x-25)-30(x-25)=0$

$(x-25)(x-30)=0$

$x-25=0$ or $x-30=0$

$x=25$ or $x=30$

The number of toys produced on that day was 25 or 30.