The school store opened on the first day of scroll with 56 notebook and 24 pencils. Within two days it sold all of these items. On the first day, twice as many notebooks were souls as pencils. On the second day for every 5 notebooks sold, 2 pencils were old. How many notebooks and how many pencils were sold on each day

Question

The school store opened on the first day of scroll with 56 notebook and 24 pencils. Within two days it sold all of these items. On the first day, twice as many notebooks were souls as pencils. On the second day for every 5 notebooks sold, 2 pencils were old. How many notebooks and how many pencils were sold on each day

Answer 1

To find out how many notebooks and pencils were sold on each day, let's break down the information given in the problem.

On the first day, twice as many notebooks were sold as pencils. Let's represent the number of pencils sold on the first day as 'x'. So, the number of notebooks sold on the first day would be 2x.

On the second day, for every 5 notebooks sold, 2 pencils were sold. Let's assume the number of notebooks sold on the second day is 'y'. So, the number of pencils sold on the second day would be (2/5)y.

Now, let's calculate the total number of notebooks and pencils that were sold:

On the first day: Number of notebooks sold + Number of pencils sold = 2x + x = 3x
On the second day: Number of notebooks sold + Number of pencils sold = y + (2/5)y = (7/5)y

According to the problem, the total number of notebooks sold is 56, and the total number of pencils sold is 24. So, we can set up the following equations:

3x + (7/5)y = 56 (equation 1)
x + (2/5)y = 24 (equation 2)

To solve this system of equations, we can use substitution or elimination method.

If we multiply equation 2 by 5, we get:

5x + 2y = 120 (equation 3)

Now we can solve this system of equations (equations 1 and 3) using the elimination method.

Multiply equation 1 by 5:

15x + (35/5)y = 280

Subtract equation 3 from the above equation:

15x + (35/5)y - (5x + 2y) = 280 - 120
10x + (25/5)y = 160
10x + 5y = 160

Divide the above equation by 5:

2x + y = 32 (equation 4)

We now have a new equation (equation 4) that relates x and y. We can use this equation to solve for either variable.

Since equation 4 has only two variables (x and y), we need additional information to solve for x and y specifically. Could you please provide any additional information?