Mrs. Cho is bringing cans and bottles to a recycling center. For a type A can, she gets 5 cents. For type B can, she gets 10 cents. She knows that she has redeemed at least 11 cans but has no more than 95cents. What are the possible combinations?

One possible solution be -

Mrs. Cho is brings type A cans = 10

Mrs. Cho is brings type B cans =  3

Step-by-step explanation:

Let us assume

Mrs. Cho is brings type A cans = x

Mrs. Cho is brings type B cans = y

As given,

For a type A can, she gets 5 cents and For type B can, she gets 10 cents

As we have x Type A cans and y Type B cans

And she has redeemed  no more than 95 cents

⇒5x + 10y ≤ 95           .......(1)

Also given,

She has redeemed at least 11 cans

⇒ x + y ≥ 11           ........(2)

∴ we get

x + y ≥ 11  

5x + 10y ≤ 95

By applying graphical method , we get

The equations become

5x + 10y = 95       .......(1)

x + y = 11              ........(2)

Table for equation (1) -

x                   y

0                    9.5

19                    0

Table for equation (2) -

x               y

0              11

11              0

The two lines intersect at point (3, 8)

So the possible combinations are in the shaded portion which is covered by both the lines.

One possible combination is ( 10, 3) which lies in the shaded region.

So,

Mrs. Cho is brings type A cans = x = 10

Mrs. Cho is brings type B cans = y = 3

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation: