Question

Amanda went to the store to purchase ink pens. She found three kinds of pens. The first cost $4 each; the price of the second kind was 4 for $1; and the cost for the third kind was 2 for $1. She bought 20 pens and she bought at least one of each kind. (It is possible to buy only 1 of the pens that are "4 for $1" or "2 for $1".) The cost was $20. When she got back to her office, Amanda decided to turn this into a math problem for me. She asked: how many of each kind did I buy?

Answer 1

Answer:

Amanda buy first kind of pen = 3

Amanda buy second kind of pen = 2

Amanda buy third kind of pen = 15

Step-by-step explanation:

Given - Amanda went to the store to purchase ink pens. She found three    

            kinds of pens. The first cost $4 each; the price of the second kind

            was 4 for $1; and the cost for the third kind was 2 for $1. She bought

            20 pens and she bought at least one of each kind. (It is possible to

            buy only 1 of the pens that are "4 for $1" or "2 for $1".) The cost was

            $20.

To find - When she got back to her office, Amanda decided to turn this into

              a math problem for me. She asked: how many of each kind did I

              buy?

Proof -

Let Amanda buy first kind of pen = x

                      second kind of pen = y

                           third kind of pen = z

As given,

She bought total pen = 20

⇒x + y + z = 20          ...............(1)

Now,

As given,

cost for first kind pen = $4 for 1 pen

As she bought x pens of first kind , so

Cost of x pens of first kind = $4x

Now,

The price of the second kind was 4 for $1

⇒Cost of second kind = $$\frac{1}{4}$ for 1 pen

As she bought y pens of send kind , so

Cost of y pens of second kind = $$\frac{1}{4}y$

Now,

The price of the third kind was 2 for $1

⇒Cost of third kind = $$\frac{1}{2}$ for 1 pen

As she bought z pens of send kind , so

Cost of z pens of third kind = $$\frac{1}{2}z$

Now,

As given, The cost was $20

⇒4x + $\frac{1}{4}y$ + $\frac{1}{2}z$ = 20

⇒16x + y + 2z = 80             .....................(2)

∴ we get 2 equations

x + y + z = 20                   .....................(1)

16x + y + 2z = 80             .....................(2)

Now,

Subtract equation (1) from equation (2) , we get

16x + y + 2z  - ( x+ y + z )= 80 - 20

⇒16x + y + 2z - x - y - z = 60

⇒15x + z = 60

⇒z = 60 - 15x

Now,

Put the value of z in equation (1) , we get

x + y + 60 - 15 x = 20

⇒ y - 14x = 20 - 60

⇒y - 14x = -40

⇒14x - y = 40

⇒y = 14x - 40

Now,

we get

z = 60 - 15x

y = 14x - 40

As given

she bought at least one of each kind

it means x > 1, y > 1, z > 1

Now,

If x = 1, then y = 14 - 40 = -26

Not possible

If x = 2 , then y = 14(2) - 40 = -12

Not possible

If x = 3, then y = 14(3) - 40 = 2 and z = 60 - 15(3) = 15

Possible.

If x = 4,  then y = 14(4) - 40 = 16 and z = 60 - 15(4) = 0

Not Possible.

If x = 5, then y = 14(5) - 40 = 30 and z = 60 - 15(5) = -15

Not Possible.

∴ we get

x = 3, y = 2, z = 15

Amanda buy first kind of pen = x = 3

Amanda buy second kind of pen = y = 2

Amanda buy third kind of pen = z = 15