Question

The cost of purchasing a bag of rice is partly constant and partly varies inversely as the square root of the number of people demanding the bag. when the cost was 100 naira the number of people were 36 when the cost was 150 naira the number of people were 144. Find

1. The cost when the number of people were 225.
The number of people when the cost was 200 naira

Answer 1

(a) 160 Naira

(b) Undefined

Explanation:

Given

Let:

`y \to` cost of bag of rice

`x \to` people demanding the bag

So, we have:

`y\ = (k)/(\sqrt x) + c` ---- The variation

`y = 100; x = 36`

`y = 150; x = 144`

We have:

`y\ = (k)/(\sqrt x) + c`

When:
`y = 100; x = 36`

`100 = \frac{k}{\sqrt {36}} + c`

`100 = (k)/(6) + c` -- (1)

When:
`y = 150; x = 144`

`150 = \frac{k}{\sqrt {144}} + c`

`150 = (k)/(12) + c`--- (2)

Subtract (1) from (2)

`150 - 100 = (k)/(12) - (k)/(6) + c - c`

`50 = (k)/(12) - (k)/(6)`

Multiply through by 12

`600 = k - 2k`

`600 = -k`

`k = -600`

To solve for x, we have:

`100 = (k)/(6) + c` -- (1)

This gives:

`100 = (-600)/(6) + c`

`100 = -100 + c`

`c = 100 + 100`

`c = 200`

So, the equation is:

`y\ = (k)/(\sqrt x) + c`

`y = -(600)/(\sqrt x) + 200`

Solving (1): y; when x = 225

We have:

`y = -(600)/(\sqrt x) + 200`

`y = -\frac{600}{\sqrt {225}} + 200`

`y = -(600)/(15) + 200`

`y = -40 + 200`

`y = 160`

Solving (2): x; when x = 200

We have:

`y = -(600)/(\sqrt x) + 200`

`200 = -(600)/(\sqrt x) + 200`

Collect like terms

`(600)/(\sqrt x) = 200 - 200`

`(600)/(\sqrt x) =0`

Cross multiply

`600 =0 * \sqrt x`

`600 =0`

x is undefined