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juin [17]
3 years ago
5

The cost of purchasing a bag of rice is partly constant and partly varies inversely as the square root of the number of people d

emanding 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
Mathematics
1 answer:
algol [13]3 years ago
7 0

Answer:

(a) 160 Naira

(b) Undefined

Step-by-step explanation:

Given

Let:

y \to cost of bag of rice

x \to people demanding the bag

So, we have:

y\ = \frac{k}{\sqrt x} + c ---- The variation

y = 100; x = 36

y = 150; x = 144

We have:

y\ = \frac{k}{\sqrt x} + c

When: y = 100; x = 36

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

100 = \frac{k}{6} + c -- (1)

When: y = 150; x = 144

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

150 = \frac{k}{12} + c--- (2)

Subtract (1) from (2)

150 - 100 = \frac{k}{12} - \frac{k}{6} + c - c

50 = \frac{k}{12} - \frac{k}{6}

Multiply through by 12

600 = k - 2k

600 = -k

k = -600

To solve for x, we have:

100 = \frac{k}{6} + c -- (1)

This gives:

100 = \frac{-600}{6} + c

100 = -100 + c

c = 100 + 100

c = 200

So, the equation is:

y\ = \frac{k}{\sqrt x} + c

y = -\frac{600}{\sqrt x} + 200

Solving (1): y; when x = 225

We have:

y = -\frac{600}{\sqrt x} + 200

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

y = -\frac{600}{15} + 200

y = -40 + 200

y = 160

Solving (2): x; when x = 200

We have:

y = -\frac{600}{\sqrt x} + 200

200 = -\frac{600}{\sqrt x} + 200

Collect like terms

\frac{600}{\sqrt x} = 200 - 200

\frac{600}{\sqrt x} =0

Cross multiply

600 =0 * \sqrt x

600 =0

<em>x is undefined</em>

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