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Preeti Dabral 1 week, 5 days ago

**Given : **Three friends Ravi, Raju and Rohit were doing buying and selling of stationery items in market.

**To Find :** Value of x , y and z

**Solution:**

The price of per dozen of Pen, Notebook and toys are Rupees x. y and z respectively

Ravi purchases 4 dozens of notebooks and sells 2 dozens of pens and 5 dozens of toys and earn 1500

=> 2x - 4y + 5z = 1500

Raju purchases 2 dozens of toy and sells 3 dozens of pens and 1 dozen of notebooks and earn 100

=> 3x + y - 2z = 100

Rohit purchases one dozen of pens and sells 3 dozens of notebooks and one dozen of toys and earn 400

-x + 3y + z = 400

2x - 4y + 5z = 1500 Eq1

3x + y - 2z = 100 Eq2

-x + 3y + z = 400 Eq3

Eq1 + 2Eq3

=> 2y + 7z = 2300

Eq2 + 3Eq3

=> 10y + z = 1300

2y + 7z = 2300 => 10y + 35z = 11500

10y + z = 1300

=> 34z = 10200

=> z = 300

10y + z = 1300

=> y = 100

on solving further** x= 200 , y = 100 and z = 300**

Posted by Sweety Rani 1 week, 5 days ago

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Preeti Dabral 1 week, 5 days ago

- (a) {tex}\frac{dx}{dt}{/tex} = k( 200,000 - x)
- (c) ₹ 1,55,555
- (d) ₹ 180,246
- (a) 200,000 - 150,000{tex}\left(\frac23\right)^{\mathrm t}{/tex}
- (b) {tex}\frac23{/tex}

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Preeti Dabral 1 week, 3 days ago

- (c) Let C(x) be the maintenance cost function, then C(x) = 5000000 + 160x - 0.04x
^{2} - (b) We have, C(x) = 5000000 + 160x - 0.04x
^{2}

Now, C{x) = 160 - 0.08x

For maxima/minima, put C'(x) = 0

{tex}\Rightarrow{/tex} 160 = 0.08x

{tex}\Rightarrow{/tex} x = 2000 - (b) Clearly, from the given condition we can see that we only want critical points that are in the interval [0, 4500]

Now, we have C(0) = 5000000

C(2000) = 5160000 and C(4500) = 4910000

{tex}\therefore{/tex} Maximum value of C(x) would be ₹5160000 - (a) The complex must have 4500 apartments to minimise the maintenance cost.
- (a) The minimum maintenance cost for each apartment woud be ₹1091.11

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