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Mathematical Equation Solving

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Mathematical Equation Solving

 

 

Table of Contents

Q 1(a). 3

Q 1(b). 3

Q 2. 3

Q 3. 3

Q 4. 4

Q 5. 4

 

Q 1(a)

 

Selling Price : $ 11 per shaver.

Variable Cost per shaver = $ 3

Fixed Cost = $ 100000

In order to break even,

R(X) – C(X) should be equal to zero.

Hence, the equation would be = (11x – 3x – 100000) should be equal to zero.

Solving for X, we get 12,500 units.

 

Q 1(b)

 

If the company sells 15000 units, then it is higher than the break even point of 12,500 units. This would result in a profit.

Hence, Profit = ((11*15000) – (3*15000) – 100000) = $ 20000

 

Q 2

 

Jenny receives $ 8/hour plus 4% sales commission

Masur receives $ 10/hour plus 8% sales commission in excess of $1000.

Both of them work 8-hour days.

We assume X to be the sales amount for which they both earn the same daily amounts.

Hence the equation would be ((8*8) + 0.04X) – ((8*10) + (0.08(X – 1000))) = 0

Solving for X we would get sales of $ 1600 which would be required for both Jenny and Masur to earn the same amount ($ 128)

 

Q 3

 

Charges for Plan I are $ 12/day and $ 0.12/mile.

Charges for Plan II are $ 30/day and no charges for miles.

If 300 miles were to be driven, then the costs for the respective plans would be :

Plan I : $ (12 + (0.12*300)) = $ 48

Plan II : $ 30

Hence, Plan II would be better.

We assume X to be the mileage for which both rates the same. The mileage at which both rates would be equal can be showcased by way of the following equation:

(12 + 0.12X) – 30 = 0

Solving for X, we would get 150.

Hence at 150 miles, the costs for both plans would be the same.

 

Q 4

 

Fixed Cost : $ 35,000

Variable Cost : $ 70/racket

Selling Price : $ 90/racket

Number of Units = 2,500

Hence, Profit = ((2500 * 90) – (70 * 2500) – 35000) = $ 15000.

 

Q 5

 

R = 225X

C = 75X + 600

For Breakeven, R – C should be equal to 0.

Hence,

225 X – (75X + 600) = 0

Solving for X, we get 4.

Hence on selling 4 items, the company will break even.

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