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# Mathematical Equation Solving ## Mathematical Equation Solving

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. 