Two companies are competing against one another for a contract to

Question

Two companies are competing against one another for a contract to provide basic vehicle maintenance for another company that maintains a fleet of 300 cars for its sales agents. Jack's Tire and Lube offers on-site maintenance for $42.00 per car.

Care includes basic oil-change services, changing wiper blades and filters as needed, and checking and filling all the car's fluids and tires every 3,000 miles. Major repairs are then referred to a partner company that offers a 20% discount to Jack's clients.

Sam's Service Express offers off-site maintenance for $38.00 per car. Care includes basic oil-changing services, changing wiper blades and filters as needed, and checking and filling all the car's fluids and tires every 3,000 miles. Major repairs are completed by Sam's company at the site where basic maintenance is performed. Sam's offers a 10% discount to contract clients.

Each company requires that a minimum of 100 cars be put on their service. The CEO who must arrange the contracts can see benefits in both service agreements. He has exactly $10,000 to spend on car service contracting. He wants to ensure that the older cars in the fleet are on Jack's service plan. Roughly one-third of the cars in the fleet are more than 2 years old and have high mileage.

How many cars in his fleet should be put on each plan to spend the $10,000 budget exactly? Make only one selection in each column.

Sajjad1994 · Answer

Official Explanation

The correct answer is 125 cars for each company.

More than one-third of the cars in the fleet must be covered by Jack's service plan. This is the only qualifier for how many cars should be on each plan. So at least 100 cars should be on Jack's plan.

Set up an equation for the total cost of the cars on each service plan, which must equal exactly $10,000 : 38$ + 42J = 10,000. Start with the lowest number, and plug in 100 for S. This equals 38(100) + 42J = 10,000. In this case, J would equal a decimal number, 147.62.

Plug in the next highest number, 125, for $ : 38(125) + 427 = 10,000. in this case, J would also equal 125. Both companies would service the same number of cars.

Purnank · Answer

42x + 38y = 10000
Plug values 
x= 125 and y=125 will satisfy.

KarishmaB · Answer

The list of care items and major repairs discounts are given to give a longer question stem. They play no role in the question. After all, we don't know how much major repairs will cost. You are given that he has $10,000 and must spend it EXACTLY on two items - one costing $42 per car and other costing $38 per car. This should make you think of linear equations in 2 variables. 100 cars are old and must go to Jack ($42). 42*J + 38*S = 10,000 Simplifying, we get 21J + 19S = 5000 Using hit and trial to solve this will be complicated here so let's get an approximate value first. J must be at least 100. Put J = 100 and we get 19S = 2900. Upon using the calculator, we see that S is a bit more than 152. 42J = 21*100 = 2100 19S = 19*152 = 2888 This is 12 less than 2900 so we should move 6 cars to J so that the amount spent would be exactly 5000. Hence J = 106 and S = 146 is the first solution for this equation. Next solution will be 106 + 19 = 125 and 146 - 21 = 125. This is available in options. Select 125 for both. Integer Solutions to equation in 2 variables is discussed here: https://anaprep.com/algebra-integer-sol ... variables/

Varshakash · Answer

Thanks Karishma...It helped

Jack's	Sam's	Contracted cars
		100
		125
		130
		155
		180
		237

Two companies are competing against one another for a contract to

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