Question

In front of a store, there is a row of parking spaces. Cars park parallel to one another, with the front of each car facing the store. Currently there are 10 compact spaces and 12 full size spaces. The store owners think they can repaint in the same space to fit 16 compact spaces and 9 full size spaces. The width of each type of space will not change. Are the store owners correct? Explain.

Answer 1

The width used for the car spaces are taken as a multiples of the width of

the compact car spaces.

Correct response:

• The store owners are incorrect

Methods used to obtain the above response

Let x represent the width of the cars parked compact, and let a·x represent the width of cars parked in full size spaces.

We have;

Initial space occupied = 10·x + 12·(a·x) = x·(10 + 12·a)

New space design = 16·x + 9×(a·x) = x·(16 + 9·a)

When the dimensions of the initial and new arrangement are equal, we have;

10 + 12·a = 16 + 9·a

12·a - 9·a = 16 - 10 = 6

3·a = 6

a = 6 ÷ 3 = 2

a = 2

Whereby the factor a < 2, such that the width of the full size space is less than twice the width of the compact spaces, by testing, we have;

10 + 12·a < 16 + 9·a

Which gives;

x·(10 + 12·a) < x·(16 + 9·a)

Therefore;

The initial total car park space is less than the space required for 16

compact spaces and 9 full size spaces, therefore; the store owners are

incorrect.

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