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
<h3 /><h3>Methods used to obtain the above response</h3>
Let <em>x</em><em> </em>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 <em>a</em> < 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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Answer:
6-13=7 , 7-13=6
6+1=7,7-1=6,
Step-by-step explanation:
