Trucks

[heshamelaziry]

One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50 percent of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

A: 18
B: 16
C: 12
D: 8
E: 4

I do not now my way around this one !

[grockit_jake]

We know that there are 20 total trucks, so the number at the lot on Saturday is (1/2 the amount rented) + (20 - amount rented).

Algebraically it's:

1/2R + (20 - R) >= 12
-1/2R + 20 >= 12
1/2R <= 8
R <= 16

Or, you can look for the pattern with the answer choices:
10 rented : 5 + 10 = 15
12 rented : 6 + 8 = 14
14 rented : 7 + 6 = 13
16 rented : 8 + 4 = 12.

Notice all the upward and downward patterns...

[antondesh]

I think of it like this:

We start with 20 trucks and x trucks were rented out:
so 20 - x


Then half of the rented trucks were returned at the end of the week:
so 20 - x + x/2


There are at least 12 trucks on the lot at the end of the week.
so 20 - x + x/2 => 12

Solving this equation, we get 16.

For me, the hardest thing in problems like this, is to follow what's happening. So it helps to break it down into a series of steps. Hope this helps!

[heshamelaziry]

We know that there are 20 total trucks, so the number at the lot on Saturday is (1/2 the amount rented) + (20 - amount rented).

Algebraically it's:

1/2R + (20 - R) >= 12
-1/2R + 20 >= 12
1/2R <= 8
R <= 16

Or, you can look for the pattern with the answer choices:
10 rented : 5 + 10 = 15
12 rented : 6 + 8 = 14
14 rented : 7 + 6 = 13
16 rented : 8 + 4 = 12.

Notice all the upward and downward patterns...

How were you able to make reason of adding (20-x)? Could you give any simpler example that is similar to the reasoning of this problem ?

Appreciated

[Stuart@KaplanGMAT]

We know that there are 20 total trucks, so the number at the lot on Saturday is (1/2 the amount rented) + (20 - amount rented).

Algebraically it's:

1/2R + (20 - R) >= 12
-1/2R + 20 >= 12
1/2R <= 8
R <= 16

Or, you can look for the pattern with the answer choices:
10 rented : 5 + 10 = 15
12 rented : 6 + 8 = 14
14 rented : 7 + 6 = 13
16 rented : 8 + 4 = 12.

Notice all the upward and downward patterns...

How were you able to make reason of adding (20-x)? Could you give any simpler example that is similar to the reasoning of this problem ?

Appreciated

Just think of his original equation as:

the number at the lot on Saturday is (1/2 the number rented) + (the number not rented)

Since there are 20 trucks to start, if we call the number rented "x", the number not rented is "20-x".

[vivek.kapoor83]

solving backwards took me less than 15 sec to figure out the ans and i think that is the fastest approach in this ques.

[Simon]

All the above were good explanations
.
Another way to look at this problem is as follows.

Monday we had 20 and Saturday they were reduced to 12, so 8 were missing.

The question states that 50% were returned, so 50% should be outside.

If 50% (outside) were 8, then 100% outside or once outside must have been 16.

The problem with this approach is that we don't take into consideration the at least 12 were out, but this should give sufficient understanding to solve/proceed further.

The above method is a conceptual approach and it does take a bit longer than the algebraic approach. In case one gets lost using a method, one can always try the other.

[Simon]

Also, according to the algebraic approach

20-x+ x/2 = >12

We know from the question that 8 is still rented, so from the above equation x/2 = 8. Solve for x =16.