Problem Solving

Wasn't sure how to tackle this problem, espeically within less time.

I saw a similar problem to this one on the real GMAT and it totally had me stumped!!

id go for 1.4x

worked it out this way:
from nov thru feb,the total prodn is 4x

then,from march onwards,the shipment per month is:
March=4x-(x/2) = 3.5x
april = 3.5x - (x/2) = 3.0x
and so on till october..when it will be zero.(I.E the prodn inventory = supply;no opening stock in november again!)

multiply each of these monthly inventory by 0.1 and,
if youadd all these..it comes to 1.4 x,the answer.

is it right?

as per me:
it would be 1.6x as the storage costs are paid only for march to oct, so the costs for the whole yr would also be no more than that.

in march, cost would be on 7x/2 rakes and every month x/2 rakes would dwindle from the pile
so that => 0.1*x (3.5 +3+ 2.5...... + 0.5) = 0.1*x [ 4*(3.5 + 0.5)]
= 0.1x * 16 = 1.6x

0.1*x (3.5 +3+ 2.5 + 2 + 1,5+1 + 0.5)=1,4x

C

OA?

Could this problem have also been solved by picking numbers?

is the answer C?

The answer is definitely C i.e 1.40 x

4x rakes total and x/2 shipped each month


Month Rakes Left
Mar 7x/2 *. 10
Apr 6x/2 * .10
May 5x/2 * .10
Jun 4x/2 * .10
Jul 3x/2 * .10
Aug 2x/2 * .10
Sep x/2 * .10
Oct 0 (since x/2 left from Sep shipped)

Therefore total

.10(7x+6x+5x+4x+3x+2x+x) / 2
= .05*28x
=1.40 x

ildude02 wrote:Wasn't sure how to tackle this problem, espeically within less time.

Answer is 1.4x

Answer is indeed 1.4x.

I later solved the problem by picking numbers(I choose value of 4 for X ) and I found that to be easier instead of calculating basing it off X.

here it is .

Hi lildude02,

Can you explain how u solved this problem picking numbers? I'm still lost on this problem!! thanks!

let's say X=10, then x/2=10/2=5

Nov. to Feb. (11, 12, 1, 2)=4*10=40

3. 40-5=35*0.1=3.5
4. 35-5=30*0.1=3.0 (you can see a pattern.)
5. 2.5
6. 2.0
7. 1.5
8. 1.0
9. 0.5
10. 0.0

3.5+3+2.5+2+1.5+1+0.5=14

14=10y
y=1.4

Ahhh that makes sense now....thanks for the explanation EricLien!!