Problem Solving

Question #19 from the mathematics section on Testprep. I know its not an official source so the problem could be the wording, but it could also be my approach. Any help appreciated

19. If edging cost $2.32 per 12-inch stone, and you want a double layer of edging around your flower bed that is 6 yards by 1 yard. How much will edging your flower bed cost?

A. $32.48
B. $64.96
C. $97.44
D. $129.92
E. $194.88




My Mspaint interpretation lol

The internal dimensions of the flower bed, converted into feet to adjust for the 12 inch blocks in which the stones are sold, are 18x3 feet. So we would need 18 stones for each long side, and 3 for each short side.

18 + 18 + 3 +3 = 42 stones

Now we would also need 4 stones to cover the corners to completely enclose the edging.

42 + 4 = 46 stones

They call for a double layer so we double our 46 and get 92 stones

92 x ($2.32) = $213.44 , obviously not one of the answer choices.

Can someone help me and tell me what im doing wrong?

What's unclear from the question is whether the "layer" includes the corner-stones or not. The 2nd layer will be larger than the 1st since it goes around the outside of the 1st layer, not the outside of the flower bed. I think that is where your confusion lies.

parkman wrote:Question #19 from the mathematics section on Testprep. I know its not an official source so the problem could be the wording, but it could also be my approach. Any help appreciated

19. If edging cost $2.32 per 12-inch stone, and you want a double layer of edging around your flower bed that is 6 yards by 1 yard. How much will edging your flower bed cost?

A. $32.48
B. $64.96
C. $97.44
D. $129.92
E. $194.88




My Mspaint interpretation lol

The internal dimensions of the flower bed, converted into feet to adjust for the 12 inch blocks in which the stones are sold, are 18x3 feet. So we would need 18 stones for each long side, and 3 for each short side.

18 + 18 + 3 +3 = 42 stones

Now we would also need 4 stones to cover the corners to completely enclose the edging.

42 + 4 = 46 stones

They call for a double layer so we double our 46 and get 92 stones

92 x ($2.32) = $213.44 , obviously not one of the answer choices.

Can someone help me and tell me what im doing wrong?

Hi!

The question definitely isn't worded clearly, since the common interpretation would be surrounding the flower bed with stones; however, by reverse engineering the choices it's clear that's not the case.

So, ignoring "filling in the corners" but using the rest of your calculations:

18 + 18 + 3 + 3 = 42 stones

Another identical layer is another 42 stones

84 * 2.32 = 194.88 which is choice (E) and I assume the accredited answer.

As an aside, if we were filling in the corners your original calculations would still be off by a bit; everything was perfect until the last step. While it's true that closing the first layer would require only 1 stone per corner, closing the second layer actually requires 3 stones per corner (draw it out and you'll see that's the case).

So, the correct number of stones would be:

42 + 42 + 4 + 12 = 100

and the answer would be 100*2.32 = $232

Not only would that be a more common interpretation of the question, but it would also be much more "GMATesque", since you don't have to do any complicated multiplication.

I actually interpreted the double layer to mean a full second surrounding of the flowerbed initially, and got the 100 stones necessary, but that number was way off so i then stacked them to get a smaller number, which was also way off.

Question for Stuart since you helped me with my last question that was curiously worded. I have since switched from the Peterson preperation material I was using to the OG as suggested. I have also read Twinspinner's post and TestPrep was a site he used in his studies. This question seems curiously worded, do you suggest I just take the offerings of the free sites with a grain of salt? I guess its not completely bad that I am getting thrown off by wording rather than fundamentals, but it could be time spent elsewhere.

THanks

Gotcha! Yeah, this question definitely seems a bit suspect to me. I would be really suspicious of "free" sites. There is so much good GMAT-prep material out there, why settle for second rate?

Once you finish with the OG and need more material, I suggest the Kaplan Quant Workbook for more overview, then the MGMAT Number Properties guide. Avoid the 1000 series and anything "free."

I also recommend Grockit (of course ). Our questions are mocked from OG materials. BTG also has a wealth of good info sorted by skill tag in the Library.

parkman wrote:I actually interpreted the double layer to mean a full second surrounding of the flowerbed initially, and got the 100 stones necessary, but that number was way off so i then stacked them to get a smaller number, which was also way off.

Question for Stuart since you helped me with my last question that was curiously worded. I have since switched from the Peterson preperation material I was using to the OG as suggested. I have also read Twinspinner's post and TestPrep was a site he used in his studies. This question seems curiously worded, do you suggest I just take the offerings of the free sites with a grain of salt? I guess its not completely bad that I am getting thrown off by wording rather than fundamentals, but it could be time spent elsewhere.

THanks

Hi!

Vivian's advice is dead on; the old adage "you get what you pay for" definitely applies to most free GMAT questions available on the web. Probably the biggest advantage of going with "brand name" materials is that you know that the top prep companies didn't get to where they are today by putting out anything less than top rate materials; I know that Kaplan invests a lot of money into our curriculum and question development and I'd be amazed if the same weren't true of the other reliable brands.

The big downside to sticking only to the OG is that, while the OG is a very valuable source of questions, it's a horrible source of explanations (especially for quant). There's pretty much no way any really high scorer could use the OG approach to math and not run out of time.