With BBQ season nearing, it’s time for hotdogs and burgers. Well, sort of. Try the problem below.

A concession stand sells either hotdogs for $1.75 each or hamburgers for $4 each. If Charlie buys a total of 9 items from the concession stand for a total of 27 dollars, then how many hotdogs did he buy?

(A) 3

(B) 4

(C) 6

(D) 7

(E) 8

On the GRE, you will most likely see a long, tedious word problem. Granted, this problem wasn’t too bad. The question is, how did you go about trying to solve it?

Many begin trying to turn the information from the problem into an equation. Some are very adept at this and get the answer quickly. But, most students I’ve seen struggle as soon as they start thinking about the above problem algebraically. So, I want you to try something that may seem counterintuitive. I want you to turn off the algebra part of your brain. Instead, I want you to work with the answer choices by plugging them back into the problem.

Let’s start with answer choice (C). The reason we start with this answer is because it is in the middle. If it turns out that the total price is less than 27 dollars, then we will need to plug in a lower number. The reason we would choose a lower number is because hotdogs are cheaper; therefore, the fewer hotdogs we have the more burgers we have, and this brings up the price.

Plugging in 6 we get $1.75 x 6 = $10.50. That leaves us 3 hamburgers at $4 each = $12. Adding the two together gives us $22.50. That amount is too low, so we can eliminate answer choices (D) and (E), as well, because they will increase the number of hotdogs and lower the overall price.

Plugging in (B) 4, we get $7 on hotdogs. That leaves us 5 hamburgers or $4 x 5 = $20. Add up $7 + $20 and we get $27, which is the correct answer.

In the next segment on plugging in, I will show you a very different type of problem in which plugging in can be equally effective.

27 is close to a multiple of 4 in 24. However, 24/4 = 8, and 24 +1.75 = 25.75, not enough. However, if there are 20 burgers (4×5 = 20), then 4 hotdogs (since there are 9 total items) and 1.28 x4 = 7.00. 20 +7 = 27. Answer is B. I hadn’t looked at the answer choices until the end.

Hi Chris! The entire Clemmonsdogpark course has been outstanding and really helpful for me! I just had a suggestion. Another trick that can be used here is that as 27 is an integer and 4 multiplied with any no. of items will also be an integer so even (1.75)*(no. of HDs) should be an integer so answer will be either 4 or 8! But I think to use this trick you need to be fast with your calculations!

That’s exactly how the GRE wants you to think–your logic saved a good minute with this problem. And it doesn’t take long at all, since if you assume the number of hotdogs is 8 and you round up 1.75 to 2, you still get 16+4, which is way short of 27. Therefore the answer has to be 4.

since he spent $ 27, 1.75HD + 4 HB = 27————1 since the total amount of stuffs he bought = 9 HD + HY= 9 ———2

solving simultaneously….i get 3.9

The plug in method seems alot faster. But under timed condition the brain will always choose the method its accustomed to and thats the way we learnt it in school. It will require alot of practise to reverse it..The plug in method is definitely better for the GRE.

I think you are right – it does take a lot of unlearning. But if you practice without using a pencil, thus forcing yourself to forgo all the algebraic steps, you will be prone to plugging in, thereby making it more second-nature. And as you noted, the plugging in method is definitely better on the GRE :).

this Clemmonsdogpark blog is very helpful and effective. I need one suggestion. I am going to take the GRE at Nov 10,2011. But should I try to adopt any new rule like plugging in. Is it frutfully works on in such practicing such a short time??? Kindly give me a suggestion.

I would definitely try plugging-in. As long as you have plenty of problems to practice with you should be able to get comfortable with plugging over the course of a few days. Clemmonsdogpark has plenty of problems in which plugging in works well, as do basically every prep book out there. So I think you should definitely have a lot of resources to practice from. If after a week, plugging in just isn’t working for you, then I would stick to the formal algebraic approach. If not, def. learn to plug-in. It will help you plenty test day.

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Hotdog = $1.75

Burgers = $4

27 is close to a multiple of 4 in 24. However, 24/4 = 8, and 24 +1.75 = 25.75, not enough. However, if there are 20 burgers (4×5 = 20), then 4 hotdogs (since there are 9 total items) and 1.28 x4 = 7.00. 20 +7 = 27. Answer is B. I hadn’t looked at the answer choices until the end.

That’s not to say I’m against using the

Hi Chris!

The entire Clemmonsdogpark course has been outstanding and really helpful for me!

I just had a suggestion. Another trick that can be used here is that as 27 is an integer and 4 multiplied with any no. of items will also be an integer so even (1.75)*(no. of HDs) should be an integer so answer will be either 4 or 8! But I think to use this trick you need to be fast with your calculations!

Thanks

That’s awesome!!

That’s exactly how the GRE wants you to think–your logic saved a good minute with this problem. And it doesn’t take long at all, since if you assume the number of hotdogs is 8 and you round up 1.75 to 2, you still get 16+4, which is way short of 27. Therefore the answer has to be 4.

Again, good job :)!

Hi chris, here is how i solved this question

Hot dogs (HD)= $1.75 each

Hamburgers (HB) = $4 each

since he spent $ 27,

1.75HD + 4 HB = 27————1

since the total amount of stuffs he bought = 9

HD + HY= 9 ———2

solving simultaneously….i get 3.9

The plug in method seems alot faster. But under timed condition the brain will always choose the method its accustomed to and thats the way we learnt it in school. It will require alot of practise to reverse it..The plug in method is definitely better for the GRE.

I think you are right – it does take a lot of unlearning. But if you practice without using a pencil, thus forcing yourself to forgo all the algebraic steps, you will be prone to plugging in, thereby making it more second-nature. And as you noted, the plugging in method is definitely better on the GRE :).

nice trick 😀 ….. i hope this may work in my exam as well

Definitely – plugging in will save you a lot of time and confusion. Remember to practice it as much as possible before the actual exam.

this Clemmonsdogpark blog is very helpful and effective. I need one suggestion. I am going to take the GRE at Nov 10,2011. But should I try to adopt any new rule like plugging in. Is it frutfully works on in such practicing such a short time??? Kindly give me a suggestion.

Hi Zafrin,

I would definitely try plugging-in. As long as you have plenty of problems to practice with you should be able to get comfortable with plugging over the course of a few days. Clemmonsdogpark has plenty of problems in which plugging in works well, as do basically every prep book out there. So I think you should definitely have a lot of resources to practice from. If after a week, plugging in just isn’t working for you, then I would stick to the formal algebraic approach. If not, def. learn to plug-in. It will help you plenty test day.