These 10 questions should take 12 minutes?

[Jamin2112]

Let me know how long it takes you to solve these 10 questions that are supposedly representative of what 10 Quantitative Reasoning questions on the GRE would be like: http://www.majortests.com/gre/problem_solving_test03. The first time I took the GRE (2 years ago), I remember most of the QR questions being no-brainers. Maybe I need to get back in the groove, or maybe I've become stupider.

 

[verty]

9m20 here, though for some reason I messed up on one question. I hate those "choose all the numbers that cannot be..." questions, one has to be so careful with them. The hardest question was the addition one, it had a little trick.

 

[Jamin2112]

9m20 here, though for some reason I messed up on one question. I hate those "choose all the numbers that cannot be..." questions, one has to be so careful with them. The hardest question was the addition one, it had a little trick.

That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.

 

[WWGD]

That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.

But D has to be 1 , since B+C +k =4 , where k=0 (if you don't carry) or k=1 ( if you carry), so B must be 8 or 9, so you must carry a 1 , which is D; you cannot have 50-something + a 2-digit number be 40-something; if 5+B+k =4, then B is either 8, or 9, so you carry a 1.

 

[jbunniii]

I got all ten within 12 minutes, but I had to spend several minutes trying to interpret what the 5A + BC = D43 question really meant. The way it is worded, one solution should be A=0, C=3, B=9, D=1. Then all the digits being summed (i.e., 5, A, B, and C) are distinct. But 13 wasn't one of the answers. Then I realized that by "all the digits in the sum," they must have meant that all of 5, A, B, C, D, 4, 3 must be distinct, and I got 22 as an answer under that interpretation. Also, they didn't specify base 10, so the question as written is ill posed.

 

[D H]

That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.

Sure you can. The only way to get D=0 is if B is -1 (or -2 in the case A+C=13), and certainly neither -1 nor -2 counts as a digit.

 

[Matterwave]

I got all 10 in 13 minutes and some seconds. I made a dumb error on the digits one and added the wrong digits up at the end after figuring out it could be 57+86=143 (I added 7,8,6 and 3...instead of 7,8,6 and 1) so I got none of the answers...took me 2 minutes to figure out my error! lol

 

[Jamin2112]

Sure you can. The only way to get D=0 is if B is -1 (or -2 in the case A+C=13), and certainly neither -1 nor -2 counts as a digit.

I need to come up with a more organized structure of how I approach those problems to solve them efficiently. Right now I'm using a "greedy" approach: try to search for the first digit whose value I know; narrow it down from there.

 

[D H]

That problem isn't that hard. You know A+C is either 3 or 13. If A+C=3, that means B=9, D=1, and the sum is 13. If A+C=13, that means B=8, D=1, and the sum is 22. The first solution isn't on the list, but the second is.

The problem would have been trickier if the problem was 5A+BC=D40 instead of 5A+BC=D43 and if both 10 and 19 were listed as choices. With this change, A+C is 0 or 10. A+C=0 means B=9, D=1, and the sum is 10. A+C=10 means B=8, D=1 and the sum is 19. Both solutions are listed! Which one is wrong? A+C=0 isn't valid because that means A=C=0, and this violates the clause that A,B,C, and D are different digits.

 

[cjl]

That problem isn't that hard. You know A+C is either 3 or 13. If A+C=3, that means B=9, D=1, and the sum is 13. If A+C=13, that means B=8, D=1, and the sum is 22. The first solution isn't on the list, but the second is.

The problem would have been trickier if the problem was 5A+BC=D40 instead of 5A+BC=D43 and if both 10 and 19 were listed as choices. With this change, A+C is 0 or 10. A+C=0 means B=9, D=1, and the sum is 10. A+C=10 means B=8, D=1 and the sum is 19. Both solutions are listed! Which one is wrong? A+C=0 isn't valid because that means A=C=0, and this violates the clause that A,B,C, and D are different digits.

But if A+C = 3, as you point out, D = 1, B = 9, but then you're stuck with A+C = 3 where A and C must be distinct digits not equal to 1 or 9 (or 5 or 4 or 3). This is not possible - the only A and C such that A+C = 3 are A = 1 and C = 2 (or A = 2 and C = 1). D is already equal to 1 though, so this fails the criterion that all the digits in the sum must be different. Thus, A+C = 13 is the only valid answer.

 

[D H]

But if A+C = 3, as you point out, D = 1, B = 9, but then you're stuck with A+C = 3 where A and C must be distinct digits not equal to 1 or 9 (or 5 or 4 or 3). This is not possible - the only A and C such that A+C = 3 are A = 1 and C = 2 (or A = 2 and C = 1). D is already equal to 1 though, so this fails the criterion that all the digits in the sum must be different. Thus, A+C = 13 is the only valid answer.

Who says A and C must be between 1 and 9? Zero is a digit. There's nothing wrong with A or C being zero. If the question had both 22 and 13 as answers and had I checked 13, I would most certainly protest that selection as being marked wrong.

 

[Medicol]

it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet

 

[Medicol]

Who says A and C must be between 1 and 9? Zero is a digit. There's nothing wrong with A or C being zero. If the question had both 22 and 13 as answers and had I checked 13, I would most certainly protest that selection as being marked wrong.

A+C can not be 3 (3 is present in the sum, D is 1 already), so A or C can not be either 0,1,2,3,4,5

 

[johnqwertyful]

it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet

$\sqrt{5}$ percent of $5\sqrt{5}$ is $\frac{\sqrt{5}}{100}5\sqrt{5}$

 

[D H]

it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet

The expression "a percent of b" means a·b/100. So in this case, with $a=\sqrt 5$ and $b=5\sqrt 5$, the answer is $\frac {(\sqrt 5)\cdot(5\sqrt 5)}{100} = \frac{5\cdot 5}{100}=\frac {25}{100} = \frac 1 4$.

A+C can not be 3 (3 is present in the sum, D is 1 already), so A or C can not be either 0,1,2,3,4,5

Sure they can. The question does not say that all of the digits in the problem are different from one another. It only says that "A,B,C and D represent different digits, and all the digits in the sum are different." The constraint against all of the digits in the sum being different just tells us that D is neither 3 nor 4, but we already know D has to be 1. The solution A=0, B=9, C=3, D=1 satisfies the sum problem and the constraints. A, B, C, and D are different digits. The only reason this is not the correct answer is because it is not one of the proffered choices.

This is somewhat common on multiple choice questions such as the GRE. There might well be multiple solutions to a problem, but only one will be listed.

 

[Medicol]

Thank you a lot.
Yes it is likely an issue with wording and comma; I think "all the digits in the sum are different" is also applied to 5A+BC=D43 not only to A+B+C+D

 

[OmCheeto]

I only got 6 correct. Were we not supposed to do these in our head? I think I probably would have gotten most all correct if I'd just written them down.

I missed:
#2: Didn't read the question correctly. Picked the closest answer to "5 is what percent of 25" = 25. Ha!
#5: Can't divide in my head anymore. By this point I was wondering how anyone could do this quiz in 10 minutes. 668 / 4 = 100 with 268 left over. 268 / 4 = 60 with 28 left over. 28? 28 doesn't look divisible by 4. :tongue:
#8: The carry threw me off by one.
#10: Word problems... Blech! I totally guessed on that one.

 

[Modest Learner]

Got 7 correct in 9 minutes 48 seconds.

The 8th one was tricky.

 

[Modest Learner]

I got all ten within 12 minutes, but I had to spend several minutes trying to interpret what the 5A + BC = D43 question really meant. The way it is worded, one solution should be A=0, C=3, B=9, D=1. Then all the digits being summed (i.e., 5, A, B, and C) are distinct. But 13 wasn't one of the answers. Then I realized that by "all the digits in the sum," they must have meant that all of 5, A, B, C, D, 4, 3 must be distinct, and I got 22 as an answer under that interpretation. Also, they didn't specify base 10, so the question as written is ill posed.

That's exactly where I went wrong.

 

[mfb]

10 correct, 4:45.

I missed the "A or C = 0" option, that saved some time.

#5 can be solved as "6^5 - 6^4 = 6^4 (6-1) = 6^4 * 5" then you don't have to divide or multiply anything (because the right answer is given as 6^4 as well).

 

[cjl]

The expression "a percent of b" means a·b/100. So in this case, with $a=\sqrt 5$ and $b=5\sqrt 5$, the answer is $\frac {(\sqrt 5)\cdot(5\sqrt 5)}{100} = \frac{5\cdot 5}{100}=\frac {25}{100} = \frac 1 4$.


Sure they can. The question does not say that all of the digits in the problem are different from one another. It only says that "A,B,C and D represent different digits, and all the digits in the sum are different." The constraint against all of the digits in the sum being different just tells us that D is neither 3 nor 4, but we already know D has to be 1. The solution A=0, B=9, C=3, D=1 satisfies the sum problem and the constraints. A, B, C, and D are different digits. The only reason this is not the correct answer is because it is not one of the proffered choices.

This is somewhat common on multiple choice questions such as the GRE. There might well be multiple solutions to a problem, but only one will be listed.

I interpret the statement "all of the digits in the sum are different" to mean that all of the digits in the problem are different, not simply that D is nether 4 nor 3. Stating that D is neither 4 nor 3 does not provide any information at all, since neither of those was a possibility from the start, so the only reasonable explanation for the introduction of that constraint is that it meant that all of the digits in the entirety of the problem (and because the problem is a summation, you could reasonably describe the problem as "the sum") are different. This also prevents A or C from being zero, since that would then require that the other one be 3, which violates the requirement that every digit is different.

 

[D H]

Arguing whether 13 is a valid solution is a bit silly because 13 was not on the list of answers.

 

[pbuk]

It only says that "A,B,C and D represent different digits, and all the digits in the sum are different."

It doesn't "only" say that, it says "In the following correctly worked addition sum, A,B,C and D represent different digits, and all the digits in the sum are different. What is the sum of A,B,C and D?"

There are two separate sentences. The first sentence refers to a "correctly worked addition sum" which is clearly the vertical sum including the digits 3, 4 and 5, and then states that all the digits in the sum (i.e. the sum which has already been referred to, not some other sum which has not yet been mentioned) are different. The second sentence then introduces a second sum - perhaps this is designed to be confusing, but by no correct interpretation can this new sum become the object of the preceding sentence.

Just under 7 minutes.

 

[Jamin2112]

Glad I'm not the only one who finds these easy to make mistakes on

 

[D H]

I didn't make a mistake. I got that one correct. (I got them all correct.) I saw that there were two solutions to that problem, solved for both, and saw that only one of my two solutions was on the list. Problem solved.

 

[Fredrik]

It took me about 14 minutes 30 seconds. I'm not fast with these things, but at least I got them all right. I think I got overconfident after doing the first four in less than 4 minutes, and slowed down a bit. I don't think I would make that mistake twice.

I had completed 8 questions when the clock passed 12 minutes.

Are you allowed to use a pen and paper? I did for the equation in question 10. I can do these things without a pen, but then I need more time.

10 correct, 4:45.

Whoa. That's crazy fast.

#5: Can't divide in my head anymore. By this point I was wondering how anyone could do this quiz in 10 minutes. 668 / 4 = 100 with 268 left over. 268 / 4 = 60 with 28 left over. 28? 28 doesn't look divisible by 4. :tongue:

I lost some time on this one too, but after a while I realized that all you have to do to check if something is divisible by 4 is to divide by 2 and see if the result is even or odd.

 

[OmCheeto]

...
after a while I realized that all you have to do to check if something is divisible by 4 is to divide by 2 and see if the result is even or odd.

I figured that out after the quiz was over. :grumpy:

All those simple math tricks we knew when we were 10 years old...

All gone, like the rest of my marbles...

 

[Medicol]

...
Are you allowed to use a pen and paper?...

Try out several more free tests there, you'll not be able to solve some problems without them.

Some of the companies I took the IQ tests as what they called them used these similar problems which are a bit selectively harder. It took me a long time to solve 10. They evaluated me as an idiot of course.

 

[Matterwave]

10 correct, 4:45.

I missed the "A or C = 0" option, that saved some time.

#5 can be solved as "6^5 - 6^4 = 6^4 (6-1) = 6^4 * 5" then you don't have to divide or multiply anything (because the right answer is given as 6^4 as well).

dang, you are supa fast

 

[mfb]

To test if a number is divisible by 4, you just need the last two digits - 100 is divisible by 4, so is every multiple of it. Therefore, 75654 is divisible by 4 if and only if 54 is.

20/4=5, so we can do the same trick again: At the next to last integer we just have to check if it is even or odd.
If it is even, we can remove it as multiple of 20, and the number is a multiple of 4 if the last digit is 0, 4 or 8.
If it is odd, we can replace it by 1, and the number is a multiple of 4 if the last digit is 2, or 6.

Examples:
647743534 -> 3 is odd, check if 14 is divisible by 4 -> no -> done.
762 -> 6 is even, check if 2 is divisible by 2 -> no -> done

There are tons of those tricks and they really make those tests and also real scientific problems easier.