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Jamin2112

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- Thread starter Jamin2112
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Jamin2112

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- #2

verty

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- #3

Jamin2112

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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.

- #4

WWGD

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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.

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- #6

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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.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.

- #7

Matterwave

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- #8

Jamin2112

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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.

- #9

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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.

- #10

cjl

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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.

- #11

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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.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.

- #12

Medicol

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

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- #13

Medicol

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

- #14

johnqwertyful

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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}##

- #15

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The expression "

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.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

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.

- #16

Medicol

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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

- #17

OmCheeto

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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.

- #18

Modest Learner

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Got 7 correct in 9 minutes 48 seconds.

The 8th one was tricky.

The 8th one was tricky.

- #19

Modest Learner

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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.

- #20

mfb

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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).

- #21

cjl

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The expression "apercent ofb" meansa·b/100. So in this case, with [itex]a=\sqrt 5[/itex] and [itex]b=5\sqrt 5[/itex], the answer is [itex]\frac {(\sqrt 5)\cdot(5\sqrt 5)}{100} = \frac{5\cdot 5}{100}=\frac {25}{100} = \frac 1 4[/itex].

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.

- #22

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

- #23

pbuk

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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.

- #24

Jamin2112

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

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- #26

Fredrik

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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.

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.

Whoa. That's crazy fast.10 correct, 4:45.

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.#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:

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- #27

OmCheeto

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...

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

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

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

- #28

Medicol

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

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

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.

- #29

Matterwave

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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

- #30

mfb

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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.

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