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In the below addition A, B, C, D, E, F, and G represent the
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Updated on: 29 Dec 2013, 04:21
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In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals Attachment:
findginsum_figure.png [ 833 Bytes  Viewed 21459 times ]
A. 2 B. 3 C. 4 D. 5 E. 6 How to tackle such questions? Do we have more question like this?
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Originally posted by ROckHIsT on 29 Dec 2013, 04:09.
Last edited by Bunuel on 29 Dec 2013, 04:21, edited 1 time in total.
Renamed the topic and edited the question.




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Re: In the below addition A, B, C, D, E, F, and G represent the
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Updated on: 15 May 2014, 13:36
OK, this is what I did First we know that E must be 1. Then we know that 0 is to be in either F and G, since they are all different numbers. Since it does not appear in one of the answer choices then F must be zero. Now, for F to be zero the sum of the two digits A and C must add to 10. Therefore we have 6 and 4. Then B and D will be 3 and 2 , or 2 and 3 giving a sum of G=5 Therefore B is equal to 5 Hope I made myself clear Cheers! J
Originally posted by jlgdr on 22 Mar 2014, 13:26.
Last edited by jlgdr on 15 May 2014, 13:36, edited 1 time in total.




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Re: In the below addition A, B, C, D, E, F, and G represent the
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29 Dec 2013, 04:25
rgyanani wrote: In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals Attachment: findginsum_figure.png A. 2 B. 3 C. 4 D. 5 E. 6 How to tackle such questions? Do we have more question like this? Similar questions to practice: inthecorrectlyworkedadditionproblemaboveabcd128953.htmltoughtrickysetofproblems85211.html#p638336inthecorrectlyworkedadditionproblemshownwherethe54154.htmlintheadditionshownaboveabcanddrepresent161656.htmlHope it helps
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Re: In the below addition A, B, C, D, E, F, and G represent the
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26 Mar 2014, 06:29
I started with F has to be 0 because if not it would result in a scenario where two of the numbers are the same. So A+C must be greater than or equal to 10. The only combination of this is 4+6. This means that A+C=10. Now you have A=4, C=6, and F=0. If A and C are 4 and 6 then E must be 1. You're left with 2,3, and 5. G is the sum of B and D, therefore it must be 5.



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Re: In the below addition A, B, C, D, E, F, and G represent the
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27 Mar 2014, 01:33
ROckHIsT wrote: In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals Attachment: findginsum_figure.png A. 2 B. 3 C. 4 D. 5 E. 6 How to tackle such questions? Do we have more question like this? Sol: Here we are adding 2 nos and get a 3 digit number When you look at the given nos we see that E has to be 1 only Now let's do back calculation and see where end up Consider G= 2 then we can have possible values of B&D (1,1 or0,2 or 2,0) But then each digit can have only 1 value So ignore G Consider G= 3 then we have B&D (1,2 or 2,1 or 3,0) Now 1,2 and 2,1 can be ruled out cause we said E has to be 1 then we have a case 3,0 but then A=3 and B=0 then G=3 which is not possible Consider G=4 then we have B&D (2,2 or 1,3 or 3,1 or 4,0) Consider 4,0 again if A=4 then B=0 and G=4 which is not possible as each Alphabet represent only single digit Consider G=5 then we have possible values of B&D (2,3 or 5,0 or 4,1 or1,4) Only case is 2,3 Keep G=5 and the only possible nos 62+43= 105 We can stop here but just to check Consider G=6 then we have (3,3 or 4,2 or 1,5 or 6,0) Now 1,5 and 6,0 can be ruled out. Consider 4,2 X4+Y2= 1A6 Which is impossible because we will need bigger nos let's say 7 8 and 9 as possible values of X and Y but these values are restricted from Q stem. So ans has to be D
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Re: In the below addition A, B, C, D, E, F, and G represent the
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10 May 2014, 13:44
If we start analyzing from E & F it will far more easier to get to the answer. My 3 step approach is below,
Step1
First E cannot be 0 thus A+C = 10/11 only then E will have 1 and F will have 0 or 1. Therefore A & C needs to be either 6 + 4 or 5 + 6.
Step2
If A & C is 5 & 6 then E=1 & F=1 but given that each variable has a different value, thus A & C will be 6 & 4. A=6 B + C=4 D  1 0 G
Step3
Now we have left with B, D & G and have 2, 3 & 5 in hand. If we put B = 2 & D = 5 then G = 7 but 7 is not in our list. Thus this combination is invalid. If we put B = 3 & D = 5 then G = 8 but 8 is not in our list. Thus this combination is invalid.
Therefore B will be 3 & D will be 2 and G will be 5.
A=6 B=3 + C=4 D=2  E=1 F=0 G=5



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In the below addition A, B, C, D, E, F, and G represent the
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22 Dec 2014, 11:07
What I did was actually I dont know what it is called , please tell me if any shorter approach is available A+C can be 10(6+4) or 11(6+5), but if A+C=11 then one number will be left unassigned due to repetition of 1 So, A+C=10 (6+4) Now , numbers left are 3,5,2 (0,1,4,6 already used) 3+2=5 And here is your answer 105
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In the below addition A, B, C, D, E, F, and G represent the
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22 Dec 2014, 19:13
EFG is a threedigit number, hence E will be equal to 1 having in mind that it cannot be any other number because addition of two twodigit numbers never yield more than 100. So, A+C must be equal 10 (note that numbers cannot be used more than once, thus 11 would not work).
In order to get get G, for B+D we must use numbers that are left: 2,3, and 5. The only possibility is 2+3=5. So 62+43=105. Answer D



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Re: In the below addition A, B, C, D, E, F, and G represent the
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22 Dec 2014, 21:55
ROckHIsT wrote: In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals Attachment: findginsum_figure.png A. 2 B. 3 C. 4 D. 5 E. 6 How to tackle such questions? Do we have more question like this? A little bit of hit and trial, a little bit of logic... E can only be 1: No two 2 digit numbers can give us a sum of 200 or more. The sum must be in the 100  199 range only. The largest digits are 5 and 6 which add to give 11. With a carryover, the sum can be 12 to give a maximum value of 2 to F. But no two of the rest of the numbers (0, 2, 3, 4) can be added to give us a carryover. So F cannot be 2 and it cannot be 1 because E is 1. So F must be 0. A and C can be 6 and 4 to give a sum of 10 leaving us with 2, 3, and 5  perfect. B and D must be 2 and 3 and G must be 5. Answer (D)
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Re: In the below addition A, B, C, D, E, F, and G represent the
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23 Dec 2014, 00:13
we have
AB CD EFG
1) A+C=10 to get E, it cannot be 11 because we get E=F or 12 because A=C. So, E=1, F=0
2) To get F=0 we have only one pair 4+6, so A=4, C=6
3) So, we have 0,1,4,6. Only 2,3,5 rest, so only possible G is 5
D



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In the below addition A, B, C, D, E, F, and G represent the
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23 Dec 2014, 00:37
Refer diagram below: Attachment:
findginsum_figure.png [ 4.64 KiB  Viewed 19494 times ]
With the given set of numbers & \(E \neq{0}\), the value of E can be maximum "1" For E = 1, A & C should add up to give "0" and "carry 1" With all the above numbers "occupied", 2 & 3 remain to give 5 These are the 4 combinations possible Answer = D = 5
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Re: In the below addition A, B, C, D, E, F, and G represent the
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19 Mar 2018, 09:03
ROckHIsT wrote: In the below addition A, B, C, D, E, F, and G represent the digits 0, 1, 2, 3, 4, 5 and 6. If each variable has a different value, and E ≠ 0, then G equals Attachment: findginsum_figure.png A. 2 B. 3 C. 4 D. 5 E. 6 How to tackle such questions? Do we have more question like this? I approached as follows:  From the fact presented in the question, E has to be 1 as no 2digit # sums to 200 or more.  Next, let's make G to have the maximum value, G cannot be 6 because no remaining digits sums to give a value of 6. So G should be 5 accordingly B and D can be 3 and 2  Next, I have numbers remaining 0, 4 and 6. I know A+C should give me a 2digit number then only E will have the value 1. Therefore the only way to have it is F=0 and A and C can have a value of 4 and 6. This gives the value of G=5. The answer is (D).
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Re: In the below addition A, B, C, D, E, F, and G represent the &nbs
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