In the below addition A, B, C, D, E, F, and G represent the

Question

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

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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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Official Answer

D

jlgdr · Accepted Answer

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

KarishmaB · Answer

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)

Bunuel · Answer

Similar questions to practice: in-the-correctly-worked-addition-problem-above-a-b-c-d-128953.html tough-tricky-set-of-problems-85211.html#p638336 in-the-correctly-worked-addition-problem-shown-where-the-54154.html in-the-addition-shown-above-a-b-c-and-d-represent-161656.html Hope it helps

hiscock90 · Answer

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.

WoundedTiger · Answer

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

DoNow · Answer

If we start analyzing from E & F it will far more easier to get to the answer. My 3 step approach is below,-

Step-1

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.

Step-2

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

Step-3

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

veergmat · Answer

What I did was actually I dont know what it is called :roll:

, 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

viktorija · Answer

EFG is a three-digit number, hence E will be equal to 1 having in mind that it cannot be any other number because addition of two two-digit 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

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

Temurkhon · Answer

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

PareshGmat · Answer

Refer diagram below: findginsum_figure.png With the given set of numbers & E eq{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

ujjwal80 · Answer

I approached as follows:
- From the fact presented in the question, E has to be 1 as no 2-digit # 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 2-digit 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).

BrentGMATPrepNow · Answer

If we add two 2-digit numbers and the sum is a 3-digit number, then the 3-digit number must start with a 1. 
So,   E = 1

In order for the sum to be a 3-digit number, A+C must be greater than 9
So, we have two options: 
EITHER A and C are 5 and 6, OR A and C are 4 and 6
If A and C are 5 and 6, then F = 1, but we already know that E = 1
So, it MUST be the case that   A and C are 4 and 6  , which means   F = 0

The three digits unaccounted for are 2, 3 and 5
Since we can see that B + D = G, it must be the case that  G = 5

Answer: D

Cheers,
Brent

hudacse6 · Answer

In order for the sum to be a 3-digit number, A+C must be greater than 9 -------what do you mean by that? it would be helpful for me if you clear it, thanks.

GMATPrepNow

BrentGMATPrepNow · Answer

If A+C is not greater than 9, then the sum of the two 2-digit numbers will not be a 3-digit number. 
For example,  2 5 +  4 1 = 66 (not a 3-digit number)
And  5 1 +  1 7 = 68 (not a 3-digit number)

Conversely,  6 1 +  7 3 = 134 (a 3-digit number)

does that help?

Kinshook · Answer

B + D = G; Since 5+6 = 11 but G<>1; 4+6 = 10 but G<>0 A + C = 10 E + F E =1 A + C = 10 + F 4 + 6 = 10 A = 4; C = 6; F=0 or A=6; C=4; F=0 Remaining digits = 2,3,5 2 + 3 = 5 B = 2; D = 3; G = 5 or B=3; D=2; G=5 G = 5 IMO D

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