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

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In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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Updated on: 29 Dec 2013, 04:21
1
22
00:00

Difficulty:

35% (medium)

Question Stats:

76% (02:42) correct 24% (02:44) wrong based on 451 sessions

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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 30257 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  [#permalink]

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Updated on: 15 May 2014, 13:36
7
1
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

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.
##### General Discussion
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Re: In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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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:
tough-tricky-set-of-problems-85211.html#p638336

Hope it helps
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Re: In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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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  [#permalink]

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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  [#permalink]

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10 May 2014, 13:44
3
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
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In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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22 Dec 2014, 11:07
1
1
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
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In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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22 Dec 2014, 19:13
1
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.
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Re: In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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22 Dec 2014, 21:55
2
1
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.

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Re: In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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23 Dec 2014, 00:13
1
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  [#permalink]

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23 Dec 2014, 00:37
2
1
Refer diagram below:

Attachment:

findginsum_figure.png [ 4.64 KiB | Viewed 28232 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

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Re: In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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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 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).
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Re: In the below addition A, B, C, D, E, F, and G represent the  [#permalink]

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10 May 2019, 05:18
1
Top Contributor
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

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

Cheers,
Brent
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Re: In the below addition A, B, C, D, E, F, and G represent the   [#permalink] 10 May 2019, 05:18
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