Author 
Message 
TAGS:


Director
Joined: 07 Jan 2018
Posts: 616
Followers: 7
Kudos [?]:
569
[0], given: 88

In the following correctly worked addition sum, A,B,C and D [#permalink]
11 Mar 2018, 08:33
Question Stats:
66% (01:25) correct
33% (01:18) wrong based on 6 sessions
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? Attachment:
shot24.jpg [ 3.9 KiB  Viewed 681 times ]
A) 23 B) 22 C) 18 D) 16 E) 14




Manager
Joined: 15 Jan 2018
Posts: 147
GMAT 1: Q V
Followers: 3
Kudos [?]:
184
[1]
, given: 0

Re: In the following correctly worked addition sum, A,B,C and D [#permalink]
12 Mar 2018, 23:02
1
This post received KUDOS
You can solve this algebraically by making a lot of equations, i.e. 50 + A + 10B + C = 100D + 43, etc. But I'd avoid that mess. When you see problems like this it's usually easier to use logic or just try digits until you figure out the logic. D looks like the easiest. It has to be 1, since there's no way you can add a 5 and a single digit to make 20 or above, which would get you a 2 or above in the hundreds place. So D is 1. Since we need a 1 for D and a 4 in the tens place, we need to add 5 and B to get 14. Thus, B is 9. Great. So what are A and C? They need to add to 3 so they must be 1 and 2 or 0 and 3. However, the problem says that all digits are different and we've already used a 1 and a 3. So both options are out. What's going on? Another way of making that 3 would be to make A and C add to 13. However, then a 1 would carry over so what we concluded in the last paragraph would need to be amended a bit. No big deal. So the carried over 1 and 5 and B need to add to get 14. So B is actually 8. Doublechecking to ensure that A and C add to 13 but don't duplicate anything: 9 and 4 don't work because of the 4, 8 and 5 don't work because of the 5, but 7 and 6 would work because they add to 13 and haven't duplicated anything. So A and C make 13, B is 8, and D is 1, adding to a total of 22. So B is the answer.
_________________
    
Need help with GRE math? Check out our groundbreaking books and app.
Last edited by SherpaPrep on 03 Apr 2018, 10:01, edited 1 time in total.



Manager
Joined: 15 Feb 2018
Posts: 53
Followers: 1
Kudos [?]:
18
[0], given: 33

Re: In the following correctly worked addition sum, A,B,C and D [#permalink]
13 Mar 2018, 23:01
I don't understand why D is 1. SherpaPrep wrote: You can solve this algebraically by making a lot of equations, i.e. 50 + A + 10B + C = 100D + 43, etc. But I'd avoid that mess. When you see problems like this it's usually easier to use logic or just try digits until you figure out the logic.
D looks like the easiest. It has to be 1, since there's no way you can add a 5 and a single digit to make 20 or above, which get you a 2 or above in the hundreds place. So D is 1.
Since we need a 1 for D and a 4 in the tens place, we need to add 5 and B to get 14. Thus, B is 9.
Great. So what are A and C? They need to add to 3 so they must be 1 and 2 or 0 and 3. However, the problem says that all digits are different and we've already used a 1 and a 3. So both options are out. What's going on? Another way of making that 3 would be to make A and C add to 13. However, then a 1 would carry over so what we concluded in the last paragraph would need to be amended a bit. No big deal. So the carried over 1 and 5 and B need to add to get 14. So B is actually 8.
Doublechecking to ensure that A and C add to 13 but don't duplicate anything: 9 and 4 don't work because of the 4, 8 and 5 don't work because of the 5, but 7 and 6 would work because they add to 13 and haven't duplicated anything.
So A and C make 13, B is 8, and D is 1, adding to a total of 22. So B is the answer.



Director
Joined: 07 Jan 2018
Posts: 616
Followers: 7
Kudos [?]:
569
[2]
, given: 88

Re: In the following correctly worked addition sum, A,B,C and D [#permalink]
15 Mar 2018, 08:57
2
This post received KUDOS
gremather wrote: I don't understand why D is 1. SherpaPrep wrote: You can solve this algebraically by making a lot of equations, i.e. 50 + A + 10B + C = 100D + 43, etc. But I'd avoid that mess. When you see problems like this it's usually easier to use logic or just try digits until you figure out the logic.
D looks like the easiest. It has to be 1, since there's no way you can add a 5 and a single digit to make 20 or above, which get you a 2 or above in the hundreds place. So D is 1.
Since we need a 1 for D and a 4 in the tens place, we need to add 5 and B to get 14. Thus, B is 9.
Great. So what are A and C? They need to add to 3 so they must be 1 and 2 or 0 and 3. However, the problem says that all digits are different and we've already used a 1 and a 3. So both options are out. What's going on? Another way of making that 3 would be to make A and C add to 13. However, then a 1 would carry over so what we concluded in the last paragraph would need to be amended a bit. No big deal. So the carried over 1 and 5 and B need to add to get 14. So B is actually 8.
Doublechecking to ensure that A and C add to 13 but don't duplicate anything: 9 and 4 don't work because of the 4, 8 and 5 don't work because of the 5, but 7 and 6 would work because they add to 13 and haven't duplicated anything.
So A and C make 13, B is 8, and D is 1, adding to a total of 22. So B is the answer. In this problem we are adding a pair of two digit numbers and our answer is a three digit number so we can safely assume that \(5+B\) is 10 or more. At this point D can be anything however we must realize that B can acquire a maximum value of \(9\) and \(5+9 = 14\). Also we must take into account that we may have to add any carry overs from the summation of A and C. When two single digit are added they can never exceed \(18\) as such Our previous sum of \(5\) and \(9\) can have only \(1\) as carry over so \(5+B\) can at maximum become \(6+B\) and If summation of two single digit number cannot exceed \(18\) so D must be 1
_________________
This is my response to the question and may be incorrect. Feel free to rectify any mistakes




Re: In the following correctly worked addition sum, A,B,C and D
[#permalink]
15 Mar 2018, 08:57





