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8. In the above correctly worked addition sum, A,B,C and D

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by Anju@Gurome » Wed Apr 24, 2013 4:07 am
varun289 wrote:5A
BC
-----
D43

In the above 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?
5A and BC are two 2-digit integers, where A, B, and C are different digits, and none of them are equal to 3 or 4 or 5.
So, the maximum value of the sum of 5A and BC is (98 + 57) = 155
So, possible values of D are 0 and 1.
But the sum cannot be 43 as it must be greater than 5A.
So, D = 1

Now, units digit of (A + C) is 3.
So, possible values of (A, C) are : (0, 3), (1, 2), (4, 9), (5, 8), and (6, 7)
As none of A and C can be equal to 1 or 3 or 4 or 5, only possible pair is (6, 7)

Now, (A + C) will result in a carry of 1.
So, (5 + B + 1) = 14 ---> B = 8

Hence, (A + B + C + D) = (6 + 8 + 7 + 1) = 22

The correct answer is B.
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by GMATGuruNY » Wed Apr 24, 2013 5:05 am
varun289 wrote:5A
BC
-----
D43

8. In the above 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?

A. 23
B. 22
C. 18
D. 16
E. 14
On the GMAT, D43 would represent a 3-digit integer, implying that D>0.
Since 99+99 = 198, D43 < 198.
Thus, D43 = 143, and D=1.
Since no digit can be repeated, 5A = 50, 52, 56, 57, 58, or 59.

Case 1: 5A = 50
Here, BC = 143-50 = 93.
Not viable, since 3 cannot appear in both 143 and 93.

Case 2: 5A = 52
Here, BC = 143-52 = 91.
Not viable, since 1 cannot appear in both 143 and 91.

Case 3: 5A = 56
Here, BC = 143-56 = 87.
This works: A=6, B=8, C=7, and D=1, with no digit appearing more than once.

Thus, A+B+C+D = 6+8+7+1 = 22.

The correct answer is B.
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