# In the Addition Problem Shown Below, Distinct Letters of the Alphabet Are Used to Represent the Digits from 0 to 9. E and D Are Both Less than 5 with D > Ewhich of the Following Letters of the Alphabe - Logical Reasoning

MCQ

Solve the following question and mark the most appropriate option.
In the addition problem shown below, distinct letters of the alphabet are used to represent the digits from 0 to 9. E and D are both less than 5 with D > E so that the sum of D and E is the average of W and T.

 W O R L D + T R A D E C E N T E R

Which of the following letters of the alphabet represents the average of the first nine natural numbers?

• E

• E

• E

• T

• T

• T

• W

• W

• W

• L

• L

• L

#### Solution 1

Since W + T = (CE), C must be 1.
Since D and E are both less than 5, they can only take values 0, 2, 3 or 4 (1 has already been used)
If either of D or E is 0, then D + E must yield a D or an E.
But in the above example D + E = R, neither D nor E can be 0.
So (D, E) = (3, 2), (4, 2) or (4, 3) and the corresponding values of R will be 5, 6 or 7.
Since R is the average of W and T, the corresponding values of W + T will be 10, 12 or 14.
Since E can only take values 2 or 3, and the sum W + T ends in 0, 2 or 4, E must be 2.

The first 2 digits of the answer are 1 and 2 respectively.
Which means W + T = 12, and therefore D + E = R, must be 6.
We can conclude that (D, E) has to be (4, 2).

 W O 6 L 4 + T 6 A 4 2 1 2 N T 2 6

Now L + 4 must be 12 so that L is 8 and 1 is carried over.

 W O 6 8 4 + T 6 A 4 2 1 2 N T 2 6

The digits remaining are 0, 3, 5, 7 and 9.
A total of 12 can be obtained as 5 + 7 or 3 + 9

Suppose (W, T) is (3, 9) then 6 + 1 + A = 9 yields A = 2 which is not possible
Suppose (W, T) is (9, 3) then 6 + 1 + A = 13 yields A = 6 which is not possible

So (W, T) is (7, 5) or (5, 7)
Suppose (W, T) is (7, 5) then 6 + 1 + A = 15 yields A = 8 which is not possible

Therefore (W, T) must be (5, 7) and 6 +1 + A = 7, yields A = 0.
The only 2 digits now remaining are 3 and 9.
Hence O has to be 3 and N has to be 9.

 5 3 6 8 4 + 7 6 0 4 2 1 2 9 7 2 6

W = 5, is the average of 1 to 9

#### Solution 2

Since W + T = (CE), C must be 1.
Since D and E are both less than 5, they can only take values 0, 2, 3 or 4 (1 has already been used)
If either of D or E is 0, then D + E must yield a D or an E.
But in the above example D + E = R, neither D nor E can be 0.
So (D, E) = (3, 2), (4, 2) or (4, 3) and the corresponding values of R will be 5, 6 or 7.
Since R is the average of W and T, the corresponding values of W + T will be 10, 12 or 14.
Since E can only take values 2 or 3, and the sum W + T ends in 0, 2 or 4, E must be 2.

The first 2 digits of the answer are 1 and 2 respectively.
Which means W + T = 12, and therefore D + E = R, must be 6.
We can conclude that (D, E) has to be (4, 2).

 W O 6 L 4 + T 6 A 4 2 1 2 N T 2 6

Now L + 4 must be 12 so that L is 8 and 1 is carried over.

 W O 6 8 4 + T 6 A 4 2 1 2 N T 2 6

The digits remaining are 0, 3, 5, 7 and 9.
A total of 12 can be obtained as 5 + 7 or 3 + 9

Suppose (W, T) is (3, 9) then 6 + 1 + A = 9 yields A = 2 which is not possible
Suppose (W, T) is (9, 3) then 6 + 1 + A = 13 yields A = 6 which is not possible

So (W, T) is (7, 5) or (5, 7)
Suppose (W, T) is (7, 5) then 6 + 1 + A = 15 yields A = 8 which is not possible

Therefore (W, T) must be (5, 7) and 6 +1 + A = 7, yields A = 0.
The only 2 digits now remaining are 3 and 9.
Hence O has to be 3 and N has to be 9.

 5 3 6 8 4 + 7 6 0 4 2 1 2 9 7 2 6

W = 5, is the average of 1 to 9

#### Solution 3

Since W + T = (CE), C must be 1.
Since D and E are both less than 5, they can only take values 0, 2, 3 or 4 (1 has already been used)
If either of D or E is 0, then D + E must yield a D or an E.
But in the above example D + E = R, neither D nor E can be 0.
So (D, E) = (3, 2), (4, 2) or (4, 3) and the corresponding values of R will be 5, 6 or 7.
Since R is the average of W and T, the corresponding values of W + T will be 10, 12 or 14.
Since E can only take values 2 or 3, and the sum W + T ends in 0, 2 or 4, E must be 2.

The first 2 digits of the answer are 1 and 2 respectively.
Which means W + T = 12, and therefore D + E = R, must be 6.
We can conclude that (D, E) has to be (4, 2).

 W O 6 L 4 + T 6 A 4 2 1 2 N T 2 6

Now L + 4 must be 12 so that L is 8 and 1 is carried over.

 W O 6 8 4 + T 6 A 4 2 1 2 N T 2 6

The digits remaining are 0, 3, 5, 7 and 9.
A total of 12 can be obtained as 5 + 7 or 3 + 9

Suppose (W, T) is (3, 9) then 6 + 1 + A = 9 yields A = 2 which is not possible
Suppose (W, T) is (9, 3) then 6 + 1 + A = 13 yields A = 6 which is not possible

So (W, T) is (7, 5) or (5, 7)
Suppose (W, T) is (7, 5) then 6 + 1 + A = 15 yields A = 8 which is not possible

Therefore (W, T) must be (5, 7) and 6 +1 + A = 7, yields A = 0.
The only 2 digits now remaining are 3 and 9.
Hence O has to be 3 and N has to be 9.

 5 3 6 8 4 + 7 6 0 4 2 1 2 9 7 2 6

W = 5, is the average of 1 to 9

Concept: Alphabet and Numeric Test
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