DILR Test 1 Practice

F + F = F.

This implies that F has to be 0. Since all the digits are less than 10, the maximum sum of any two letters will be 17 (i.e. 9 + 8).
Therefore, a maximum of 1 can be carried over to the digits place on the left. In the ten thousands place, either the units digit of H + H = F or the units digit of H + H + 1 = F.
It cannot be H + H + 1 because 2H + 1 is an odd number and the units digit of that cannot be 0. From this, we can infer that the units digit of 2H must be 0.
This implies that H = 5. The sum of both the 6-digit numbers is a 7-digit number. This implies that the leftmost digit in the 7-digit number has to be 1. Therefore. A = 1.
Since H = 5, H + H = 10. This means that B + A + 1 = AA = 10A + A. A = 1. Therefore. B = 9.
In the hundreds place the units place of the sum A + F is C. Now A = 1 and F = 0.
But C cannot be 1. Hence, C has to be A + 1, i.e. C = 2. Therefore, G + K should be greater than 10.
The units digit of the sum of G + K is 1. This implies that G and K are either 3 and 8 or 4 and 7. not necessarily in that order.
We now have the following:

Since G is a single digit, J + 1 is less than 9.
If J = 3. G = 4 and K = 7. In that case. D and E will be 6 and 8. not necessarily in that order.
If j = 4. G = 5. which is not possible because H = 5 and each letter has a distinct number.
If J = 6. G = 7 and K = 4. In that case. D and E will be 3 and 8. not necessarily in that order.
If j = 7. G = 8 and K = 3. In that case, D and E will be 4 and 6 not necessarily in that order.
If J = 8, G = 9, which is not possible because B = 9 and each letter has a distinct number.

F + F = F.

This implies that F has to be 0. Since all the digits are less than 10, the maximum sum of any two letters will be 17 (i.e. 9 + 8).
Therefore, a maximum of 1 can be carried over to the digits place on the left. In the ten thousands place, either the units digit of H + H = F or the units digit of H + H + 1 = F.
It cannot be H + H + 1 because 2H + 1 is an odd number and the units digit of that cannot be 0. From this, we can infer that the units digit of 2H must be 0.
This implies that H = 5. The sum of both the 6-digit numbers is a 7-digit number. This implies that the leftmost digit in the 7-digit number has to be 1. Therefore. A = 1.
Since H = 5, H + H = 10. This means that B + A + 1 = AA = 10A + A. A = 1. Therefore. B = 9.
In the hundreds place the units place of the sum A + F is C. Now A = 1 and F = 0.
But C cannot be 1. Hence, C has to be A + 1, i.e. C = 2. Therefore, G + K should be greater than 10.
The units digit of the sum of G + K is 1. This implies that G and K are either 3 and 8 or 4 and 7. not necessarily in that order.
We now have the following:

Since G is a single digit, J + 1 is less than 9.
If J = 3. G = 4 and K = 7. In that case. D and E will be 6 and 8. not necessarily in that order.
If j = 4. G = 5. which is not possible because H = 5 and each letter has a distinct number.
If J = 6. G = 7 and K = 4. In that case. D and E will be 3 and 8. not necessarily in that order.
If j = 7. G = 8 and K = 3. In that case, D and E will be 4 and 6 not necessarily in that order.
If J = 8, G = 9, which is not possible because B = 9 and each letter has a distinct number.

F + F = F.

This implies that F has to be 0. Since all the digits are less than 10, the maximum sum of any two letters will be 17 (i.e. 9 + 8).
Therefore, a maximum of 1 can be carried over to the digits place on the left. In the ten thousands place, either the units digit of H + H = F or the units digit of H + H + 1 = F.
It cannot be H + H + 1 because 2H + 1 is an odd number and the units digit of that cannot be 0. From this, we can infer that the units digit of 2H must be 0.
This implies that H = 5. The sum of both the 6-digit numbers is a 7-digit number. This implies that the leftmost digit in the 7-digit number has to be 1. Therefore. A = 1.
Since H = 5, H + H = 10. This means that B + A + 1 = AA = 10A + A. A = 1. Therefore. B = 9.
In the hundreds place the units place of the sum A + F is C. Now A = 1 and F = 0.
But C cannot be 1. Hence, C has to be A + 1, i.e. C = 2. Therefore, G + K should be greater than 10.
The units digit of the sum of G + K is 1. This implies that G and K are either 3 and 8 or 4 and 7. not necessarily in that order.
We now have the following:

Since G is a single digit, J + 1 is less than 9.
If J = 3. G = 4 and K = 7. In that case. D and E will be 6 and 8. not necessarily in that order.
If j = 4. G = 5. which is not possible because H = 5 and each letter has a distinct number.
If J = 6. G = 7 and K = 4. In that case. D and E will be 3 and 8. not necessarily in that order.
If j = 7. G = 8 and K = 3. In that case, D and E will be 4 and 6 not necessarily in that order.
If J = 8, G = 9, which is not possible because B = 9 and each letter has a distinct number.

F + F = F.

This implies that F has to be 0. Since all the digits are less than 10, the maximum sum of any two letters will be 17 (i.e. 9 + 8).
Therefore, a maximum of 1 can be carried over to the digits place on the left. In the ten thousands place, either the units digit of H + H = F or the units digit of H + H + 1 = F.
It cannot be H + H + 1 because 2H + 1 is an odd number and the units digit of that cannot be 0. From this, we can infer that the units digit of 2H must be 0.
This implies that H = 5. The sum of both the 6-digit numbers is a 7-digit number. This implies that the leftmost digit in the 7-digit number has to be 1. Therefore. A = 1.
Since H = 5, H + H = 10. This means that B + A + 1 = AA = 10A + A. A = 1. Therefore. B = 9.
In the hundreds place the units place of the sum A + F is C. Now A = 1 and F = 0.
But C cannot be 1. Hence, C has to be A + 1, i.e. C = 2. Therefore, G + K should be greater than 10.
The units digit of the sum of G + K is 1. This implies that G and K are either 3 and 8 or 4 and 7. not necessarily in that order.
We now have the following:

Since G is a single digit, J + 1 is less than 9.
If J = 3. G = 4 and K = 7. In that case. D and E will be 6 and 8. not necessarily in that order.
If j = 4. G = 5. which is not possible because H = 5 and each letter has a distinct number.
If J = 6. G = 7 and K = 4. In that case. D and E will be 3 and 8. not necessarily in that order.
If j = 7. G = 8 and K = 3. In that case, D and E will be 4 and 6 not necessarily in that order.
If J = 8, G = 9, which is not possible because B = 9 and each letter has a distinct number.

The distribution of subjects based on the information given is as follows:

The only female student is P.

The distribution of subjects based on the information given is as follows:

The only female student is P.

The distribution of subjects based on the information given is as follows:

The only female student is P.

The distribution of subjects based on the information given is as follows:

The only female student is P.

The distribution of subjects based on the information given is as follows:

The only female student is P.

This question actually expects the test takers to identify the number of pairs of letters in a word in which the number of letters between the first letter and the second letter of a pair must have the same number of letters as in the English alphabet. For example, the positional difference between E and I in English alphabet is three (F,G and H are between them); the positional difference between E and I in the word CHILDREN is also three (L,D and R are between them). Hence, (E, I) is one such pair. Similarly, in the word ‘CHILDREN’, the other pairs can be identified as below: • (H, I): The number of letter between ‘H’ and ‘I’ in English alphabet is zero as there is no letter between ‘H’ and ‘I’. Similarly in the word ‘CHILDREN’ the number of letter between ‘H’ and ‘I’ is also zero as there is no letter between ‘H’ and ‘I’. • (H, N): The number of letter between ‘H’ and ‘N’ in English alphabet is five as there are five letters between ‘H’ and ‘N’. Similarly in the word ‘CHILDREN’ the number of letter between ‘H’ and ‘N’ is also five as there are five letters between ‘H’ and ‘N’. (I, N) : The number of letter between ‘I’ and ‘N’ in English alphabet is four as there are four letters between ‘I’ and ‘N’. Similarly in the word ‘CHILDREN’ the number of letter between ‘I’ and ‘N’ is also four as there are four letters between ‘I’ and ‘N’. Therefore, there are four such pairs – (E, I), (H, I), (H, N) and (I, N). The correct answer is C.