CAT 2025 Lesson : Numerical Reasoning - Case 8

Read the following and answer the questions that follow

The following table represents addition of two six-digit numbers given in the first and the second rows, while the sum is given in the third row. In the representation, each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 has been coded with one letter among A, B, C, D, E, F, G, H, J, K, with distinct letters representing distinct digits.



[CAT 2019 S1]

26) Which digit does the letter A represent?

Answer:

27) Which digit does the letter B represent?

Answer:

28) Which among the digits 3, 4, 6 and 7 cannot be represented by the letter D?

Answer:

29) Which among the digits 4, 6, 7 and 8 cannot be represented by the letter G?

Answer:

Solution

26)


As stated, Digits from 0 to 9 are coded to represent two six-digit numbers and added. In addition, remember to consider the carry up(For ex: 8 +7 =15, here the carry up is 1).

Looking at the table, A & F are the most repeating letters.

The unit's digit of the numbers are added to get the same digit.
F + F = F which gives a hint that F is an even number. Write down all possible combinations.

0 + 0 = 0
2 + 2 = 4
4 + 4 = 8
6 + 6 =12
8 + 8 =16

Hence F = 0 $\longrightarrow$(1)is the only possible digit. Fill F in the table.



The digit H is interesting.
H + H = F

The only possible digit satisfying the equation is 5 + 5 = 10. Hence H = 5.$\longrightarrow$(2)



Consider B + A = A and the seventh letter is also A. Here the Seventh digit is the carry up, The maximum possible single digit addition carry up is 1( 9 + 9 =18). Hence A = 1.$\longrightarrow$(3)

Answer: 1

27)


As stated, Digits from 0 to 9 are coded to represent two six-digit numbers and added. In addition, remember to consider the carry up(For ex: 8 +7 =15, here the carry up is 1).

Looking at the table, A & F are the most repeating letters.

The unit's digit of the numbers are added to get the same digit.
F + F = F which gives a hint that F is an even number. Write down all possible combinations.

0 + 0 = 0
2 + 2 = 4
4 + 4 = 8
6 + 6 =12
8 + 8 =16

Hence F = 0 $\longrightarrow$(1) is the only possible digit. Fill F in the table.



The digit H is interesting.
H + H = F

The only possible digit satisfying the equation is 5 + 5 = 10. Hence H = 5.$\longrightarrow$(2)



Consider B + A = A and the seventh letter is also A. Here the Seventh digit is the carry up, The maximum possible single digit addition carry up is 1(9 + 9 =18). Hence A = 1$\longrightarrow$(3)

B + A = A
from 1.

– + 1 + (1)= 11
9 + 1 + (1)= 11

Hence B = 9$\longrightarrow$(4)

Answer: 9

28)


As stated, Digits from 0 to 9 are coded to represent two six-digit numbers and added. In addition, remember to consider the carry up( For ex: 8 +7 =15, here the carry up is 1).

Looking at the table, A & F are the most repeating letters.

The unit's digit of the numbers are added to get the same digit.
F + F = F which gives a hint that F is an even number. Write down all possible combinations.

0 + 0 = 0
2 + 2 = 4
4 + 4 = 8
6 + 6 =12
8 + 8 =16

Hence F = 0 $\longrightarrow$(1) is the only possible digit. Fill F in the table.



The digit H is interesting.
H + H = F

The only possible digit satisfying the equation is 5 + 5 = 10. Hence H = 5.$\longrightarrow$(2)



Consider B + A = A and the seventh letter is also A. Here the Seventh digit is the carry up, The maximum possible single digit addition carry up is 1( 9 + 9 =18). Hence A = 1$\longrightarrow$(3).

B + A = A
from 1.

– + 1 + (1)= 11
9 + 1 + (1)= 11

Hence B = 9$\longrightarrow$(4).

G + K = A . Here G + K = 11 is the only possible equation which gives a carry up 1.



Consider A + F = C
Substitute values of A & F in the above equation.

1 + 0 + (1) = 2
Hence C = 2$\longrightarrow$(5).



Frame equations to solve the rest.



G + K = 11
K = 11 - G $\longrightarrow$(6)
1 + J +(0) = G
G = J +1 $\longrightarrow$(7)

The left over digits are 3,4,6,7 & 8 for letters J, G, K, D & E

Substitute values for J.
(i) J = 3, Then G = 4 & K = 7 $\longrightarrow$(8)
(ii) J = 4, Then G = 5 is rejected as H = 5
(iii) J = 6, Then G = 7 & K = 4 $\longrightarrow$(9)
(iv) J = 7, Then G = 8 & K= 3 $\longrightarrow$(10)

From 8, 9 & 10
D & E can either be represented by 3,4,6 & 8

Hence D cannot be represented by 7.

Answer: 7

29)


As stated, Digits from 0 to 9 are coded to represent two six-digit numbers and added. In addition, remember to consider the carry up( For ex: 8 +7 =15, here the carry up is 1).

Looking at the table, A & F are the most repeating letters.

The unit's digit of the numbers are added to get the same digit.
F + F = F which gives a hint that F is an even number. Write down all possible combinations.

0 + 0 = 0
2 + 2 = 4
4 + 4 = 8
6 + 6 =12
8 + 8 =16

Hence F = 0 $\longrightarrow$(1) is the only possible digit. Fill F in the table.



The digit H is interesting.
H + H = F

The only possible digit satisfying the equation is 5 + 5 = 10. Hence H = 5.$\longrightarrow$(2)



Consider B + A = A and the seventh letter is also A. Here the Seventh digit is the carry up, The maximum possible single digit addition carry up is 1( 9 + 9 =18). Hence A = 1$\longrightarrow$(3).

B + A = A
from 1.

– + 1 + (1)= 11
9 + 1 + (1)= 11

Hence B = 9$\longrightarrow$(4).

G + K = A . Here G + K = 11 is the only possible equation which gives a carry up 1.



Consider A + F = C
Substitute values of A & F in the above equation.

1 + 0 + (1) = 2
Hence C = 2$\longrightarrow$(5).



Frame equations to solve the rest.



G + K = 11
K = 11- G $\longrightarrow$(6)
1 + J +(0) = G
G = J +1 $\longrightarrow$(7)

The left over digits are 3,4,6,7 & 8 for letters J, G, K, D & E

Substitute values for J .
(i) J = 3, Then G = 4 & K = 7 $\longrightarrow$(8)
(ii) J = 4, Then G = 5 is rejected as H = 5
(iii) J = 6, Then G = 7 & K = 4 $\longrightarrow$(9)
(iv) J = 7, Then G = 8 & K= 3 $\longrightarrow$(10)

From 8, 9 & 10
D & E can either be represented by 3,4,6 & 8

Hence G cannot be represented by 6.

Answer: 6

Answer:
26) 1
27) 9
28) 7
29) 6