Practice Question - 32 (Quant Based DI) | 100 DILR Questions for CAT Preparation PDF Download

The value of F can only be 0 as F+F=F can only hold if F=0.

Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)

So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1  => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:

The letter A represents 1.

The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11  => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1  => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:

The letter B represents 9.

The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1 => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:

In all possible cases 7 is already represented by a letter other than D. Hence 7 is the answer.

The value of F can only be 0 as F+F=F can only hold if F=0.
Also, A can only be 1(in the second column) because to get a carry of more than 1, B has to be a double-digit number which is not possible. (A carry is a digit that is transferred from one column of digits to another column of more significant digits.)
So the data can be tabulated as follows:

Since the last row in the third column is 0, the carry to the second column must have been 1, Hence B+1+1=11 => B=9
In the 4th column, H+H = 10 since a carry 1 has gone to the 3rd column. Hence H=5.
G+K must be 11 and the carry 1 goes to the next column, so C=1+1=2.
Now, G,K can take values (3,8), (4,7) and (5,6) in any order.
From 5th column G=J+1  => J=G-1
Case: G=3 and K=8, here J =2 which is not possible as C =2
Case: G=8 and K=3, J=7, a possible case.
Case: G=4 and K=7, J=3 possible
Case: G=7 and K=4, J=6 possible
Case: G=5 and K=6, J=4 not possible as H =5.
Case: G=6 and K=5, J=5 both J and K are same, not possible.
Hence the cases can be tabulated as follows:

From the table it is clear that 6 cannot be represented by G.