1
Question
Column I Column II T A P E m o s j S O U P i h h p o k T I E n i s SAY nkpl T R E A T m s r Y E A R h r p S I P n h p k T Y R E k h i p h
Find the codes of the following words with the help of above table
(1) S T A R (2)R O P E (3)T A S T Y (4) P E R S U E (5) S U R P R I S E
| Column I | Column II |
| T A P E | m o s j |
| S O U P | i h h p o k |
| T I E | n i s |
| SAY | nkpl |
| T R E A T | m s r |
| Y E A R | h r p |
| S I P | n h p k |
| T Y R E | k h i p h |
(1) S T A R (2)R O P E (3)T A S T Y (4) P E R S U E (5) S U R P R I S E
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Solution
The correct option is A 0
In the above problem the only six letters word is R O T A T E so its code is I h h p o k since it is the only six letters code Further T is repeated in the word and h is repeated in its code so the code of T is h
the only five letter word is T R E A T so its code is k h l p h
In the two words T R E A T and R O T A T E the letters T R E A and T are common in R O T A T E O is the extra letters Hence the code for O is o
T I E is the only three letter word containing T since we know the code of T is h we can conclude that the code of the word T I E is hrp both the word T I E and S I P have I in common similarly both the codes h r p and m s r have r in common this implies that the code of the word S I P is m s r and the code of I is r now we know the codes of the letters T and I in the word T I E the only letter left out is E we can easily conclude that the code of E is P
the only three letter word left out now is S A Y so the code of S A Y is n l s
in S A Y and S I P the common letter is S and the common code in their codes is s this implies that s is the code of S
We know that the code for O is o the only four letter word in which we find O is S O U P hence the code of S O U P is m o s j
Both the words T A P E and S O U P have P thus the code of T A P E ks p m l h and the code of the letter p is m (m is common in p m l h and m o i j )
Out of the two remaining words T Y R E and Y E A R first word contains T we know the code of T is h so the code of T Y R E is n h p k thus the code of Y E A R is n k p l further these.two words have the letters Y R and E in common and their codes have the codes n k and p in common so the code of A is L
Comparing the codes of S A Y (n l s) and T Y R E ( n g p k) we can conclude that the code of Y is n Now we know the codes of Y is n Now we know the codes of T Y and E in the word T Y R E and they are h and p respectively this implies that the code of R is k
Now consider the word S O U P the code of which is m o s j we already know the codes of S O and P they are s o and m respectively. Hence the cide of U is j
on rearranging the words along with their appropriate codes the table appears as folows.
In the above problem the only six letters word is R O T A T E so its code is I h h p o k since it is the only six letters code Further T is repeated in the word and h is repeated in its code so the code of T is h
the only five letter word is T R E A T so its code is k h l p h
In the two words T R E A T and R O T A T E the letters T R E A and T are common in R O T A T E O is the extra letters Hence the code for O is o
T I E is the only three letter word containing T since we know the code of T is h we can conclude that the code of the word T I E is hrp both the word T I E and S I P have I in common similarly both the codes h r p and m s r have r in common this implies that the code of the word S I P is m s r and the code of I is r now we know the codes of the letters T and I in the word T I E the only letter left out is E we can easily conclude that the code of E is P
the only three letter word left out now is S A Y so the code of S A Y is n l s
in S A Y and S I P the common letter is S and the common code in their codes is s this implies that s is the code of S
We know that the code for O is o the only four letter word in which we find O is S O U P hence the code of S O U P is m o s j
Both the words T A P E and S O U P have P thus the code of T A P E ks p m l h and the code of the letter p is m (m is common in p m l h and m o i j )
Out of the two remaining words T Y R E and Y E A R first word contains T we know the code of T is h so the code of T Y R E is n h p k thus the code of Y E A R is n k p l further these.two words have the letters Y R and E in common and their codes have the codes n k and p in common so the code of A is L
Comparing the codes of S A Y (n l s) and T Y R E ( n g p k) we can conclude that the code of Y is n Now we know the codes of Y is n Now we know the codes of T Y and E in the word T Y R E and they are h and p respectively this implies that the code of R is k
Now consider the word S O U P the code of which is m o s j we already know the codes of S O and P they are s o and m respectively. Hence the cide of U is j
on rearranging the words along with their appropriate codes the table appears as folows.
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