type 10 conditional coding & decoding Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (e)

When condition is not applied, the coding is done as follows.

3 → * 6 → F 4 → A 8 → K 1 → % 9 → M

But here the first and the last digits are odd, therefore condition (i) is applied here.

As condition (i) is applied here 3 and 9 will be coded as X.

3 → X 6 → F 4 → A 8 → K 1 → % 9 → X

Hence code for 364819 ⇒ XFAK%X

Answer: (a)

When condition is not applied, the coding is done as follows.

5 → B 4 → A 6 → F 8 → K 3 → * 9 → 1

But here the first and the last digits are odd, therefore condition (i) is applied here.

As condition (i) is applied here 5 and 9 will be coded as X. 5 → X 4 → A 6 → F 8 → K 3 → * 9 → X , X A F K * X.

Hence code for 546839 ⇒ XAFK*X

Answer: (c)

Here, none of the condition is applied,

so the coding will be done as follows.

7 → E 6 → F 5 → B 0 → R 8 → K 2 → @

Hence, code for 765082 ⇒ EFBRK@.

Answer: (b)

When condition is not applied, the coding is done as follows.

7 → E 1 → % 3 → * 5 → B 4 → A 0 → R

But here the last digit is 0, therefore condition (iii) is applied here.

As condition (iii) is applied here,

0 will be coded as # 7 → E 1 → % 3 → * 5 → B 4 → A 0 → #

Hence, code for 713540 ⇒ E%*BA#.

Answer: (b)

When the condition is not applied, the coding is done as follows.

4 → A, 8 → K, 7 → E, 6 → F, 9 → M, 2 → @.

But here the first and the last digits are even, therefore condition (ii) is applied here.

As condition (ii) is applied here, 4 and 2 will be coded as &.

4 → &, 8 → 8, 7 → E, 6 → F, 9 → M, 2 →&.

Hence, code for 487692 ⇒ &KEFM&.