Enroll to CMAT 2023 Crash Course<\/a><\/p>\n Instructions<\/b><\/p>\n <p “=””>In each of the questions below, a group of numerals is given followed by four groups of symbol\/letter combinations numbered a:, b:, c: and d:. Numerals are to be coded as per the codes and conditions given below. You have to nd out which of the combinations a:, b:, c: and d: is correct and indicate your answer accordingly. If none of the four combinations represent the correct code, mark e: as your answer. Question 1:\u00a0<\/b>713540<\/p>\n a)\u00a0X%*BA#<\/p>\n b)\u00a0E%*BA#<\/p>\n c)\u00a0E%*BAR<\/p>\n d)\u00a0X%*BAR<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0713540<\/p>\n The last digit is ‘0’, hence condition (iii) is applied, and ‘0’ is coded as = ‘#’<\/p>\n Thus,\u00a0713540\u00a0:\u00a0E%*BA#<\/strong><\/p>\n => Ans – (B)<\/p>\n Question 2:\u00a0<\/b>487692<\/p>\n a)\u00a0$KEFM@<\/p>\n b)\u00a0AKEFM@<\/p>\n c)\u00a0AKEFM$<\/p>\n d)\u00a0$KEFM$<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number – 487692<\/p>\n The first and last digits are even, hence condition (ii) is followed and both ‘4’ and ‘2’ are coded as = ‘$’<\/p>\n Thus, 487692\u00a0: $KEFM$<\/strong><\/p>\n => Ans – (D)<\/p>\n Question 3:\u00a0<\/b>364819<\/p>\n a)\u00a0*FAK%X<\/p>\n b)\u00a0XFAK&M;<\/p>\n c)\u00a0*FAK%M<\/p>\n d)\u00a0*EAK%X<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0364819<\/p>\n The first and last digits are odd, hence condition (i) is followed and both ‘3’ and ‘9’ are coded as = ‘X’<\/p>\n Thus,\u00a0364819\u00a0: XFAK%X<\/strong><\/p>\n => Ans – (E)<\/p>\n Question 4:\u00a0<\/b>765082<\/p>\n a)\u00a0EFB#K@<\/p>\n b)\u00a0XFBRIK@<\/p>\n c)\u00a0EFBRK@<\/p>\n d)\u00a0EFBR#K<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0765082<\/p>\n None of the condition follows, all the digits are coded as given.<\/p>\n Thus,\u00a0765082\u00a0:\u00a0EFBRK@<\/strong><\/p>\n => Ans – (C)<\/p>\n Question 5:\u00a0<\/b>546839<\/p>\n a)\u00a0XAFK*X<\/p>\n b)\u00a0XAFK*M<\/p>\n c)\u00a0BAFK*X<\/p>\n d)\u00a0BAFK*M<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0546839<\/p>\n The first and last digits are odd, hence condition (i) is followed and both ‘5’ and ‘9’ are coded as = ‘X’<\/p>\n Thus,\u00a0546839\u00a0:\u00a0XAFK*X<\/strong><\/p>\n => Ans – (A)<\/p>\n Instructions<\/b><\/p>\n Study the following information to answer the given questions. In a certain code language, `economy and work related’ is written as `oj my bx st’ `work and employment today’ is written as `pk bx oj dy’ `employment for growth only’ is written as ‘el pk fd zn’ `growth is related today’ is written as ‘el dy gm my’ (All codes are two letter codes only)<\/p>\n Question 6:\u00a0<\/b>What is the code for ‘growth today’ in the given code language ?<\/p>\n a)\u00a0fd el<\/p>\n b)\u00a0dy fd<\/p>\n c)\u00a0pk dy<\/p>\n d)\u00a0dy el<\/p>\n e)\u00a0an fd<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The common words in first two statements are ‘work’ and ‘and’ coded as = ‘oj’ or ‘bx’<\/p>\n The common word in first and last statements is ‘related’ coded as = ‘my’<\/p>\n => Only word left in first statement is ‘economy’ coded as = ‘st’<\/p>\n The common word in second and third statements is ’employment’ coded as = ‘pk’<\/p>\n => Only word left in second statement is ‘today’ coded as = ‘dy’<\/p>\n The common common word in last two statements is ‘growth’ coded as = ‘el’<\/p>\n => Only words left in third statement are ‘for’ and ‘only’ coded as = ‘fd’ or ‘zn’<\/p>\n => Only word left in last statement is ‘is’ coded as = ‘gm’<\/p>\n Thus,\u00a0the code for ‘growth today’\u00a0=\u00a0dy el<\/strong><\/p>\n => Ans – (D)<\/p>\n Question 7:\u00a0<\/b>Which of the following additional statements is required to definitely find the code of `and’ in the given code language ?<\/p>\n a)\u00a0‘work and prosper now’ is written as `bx yp jn oj’<\/p>\n b)\u00a0‘work today also important’ is written as ‘Iv en oj dy’<\/p>\n c)\u00a0No additional statement is required to find the code<\/p>\n d)\u00a0`and more work today’ is written as `z1 oj dy bx’<\/p>\n e)\u00a0‘related only for employment’ is written as ‘my zn fd pk’<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The common words in first two statements are ‘work’ and ‘and’ coded as = ‘oj’ or ‘bx’<\/p>\n The common word in first and last statements is ‘related’ coded as = ‘my’<\/p>\n => Only word left in first statement is ‘economy’ coded as = ‘st’<\/p>\n The common word in second and third statements is ’employment’ coded as = ‘pk’<\/p>\n => Only word left in second statement is ‘today’ coded as = ‘dy’<\/p>\n The common common word in last two statements is ‘growth’ coded as = ‘el’<\/p>\n => Only words left in third statement are ‘for’ and ‘only’ coded as = ‘fd’ or ‘zn’<\/p>\n => Only word left in last statement is ‘is’ coded as = ‘gm’<\/p>\n Thus,\u00a0to definitely find the code of ‘and’, we need a statement that contains only one of ‘work’ or ‘and’<\/p>\n => Ans – (B)<\/p>\n Question 8:\u00a0<\/b>What may be the code for `economy is boosting’ in the given code language ?<\/p>\n a)\u00a0gm rc st<\/p>\n b)\u00a0zn gm st<\/p>\n c)\u00a0ye st el<\/p>\n d)\u00a0cp st rc<\/p>\n e)\u00a0st bx gm<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The common words in first two statements are ‘work’ and ‘and’ coded as = ‘oj’ or ‘bx’<\/p>\n The common word in first and last statements is ‘related’ coded as = ‘my’<\/p>\n => Only word left in first statement is ‘economy’ coded as = ‘st’<\/p>\n The common word in second and third statements is ’employment’ coded as = ‘pk’<\/p>\n => Only word left in second statement is ‘today’ coded as = ‘dy’<\/p>\n The common common word in last two statements is ‘growth’ coded as = ‘el’<\/p>\n => Only words left in third statement are ‘for’ and ‘only’ coded as = ‘fd’ or ‘zn’<\/p>\n => Only word left in last statement is ‘is’ coded as = ‘gm’<\/p>\n Thus,\u00a0the code for `economy is boosting’\u00a0may be =\u00a0gm rc st<\/strong><\/p>\n => Ans – (A)<\/p>\n Question 9:\u00a0<\/b>What does the code `pk’ stand for in the given code language ?<\/p>\n a)\u00a0growth<\/p>\n b)\u00a0employment<\/p>\n c)\u00a0only<\/p>\n d)\u00a0economy<\/p>\n e)\u00a0today<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The common words in first two statements are ‘work’ and ‘and’ coded as = ‘oj’ or ‘bx’<\/p>\n The common word in first and last statements is ‘related’ coded as = ‘my’<\/p>\n => Only word left in first statement is ‘economy’ coded as = ‘st’<\/p>\n The common word in second and third statements is ’employment’ coded as = ‘pk’<\/p>\n => Only word left in second statement is ‘today’ coded as = ‘dy’<\/p>\n The common common word in last two statements is ‘growth’ coded as = ‘el’<\/p>\n => Only words left in third statement are ‘for’ and ‘only’ coded as = ‘fd’ or ‘zn’<\/p>\n => Only word left in last statement is ‘is’ coded as = ‘gm’<\/p>\n Thus,\u00a0the code `pk’ stands for\u00a0=\u00a0employment<\/strong><\/p>\n => Ans – (B)<\/p>\n Question 10:\u00a0<\/b>If the code for ‘related people only’ is ‘xd my fd’ then what may be the code for ‘people for decision’ in the given code language ?<\/p>\n a)\u00a0to xd my<\/p>\n b)\u00a0zn xd fd<\/p>\n c)\u00a0zn xd dy<\/p>\n d)\u00a0zn kz xd<\/p>\n e)\u00a0kz fd xd<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The common words in first two statements are ‘work’ and ‘and’ coded as = ‘oj’ or ‘bx’<\/p>\n The common word in first and last statements is ‘related’ coded as = ‘my’<\/p>\n => Only word left in first statement is ‘economy’ coded as = ‘st’<\/p>\n The common word in second and third statements is ’employment’ coded as = ‘pk’<\/p>\n => Only word left in second statement is ‘today’ coded as = ‘dy’<\/p>\n The common common word in last two statements is ‘growth’ coded as = ‘el’<\/p>\n => Only words left in third statement are ‘for’ and ‘only’ coded as = ‘fd’ or ‘zn’<\/p>\n => Only word left in last statement is ‘is’ coded as = ‘gm’<\/p>\n If the code for ‘related people only’ is ‘xd my fd’, => code for ‘people’ = ‘xd’<\/p>\n Thus,\u00a0code for ‘people for decision’\u00a0=\u00a0zn kz xd<\/strong><\/p>\n => Ans – (D)<\/p>\n Instructions<\/b><\/p>\n <p “=””>In each question below is given a group of numbers\/symbols followed by live combinations of letter codes numbered a: b:c: d: and e:.You have to find out which of the combinations correctly represents the group of numbers\/symbols based on the following coding system and the conditions and mark the number of that combination as your answer : Question 11:\u00a0<\/b>7\u00a98$#&<\/p>\n a)\u00a0TGYDPC<\/p>\n b)\u00a0TGCPDC<\/p>\n c)\u00a0TGCDPC<\/p>\n d)\u00a0TGCPPC<\/p>\n e)\u00a0TSCDPC<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a07\u00a98$#&<\/p>\n The third element is an even number ‘8’ and the last is a symbol ‘&’, hence condition (i) is applied and ‘8’ is coded as = ‘C’<\/p>\n =\u00a0TGCDPC<\/p>\n => Ans – (C)<\/p>\n Question 12:\u00a0<\/b>3+@5#7<\/p>\n a)\u00a0HRLAPT<\/p>\n b)\u00a0LRAKPT<\/p>\n c)\u00a0LRHAPT<\/p>\n d)\u00a0LAHRPT<\/p>\n e)\u00a0LRHATP<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a03+@5#7<\/p>\n There \u00a0are no even numbers, hence condition (iii) is applied, thus ‘L’ and ‘H’ are interchanged.<\/p>\n =\u00a0LRHAPT<\/p>\n => Ans – (C)<\/p>\n Question 13:\u00a0<\/b>@7#29%<\/p>\n a)\u00a0LOWPEV<\/p>\n b)\u00a0LTPWEV<\/p>\n c)\u00a0LOPVEW<\/p>\n d)\u00a0LOPWEV<\/p>\n e)\u00a0LOPWEG<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0@7#29%<\/p>\n The odd number ‘7’ is immediately\u00a0preceded as well as immediately followed by symbols, hence condition (ii) is applied and ‘7’ is coded as = ‘O’<\/p>\n =\u00a0LOPWEV<\/p>\n =>\u00a0Ans – (D)<\/p>\n Question 14:\u00a0<\/b>2$%53+<\/p>\n a)\u00a0WDVHAR<\/p>\n b)\u00a0WVDAHR<\/p>\n c)\u00a0WDVAHG<\/p>\n d)\u00a0DVVVAHR<\/p>\n e)\u00a0WDVAHR<\/p>\n 14)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a02$%53+<\/p>\n None of the condition is followed, hence the code is<\/p>\n =\u00a0WDVAHR<\/p>\n => Ans – (E)<\/p>\n Question 15:\u00a0<\/b>%96&4$<\/p>\n a)\u00a0VESCZD<\/p>\n b)\u00a0VEDCZD<\/p>\n c)\u00a0EVDCZD<\/p>\n d)\u00a0VEDZCD<\/p>\n e)\u00a0VEDAZD<\/p>\n 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0%96&4$<\/p>\n The third element is an even number ‘6’ and the last is a symbol ‘$’, hence condition (i) is applied and ‘6’ is coded as = ‘D’<\/p>\n =\u00a0VEDCZD<\/p>\n => Ans – (B)<\/p>\n Instructions<\/b><\/p>\n In each question below is given a group of letters followed by five combinations of number\/symbol codes numbered a:, b:, c:, d: and e:. You have to find out which of the combinations correctly represents the group of letters based on the following coding system and the conditions and mark the number of that combination as your answer.<\/p>\n Conditions : Question 16:\u00a0<\/b>AUWDBE<\/p>\n a)\u00a095\u03b2992<\/p>\n b)\u00a025\u03b2693<\/p>\n c)\u00a03596\u03b22<\/p>\n d)\u00a035\u03b2629<\/p>\n e)\u00a035\u03b2692<\/p>\n 16)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n AUWDBE<\/p>\n As we can see, none of the conditions will apply in this case. Using the table, the code for AUWDBE is 35\u03b2692. Option E is the right answer.<\/p>\n Question 17:\u00a0<\/b>RPCUDH<\/p>\n a)\u00a044*568<\/p>\n b)\u00a0#4*568<\/p>\n c)\u00a0#46*58<\/p>\n d)\u00a0#4*56#<\/p>\n e)\u00a084*56#<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n RPCUDH<\/p>\n The string contains only one vowel. Hence, condition 2 will apply here. The codes for the first and last letters must be interchanged.<\/p>\n #4*568<\/p>\n Option B is the right answer.<\/p>\n Question 18:\u00a0<\/b>UTDARM<\/p>\n a)\u00a0@@638@<\/p>\n b)\u00a0@@6$83<\/p>\n c)\u00a0$@638@<\/p>\n d)\u00a0@@638$<\/p>\n e)\u00a0$638$<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n UTDARM<\/p>\n Here, the first letter is a vowel and the last letter is a consonant. Condition 3 will apply here. U must also take the code for T.<\/p>\n @@638$<\/p>\n Option D is the right answer.<\/p>\n Question 19:\u00a0<\/b>HEQAFK<\/p>\n a)\u00a0#3\u00a93%7<\/p>\n b)\u00a0#2\u00a92%7<\/p>\n c)\u00a0#2\u00a937%<\/p>\n d)\u00a072\u00a93%7<\/p>\n e)\u00a0#3\u00a92%7<\/p>\n 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n HEQAFK<\/p>\n Here, the second and fourth letters are vowels. Hence, condition 1 will apply. Second and fourth letters will take the code for A.<\/p>\n #3\u00a93%7<\/p>\n Hence, option A is the right answer.<\/p>\n Question 20:\u00a0<\/b>MFQBEK<\/p>\n a)\u00a07%92\u00a9$<\/p>\n b)\u00a07%\u00a992$<\/p>\n c)\u00a0$%\u00a9927<\/p>\n d)\u00a07%\u00a929$<\/p>\n e)\u00a07%\u00a9927<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n First, let us find out the additional conditions that will apply to MFQBEK.<\/p>\n Only condition 2 will apply since the string contains only 1 vowel.<\/p>\n M – $<\/p>\n F – %<\/p>\n Q – \u00a9<\/p>\n B – 9<\/p>\n E – 2<\/p>\n K – 7<\/p>\n The first and last characters are to be interchanged. Hence the correct answer is 7%\u00a992$. Option B is the right answer.<\/span><\/p>\n
\n<p “=””><img “=”” class=”img-responsive” src=”https:\/\/cracku.in\/media\/uploads\/73576.png”\/>Following conditions apply :
\n(i) If the first digit as well as the last digit is odd, both are to be coded as \u2018X\u2019.
\n(ii) If the first digit as well as the last digit is even, both are to be coded as \u2018$\u2019.
\n(iii) If the last digit is \u20180\u2019, it is to be coded as \u2018#\u2019.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/68302.png\")
Conditions :<\/strong>
\n(i) If the third element is an even number and the last element is a symbol then that even number is to be coded as the code for the symbol.
\n(ii) If an odd number is immediately preceded as well as immediately followed by symbols then the odd number is to be coded as \u2018O\u2019.
\n(iii) If there are no even numbers then the codes of the first and the third elements have to be interchanged.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/tabl.JPG\")
<\/p>\n
\n(i) If both the second and the fourth- elements are vowels, both these are to be coded as the code for the fourth vowel.
\n(ii) If the group of letters contains only one vowel, the codes for the first and the last letters are to be interchanged.
\n(iii) If first element is a vowel and the last a consonant, then that vowel is to be coded as the code for the letter following it.<\/p>\n