Download Coding Decoding Questions for MAH-CET <\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Instructions<\/b><\/p>\n Read the following information and answer the given questions.Symbolise the given number and symbol. Question 1:\u00a0<\/b>Which of the following is symbol of 673258?<\/p>\n a)\u00a0%+?*$!<\/p>\n b)\u00a0@+?*$!<\/p>\n c)\u00a0@?+$*!<\/p>\n d)\u00a0%+?*$\u00a9<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0673258<\/p>\n The number ends by an even number, thus condition (ii) is applied and the even number is coded as =\u00a0\u00a9<\/p>\n Thus,\u00a0673258\u00a0:\u00a0%+?*$\u00a9<\/strong><\/p>\n => Ans – (D)<\/p>\n Question 2:\u00a0<\/b>Which of the following is the symbol of 236475?<\/p>\n a)\u00a0?*%&x;@<\/p>\n b)\u00a0\u00a9 ?%&x;@<\/p>\n c)\u00a0x?%&+@<\/p>\n d)\u00a0\u00a9 ?%&+$<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0236475<\/p>\n None of the conditions are applied, the codes are written as given.<\/p>\n Thus,\u00a0236475\u00a0:\u00a0*?%&+$<\/strong><\/p>\n => Ans – (E)<\/p>\n Question 3:\u00a0<\/b>Which of the following will be the symbol of 178524?<\/p>\n a)\u00a0@+!$*\u00a9<\/p>\n b)\u00a0@x+!$\u00a9<\/p>\n c)\u00a0@+!$*&<\/p>\n d)\u00a0*+!$&\u00a9<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0178524<\/p>\n The number begins by odd number and ends with an even number, thus both conditions are applied.<\/p>\n Thus,\u00a0178524\u00a0:\u00a0@+!$*\u00a9<\/strong><\/p>\n => Ans – (A)<\/p>\n Question 4:\u00a0<\/b>Which of the following will be the symbol of 178524?<\/p>\n a)\u00a0@+!$*\u00a9<\/p>\n b)\u00a0@x+!$\u00a9<\/p>\n c)\u00a0@+!$*&<\/p>\n d)\u00a0*+!$&\u00a9<\/p>\n e)\u00a0None of these<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a0178524<\/p>\n The number begins by odd number and ends with an even number, thus both conditions are applied.<\/p>\n Thus,\u00a0178524\u00a0:\u00a0@+!$*\u00a9<\/strong><\/p>\n => Ans – (A)<\/p>\n Question 5:\u00a0<\/b>Which of the following will be the symbol of 25486 ?<\/p>\n a)\u00a0*$ &!\u00a3<\/p>\n b)\u00a0*$&!\u00a9<\/p>\n c)\u00a0\u00a9 $&!%<\/p>\n d)\u00a0*$!&%<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number –\u00a025486<\/p>\n The number ends by an even number, thus condition (ii) is applied and the even number is coded as =\u00a0\u00a9<\/p>\n Thus,\u00a025486\u00a0:\u00a0*$&!\u00a9<\/strong><\/p>\n => Ans – (B)<\/p>\n Instructions<\/b><\/p>\n <\/p>\n CONDITIONS:<\/p>\n I. if the first as well the last digit is odd ,both are coded as \u00a9 Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>468910<\/p>\n a)\u00a0$HJ@KL<\/p>\n b)\u00a0LHJ@K$<\/p>\n c)\u00a0\u2605HJ@K$<\/p>\n d)\u00a0$HJ@K\u2605<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0: 468910<\/p>\n Since the last digit is ‘0’, rule (iii) will be applied and 0 is coded as =\u00a0\u2605<\/p>\n => Code for 468910 =\u00a0$HJ@K\u2605<\/p>\n => Ans – (D)<\/p>\n Question 7:\u00a0<\/b>592476<\/p>\n a)\u00a0H@Q$%T<\/p>\n b)\u00a0\u00a9@Q$%T<\/p>\n c)\u00a0H@Q$%\u00a9<\/p>\n d)\u00a0\u00a9@Q$%\u00a9<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0592476<\/p>\n Since, both the first and last digits are neither odd nor even. Also, last digit is not ‘0’. Thus. no rule is applied.<\/p>\n Code for\u00a0592476 = T@Q$%H<\/p>\n => Ans – (E) : none of these<\/p>\n Question 8:\u00a0<\/b>875306<\/p>\n a)\u00a0J%T#LH<\/p>\n b)\u00a0H%T#LH<\/p>\n c)\u00a0H%T#LJ<\/p>\n d)\u00a0J%T#LJ<\/p>\n e)\u00a0NONE OF THESE<\/p>\n 8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0875306<\/p>\n Since, the first and the last digits are even and the last digit is not ‘0’, thus rule (ii) will be applied and codes for 8 and 6 are swapped.<\/p>\n => Code for\u00a0875306 =\u00a0H%T#LJ<\/p>\n => Ans – (C)<\/p>\n Question 9:\u00a0<\/b>364279<\/p>\n a)\u00a0\u00a9H$Q%\u00a9<\/p>\n b)\u00a0#H$Q%@<\/p>\n c)\u00a0@H$Q%#<\/p>\n d)\u00a0#H$Q%&<\/p>\n e)\u00a0None of these<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0364279<\/p>\n Since, the first and the last digits are both odd, thus rule (i) is applied and both are coded as\u00a0\u00a9<\/p>\n => Code for\u00a0364279 =\u00a0\u00a9H$Q%\u00a9<\/p>\n => Ans – (A)<\/p>\n Question 10:\u00a0<\/b>270514<\/p>\n a)\u00a0\u2605%LTK<\/p>\n b)\u00a0$%LTKQ<\/p>\n c)\u00a0Q%LTK$<\/p>\n d)\u00a0$%\u2605TKQ<\/p>\n e)\u00a0None of these<\/p>\n 10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0270514<\/p>\n Since, the first and the last digits are even and the last digit is not ‘0’, thus rule (ii) will be applied and codes for 2 and 4 are swapped.<\/p>\n => Code for\u00a0270514 =\u00a0$%LTKQ<\/p>\n => Ans – (B)<\/p>\n Instructions<\/b><\/p>\n Study the following information carefully and answer the questions given below: Question 11:\u00a0<\/b>What is the code for \u2018available\u2019 in the given code language?<\/p>\n a)\u00a0either \u2018pz\u2018 or \u2018fo\u2019<\/p>\n b)\u00a0either `nb\u2018 or `mk\u2018<\/p>\n c)\u00a0li<\/p>\n d)\u00a0hu<\/p>\n e)\u00a0Other than those given as options<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n From statement 1 & 2, common word is ‘few’ coded as ‘gs’.<\/p>\n Similarly, from 2\u00a0& 3, code for ‘fertilizer = nb’<\/p>\n Now, in statement 2, two words are left, i.e. ‘products’ and ‘available’ coded as ‘fo’ or ‘pz’ interchangeably.<\/p>\n Thus, code for available = ‘pz’ or ‘fo’<\/p>\n => Ans – (A)<\/p>\n Question 12:\u00a0<\/b>If \u2018waste management techniques\u2019 is coded as \u2018ax da mk\u2019 in the given code language, then how will \u2018farming fertilizer management\u2019 be coded as?<\/p>\n a)\u00a0ax nb cr<\/p>\n b)\u00a0liaxpt<\/p>\n c)\u00a0gs nb li<\/p>\n d)\u00a0nb ax li<\/p>\n e)\u00a0Other than those given as options<\/p>\n 12)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n From statement 1 & 2, common word is ‘few’ coded as ‘gs’.<\/p>\n From 1\u00a0& 4, code for ‘farming = li’<\/p>\n From 1\u00a0& 3, code for ‘organic = cr’<\/p>\n From above statements, in statement 1, code for ‘techniques = da’<\/p>\n Similarly, from 2\u00a0& 3, code for ‘fertilizer = nb’<\/p>\n From 3 & 4, code for ‘waste = mk’<\/p>\n Now, it is given that code for ‘waste management techniques’ = ‘ax da mk’<\/p>\n => Since, we already know codes for waste and technique, => code for ‘management = ax’<\/p>\n Thus, code for ‘farming fertilizer management’ is any combination of ‘li nb ax’<\/p>\n => Ans – (D)<\/p>\n Question 13:\u00a0<\/b>In the given code language, what does the code \u2018yu. stand for?<\/p>\n a)\u00a0farming<\/p>\n b)\u00a0techniques<\/p>\n c)\u00a0either \u2018of\u2019 or \u2018disposal\u2019<\/p>\n d)\u00a0waste<\/p>\n e)\u00a0either \u2018into\u2019 or \u2018few\u2019<\/p>\n 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n From statement 1 & 4, common word is ‘farming’ coded as ‘li’.<\/p>\n Similarly, from 3 & 4, code for ‘waste = mk’<\/p>\n Now, in statement 4, two words are left, i.e. ‘disposal’ and ‘of’ coded as ‘hu’ or ‘yu’ interchangeably.<\/p>\n Thus, the code ‘yu’ stans for = ‘disposal’ or ‘of’<\/p>\n => Ans – (C)<\/p>\n Question 14:\u00a0<\/b>What is the code for \u2018organic\u2019 in the given code language?<\/p>\n a)\u00a0gs<\/p>\n b)\u00a0cr<\/p>\n c)\u00a0da<\/p>\n d)\u00a0pt<\/p>\n e)\u00a0Other than those given as options<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n From statements 1 and 3, the common word is organic.<\/p>\n Thus, organic is coded as = ‘cr’<\/p>\n => Ans – (B)<\/p>\n Question 15:\u00a0<\/b>What will be the code for \u2018few waste\u2019 in the given code language?<\/p>\n a)\u00a0gs li<\/p>\n b)\u00a0pt da<\/p>\n c)\u00a0da mk<\/p>\n d)\u00a0Other than those given as options<\/p>\n e)\u00a0mk gs<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n From statement 1 & 2, common word is ‘few’ coded as ‘gs’.<\/p>\n Similarly, from 3 & 4, code for ‘waste = mk’<\/p>\n Thus, code for ‘few waste’ is the combination of = ‘mk gs’<\/p>\n => Ans – (E)<\/p>\n
\nNumber \u00a0 \u00a01 \u00a0 2 \u00a0 3 \u00a0 4 \u00a0 \u00a05 \u00a0 6 \u00a07 \u00a0 8 \u00a0 \u00a09
\nSymbol \u00a0 \u00a0 x \u00a0 \u00a0* \u00a0 ? \u00a0 \u00a0& \u00a0 $ \u00a0% \u00a0+ \u00a0 ! \u00a0 \u00a0 \u00a3
\nConditions:
\nI. if any number begins by the odd number then the odd number symbolised by @
\nII. if any number ends by even number then the even number symbolised as \u00a9<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/DeepinScreenshot20171024162118.png\")
<\/figure>\n
\nIi. if the first as well as last digit is even ,thier codes are to be swapped
\nIii. if \u2018O\u2019 is the last digit ,it is to be code as \u2605<\/p>\n
\nIn a certain code language,
\n\u2018few organic farming techniques\u2019 is written as \u2018Ii gs da Cr\u2019
\n\u2018fertilizer products few available\u2019 Is written \u2018fo pz nb gs\u201d
\n\u2018organic waste into fertilizer\u2019 is written \u2018nb Cr pt mk\u2019
\n\u2018disposal of farming waste\u2019 is written as \u2018hu mk li yu\u2019
\n(All codes are two letter codes only)<\/p>\n