Here you can download a free Series Completion questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Series Completion of answers for the given questions. These questions will help you to make practice and solve the Series Completion<\/strong>\u00a0questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Series Completion MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Series Completion Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Instructions<\/b><\/p>\n Answer the following questions based on the alphabet given below : Question 1:\u00a0<\/b>If the alphabets are written in reverse order after interchanging alphabets from \u2018D to L\u2019 with those from \u2018R to Z\u2019 respectively, which letter would be midway between W and E in the new order ?<\/p>\n a)\u00a0Z<\/p>\n b)\u00a0N<\/p>\n c)\u00a0L<\/p>\n d)\u00a0There is no such letter<\/p>\n e)\u00a0None of these<\/p>\n 1)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Alphabets\u00a0:\u00a0A B C D E F G H I J K L M N O P Q R S T U V W X Y Z<\/p>\n After interchanging alphabets from \u2018D to L\u2019 with those from \u2018R to Z\u2019 respectively, then\u00a0:<\/p>\n = A B C R S T U V W X Y Z M N O P Q D E F G H I J K L<\/p>\n Letter which is midway between W and E =\u00a0N<\/strong><\/p>\n => Ans – (B)<\/p>\n Question 2:\u00a0<\/b>What will come in the place of question (?) mark in the following series ? a)\u00a0GIHPSUW<\/p>\n b)\u00a0GIPQSUW<\/p>\n c)\u00a0GIKPSUW<\/p>\n d)\u00a0GIJPSUV<\/p>\n e)\u00a0None of these<\/p>\n 2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series\u00a0:\u00a0GPW, GPUW, GIPUW, GIPSUW, ?<\/p>\n We are adding one letter from the back\u00a0<\/span>series, and similarly, adding one letter\u00a0<\/span>from the front side.\u00a0<\/span><\/p>\n Eg\u00a0:- In 2nd term, U is added at the second last position and in the next term after that I is added at 2nd position from front. Thus, in the missing term, a new letter is added at 3rd position from front.<\/p>\n Thus, missing term =\u00a0GIKPSUW<\/strong><\/span><\/p>\n => Ans – (C)<\/p>\n Question 3:\u00a0<\/b>If the letters of the word PRODUCTIVE which are at the odd numbered position in the English alphabet are picked up and rearranged in alphabetical order from left and if they are now substituted by Z, Y, X and so on, beginning from left which letter will get substituted by W ?<\/p>\n a)\u00a0No letter will get substituted by W<\/p>\n b)\u00a0O<\/p>\n c)\u00a0E<\/p>\n d)\u00a0C<\/p>\n e)\u00a0None of these<\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word –\u00a0PRODUCTIVE<\/p>\n Letters at odd numbered position in English alphabet = O,U,C,I,E<\/p>\n Arranging letters in alphabetical order = CEIOU<\/p>\n Substituting the letters, we get : ZYXWV<\/p>\n Thus, O\u00a0will get substituted by W.<\/p>\n => Ans – (B)<\/p>\n Question 4:\u00a0<\/b>In the following set of numbers if \u20181\u2019 is added to the last digit of and then the order of digits is reversed ,which of the following will be the fourth if arranged in the ascending order 567 284 696 865 738<\/p>\n a)\u00a0567<\/p>\n b)\u00a0284<\/p>\n c)\u00a0696<\/p>\n d)\u00a0865<\/p>\n e)\u00a0738<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Numbers\u00a0: 567 284 696 865 738<\/p>\n If\u00a0\u20181\u2019 is added to the last digit, then numbers :\u00a0568 , 285 , 697 , 866 , 739<\/p>\n Now, order of digits is reversed, = 865 , 582 , 796 , 668 , 937<\/p>\n After arranging in ascending order = 582 , 668 , 796 , 865 , 937<\/p>\n Thus, 4th from left = $865\\equiv567$<\/p>\n => Ans – (A)<\/p>\n Question 5:\u00a0<\/b>How many such pairs of letters are there in the word PHYSICAL each of which has as many letters between them in the word as apperar in the English alphabets.<\/p>\n a)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘PHYSICAL’<\/p>\n There are 2 pairs of letters which have as many letters between them in the word as they have between them in the English alphabetical series<\/p>\n = (P,S) , (I,L)<\/p>\n => Ans – (C)<\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enroll to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>Four of the following are like in certain way on the basis of their positions in English alphabets and so form a group.which of the following does not belong to that group ?<\/p>\n a)\u00a0HJG<\/p>\n b)\u00a0PQN<\/p>\n c)\u00a0DEB<\/p>\n d)\u00a0TUR<\/p>\n e)\u00a0KLI<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n (A) : H\u00a0(+2<\/strong> letters) = J (-3 letters) = G<\/p>\n (B) : P (+1 letter) = Q (-3 letters) = N<\/p>\n (C) : D\u00a0(+1 letter) = E (-3 letters) = B<\/p>\n (D) : T\u00a0(+1 letter) = U (-3 letters) = R<\/p>\n (E) : K\u00a0(+1 letter) = L (-3 letters) = I<\/p>\n => Ans – (A)<\/p>\n Question 7:\u00a0<\/b>What should come next in the following letter series ? a)\u00a0JZ<\/p>\n b)\u00a0IZ<\/p>\n c)\u00a0IY<\/p>\n d)\u00a0KA<\/p>\n e)\u00a0LZ<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series :\u00a0AN CQ ET GW ?<\/p>\n The pattern followed in each letter of the terms is\u00a0:<\/p>\n 1st letter : A (+2 letters) = C\u00a0(+2 letters) =\u00a0E\u00a0(+2 letters) =\u00a0G\u00a0(+2 letters) =\u00a0I<\/p>\n 2nd letter\u00a0: N\u00a0(+3 letters) =\u00a0Q\u00a0(+3 letters) =\u00a0T\u00a0(+3 letters) =\u00a0W\u00a0(+3 letters) =\u00a0Z<\/p>\n Thus, missing term =\u00a0IZ<\/strong><\/p>\n => Ans – (B)<\/p>\n Question 8:\u00a0<\/b>Four of the following five are alike in a certain way (based on their positions of alphabets in the English alphabetical series) and hence form a group. Which is the one that does not belong to that group?<\/p>\n a)\u00a0BFD<\/p>\n b)\u00a0EIH<\/p>\n c)\u00a0KOM<\/p>\n d)\u00a0TXV<\/p>\n e)\u00a0LPN<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n BCDEF = BFD (1st -5th-3rd)<\/p>\n KLMNO = KOM(1st -5th-3rd)<\/p>\n TUVWX = TXV\u00a0(1st -5th-3rd)<\/p>\n LMNOP = LPN(1st -5th-3rd)<\/p>\n EFGHI = it should be EIG but it is given as EIH and hence this is the odd one out<\/p>\n Question 9:\u00a0<\/b>The sum of the squares of two odd numbers is 11570. The square of the smaller number is 5329. What is the other number?<\/p>\n a)\u00a073<\/p>\n b)\u00a075<\/p>\n c)\u00a078<\/p>\n d)\u00a079<\/p>\n e)\u00a080<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the two numbers be a and b , where b is the smaller number<\/p>\n $(a)^2$ + $(b)^2$ = 11570<\/p>\n $(a)^2$ + 5329 = 11570<\/p>\n $(a)^2$ = 6241 = 79 x 79<\/p>\n a = 79<\/p>\n Hence option D is correct<\/p>\n Question 10:\u00a0<\/b>If the digits in the number 79246358 are arranged in descending order from left to right, what will be the difference between the digits which are third from the right and second from the left in the new arrangement?<\/p>\n a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a03<\/p>\n d)\u00a04<\/p>\n e)\u00a05<\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Number – ‘79246358’<\/p>\n If the digits are arranged in descending order from left to right,\u00a0new number = 98765432<\/p>\n Difference between the digits which are third from the right and second from the left\u00a0= $8-4=4$<\/p>\n => Ans – (D)<\/p>\n Question 11:\u00a0<\/b>How many such pairs of letters are there in the word FORMATION each of which has as many letters between them in the word (in both forward and backward directions) as they have between them in the English alphabetical order?<\/p>\n a)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘FORMATION’<\/p>\n There are 3 pairs of letters which have as many letters between them in the word as they have between them in the English alphabetical series.<\/p>\n = (O,M) , (O,N) , (F,N)<\/p>\n => Ans – (d)<\/p>\n Question 12:\u00a0<\/b>If each of the vowels in the word GOLIATHS is changed to the next letter in the English alphabetical series and each consonant is changed to the previous letter in the English alphabetical series, and then the alphabets so formed are arranged in alphabetical order from left to right, which of the following will be sixth from the left of the new arrangement thus formed?<\/p>\n a)\u00a0S<\/p>\n b)\u00a0P<\/p>\n c)\u00a0G<\/p>\n d)\u00a0J<\/p>\n e)\u00a0F<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘GOLIATHS’<\/p>\n If each vowel is changed to next letter in the English alphabetical series and each consonant is changed to previous letter in English alphabetical series, then new word – ‘FPKJBSGR’<\/p>\n Word after arranging in alphabetical order = ‘BFGJKPRS’<\/p>\n 6th from the leftt end = P<\/p>\n => Ans – (B)<\/p>\n Question 13:\u00a0<\/b>How many such pairs of letters are there in the word WONDERS, each of which has as many letters between them in the word (in both forward and backward directions) as they have between them in the English alphabetical series ?<\/p>\n a)\u00a0One<\/p>\n b)\u00a0Two<\/p>\n c)\u00a0Three<\/p>\n d)\u00a0More than three<\/p>\n e)\u00a0None<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘WONDERS’<\/p>\n There are 4, i.e. more than three pairs of letters which have as many letters between them in the word as they have between them in the English alphabetical series.<\/p>\n = (O,N) , (D,E) , (R,S) , (W,R)<\/p>\n => Ans – (D)<\/p>\n Question 14:\u00a0<\/b>Which of the following will come next in the following series ? a)\u00a00<\/p>\n b)\u00a06<\/p>\n c)\u00a09<\/p>\n d)\u00a07<\/p>\n e)\u00a04<\/p>\n 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series :\u00a00 9 0 1 9 0 1 2 9 0 1 2 3 9 0 1 2\u00a03 4 9 0 1 2 3 4 5 9 0 1 2 3 4 5<\/p>\n The pattern followed is that whole numbers are written in increasing order and at the end ‘9’ is appended.<\/p>\n Eg :- 0 9 , 01 9 , 012 9 , 0123 9 , 01234 9 , 012345 9<\/p>\n Thus, next number after 5 will be =\u00a06<\/strong><\/p>\n => Ans – (B)<\/p>\n Question 15:\u00a0<\/b>What should come in place of the question mark (?) in the following number series ? a)\u00a01376<\/p>\n b)\u00a01780<\/p>\n c)\u00a01570<\/p>\n d)\u00a01725<\/p>\n e)\u00a0None of these<\/p>\n 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern followed is\u00a0:<\/p>\n 9 $\\times1+1^2$ = 10<\/p>\n 10\u00a0$\\times2+2^2$ =\u00a024<\/p>\n 24\u00a0$\\times3+3^2$ =\u00a081<\/p>\n 81\u00a0$\\times4+4^2$ =\u00a0340<\/p>\n 340\u00a0$\\times5+5^2$ =\u00a01725<\/strong><\/p>\n => Ans – (D)<\/p>\n
\nA B C D E F G H I J K L M N O P Q R S T U V W X Y Z<\/p>\n
\nGPW, GPUW, GIPUW, GIPSUW, ?<\/p>\n
\nAN CQ ET GW ?<\/p>\n
\n0 9 0 1 9 0 1 2 9 0 1 2 3 9 0 1 2 3 4 9 0 1 2 3 4 5 9 0 1 2 3 4 5<\/p>\n
\n9 10 24 81 340 ?<\/p>\n