Download Series Completion Questions for MAH-CET PDF<\/a><\/p>\n Enroll to MAH-CET Crash Course<\/a><\/p>\n Question 1:\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 1)\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 2:\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 2)\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 Question 3:\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 3)\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 4:\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 4)\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 5:\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 5)\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 6:\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 6)\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 7:\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 7)\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 8:\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 8)\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 9:\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 9)\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 10:\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 10)\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 Take Free MAH-CET mock tests here<\/a><\/p>\n Question 11:\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 11)\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 12:\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 12)\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 Question 13:\u00a0<\/b>Which of the following will come in the place of the question mark (?) in the following series based on the English alphabetical order ? a)\u00a0VF<\/p>\n b)\u00a0EU<\/p>\n c)\u00a0UE<\/p>\n d)\u00a0EV<\/p>\n e)\u00a0VE<\/p>\n 13)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Series :\u00a0ZA BY XC DW ?<\/p>\n The pattern followed is that alphabets starting from end are written in alternate order to those of starting from top<\/p>\n = (Z)A , B(Y) , (X)C , D(W)<\/p>\n Thus, next term =\u00a0VE<\/strong><\/p>\n => Ans – (E)<\/p>\n Question 14:\u00a0<\/b>Which of the following will come next in the following series ? a)\u00a07<\/p>\n b)\u00a08<\/p>\n c)\u00a03<\/p>\n d)\u00a02<\/p>\n e)\u00a04<\/p>\n 14)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Arrangement\u00a0:\u00a09 1 9 8 2 9 8 7 3 9 8 7 6 4 9 8 7 6 5 5 9 8 7 6 5 4 6 9 8 7 6 5 4<\/p>\n The pattern followed is that in alternate groups of numbers, one digit starting from ‘1’ in increasing order are written and in the other group digits in decreasing order are written with one digit appended after each new group.<\/p>\n 9 , 1 , 9 8 , 2 , 9 8 7 , 3 , 9 8 7 6 , 4 , 9 8 7 6 5 , 5 , 9 8 7 6 5 4 , 6 , 9 8 7 6 5 4\u00a03<\/strong><\/p>\n Thus, next number = 3<\/p>\n => Ans – (C)<\/p>\n Question 15:\u00a0<\/b>How many such pairs of letters are there in the word BARGAINED, 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)\u00a0Two<\/p>\n b)\u00a0Three<\/p>\n c)\u00a0None<\/p>\n d)\u00a0One<\/p>\n e)\u00a0More than three<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘BARGAINED’<\/p>\n There are 4\u00a0pairs of letters which have 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 = (B,A) , (R,N) , (G,I) , (E,D)<\/p>\n => Ans – (E)<\/p>\n Question 16:\u00a0<\/b>The positions of how many alphabets will remain unchanged if each of the alphabets in the word PROACTIVE is arranged in alphabetical order from left to right ?<\/p>\n a)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘PROACTIVE’<\/p>\n If each of the alphabet is arranged in alphabetical order = ACEIOPRTV<\/p>\n Thus,\u00a0position of none of the alphabets will remain unchanged.<\/p>\n => Ans – (A)<\/p>\n Question 17:\u00a0<\/b>How many meaningful words can be formed with the letters LDFO, using each letter only once in each word ?<\/p>\n a)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Letters – L,D,F,O<\/p>\n Only 1 meaningful word can be formed = Fold<\/p>\n => Ans – (B)<\/p>\n Question 18:\u00a0<\/b>How many such pairs of letters are there in the word BANTER, each of which has 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)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘BANTER’<\/p>\n There are 2\u00a0pairs of letters which have 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 = (B,A) , (T,R)<\/p>\n => Ans – (C)<\/p>\n Question 19:\u00a0<\/b>How many such pairs of letters are there in the word DECAYS, 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)\u00a0None<\/p>\n e)\u00a0More than three<\/p>\n 19)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘DECAYS’<\/p>\n There are 2\u00a0pairs of letters which have 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 = (D,E) , (D,A)<\/p>\n => Ans – (B)<\/p>\n Question 20:\u00a0<\/b>If each vowel in the word BROUGHT 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 an alphabetical order from left to right, which of the following will be fourth from the left ?<\/p>\n a)\u00a0V<\/p>\n b)\u00a0P<\/p>\n c)\u00a0G<\/p>\n d)\u00a0Q<\/p>\n e)\u00a0S<\/p>\n 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Word – ‘BROUGHT’<\/p>\n If each vowel 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, new word – ‘AQPVFGS’<\/p>\n When the word is\u00a0arranged in an alphabetical order = AFGPQSV<\/p>\n 4th from left end = P<\/p>\n => Ans – (B)<\/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
\nZA BY XC DW ?<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/17107_zmfaon7.PNG\")
<\/figure>\n
\n9 1 9 8 2 9 8 7 3 9 8 7 6 4 9 8 7 6 5 5 9 8 7 6 5 4 6 9 8 7 6 5 4<\/p>\n