35 IBPS RRB Clerk Mocks @ Rs. 149<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Take a free mock test for IBPS RRB Clerk<\/a><\/p>\n Download IBPS RRB Clerk Previous Papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>What would be the rank of the word ALARM in a dictionary made up of all the possible permutations of the letters of the word?<\/p>\n a)\u00a08<\/p>\n b)\u00a016<\/p>\n c)\u00a055<\/p>\n d)\u00a012<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>If all the possible permutations of the word PAINT are written down in alphabetical order, what is the rank of the word PAINT?<\/p>\n a)\u00a055<\/p>\n b)\u00a049<\/p>\n c)\u00a072<\/p>\n d)\u00a073<\/p>\n e)\u00a061<\/p>\n Question 3:\u00a0<\/b>In approximately how many ways can the letters of the word “PERMUTATION” be arranged?<\/p>\n a)\u00a01 crore<\/p>\n b)\u00a02 crores<\/p>\n c)\u00a03 crores<\/p>\n d)\u00a04 crores<\/p>\n e)\u00a05 crores<\/p>\n Question 4:\u00a0<\/b>In how many ways can the letters of the word \u2018PERMUTATION\u2019 be arranged such that all the consonants are together?<\/p>\n a)\u00a0$6!\/2$<\/p>\n b)\u00a0$6!^2\/2$<\/p>\n c)\u00a0$5!^2\/2$<\/p>\n d)\u00a0$6!^2$<\/p>\n e)\u00a0None of the above<\/p>\n Question 5:\u00a0<\/b>If all the permutations of the letters of the word \u201cMATHS\u201d are arranged as in a dictionary, what is the rank of the word \u201cMATHS\u201d?<\/p>\n a)\u00a048<\/p>\n b)\u00a053<\/p>\n c)\u00a055<\/p>\n d)\u00a054<\/p>\n e)\u00a050<\/p>\n Question 6:\u00a0<\/b>In how many ways can the letters of the word \u2018PERMUTATION\u2019 be arranged?<\/p>\n a)\u00a011!\/2!<\/p>\n b)\u00a011!<\/p>\n c)\u00a011!*2!<\/p>\n d)\u00a011!\/(2!*2!)<\/p>\n e)\u00a0None of the above<\/p>\n Question 7:\u00a0<\/b>A committee of 2 males and 3 female professors has to be formed from a pool of 8 males and 9 female professors. How many different combinations for a committee are possible?<\/p>\n a)\u00a02352<\/p>\n b)\u00a07056<\/p>\n c)\u00a01176<\/p>\n d)\u00a01008<\/p>\n e)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>In how many ways can a team of 3 people having at least one boy can be formed from a group of 4 boys and 4 girls ?<\/p>\n a)\u00a0520<\/p>\n b)\u00a0500<\/p>\n c)\u00a0510<\/p>\n d)\u00a0592<\/p>\n e)\u00a0536<\/p>\n Question 9:\u00a0<\/b>In how many different ways can the letters of the word \u2018AUSTRALIA\u2019 such that all the vowels are together ?<\/p>\n a)\u00a02400<\/p>\n b)\u00a03600<\/p>\n c)\u00a01800<\/p>\n d)\u00a03000<\/p>\n e)\u00a04200<\/p>\n Question 10:\u00a0<\/b>In how many ways can a team of 5 people having at least one boy can be formed from a group of 6 boys and 6 girls ?<\/p>\n a)\u00a0120<\/p>\n b)\u00a0800<\/p>\n c)\u00a0720<\/p>\n d)\u00a0792<\/p>\n e)\u00a0786<\/p>\n Free Mock Test for IBPS RRB Clerk<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/p>\n Question 11:\u00a0<\/b>What is the probability of selecting 2 red balls and 1 blue ball in the particular given order one at a time from a bag consisting of 6 red balls and 3 blue balls ?<\/p>\n a)\u00a05\/26<\/p>\n b)\u00a05\/28<\/p>\n c)\u00a05\/21<\/p>\n d)\u00a015\/28<\/p>\n e)\u00a05\/7<\/p>\n Question 12:\u00a0<\/b>If a number is randomly selected from the first 50 natural numbers then what is the probability that it is divisible by either 4 or 5 ?<\/p>\n a)\u00a00.33<\/p>\n b)\u00a00.40<\/p>\n c)\u00a00.50<\/p>\n d)\u00a00.25<\/p>\n e)\u00a00.55<\/p>\n Question 13:\u00a0<\/b>If all the words that are formed by using the letters in the word \u201cALIEN\u201d are arranged as listed in the dictionary then how many words are listed before \u201cILANE\u201d ?<\/p>\n a)\u00a064<\/p>\n b)\u00a060<\/p>\n c)\u00a061<\/p>\n d)\u00a062<\/p>\n e)\u00a063<\/p>\n Question 14:\u00a0<\/b>There are 15 chairs in a row and 5 people are to be seated in them such that odd number of chairs should be present between any two people.In how many can it be done ?<\/p>\n a)\u00a070<\/p>\n b)\u00a077<\/p>\n c)\u00a074<\/p>\n d)\u00a072<\/p>\n e)\u00a075<\/p>\n Question 15:\u00a0<\/b>There are 4 male and 4 female badminton players and if 2 teams comprising of 1 male and 1 female has to be made then in how many ways can it be done ?<\/p>\n a)\u00a018<\/p>\n b)\u00a080<\/p>\n c)\u00a0144<\/p>\n d)\u00a036<\/p>\n e)\u00a072<\/p>\n Question 16:\u00a0<\/b>A sum of Rs 80 has to be made by using Rs 1 or Rs 5 coins.In how many different ways can it be done ?<\/p>\n a)\u00a010<\/p>\n b)\u00a017<\/p>\n c)\u00a014<\/p>\n d)\u00a015<\/p>\n e)\u00a016<\/p>\n Question 17:\u00a0<\/b>A box contains 5 blue pens, 6 green pens and 10 black pens. The number of ways in which 3 pens can be selected such that at least 2 pens are of the same colour is<\/p>\n a)\u00a01000<\/p>\n b)\u00a01300<\/p>\n c)\u00a01330<\/p>\n d)\u00a01030<\/p>\n e)\u00a01310<\/p>\n Question 18:\u00a0<\/b>Find the number of rectangles in the following diagram:<\/p>\n a)\u00a0225<\/p>\n b)\u00a0315<\/p>\n c)\u00a0360<\/p>\n d)\u00a0441<\/p>\n e)\u00a0482<\/p>\n Question 19:\u00a0<\/b>A committee of 5 members is to be formed from 4 men and 4 women. If the committee must have at least 3 men, the number of ways in which the committee can be formed is<\/p>\n a)\u00a024<\/p>\n b)\u00a032<\/p>\n c)\u00a036<\/p>\n d)\u00a040<\/p>\n e)\u00a028<\/p>\n Question 20:\u00a0<\/b>Letters of the word DIRECTOR are arranged in such a way that all the vowel come together .Find the No of ways making such arrangement?<\/p>\n a)\u00a04320<\/p>\n b)\u00a0720<\/p>\n c)\u00a02160<\/p>\n d)\u00a0120<\/p>\n e)\u00a0None of these<\/p>\n Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The alphabetical order of the letters is AALMR.<\/p>\n Number of words beginning with AA = 3! = 6 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n All the words starting with A are ahead of PAINT. 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n There are 11 letters in the word PERMUTATION of which T repeats two times. So, the number of ways in which the word can be arranged is 11!\/2 = 1,99,58,400<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Consonants in the word \u2018PERMUTATION\u2019 are P, R, M, T, T, N Let the consonants be one unit. So, total number of units = 5 + 1 = 6 So, total number of ways of arrangement = $6!^2\/2$<\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The alphabetical order of letters is A, H, M, S, T. So, the rank of the word \u201cMATHS\u201d is 24 + 24 + 5 = 53<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The total number of letters in the word \u2018Permutation\u2019 is 11. Out of these, the letter \u2018t\u2019 occurs twice. 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Number of ways of choosing 2 males from 8 males is $^8C_2$ 8)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Having at least one boy can be calculated by deleting the case of no boys from the total number of cases i.e 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n In AUSTRALIA we have 5 vowels i.e AAAIU and 4 other letters STRL and so assuming all the vowels as single letter we have total of 5 letters to arrange and the internal arrangement of 5 vowels i.e vowels AAAIU can be arranged in 5!\/3! Ways 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Having at least one boy can be calculated by deleting the case of no boys from the total number of cases i.e 11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given we have 6 red balls and 3 blue balls 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n All the multiples of 4 till 50 are 12 i.e from 4,8\u2026\u2026.48 13)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Total 5! words can be formed using the 5 letters and so by taking A as the first letter is 4!=24 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n In the given condition if all the 5 people are seated in either odd places or even places then the number of chairs between them will be odd and so we have odd places 8 starting from 1,3\u2026.15 and even places 7 starting from 2,4…14 and so filing 5 persons in 8 chairs=8C5 =(8*7*6)\/(3*2)=56 15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Two teams are to be selected so 4 players i.e 2 males and 2 females are to be selected. 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n let the number of Rs 1 coins be x 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Number of ways in which at least 2 pens of the same colour can be selected = Total number of ways in which 3 pens can be selected – Number of ways in which 3 pens of different colour can be selected.<\/p>\n There are 5+6+10 = 21 pens in total. Therefore, Number of ways in which at least 2 pens of the same colour can be selected = 1330 – 300 = 1030. Therefore, option D is the right answer.<\/p>\n 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n There are 7 horizontal lines and 6 vertical lines in the diagram.<\/p>\n For a rectangle to be formed, 2 out of these 7 horizontal lines and 2 out of these 6 vertical lines must be selected.<\/p>\n Therefore, total number of rectangles = 7C2*6C2 = 21*15 = 315. 19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The committee can have either 3 men or 4 men. If 4 men are to be selected, the committee can be formed in 4C4*4C1 = 1*4 = 4 ways. 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Word – DIRECTOR<\/p>\n So \u201cI,E,O\u201d are there are 3! ways to arrange the vowels<\/p>\n Now \u201cD,R,C,T,R\u201d are the remaining alphabets ,<\/p>\n Condition is that the vowels should always be together so we can assume the vowels as a single alphabet\/unit say \u201cX\u201d (\u2018X\u2019=\u2019I,E,O\u2019) so now we have a new word – \u201cD,R,C,T,R,X\u201d<\/p>\n Possible arrangements for this word = 6!<\/p>\n Thus total number of ways to rearrange DIRECTOR with vowels grouped together = (Possible arrangements of \u2018DRCTRX\u2019) $\\times$ (Possible arrangements of vowels)<\/p>\n = 6! $\\times$ 3! = $720 \\times 6 = 4320$<\/p>\n => Ans – (A)<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/board_ttxR89B.JPG\")
<\/figure>\n
\nNext would come ALAMR and then ALARM.
\nHence, rank would be 8.<\/p>\n
\nThe number of such words are 4! = 24
\nSimilarly, all the words starting with I are ahead of PAINT.
\nThe number of such words are 4! = 24
\nSimilarly, all the words starting with N are ahead of PAINT.
\nThe number of such words are 4! = 24
\nThe first word starting with P and using these letters is PAINT.
\nHence, its rank is 24+24+24+1 = 73<\/p>\n
\nVowels are E, U, A, I, O<\/p>\n
\nNumber of ways in which these 6 units can be arranged = 6!
\nNumber of ways in which the consonants can be arranged among themselves = 6!\/2!<\/p>\n
\nNumber of words whose first letter is A = 4! = 24
\nNumber of words whose first letter is H = 4! = 24
\nThe next few words will be MAHST, MAHTS, MASHT, MASTH, MATHS.<\/p>\n
\nSo, the number of arrangements of the word = 11!\/2!<\/p>\n
\nNumber of ways of choosing 3 females from 9 females is $^9C_3$
\nTotal number of ways = $^8C_2 \\times ^9C_3 = 2352$<\/p>\n
\nTotal number of boys and girls=12
\nFrom 12 we have to select 5 ie 12C5=792
\nNo of cases having no boys i.e 6C5=6
\nSo number of cases having at least 1 boy=792-6=786<\/p>\n
\n=20 ways
\nAll the 5 letters can be arranged in 5! ways therefore
\nTotal ways=5!*(5!\/3!)
\n=2400 ways<\/p>\n
\nTotal number of boys and girls=12
\nFrom 12 we have to select 5 ie 12C5=792
\nNo of cases having no boys i.e 6C5=6
\nSo number of cases having at least 1 boy=792-6=786<\/p>\n
\nSo the required probability is $\\dfrac{6_{C_{1}}}{9_{C_1}} \\times \\dfrac{5_{C_{1}}}{8_{C_1}} \\times \\dfrac{3_{C_{1}}}{7_{C_1}}$
\n=(6*5*3)\/(9*8*7)
\n=5\/28<\/p>\n
\nAll the multiples of 5 till 50 are 10 i.e from 5,10\u2026.50
\nIn the above multiples there are few numbers which are common multiples of 4 and 5 and so they are repeated twice so they are to be removed
\nLCM of 4 and 5 is 20
\nMultiples of 20 are 20 and 40 which lie below 50
\nTherefore total numbers are 12+10-2
\n=22-2
\n=20
\nProbability of selecting a number which is either divisible by 4 or 5 from first 50 natural numbers is 20\/50
\n=2\/5
\n=0.4<\/p>\n
\nBy taking E as first letter we have 4! words=24
\nBy taking I as first letter and A as the second letter we have 3! Words
\nBy taking I as first letter and E as the second letter we have 3! Words
\nBy taking I as first letter and L as second letter and A as third letter we have 2 words
\nAnd the second word is ILANE
\nSo 24+24+6+6+1=61 words come before the word ILANE<\/p>\n
\n5 persons sitting in 7 places i.e i 7C5 ways=(7*6)\/(1*2)=21
\nSo total number of ways=56+21=77 ways<\/p>\n
\nFirst select 2 males out of four i.e 4C2=(4*3)\/2 =6 ways
\nthen select 2 females out of four i.e 4C2=(4*3)\/2 =6 ways
\nThen first male can be paired with two females and second male can be paired with only one female.
\nTherefore required ways=6*6*2*1
\n=72 ways<\/p>\n
\nAnd the number of Rs 2 coins be y
\nTherefore x+5y=80
\nSo for every x there should be one y and vice versa
\nY take values from y=0 to y=16
\nI.e for every y from 0 to 16,we will have x
\nSo for y=0 we have x=80
\nFor y=1 we have x=75
\nSo on
\nTherefore 16+1=17 ways are possible<\/p>\n
\nTotal number of ways in which 3 pens can be selected = 21C3 = (21*20*19)\/(1*2*3) = 1330 ways.
\nNumber of ways in which 3 pens of different colour can be selected = 5*6*10 = 300 ways<\/p>\n
\nTherefore, option B is the right answer.<\/p>\n
\nIf 3 men are to be selected, the number of ways in which the committee can be formed is 4C3 * 4C2
\n= 4*6 = 24 ways.<\/p>\n
\nTherefore, the total number of ways in which the committee can be formed is 24+4 = 28 ways. Hence, option E is the right answer.<\/p>\n