CGI Group Aptitude Interview Questions & Answers

  • Question:Find The Selling Price Of A Scooter Toy Which Is Bought For Rs. 1200 And Sold At Profit Of 30 %?

    Answer :

    Given, Cost price = Rs. 1200

    Profit = 30%

    Selling price = {[100 + gain%] / 100} * Cost price

    = [130 / 100] * 1200

    = 130 * 12

    = 1560

    Therefore, Selling price = 1560.

  • Question:A Man Sold An Item For Rs. 1500 At A Loss Of 25%. What Will Be The Selling Price Of Same Item If He Sells It At A Profit Of 20 %?

    Answer :

    S.P = 75 % of CP

    => 75 x CP /100= 1500

    => CP = 2000

    20 % of CP = (20/100) x 2000 = 400

    SP = 2000 + 400 = 2400.

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  • Question:Ramesh Bought 10 Cycles For Rs. 500 Each. He Spent Rs. 2,000 On The Repair Of All Cycles. He Sold Five Of Them For Rs. 750 Each And The Remaining For Rs. 550 Each. Then The Total Gain Or Loss % Is?

    Answer :

    Total CP = Rs. (500 X 10 + 2000) = Rs. 7000

    SP = Rs. (5 X 750 + 5 X 550) = Rs. 6500

    Loss = CP – SP = 7000 – 6500 = 500

    Loss Percent = 500/7000 X 100 = 50/7.

  • Question:A, B And C Entered Into Partnership In Business A Got 3/5 Of The Profit And B And C Distributed The Remaining Profit Equally. If C Got Rs. 400 Less Than A, The Total Profit Was?

    Answer :

    Let x be the total profit

    A’s share in profit = Rs. 3x/5

    Remaining Profit = x – (3x/5) = 2x/5

    So, B’s share in profit = Rs. x/5

    C’s share in profit = Rs. x/5

    Given,(3x/5 – x/5) = 400

    => 2x/5 = 400

    => x = (400×5)/ 2

    => x = Rs. 1000

    Therefore, Total Profit = x =Rs. 1000.

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  • Question:If A Shirt Costs Rs. 64 After 20% Discount Is Allowed, What Was Its Original Price In Rs?

    Answer :

    Let the price be Rs 100

    After discount of 20% we get = 100-20 = 80

    shirt costs Rs. 64

    Let x be the cost price of the shirt,

    x * 80/100 = 64

    x = (64 x 100) / 80 = 80

    Original price of shirt in Rs. 80.

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  • Question:If There Is A Profit Of 20% On The Cost Price Of An Article, The Percentage Of Profit Calculated On Its Selling Price Will Be?

    Answer :

    Let the Cost Price be Rs. 100,

    Sincethere is a profit of 20% on the Cost Price,

    then Selling Price = C.P + 20% of C.P

    = 100 + 20

    = Rs. 120

    =>Selling Price =Rs. 120

    Gain = SP – CP 

    = 120 – 100

    = 20

    Gain % on S.P = (Gain / S.P) * 100%

    = (20/120) x 100

    = 50/3%.

  • Question:An Article Was Sold At A Loss Of 5%.if It Were Sold For Rs. 30 More ,the Gain Would Have Been 1.25%. The Cost Price Of The Article Is?

    Answer :

    Let, cost price of the article = Rs.100x

    Then selling price = 5% loss of Cost price

    = C.P – loss

    = 100x – (5/100)*100x

    = 100x – 5x

    = 95x

    => selling price = 95x

    But if he sold the product for Rs.30 more, ==> his profit is 1.25%.

    In this case his selling price = 100x + (1.25/100) * 100x

    = 100x + 1.25x

    = 101.25x

    => selling price = 101.25x

    Difference in two selling prices = Rs.30

    => 101.25x – 95x = Rs.30

    => 6.25x = Rs.30

    => x = Rs.30 / 6.25

    => x = Rs.4.8 —>Substitutingin cost price, we get

    Cost Price of the article = Rs. 100x = Rs. 100 * 4.8 = Rs. 480

    Therefore, Cost Price of the article = Rs. 480.

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  • Question:A Tradesman’s Prices Are 20% Above C.p. He Allows His Customers Some Discount On His Bill And Makes A Profit Of 8%. The Rate Of Discount Is?

    Answer :

    Let the Cost Price(C.P) be Rs. 100

    Given,tradesman’s prices are 20% above C.P

    => Marked Price (M.P) = 20% more than C.P

    => M.P = Rs. 120

    Given,profit = 8%

    => Selling price (S.P) = 8% more than C.P

    => S.P = Rs. 108

    Rate of Discount = {(M.P – S.P) / M.P} * 100%

    = {(120 – 108) / 120} * 100%

    = (12 / 120) * 100%

    = 10 %

    Thus,Rate of Discount = 10%.

  • Question:By Selling Sugar At Rs. 5.58 Per Kg. A Man Loses 7%. To Gain 7% It Must Be Sold At The Rate Of Rs?

    Answer :

    Cost of sugar = Rs 5.58 / kg

    His lost percent =7 %

    = 100 – 7

    = 93.

    Gain percent

    = 100+ 7

    = 107.

    So, Gain = CP * gain / 100 – loss

    = 5.58 * 107 / 100 – 7

    = 5.58 * 107 / 93

    = 597.06 / 93

    = 6.42 .per kg 

    6.42 .per kg is to be sold to gain 7 % .

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  • Question:Toffee Are Bought At A Rate Of 8 For One Rupee. To Gain 60% They Must Be Sold At?

    Answer :

    Given, Cost price (C.P) of 8 toffees = Re. 1

    Gain = 60%

    So, Selling price, (S.P) = {[100 + Gain%] / 100} * C.P

    = Rs. (160 / 100) x 1

    = Rs. 8 / 5

    For Rs. 8 / 5, toffees sold = 8

    For Re. 1, toffees sold = (8 x 5) / 8 = 5

    So, to gain 60%, toffees must be sold at 5 for Re. 1.

  • Question:A Fruit Seller Buys Lemons At 2 For A Rupee And Sells Them At 5 For Three Rupees. His Gain Percent Is?

    Answer :

    Given, Cost price (C.P) of 2 lemons = Rs. 1

    =>C.P of 1 lemon = Rs. 1/2 = Rs. 0.50

    Given, Selling price (S.P) of 5 lemons = Rs. 3

    =>S.P of 1 lemon = Rs. 3/5 = 0.60

    Gain = S.P of1 lemon -C.P of 1 lemon

    = 0.60 – 0.50

    = 0.10

    =>Gain = Rs. 0.10

    Gain % = (Gain / C.P of 1 lemon) * 100%

    = (0.10 / 0.50)* 100%

    = 20%

    Thus, Gain % = 20%.

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  • Question:A Man Sold A Chair At A Loss Of 6%. Had He Been Able To Sell It At A Gain Of 10%, It Would Have Fetched Rs. 96 More Than It Did. What Was His Cost Price?

    Answer :

    In the given problem, let C.P denote the cost price,

    then (100 +10)% of CP -(100-6) % of CP = Rs. 96

    =>(110)% of CP – (94) % of CP = Rs.96

    =>16 % of CP = 96

    => 16 / 100 = 96

    => CP = 96 x 100 / 16

    => 9600 / 16

    = 600 Rs

    Rs 600is cost price.

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  • Question:A Man Sold A Horse At A Loss Of 4%. Had He Been Able To Sell It At A Gain Of 12%, It Would Have Fetched Rs. 64 More Than It Did. What Was His Cost Price?

    Answer :

    In the given problem,

    Let C.P denote the cost price,

    Then (100+12)% of CP = (100-4) % of Cost

    => Rs. 128 (112)% of CP = (96) % of Cost = Rs.128

    16 % of CP = 128

    => CP = 128 x 100 / 16

    = 12800 / 16

    = 800.

  • Question:The Profit Earned By Selling An Article For Rs. 600 Is Equal To The Loss Incurred When The Same Article Is Sold For Rs. 400. What Should Be The Sale Price Of The Article For Making 25% Profit?

    Answer :

    Let the cost price be Rs. k

    Now, as per the question,

    600 – k = k – 400

    => 2k = 1000

    => k = 500

    Again, selling price of the article for making 25 % profit

    = (500 x 125) /200

    = 125 * 5

    = Rs. 625.

  • Question:Out Of 7 Consonants And 4 Vowels, How Many Words Of 3 Consonants And 2 Vowels Can Be Formed?

    Answer :

    Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3*4C2

    = 210. 

    Number of groups, each having 3 consonants and 2 vowels = 210. 

    Each group contains 5 letters. 

    Number of ways of arranging 5 letters among themselves = 5! = 120 

    Required number of ways = (210 x 120) = 25200.

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  • Question:A Committee Of 5 Persons Is To Be Formed From 6 Men And 4 Women. In How Many Ways Can This Be Done When At Least 2 Women Are Included ?

    Answer :

    When at least 2 women are included.

    The committee may consist of 3 women, 2 men : It can be done in  4C3*6C2ways

    or, 4 women, 1 man : It can be done in  4C4*6C1ways

    or, 2 women, 3 men : It can be done in 4C2*6C3ways.

    Total number of ways of forming the committees

    = 4C2*6C3+4C3*6C2+4C4*6C1

    = 6 x 20 + 4 x 15 + 1x 6

    = 120 + 60 + 6 =186.

  • Question:If The Letters Of The Word Sachin Are Arranged In All Possible Ways And These Words Are Written Out As In Dictionary, Then The Word ‘sachin’ Appears At Serial Number?

    Answer :

    If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

    If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

    If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

    The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

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  • Question:A College Has 10 Basketball Players. A 5-member Team And A Captain Will Be Selected Out Of These 10 Players. How Many Different Selections Can Be Made?

    Answer :

    A team of 6 members has to be selected from the 10 players.

    This can be done in 10C6 or 210 ways. 

    Now, the captain can be selected from these 6 players in 6 ways.

    Therefore, total ways the selection can be made is 210×6= 1260.

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  • Question:How Many 4-letter Words With Or Without Meaning, Can Be Formed Out Of The Letters Of The Word, ‘logarithms’, If Repetition Of Letters Is Not Allowed?

    Answer :

    ‘LOGARITHMS’ contains 10 different letters.

    Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

    = 10P4

    = 5040.

  • Question:The Indian Cricket Team Consists Of 16 Players. It Includes 2 Wicket Keepers And 5 Bowlers. In How Many Ways Can A Cricket Eleven Be Selected If We Have To Select 1 Wicket Keeper And Atleast 4 Bowlers?

    Answer :

    We are to choose 11 players including 1 wicket keeper and 4 bowlers  or, 1 wicket keeper and 5 bowlers.

    Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in 2C1*5C4*9C6= 840

    Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2C1*5C5*9C5=252

    Total number of ways of selecting the team = 840 + 252 = 1092.

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  • Question:How Many Arrangements Can Be Made Out Of The Letters Of The Word Committee, Taken All At A Time, Such That The Four Vowels Do Not Come Together?

    Answer :

    There are total 9 letters in the word COMMITTEE in which there are 2M’s, 2T’s, 2E’s.

    The number of ways in which 9 letters can be arranged = 9!2!×2!×2! = 45360

    There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6!2!×2! = 180 ways.

    In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4!2! = 12 ways.

    The number of ways in which the four vowels always come together = 180 x 12 = 2160.

    Hence, the required number of ways in which the four vowels do not come together = 45360 – 2160 = 43200.

  • Question:When Four Fair Dice Are Rolled Simultaneously, In How Many Outcomes Will At Least One Of The Dice Show 3?

    Answer :

    When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

    The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

    Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671.

  • Question:If The Letters Of The Word Chasm Are Rearranged To Form 5 Letter Words Such That None Of The Word Repeat And The Results Arranged In Ascending Order As In A Dictionary What Is The Rank Of The Word Chasm ?

    Answer :

    The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

    The first 24 of these words will start with A.

    Then the 25th word will start will CA _ _ _. 

    The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

    The next word starts with CH and then A, i.e., CHA _ _. 

    The first of the words will be CHAMS. The next word will be CHASM.

    Therefore, the rank of CHASM will be 24+6+2= 32.

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  • Question:How Many Arrangements Of The Letters Of The Word ‘bengali’ Can Be Made If The Vowels Are To Occupy Only Odd Places?

    Answer :

    There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

    There are 4 odd places and 3 even places.

    3 vowels can occupy 4 odd places in 4P3 ways and 4 constants can be arranged in 4P4ways.

    Number of words =4P3x4P4= 24 x 24 = 576.

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  • Question:How Many 7 Digit Numbers Can Be Formed Using The Digits 1, 2, 0, 2, 4, 2, 4?

    Answer :

    There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

    Number of 7 digit numbers = 7!3!×2! = 420 

    But out of these 420 numbers, there are some numbers which begin with ‘0’ and they are not 7-digit numbers.

    The number of such numbers beginning with ‘0’. 

    =6!3!×2! = 60

    Hence the required number of 7 digits numbers = 420 – 60 = 360.

  • Question:How Many Different Four Letter Words Can Be Formed (the Words Need Not Be Meaningful Using The Letters Of The Word “mediterranean” Such That The First Letter Is E And The Last Letter Is R?

    Answer :

    The first letter is E and the last one is R.

    Therefore, one has to find two more letters from the remaining 11 letters.

    Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

    The second and third positions can either have two different letters or have both the letters to be the same.

    Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

    Case 2: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.

    Total number of possibilities = 56 + 3 = 59.

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  • Question:From 5 Consonants And 4 Vowels, How Many Words Can Be Formed Using 3 Consonants And 2 Vowels ?

    Answer :

    From 5 consonants, 3 consonants can be selected in 5C3 ways.

    From 4 vowels, 2 vowels can be selected in 4C2ways.

    Now with every selection, number of ways of arranging 5 letters is 5P5ways.

    Total number of words = 5C3*4C2*5P5= 10x 6 x 5 x 4 x 3 x 2 x 1= 7200.

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  • Question:A Letter Lock Consists Of Three Rings Each Marked With Six Different Letters. The Number Of Distinct Unsuccessful Attempts To Open The Lock Is At The Most?

    Answer :

    Since each ring consists of six different letters,

    the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts,

    one of them is a successful attempt.

    Maximum number of unsuccessful attempts = 216 – 1 = 215.

  • Question:In How Many Ways Can 4 Girls And 5 Boys Be Arranged In A Row So That All The Four Girls Are Together ?

    Answer :

    Let 4 girls be one unit and now there are 6 units in all.

    They can be arranged in 6! ways.

    In each of these arrangements 4 girls can be arranged in 4! ways. 

    Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280.

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  • Question:A Team Of 8 Students Goes On An Excursion, In Two Cars, Of Which One Can Seat 5 And The Other Only 4. In How Many Ways Can They Travel?

    Answer :

    There are 8 students and the maximum capacity of the cars together is 9.

    We may divide the 8 students as follows

    Case I: 5 students in the first car and 3 in the second 

    Case II: 4 students in the first car and 4 in the second

    Hence, in Case I: 8 students are divided into groups of 5 and 3 in8C3 ways.

    Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4ways.

    Therefore,

    the total number of ways in which 8 students can travel is:

    8C3+8C4=56+70= 126.

  • Question:Consider The Word Rotor. Whichever Way You Read It, From Left To Right Or From Right To Left, You Get The Same Word. Such A Word Is Known As Palindrome. Find The Maximum Possible Number Of 5-letter Palindromes?

    Answer :

    The first letter from the right can be chosen in 26 ways because there are 26 alphabets. 

    Having chosen this, the second letter can be chosen in 26 ways 

    The first two letters can chosen in 26 x 26 = 676 ways 

    Having chosen the first two letters, the third letter can be chosen in 26 ways. 

    All the three letters can be chosen in 676 x 26 =17576 ways. 

    It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

  • Question:In A Box, There Are 5 Black Pens, 3 White Pens And 4 Red Pens. In How Many Ways Can 2 Black Pens, 2 White Pens And 2 Red Pens Can Be Chosen?

    Answer :

    Number of ways of choosing 2 black pens from 5 black pens in 5C2ways.

    Number of ways of choosing 2 white pens from 3 white pens in 3C2ways.

    Number of ways of choosing 2 red pens from 4 red pens in 4C2ways.

    By
    the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.

  • Question:Find The Number Of Subsets Of The Set {1,2,3,4,5,6,7,8,9,10,11} Having 4 Elements?

    Answer :

    Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

    We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

    This can be done in 11C4ways = 330 ways.

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  • Question:In How Many Ways Can The Letters Of The Word “problem” Be Rearranged To Make 7 Letter Words Such That None Of The Letters Repeat?

    Answer :

    There are seven positions to be filled.

    The first position can be filled using any of the 7 letters contained in PROBLEM.

    The second position can be filled by the remaining 6 letters as the letters should not repeat.

    The third position can be filled by the remaining 5 letters only and so on.

    Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

  • Question:There Are Three Rooms In A Hotel: One Single, One Double And One For Four Persons. How Many Ways Are There To House Seven Persons In These Rooms?

    Answer :

    Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room,

    Then, 7C1 x 6C2 x 4C4 

    = 7 x 15 x 1 = 105.

  • Question:Suppose You Want To Arrange Your English, Hindi, Mathematics, History, Geography And Science Books On A Shelf. In How Many Ways Can You Do It?

    Answer :

    We have to arrange 6 books.

    The number of permutations of n objects is n! = n. (n – 1) . (n – 2) … 2.1

    Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720.

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  • Question:5 Men And 4 Women Are To Be Seated In A Row So That The Women Occupy The Even Places . How Many Such Arrangements Are Possible?

    Answer :

    There are total 9 places out of which 4 are even and rest 5 places are odd.

    4 women can be arranged at 4 even places in 4! ways.

    and 5 men can be placed in remaining 5 places in 5! ways.

    Hence, the required number of permutations  = 4! x 5! = 24 x 120 = 2880.

  • Question:A Box Contains 2 White Balls, 3 Black Balls And 4 Red Balls. In How Many Ways Can 3 Balls Be Drawn From The Box, If At Least One Black Ball Is To Be Included In The Draw?

    Answer :

    We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

    Required number of ways=(3C1*6C2)+(3C2*6C1)+3C3 = (45 + 18 + 1) =64.

  • Question:There Are 5 Novels And 4 Biographies. In How Many Ways Can 4 Novels And 2 Biographies Can Be Arranged On A Shelf ?

    Answer :

    4 novels can be selected out of 5 in 5C4 ways.

    2 biographies can be selected out of 4 in 4C2 ways.

    Number of ways of arranging novels and biographies = 5C4*4C2= 30 

    After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.

    By the Counting Principle, the total number of arrangements = 30 x 720 = 21600.

  • Question:Compute The Sum Of 4 Digit Numbers Which Can Be Formed With The Four Digits 1,3,5,7, If Each Digit Is Used Only Once In Each Arrangement?

    Answer :

    The number of arrangements of 4 different digits taken 4 at a time is given by 4P4 = 4! = 24.

    All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

    Thus,each digit will occur 24/4 = 6 times in each of the position.

    The sum of digits in one’s position will be 6 x (1+3+5+7) = 96.

    Similar is the case in ten’s,hundred’s and thousand’s places.

    Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656.

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  • Question:If Repetition Of The Digits Is Allowed, Then The Number Of Even Natural Numbers Having Three Digits Is?

    Answer :

    In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

    10’s place can be filled in 10 different ways

    100’s place can be filled in 9 different ways

    There fore total number of ways = 5X10X9 = 450.

  • Question:From A Total Of Six Men And Four Ladies A Committee Of Three Is To Be Formed. If Mrs. X Is Not Willing To Join The Committee In Which Mr. Y Is A Member, Whereas Mr.y Is Willing To Join The Committee Only If Mrs Z Is Included, How Many Such Committee Are Possible?

    Answer :

    We first count the number of committee in which

    (i). Mr. Y is a member 

    (ii). the ones in which he is not

    Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.

    Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

    We can choose 1 more in5+2C1=7 ways.

    Case (ii): If Mr. Y is not a member then we left with (6+4-1) people. 

    we can select 3 from 9 in 9C3=84 ways.

    Thus, total number of ways is 7+84= 91 ways.

  • Capgemini Aptitude Interview Questions

  • Question:Find The Value Of ‘n’ For Which The Nth Term Of Two Ap’s:15,12,9…. And -15,-13,-11…… Are Equal?

    Answer :

    Given are the two AP’S:

    15,12,9…. in which a=15, d=-3………….(1) 

    -15,-13,-11….. in which a’=-15 ,d’=2…..(2) 

    now using the nth term’s formula,we get

    a+(n-1)d = a’+(n-1)d’

    substituting the value obtained in eq. 1 and 2,

    15+(n-1) x (-3) = -15+(n-1) x 2

    => 15 – 3n + 3 = -15 + 2n – 2

    => 12 – 3n = -17 + 2n

    => 12+17 = 2n+3n

    => 29=5n

    => n= 29/5.

  • Question:How Many 6-digit Even Numbers Can Be Formed From The Digits 1, 2, 3, 4, 5, 6 And 7 So That The Digits Should Not Repeat And The Second Last Digit Is Even ?

    Answer :

    Let last digit is 2

    when second last digit is 4 remaining 4 digits can be filled in 120 ways,

    similarly second last digit is 6 remained 4 digits can be filled in 120 ways.

    so for last digit = 2, total numbers=240

    Similarly for 4 and 6

    When last digit = 4, total no. of ways =240

    and last digit = 6, total no. of ways =240

    so total of 720 even numbers are possible.

  • Question:Find The Total Number Of Distinct Vehicle Numbers That Can Be Formed Using Two Letters Followed By Two Numbers. Letters Need To Be Distinct?

    Answer :

    Out of 26 alphabets two distinct letters can be chosen in 26P2 ways.

    Coming to numbers part, there are 10 ways.

    (any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit.

    Hence there are totally 10X10 = 100 ways. 

    Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

  • Question:How Many Alphabets Need To Be There In A Language If One Were To Make 1 Million Distinct 3 Digit Initials Using The Alphabets Of The Language?

    Answer :

    1 million distinct 3 digit initials are needed. 

    Let the number of required alphabets in the language be ‘n’. 

    Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

    Note distinct initials is different from initials where the digits are different. 

    For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different. 

    This n3 different initials = 1 million 

    i.e. n3=106  (1 million = 106)

      => n = 102 = 100

    Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

  • Question:Suppose You Can Travel From A Place A To A Place B By 3 Buses, From Place B To Place C By 4 Buses, From Place C To Place D By 2 Buses And From Place D To Place E By 3 Buses. In How Many Ways Can You Travel ?from A To E?

    Answer :

    The bus fromA to B can be selected in 3 ways.

    The bus from B to C can be selected in 4 ways.

    The bus from C toD can be selected in 2 ways.

    The bus fromD to E can be selected in 3 ways.

    So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72.

  • Question:There Are 2 Brothers Among A Group Of 20 Persons. In How Many Ways Can The Group Be Arranged Around A Circle So That There Is Exactly One Person Between The Two Brothers?

    Answer :

    fix one person and the brothers B1 P B2 = 2 ways to do so. 

    other 17 people= 17! 

    Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!.

  • Question:In How Many Ways Can 5 Letters Be Posted In 4 Letter Boxes?

    Answer :

    First letter can be posted in 4 letter boxes in 4 ways.

    Similarly second letter can be posted in 4 letter boxes in 4 ways and so on.

    Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024.