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Thread: TestFunda Answers - Higher Maths

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    Default TestFunda Answers - Higher Maths

    This thread contains questions, and solutions in Higher Maths Topic from TestFunda Answers.
    Last edited by TestFunda; 06-Oct-10 at 11:59 AM.

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    Post TestFunda Answers - Higher Maths

    what is the remainder when 555555......98 times    is divided by 133?

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    Post TestFunda Answers - Higher Maths

    The answer is 125
    Now what i did is very unconventional but if you are even reasonably good in calculation you can reach there very fast
    simply start deviding
    Number ----No of 5------ Remainder 
    5                    1                 5
    55                   2                55
    555                 3                23
    5555               4               102
    55555             5                94
    ..................
    ...................
    No of 5 ----------------Remainder

    6                                   14
    7                                  12
    8                                   125
    9                                  58
    10                                 53
    11                                 3
    12                                 35
    13                                 89
    14                                 97
    15                                  44
    16                                  46
    17                                  66
    18                                   0
    19     again again             5
    so we have the whole series now 98 mod 18 =8 or 5555...8 times devided by 133 remainder =125

    If someone has an easier solution than this donkey type method please post

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    Post TestFunda Answers - Higher Maths

    A coin whose faces r  marked 2,3 is thrown 5 times,what is the chance of obtaining a total of 12?
    Options
    1) 
    5/16
    2) 
    11/16
    3) 
    7/16
    4) 
    3/16

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    Post TestFunda Answers - Higher Maths


    Total possibilities = 2^5 = 32

    Total cases where sum = 12

    The only way the sum can be 12 is when the combinations are 2, 2, 2, 3 and 3

    The number of ways this can occur is the same as the number of ways in which the 3's can be placed

    3 3 _ _ _

    3 _ 3 _ _ 

    3 _ _ 3 _ 

    3 _ _ _ 3

    _ 3 3 _ _ 

    _ 3 _ 3 _

    _ 3 _ _ 3

    _ _ 3 3 _

    _ _ 3 _ 3

    _ _ _ 3 3

    This can occur in 4+3+2+1 = 10 ways

     

    So, required probability = 10/32 = 5/16

    Total possibilities = 2^5 = 32
    Total cases where sum = 12
    The only way the sum can be 12 is when the combinations are 2, 2, 2, 3 and 3
    The number of ways this can occur is the same as the number of ways in which the 3's can be placed
    3 3 _ _ _
    3 _ 3 _ _ 
    3 _ _ 3 _ 
    3 _ _ _ 3
    _ 3 3 _ _ 
    _ 3 _ 3 _
    _ 3 _ _ 3
    _ _ 3 3 _
    _ _ 3 _ 3
    _ _ _ 3 3
    This can occur in 4+3+2+1 = 10 ways
    So, required probability = 10/32 = 5/16

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    Post TestFunda Answers - Higher Maths

    The answer is 125
    Now what i did is very unconventional but if you are even reasonably good in calculation you can reach there very fast
    simply start deviding
    Number ----No of 5------ Remainder 
    5                    1                 5
    55                   2                55
    555                 3                23
    5555               4               102
    55555             5                94
    ..................
    ...................
    No of 5 ----------------Remainder

    6                                   14
    7                                  12
    8                                   125
    9                                  58
    10                                 53
    11                                 3
    12                                 35
    13                                 89
    14                                 97
    15                                  44
    16                                  46
    17                                  66
    18                                   0
    19     again again             5
    so we have the whole series now 98 mod 18 =8 or 5555...8 times devided by 133 remainder =125

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    Post TestFunda Answers - Higher Maths

    This is a question of putting 5 labels on 5 boxes such that no box has the right label.

    We can first calculate the cases where exactly 1, 2, 3, 4 and 5 boxes are labelled correctly. 

    Case 1: Exactly 5 boxes are labelled correctly
    No of ways = 1

    Case 2: Exactly 4 boxes are labelled correctly
    This case is same as the case above, because the 5th box will by default get labelled correctly.

    Case 3: Exactly 3 boxes are labelled correctly
    Consider the case where box 1, box 2 and box 3 are labelled correctly. So box 4 and 5 are labelled incorrectly. The number of ways in which this can be done is 1.

    So total number of ways for labeling exactly 3 boxes correctly = 5C3*1 = 10

    Case 4: Exactly 2 boxes are labelled correctly
    Consider the case where box 1and box 2 are labelled correctly and box 3, box 4 and box 5 are labelled wrongly. The number of ways to do this is 2 as follows.
    534, 453

    So total number of ways for labeling exactly 2 boxes correctly = 5C2*2 = 20

    Case 5: Exactly 1 box is labelled correctly
    Consider the case where box 1is labelled correctly and box 2, box 3, box 4 and box 5 are labelled wrongly. The number of ways to do this is 9 as follows.
    3254, 5234, 4253, 3524, 4523, 5423, 3452, 4532, 5432

    So total number of ways for labeling exactly 2 boxes correctly = 5C1*9 = 45

    Total no of ways in which at least 1 box is labelled correctly = 1+10+20+45 = 76

    Total no of ways in which 5 boxes can be labelled = 5! = 120

    Total no of ways in which no box is labelled correctly = 120 - 76 = 44

    Probability of labeling all boxes incorrectly = 44/120 = 11/30

    Hence, option 3. 

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    Post TestFunda Answers - Higher Maths


    Let's start with a small problem. Assume that there are 4 balls and 4 boxes. Let A be the event that ball 1 is in the correct box and similarly let B be the event that ball 2 is in the correct box. Then

     

    P(A) = 1/4, P(B) = 1/4 and

    P(A and B) = 1/4*1/3

    ...............

     

    Now use the inclusion-exclusion principle to get the probability that ball 1 or ball 2 or ball 3 or ball 4 is correctly placed.

     

    P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A and B) - P(A and C) - P(A and D) - P(B and C) - P(B and D) - P(C and D) + P(A and B and C) + P(A and B and D) + P(A and C and D) + P(B and C and D) - P(A and B and C and D) 

     

    = 4C1*1/4 - 4C2*1/4*1/3 + 4C3*1/4*1/3*1/2 - 4C4*1/4*1/3*1/2*1/1

    = 1 - ((4*3)/2)*1/4*1/3 + ((4*3*2)/(3*2))*1/4*1/3*1/2 - 1*1/4*1/3*1/2*1/1

    = 1 - 1/2! + 1/3! - 1/4!

     

    Note that it's following a pattern! For 6 balls and 6 boxes the result will be

    1 - 1/2! + 1/3! - 1/4! + 1/5! - 1/6!

     

    So number of ways of getting at least one ball in correct box = Probability * Total number of ways of placing the balls = (1 - 1/2 + 1/6 - 1/24 + 1/120 - 1/720)*6! = 455

     

    Number of ways of getting at least 2 balls in correct boxes = Number of ways of getting at least one ball in correct box - number of ways of getting exactly one ball in correct box

     

    Nubmer of ways of getting exactly one ball in correct box = 6C1 x Number of ways of getting all 5 balls in incorrect boxes = 6 x (5! - number of ways of getting at least 1 out of the 5 balls placed correctly) = 6 x (5! - 5!*(1 - 1/2! + 1/3! - 1/4! + 1/5!)) = 6 x (120 - 120(1 - 1/2 + 1/6 - 1/24 + 1/120)) = 6 x 44 = 264

     

    Number of ways of getting at least 2 balls in correct boxes = 455 - 264 = 191

     

    Required probablity = 191/720

     

    Hence, option 1.

    Let's start with a small problem. Assume that there are 4 balls and 4 boxes.
    Let A be the event that ball 1 is in the correct box and similarly let B be the event that ball 2 is in the correct box.
    Then
    P(A) = 1/4, P(B) = 1/4 andP(A and B) = 1/4*1/3...............

    Now use the inclusion-exclusion principle to get the probability that ball 1 or ball 2 or ball 3 or ball 4 is correctly placed.
    P(A or B or C or D) = P(A) + P(B) + P(C) + P(D) - P(A and B) - P(A and C) - P(A and D) - P(B and C) - P(B and D) - P(C and D) + P(A and B and C) + P(A and B and D) + P(A and C and D) + P(B and C and D) - P(A and B and C and D) 
    = 4C1*1/4 - 4C2*1/4*1/3 + 4C3*1/4*1/3*1/2 - 4C4*1/4*1/3*1/2*1/1= 1 - ((4*3)/2)*1/4*1/3 + ((4*3*2)/(3*2))*1/4*1/3*1/2 - 1*1/4*1/3*1/2*1/1= 1 - 1/2! + 1/3! - 1/4!

    Note that it's following a pattern! For 6 balls and 6 boxes the result will be
    1 - 1/2! + 1/3! - 1/4! + 1/5! - 1/6!
    So number of ways of getting at least one ball in correct box = Probability * Total number of ways of placing the balls = (1 - 1/2 + 1/6 - 1/24 + 1/120 - 1/720)*6! = 455
    Number of ways of getting at least 2 balls in correct boxes = Number of ways of getting at least one ball in correct box - number of ways of getting exactly one ball in correct box
    Number of ways of getting exactly one ball in correct box = 6C1 x Number of ways of getting all 5 balls in incorrect boxes = 6 x (5! - number of ways of getting at least 1 out of the 5 balls placed correctly) = 6 x (5! - 5!*(1 - 1/2! + 1/3! - 1/4! + 1/5!)) = 6 x (120 - 120(1 - 1/2 + 1/6 - 1/24 + 1/120)) = 6 x 44 = 264
    Number of ways of getting at least 2 balls in correct boxes = 455 - 264 = 191
    Required probability = 191/720
    Hence, option 1.

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    Post TestFunda Answers - Higher Maths

    What is the coefficient of x3y4z5 in the expansion of (xy + yz + zx)6?

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    Post TestFunda Answers - Higher Maths

    x^3y^4z^5=(xy)^3 (yz) (zx)^2

    which is same as 6C24C3=15x4=60 

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    Post TestFunda Answers - Higher Maths

    Find the no.of divisors of 1080 excluding the throughout divisors which are perfect squares?

    Options
    1) 
    28
    2) 
    29
    3) 
    30
    4) 
    31

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    Post TestFunda Answers - Higher Maths

    Option 2.
    1080=2^3*3^3*5
    So, total divisors are: 32
    Divisors which are perfect square: 2^2, 3^2, 6^2
    Answer: 29 

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    Post TestFunda Answers - Higher Maths

    The question should be which are not perfect squares

    Option 1.
    1080=2^3*3^3*5
    So, total divisors are: 32
    Divisors which are perfect square:1,  2^2, 3^2, 6^2
    Answer: 28  

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    Post TestFunda Answers - Higher Maths

    In a class of 50 students, 28 like pop music, and 20 like classical music. If 10 students like one of the two kinds of music, find the number of students who like both kinds of music?
    Options
    1) 
    8
    2) 
    5
    3) 
    7
    4) 
    4

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    Post TestFunda Answers - Higher Maths

    Either the options or the question is wrong.

    Please recheck  

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    Post TestFunda Answers - Higher Maths

    From three different softdrinks, 4 chinese dishes and 2 ice-creams, how many different meals are possible if atleast one of each of the three items is to be included depending upon no. of people likely to turn up?
    Options
    1) 
    315
    2) 
    282
    3) 
    864
    4) 
    none of these

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    Post TestFunda Answers - Higher Maths

    How many license plates involving three letters and 3 digits are there if the three letters appear together either at beginning or at end of the license? 
    Options
    1) 
    2 x 263 x 103
    2) 
    54102
    3) 
    4 x 252 x 104
    4) 
    none of these

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