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Thread: TestFunda Answers - Higher Maths

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    Post TestFunda Answers - Higher Maths


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    Post TestFunda Answers - Higher Maths

    Let us say there are only 3 questions. Thus there are 23-1 = 4 students who have done 1 or more
    questions wrongly, 23-2 = 2 students who have done 2 or more questions wrongly and 23-3 = 1 student who
    must have done all 3 wrongly. Thus total number of wrong answers = 4 + 2 + 1 = 7 = 23 - 1 = 2n - 1.
    In our question, the total number of wrong answers = 4095 = 212 - 1. Thus n = 12 is the ans

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    Post TestFunda Answers - Higher Maths

    best to keep each digit at hundredth place(other than 0 and 9 as tenth place digit is more than that at hundredth place) .
    for 1 --> total numbers will be 44
    for 2 --> total numbers will be 42
    for 3 --> total numbers will be 39
    for 4 --> total numbers will be 35
    for 5 --> total numbers will be 30
    for 6 --> total numbers will be 24
    for 7 --> total numbers will be 17
    for 8 --> total numbers will be 9.
    Hence, total will be 44+42+39+35+30+24+17+9 = 240  

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    Post TestFunda Answers - Higher Maths

    1144 = 2^3*11*13
    Now 777777.......6k times would be divisible by both 11 & 13
    and 777777..........429times = 7777.....7000(426 times 7)+777
    Now 7777....7000 wud be divisible by 8,11,13 (last 3 digits divisible by 8 too),
    So, remainder = 777

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    Post TestFunda Answers - Higher Maths

    Fx = 1! + 2! + 3! + … + x!

    What will be the value of 3.1! + 4.2! + 5.3! + 6.4! +… + 19.17! + 20.18! in terms of F19?
    Options
    1) 
    2(F19) – 19! – 1
    2) 
    (F19) + 19! – 1
    3) 
    3(F19) – 2.19!
    4) 
    none of d above

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    Post TestFunda Answers - Higher Maths

    3.1! + 4.2! + 5.3! + ..... + 20.18! = (2+1)1! + (3+1)2! + (4+1)3! + .... + (19+1)18!
                                                        [(2! + 3! + 4! + .... + 19!) + (1! + 2! + .... + 18!)]
                                                     =  [(1! + 2! + 3! + 4! + .... + 19!) - 1! + (1! + 2! + .... + 18! + 19!) -19!]
                                                     = F19 -1! + F19 -19! = 2F19 -1- 19!

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    Post TestFunda Answers - Higher Maths

    A and B play a game of dice between them. The dice consist of colors on their faces (instead of numbers). When the dice are thrown, A wins if both show the same color; otherwise B wins. One die has 4 red face and 2 blue faces. How many red and blue faces should the other die have if the both players have the same chances of winning?
    Options
    1) 
    3 red and 3 blue
    2) 
    2 red and remaining blue
    3) 
    6 red and 0 blue
    4) 
    4 red and 2 blue

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    Post TestFunda Answers - Higher Maths

    probability of wining both will be equal if total of 6 red and 6 blue ARE PRESENT

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    Post TestFunda Answers - Higher Maths

    probability that A will win is 0.5
    P(both red) +P(both blue) = A wins
    let x be no of red
    (4/6)*(x/6) + (2/6)*((6-x)/6) = 0.5
    solve for x
    x=3
    so red =3 and blue = 3  

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    Post TestFunda Answers - Higher Maths


    3.1! + 4.2! + 5.3! + ..... + 20.18! = (2+1)1! + (3+1)2! + (4+1)3! + .... + (19+1)18!
                                                        [(2! + 3! + 4! + .... + 19!) + (1! + 2! + .... + 18!)]
                                                     =  [(1! + 2! + 3! + 4! + .... + 19!) - 1! + (1! + 2! + .... + 18! + 19!) -19!]
                                                     = F19 -1! + F19 -19! = 2F19 -1- 19!  


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    Post TestFunda Answers - Higher Maths

    find the highest power of 2 in 1! + 2! + 3! + 4! + 5! + .......... + 600!
    Options
    1) 
    1
    2) 
    494
    3) 
    3.0
    4) 
    256
    Additional Notes: Q is from Arun Sharma. Answer given is c

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    Post TestFunda Answers - Higher Maths


    In A Pair regular polygons, the ratio of the measure of interior is 10:9 . then the maximum number of such pairs is ?

    Options
    1) 
    9
    2) 
    11
    3) 
    15
    4) 
    19

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    Post TestFunda Answers - Higher Maths

    A car manufacturer produces only red and blue models which come out of the final testing area completely at

    random. What are the odds that 5 consecutive cars of the same color will come through the test area at a time?
    Options
    1) 
    1in16
    2) 
    1in125
    3) 
    1in32
    4) 
    1in25

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    Post TestFunda Answers - Higher Maths

    Its very simple ;
    ( 2,2,2,3,3) thats the only case to get sum of 12 in 5 tosses .
    So probablity will be ( total favorable cases) / total number of possible outcomes.
    Total favorable cases ( 5!/3!*2!) = 10
    probability of one single outcome be it 2 or 3 , it will be 1/2 .
    So final probablity is 10* (1/2)^5 = 5/16  .

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    Post TestFunda Answers - Higher Maths

    probability of first car will be red 1/2
    Similarly probability of each car will be red = 1/2 
    Probability of all car would be red =  1/2 * 1/2 *1/2 *1/2 *1/2   = 1/32
    Similarly of all car blue = 1/32
    1/32   +   1/32=  1/16 

  18. #2718
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    Post TestFunda Answers - Higher Maths

    probability of first car will be red 1/2
    Similarly probability of each car will be red = 1/2 
    Probability of all car would be red =  1/2 * 1/2 *1/2 *1/2 *1/2   = 1/32
    Similarly of all car blue = 1/32
    1/32   +   1/32=  1/16  

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    Post TestFunda Answers - Higher Maths

    there are 2 choices everytime to select the car. Hence total cases for 5 consecutive cars would be 25 that would be 32 and there is only one case where all 5 cars would be of the same colour. Hence solution is 1/32.Similar car of other colour would be 1/32. Total is 1/16.
    Last edited by mohit311989; 09-Aug-13 at 11:47 PM.

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    Post TestFunda Answers - Higher Maths

    when 50! is divided by 16^15 what will be the remainder ?
    Options
    1) 
    1
    2) 
    2
    3) 
    3
    4) 
    0
    Additional Notes: Please provide a detailed or proper solution to the question

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