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Thread: TestFunda Answers - Higher Maths

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    Post TestFunda Answers - Higher Maths

    the series will end up  (-1)^3+(-2)^3.... when calcupated we get 2 

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    Post TestFunda Answers - Higher Maths

    What is the last non zero digit in 100!?
    please explain its method also 


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    Post TestFunda Answers - Higher Maths

    Since you only want the last digit only concentrate on the last digits which would be multiplied. This question in eeffective becomes, last digit of (9!)10.
    if you know the answer of 9! then awesome or else, break it down.
    You would know, 7! = 5040
    So, last digit = 4.
    Last digit of 8! = 2 (4*8=32)
    Last digit of 9! = 8 (9*2=18)
     Hence, last digit of 100! = last digit of 810 = last digit of (210)3 
    210 = 1024
    43 = 64
    Last non-zero digit of 100! is 4 

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    Post TestFunda Answers - Higher Maths

    Since you only want the last digit only concentrate on the last digits which would be multiplied. This question in eeffective becomes, last digit of (9!)10.
    if you know the answer of 9! then awesome or else, break it down.
    You would know, 7! = 5040
    So, last digit = 4.  
    Last digit of 8! = 2 (4*8=32)
    Last digit of 9! = 8 (9*2=18)
     Hence, last digit of 100! = last digit of 810 = last digit of (210)3 
    210 = 1024
    4= 64
    Last non-zero digit of 100! is 4

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    Post TestFunda Answers - Higher Maths


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    Post TestFunda Answers - Higher Maths



    2x+4x=240;
    x=40;
    then two whl=2*40=80
    4 whl =4*40=160;
    ans: 160/4= i-e 40  


    Read more at http://www.m4maths.com/2639-There-are-some-2-wheelers-and-4-wheelers-parked-total-number-of-wheels-present-is-240-then-how-many-4.html#F39eXTbVYGBrwWuI.99


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    Post TestFunda Answers - Higher Maths

    First Read the question carefully, According to the given question all the numbers will be decided by alok and first 2 would be put by bhanu whereas alok will put the 3rd one.
    So for getting maximum alok will tell 7 as the first Number
    two cases arise
    1) if bhanu will put this number as Z then 9, 9 will be the next 2 number. and 9+x+y-z = 20
    2) if bhanu will put this number X, then alok will tell 4 or 5 as next number.
    if bhanu will put 4 or 5 as Z then alok will put 9 as Y. So N will be
    9+x+y-z= 9+7+9-4 = 21 OR 9+7+9-5 = 20

    If bhanu will put 4 or 5 as Y then alok will put 0 as Z
    9+x+y-z = 9+7+4-0 = 20 OR 9+7+5-0 = 21

    So if both are playing with there best effort then 9+x+y-z will be 20

    20 is the correct answer  


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    Post TestFunda Answers - Higher Maths

    A and B walk up an escalator (moving staircase). The escalator moves at a constant speed. A takes 6 steps for every 4 steps of B . A gets to the top of the escalator after having taken 50 steps while B (because of his slower pace lets the escalator do a little more of the work) takes only 40 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?
    Options
    1) 

    110

    2) 
    100
    3) 
    120
    4) 
    150

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    Post TestFunda Answers - Higher Maths

    Let speed of escalator be E and total steps be X. Then i) 50 + E. 50/Sa = X where Sa is speed of A. Similarly , ii ) 40 + E.40/Sb =  X where Sb is speed of b in steps per second. Also , iii) Sa/Sb = 6/4. Subtrating (ii) from (i) and substituting  (iii) , we get X=100

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    Post TestFunda Answers - Higher Maths

    Wilson's Theorem is about primes, namely, for integer n>1, 
    (n-1)! = -1 mod n iff n is prime. 

    Since 101 is prime, we can use it, along with the fact that 
    100*99*98*97! = 100! 

    Then taking mod 101 and using Wilson's Theorem, 
    100*99*98*97! = -1 mod 101 
    (-1)(-2)(-3)97! = -1 mod 101 

    so that, multiplying both sides by -1, 
    6*97! = 1 mod 101 

    Now, we'd like to find the inverse mod 101 of 6. To do this, use the fact that 
    6*17 = 102 = 1 mod 101, to multiply both sides by 17, yielding 

    17*6*97! = 97!mod 101 = 17 mod 101 

    So it's 17    

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    Post TestFunda Answers - Higher Maths

    The number will be of the form 44k+13.
    Put k=0, rem is 13
    put k=1, rem is 24
    put k=2, rem=2
    put k=3, rem=13
    put k=4,rem=24...so cycle following
    therefore sum of distinct rems=2+24+13=39  

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    Post TestFunda Answers - Higher Maths

    The approach taken is that the limiting factor in both 35^6 and the 21^12 is 7.
    On calculating we will see that 42! has highest power of 7 as 6.
    And 76! has highest power of 7 as 11 but 77 will have highest power of 7 as 12.so we need to discard 77.
    Therefore total number of value of n possible =76-42+1=35    
    Last edited by vimbloid; 14-Nov-14 at 11:12 AM.

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    Post TestFunda Answers - Higher Maths

    100<x<200
    100/7=rem(2)...ie 98 divivisible by 7...so in 100 to 200 range lowest no divisible by 100=98+7=105
    200/7=rem(4)...i.e. 196 is the largest within 200 to be divisible by 7...so 
    now do highest-lowest/7=196-105/7=91/7=13  

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    Post TestFunda Answers - Higher Maths

    Let 'n' be the total number of participants
    Therefore, No. of men= n-2
                  & No. of women= 2
    No. of chess matches betwee men= 2 x [(n-2)C2] = (n-2)(n-3)
    {Factor 2 comes from the info that "every participant played 2 games with every other participant"}
    AND No. of chess matches betwee men and women = [2 x (n-2)] + [2 x (n-2)] = 4 x (n-2)
    Ac. to question:  No. of chess matches betwee men = 66 + No. of chess matches betwee men and women
    Or  (n-2)(n-3) = 66 + 4 x (n-2)
    Solving for n, we get n=13 

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    Post TestFunda Answers - Higher Maths


    2520= 7 x 5 x 2x2x2 x 3x3 .


    Here,


    7 is the highest prime. So counting the number of 7's in 50! we get:


    50/7 + 50/49 


    =7 + 1


    =8.



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    Post TestFunda Answers - Higher Maths

    Out of a pck of 52 card one is lost , from the remaining of the pack two cards are drawn and are found to be spade. find the chances is the spade.
    Options
    1) 
    11/50
    2) 
    11/49
    3) 
    10/49
    4) 
    10/50
    Additional Notes: Answer is giving option a. but i did not get concept, please explaine.

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    Post TestFunda Answers - Higher Maths


    Three people Kishore, Rakesh and Ajeet start their journey from Bangalore to Delhi. They

    leave Bangalore at 7 am, 11 am and 4 pm respectively. In how many hours after Ajeet starts, will

    they meet if their speeds are 25km/hr, 30km/hr and 40 km/hr respectively?

    A ) 24 hrs

    B ) 20 hrs

    C ) 15 hrs

    D ) 12 hrs

    SOLUTION

    Let they meet after x hours. Distance travelled by them should be the same

    So 40x = 30(x + 5)

    10x = 150

    x = 15 hours

    The correct

    Three people Kishore, Rakesh and Ajeet start their journey from Bangalore to Delhi. Theyleave Bangalore at 7 am, 11 am and 4 pm respectively. In how many hours after Ajeet starts, willthey meet if their speeds are 25km/hr, 30km/hr and 40 km/hr respectively?
    Options
    1) 
    24hrs
    2) 
    20hrs
    3) 
    15hrs
    4) 
    12hrs

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