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Thread: TestFunda Answers - Higher Maths

  1. #2501
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    Post TestFunda Answers - Higher Maths

    option 1..265
    classic case of dearrangements.

    nos. of ways = 6!*(1- 1/1! + 1/2! - 1/3! + 1/4! - 1/5! + 1/6!) = 265

  2. #2502
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    Post TestFunda Answers - Higher Maths

    if the sets are identical then option B


    otherwise if are considered as different like Set 1 Set 2 etc. then option D


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    Post TestFunda Answers - Higher Maths

    Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the best describes the minimum value say m of these three integers?
    Options
    1) 
    1<=m<=3
    2) 
    4<=m<=6
    3) 

    7<=m<=9

    4) 
    10<=m<=12
    5) 
    13<=m<=15

  4. #2504
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    Post TestFunda Answers - Higher Maths

    A person wrote first N consecutive numbers (natural) on a board and found their sum, Another person came and deleted the smallest number and found the sum of remaaining numbers. The process continued until N remained on the board. The average of all the sums is found to out to be 105. Find N?
    Options
    1) 
    17
    2) 
    20
    3) 
    19
    4) 
    23

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    Post TestFunda Answers - Higher Maths

    X and Y belong to the set of positive integers. The probability that a randomly chosen value of Y satisfies the equation |3X-13Y|= 2 is:
    Options
    1) 
    1/8
    2) 
    2/3
    3) 
    3/13
    4) 
    Cannot be determined

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    Post TestFunda Answers - Higher Maths

    Karan and Arjun run a 100 metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true  in connection with the second race?
    Options
    1) 
    Karan and Arun reach the finishing line simultaneously.
    2) 
    Arjun beats Karan by 1 metre.
    3) 
    Arjun beats Karan by 11 metre.
    4) 
    Karan beats Arjun by 1 metre.

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    Post TestFunda Answers - Higher Maths

    After travelling 50km, a train is meeting with an accident and travels at 3/4th of the usual speed and reaches 45 min late. Had the accident happened 10km further on it would have reached 35 min late. Find the usual speed?
    Options
    1) 
    15 km/hr
    2) 
    20 km/hr
    3) 
    25 km/hr
    4) 
    30 km/hr

  8. #2508
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    Post TestFunda Answers - Higher Maths

    Let three integers be x-1, x & x+1.
    Now acc. to question x-1 + x^2 + (x+1)^3 = (3x)^2
    Solving equation we will have x= 0,1,4.
    x cannot be 0, 1 as numbers are positive integers, so x= 4.

    So integers are 3,4,5.
    And our answer is 3.

  9. #2509
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    Post TestFunda Answers - Higher Maths

    Three wheels can complete 60, 36, 24 revolutions per minute respectively. There is a spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again:
    Options
    1) 
    5/2 sec
    2) 
    5/3 sec
    3) 
    6 sec
    4) 
    7.5 sec

  10. #2510
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    Post TestFunda Answers - Higher Maths

    option 2..20 Kph

    let total distance be 'd', original speed of train be 's' and time taken to reach d be 't'
    d/s = t

    now,
    d/s + 4(d-50)/3s = d/s + 3/4 ...............eq.1
    d/s + 4(d-60)/3s = d/s + 7/12 ...............eq.2

    solving eq.1 and eq.2...we get s = 20 Kph  

  11. #2511
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    Post TestFunda Answers - Higher Maths

    option 4...karan beats arjun by 1 meter
    from 1st race:
    Skaran / Sarjun = 10/9 ....let their speeds be 10m/s and 9 m/s respectively.
    in  2nd race Karan runs 110 mtrs to complete the race, i.e., he takes 11 seconds to complete the racing distance of 100 mtrs.
    In this 11 secs Arjun runs 9*11 = 99mtrs
    Therefore, Karan beats Arjun by 1 mtr in the 2nd race

  12. #2512
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    Post TestFunda Answers - Higher Maths

    option 2..2/3

    Upon removing the modulus sign, we get (3x-13y) = +2 and (3x-13y) = -2,
    Solving we see that y exists for positive integers (1,2,4,5,7,8,10,11,…..)   
    values like 3,6,9,12,….. cannot satisfy the equation for positive integer values of x….. hence ,probability is 2/3


     

  13. #2513
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    Post TestFunda Answers - Higher Maths

    option 1...17
    The sum of the series can be writen as 1*1 + 2*2 + 3*3 + ........ + n*n  = n*(n+1)*(2n+1)/6

    now,
    (n+1)*(2n+1)/6 = 105  {given, average of sum of n series as 105}

    solving: n = 17

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    Post TestFunda Answers - Higher Maths

    If the same 100 apples had to be divided among 8 people instead of 5 in a similar manner, then how many apples will the eighth person get, if the no.of apples the first person gets is unchanged?

    Options
    1) 
    70/3
    2) 
    65/6
    3) 
    115/3
    4) 
    Indeterminate

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    Post TestFunda Answers - Higher Maths

    is a,a,a,a,a,a,a,a,a................. an A.P? if yes then how and what will be the sum of n terms in the series and if no then why.

  16. #2516
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    Post TestFunda Answers - Higher Maths

    how it can be an A.P all terms are equal to its first term and all of them had a common difference of 0..................................................i dont think it is in progressions

    try to use A.M >= G.M
    ARITHMETIC MEAN GREATER THAN OR EQUAL TO GEOMETRIC MEAN CAN BE ONE APPROACH IF ITS SOLUTION EXIST

  17. #2517
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    Post TestFunda Answers - Higher Maths


    A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kgs. The clerk weighs the boxes in pairs. The weights obtained are 110,112, 113, 114, 115, 116, 117, 118, 120 and 121 kgs. What is the weight, in kgs, of the heaviest box?





    Options
    1) 
    60
    2) 
    62
    3) 
    64
    4) 
    cannot be determined
    Additional Notes: Explain ur method

  18. #2518
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    Post TestFunda Answers - Higher Maths

    it's in an AP with the common diff. 0. as the sum of AP is given by n/2(2a+(n-1)d= n/2(2a)=an

  19. #2519
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    Post TestFunda Answers - Higher Maths

    what is the total no. of ways of selecting twenty balls from an infinite no. of blue,green and yellow balls?
    Options
    1) 
    3^20
    2) 
    20^3
    3) 
    231
    4) 
    1771
    Additional Notes: please help why cant the ans to this ques be 3^20??

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    Post TestFunda Answers - Higher Maths

    No. of ways for finding 1st ball out of the 3 types = 3
    No. of ways for finding 2nd ball out of the 3 types = 3
    Similarly for 3rd ball=3, 4th ball=3,....
    Thus, for 20 balls = 3^20

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