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Thread: TestFunda Answers - Higher Maths

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    Post TestFunda Answers - Higher Maths


    Quark Express Couriers (QEC) sends a person on motorcycle every day to the airport to collect the courier. He reaches the airport exactly when the plane lands at the scheduled time. One day QEC came to know that the flight would be late by 50 minutes. So, the motorcyclist also started from his office 50 minutes later than the usual time. However, the plane landed earlier than was anticipated by the QEC. Immediately, the airport authorities dispatched the courier through a cyclist to QEC. The cyclist met the motorcyclist of QEC on the way after 20 minutes and handed over the courier to him. As a result, the QEC person reached office 10 minutes earlier than expected. The motorcyclist and the cyclist travel at their own uniform speeds.


    If the cyclist covered a distance of 2.5 km before he met the motorcyclist, what is the speed of the motorcyclist?


    Options
    1) 
      60 kmph
    2) 
    45 kmph
    3) 
    30 kmph
    4) 
    25 kmph
    5) 
    Cannot be determined

  3. #83
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    Post TestFunda Answers - Higher Maths

    option 3.
    Motor-cyclist is traveling 5 km less, and he is reaching 10 mins early. So, speed is 30 kmph 

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    Post TestFunda Answers - Higher Maths

    The motorcyclist reaches early by 10 minutes because he meets the cyclist mid-way.
    So, if the cyclist wasn't there, he would have used 10/2 = 5 minutes to reach the airport and 5 minutes from the airport to the place where he had met the cyclist.
    This distance = 2.5 km
    So, 2.5 km in 5 minutes
    So, 30 km in 1 hour
    So, speed = 30 kmph
    Last edited by vimbloid; 27-Oct-10 at 6:33 AM.

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    Post TestFunda Answers - Higher Maths


    (SP-CP)/CP *100% = 10%

    10SP- 10CP = CP

    So, 10SPold = 11CPold.................(A)

     

    CPnew = 0.95CPold

    SPnew = SPold + 7

    Accordingly,

    5SPnew = 6CPnew

    So, 5(SPold + 7) = 6(0.95CPold)

    So, 5SPold + 35 = 5.7CPold..................(B)

     

    Solving A and B, we get,

    5(11/10 CPold) + 35 = 5.7 CPold

    11CPold + 70 = 11.4CPold

    CPold = 700/4 = 175

    (SP-CP)/CP *100% = 10%
    10SP- 10CP = CP
    So, 10SPold = 11CPold.................(A)

    CPnew = 0.95CPold
    SPnew = SPold + 7
    Accordingly,
    5SPnew = 6CPnew
    So, 5(SPold + 7) = 6(0.95CPold)
    So, 5SPold + 35 = 5.7CPold..................(B)
    Solving A and B, we get,
    5(11/10 CPold) + 35 = 5.7CPold
    11CPold + 70 = 11.4CPold
    CPold = 700/4 = 175

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    Post TestFunda Answers - Higher Maths

    Five persons, working together, take 12 days to complete a piece of work. If the five people, instead do the work in a peculiar manner such that the first person works for two days and takes rest on the third day, the second person works for three days and takes rest for the next two days, the third person works for four days and takes rest for the next three days and so on, then in how many days will the work be completed, given all of them start the work on the same day?

    Options
    1) 
    19
    2) 
    19
    3) 
    20
    4) 
    21
    Additional Notes: aimcat04- 13

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    Post TestFunda Answers - Higher Maths

     There are m integers ni, where i = 1, 2, 3,…. m. If S = n 1 + n1n2 + n1n2n3 + n1n2n3n4+…………..…+ n1n2n3…..nm, which of the following statements is always true?
    Options
    1) 
     S is even, if m is even.
    2) 
     If there are an odd number of odd ni and m is odd, then S is even.
    3) 
     If the least value of i, for which ni is even, is even, then S is odd.
    4) 
     If the largest value of i, for which ni is odd, is odd, then S is even.

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    Post TestFunda Answers - Higher Maths


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    Post TestFunda Answers - Higher Maths

    Its time consuming one for me.I have listed the daily work and found the answer. But need to know some alternatives.

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    Post TestFunda Answers - Higher Maths


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    Post TestFunda Answers - Higher Maths


    Suppose there are 4 integers: 1,2,3 and 4

    S is even, if m is even

    Here m is even (4)

    S = 1+1*2+1*2*3+1*2*3*4 = 33

    So, S is not even

    Hence, option 1 is eliminated

     

    Suppose there are 5 integers: 1, 2, 3, 4 and 5

    So, m = 5 = odd

    Odd i = 3 ( one, three and five) = odd

    S =  1+1*2+1*2*3+1*2*3*4+1*2*3*4*5 = 153 = odd

    So, option 2 is eliminated

     

    Suppose there are 5 integers: 1, 2, 3, 4 and 5

    n1=1, n2=2, n3=3, n4=4 and n5=5

    Odd ni = n1, n3 and n5

    Largest value of i for which ni is odd = n5 = 5, which is odd

    S =  1+1*2+1*2*3+1*2*3*4+1*2*3*4*5 = 153 = odd

    So, option 4 is eliminated

     

    Suppose there are 5 integers: 1, 2, 3, 4 and 5

    n1=1, n2=2, n3=3, n4=4 and n5=5

    Even ni = n2 and n4

    Least value of i for which ni is even = 2, which is even

    S =  1+1*2+1*2*3+1*2*3*4+1*2*3*4*5 = 153 = odd

    So, the answer is option 3

    Suppose there are 4 integers: 1,2,3 and 4
    S is even, if m is even
    Here m is even (4)
    S = 1+1*2+1*2*3+1*2*3*4 = 33
    So, S is not even
    Hence, option 1 is eliminated

    Suppose there are 5 integers: 1, 2, 3, 4 and 5
    So, m = 5 = odd
    Odd i = 3 ( one, three and five) = odd
    S =  1+1*2+1*2*3+1*2*3*4+1*2*3*4*5 = 153 = odd
    So, option 2 is eliminated

    Suppose there are 5 integers: 1, 2, 3, 4 and 5
    n1=1, n2=2, n3=3, n4=4 and n5=5
    Odd ni = n1, n3 and n5
    Largest value of i for which ni is odd = n5 = 5, which is odd
    S =  1+1*2+1*2*3+1*2*3*4+1*2*3*4*5 = 153 = odd
    So, option 4 is eliminated

    Suppose there are 5 integers: 1, 2, 3, 4 and 5
    n1=1, n2=2, n3=3, n4=4 and n5=5
    Even ni = n2 and n4
    Least value of i for which ni is even = 2, which is even
    S =  1+1*2+1*2*3+1*2*3*4+1*2*3*4*5 = 153 = odd
    So, the answer is option 3

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    Post TestFunda Answers - Higher Maths

    Take total work as 60 , and form a grid to simply calculate the number of people that work in one day
    like Day 1 =5, day2=5 , day 3=4, day 4=4, day5=3,day6=2.....
    sum up the number of people to 60 , ie(5+5+4+4+3+2.....) on Day 19 sum will be excatly 60 that is the answer.Thanks! 

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    Post TestFunda Answers - Higher Maths

    Gautam gave to his brother Uttam, a chain comprising N closed links numbered from 1 to N (in that order, from one end to the other) and asked him to cut them into sets of one or more links such that he can return to Gautam any number of links that Gautam may ask for (i.e. from 1 to N).

    Cutting a link means that the link is cut in only one place. For example if there is a chain comprising 4 closed links, numbered from 1 to 4, we can cut exactly one link and obtain three sets of links.
    Caselet
       

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    Post TestFunda Answers - Higher Maths

     This morning, Gopal, the local milkman, was in a hurry to leave home for selling milk. In his hurry he asked Suppandi, his servant to pour half of the water from the water drum into the milk drum containing pure milk and then to bring him a full can of the diluted milk for selling. However, Suppandi got confused and he poured half the milk from the milk drum into the water drum and brought out a full can of the contents from the latter for his master to sell. The concentration of the milk in the can that Suppandi brought was 1/n times that intended by Gopal
    Caselet
       

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     An infinite plane has to be tiled (without gaps or overlaps) using identical tiles, each in the shape of a regular n-sided polygon. How many values are possible for n?
    Options
    1) 
    3
    2) 
    5
    3) 
    6

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    Post TestFunda Answers - Higher Maths

    Consider the initial volume of milk (in the container) to be M and that of water to be W.
    Now, on mixing half the contents of the Water container into the milk container, we get the total final volume (of the milk + water mixture) as M + W/2. The concentration of milk in this is M/(M+W/2).

    However, Suppandi commits a blunder by doing the opposite of what is asked of him. After his actions, the final volume in the container (re: water) becomes W + M/2. The concentration of milk in this is (M/2)/(W + M/2).

    on equating as per the given conditions,
    (M/2)/(W + M/2) = M/(M+W/2) * (1/n)

    Put n = 3 and get the answer.

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    Post TestFunda Answers - Higher Maths

    Refer to the previous solution. Here, the ratio M/W is given as 4.5 and the value of n is asked.

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    Post TestFunda Answers - Higher Maths

    Try rearranging the terms of the equation in the solution to the first part in such a manner that you can get n in terms of W and M. I'll give you a hint. n ε (1 , 4). Try to reason why n can't take values beyond this range.

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    Post TestFunda Answers - Higher Maths

     X, Y and Z are points on the sides AB, BC and AC of the triangle ABC, such that AX : XB = 4 : 3, BY : YC = 2 : 3 and CZ : ZA = 2 : 1. Find the ratio of the area of the triangle XYZ to that of the triangle ABC.
    Options
    1) 
    4/35
    2) 
    4/21
    3) 
    5/21
    4) 
    3/14
    Additional Notes: Any shortcut any trick for this ques

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