Discussion Room : Quant



  • ABCD is a square with side length 2 cm. It is divided into five rectangles of equal areas, as shown in the figure. The perimeter of the rectangle BEFG is

    0_1516899509808_0f9deca4-db08-4d50-a9b0-8d16877b2efc-image.png

    (a) 51/16
    (b) 36/11
    (c) 58/15
    (d) 47/13



  • Can anyone explain please.


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    @vinaycat2017

    Side length = 2 cm.
    So area of square = 4 cm^2
    Given square is divided into 5 rectangles of equal areas. So each rectangle has an area of 4/5 cm^2.
    Proceed with this understanding (i.e, if we know one side of rectangle = 2 then other side should be 2/5 to get an area of 4/5) and we get EB = 4/3 cm and BG = 3/5 cm. Figure as below

    0_1517200748962_ac81beed-447d-4d5a-8429-731d8711bd1b-image.png

    Perimeter = 58/15 cm (does it match with the OA?)





  • @0_1516904528318_QUES.png



  • @zabeer
    Sir one more
    0_1516904775903_how.png



  • @vinaycat2017

    58/15.
    One side comes out to be 4/3 and another one comes out to be 3/5



  • @vinaycat2017

    One way could be
    You take two triplets:-
    9,12,15
    5,12,13
    a=12, b=5, c=9
    Area comes out to be 84...by heron's.
    If u put these values in options...only third option would give 84 as answer.


  • Being MBAtious!


    @vinaycat2017
    Tan(A + B) = (Tan A + Tan B)/(1 - TanA * TanB)
    Tan70 = (Tan50 + Tan20)/(1 - Tan50 * Tan20)
    Tan70 - Tan70 * Tan50 * Tan20 = Tan50 + Tan20
    We know, Tan70 * Tan20 = 1
    So, Tan70 - Tan50 = Tan50 + Tan20
    Tan70 = Tan20 + 2Tan50.



  • A swimming pool is of length 50 m. A and B enter a 300 m race starting simultaneously at one end of the pool at speeds of 3 m/s and 5 m/s. How many times will they meet while travelling in opposite directions before B completes the race?
    a) Twice
    b) Thrice
    c) Once
    d) 5 times

    In every meet their combined distance travelled is 100 m.

    Let there be "a" meets.

    =>[(100a) * 5/(5+3)]=300
    Which gives a=4.8

    Though it cant be decimal.

    But Where am I proceeding wrong with this approach????could anyone help me?

    I guess the problem lies when they might be travelling in same direction ..



  • P and Q are two diametrically opposite points on a circular track. A and B start from P at speed of 20 and 40 rounds per hr respectively. C starts from Q at the same time at 10 rounds per hour. If B and C move in clockwise direction and A in anticlockwise direction ,find the time after which they meet for the third time anywhere on the track?

    Sol:--What i tried is as follows:-
    A and C
    They meet first in L/2(30)L hr
    So 3rd time they meet in L/60L+L/30L+L/30L= 5/60 hr

    A and B will meet for the third time = 3L/60L hour

    B and C :
    they meet first at L/2(40-10)=L/60L hour
    Third time they meet in:
    L/60+ 2L/30=5L/60L

    They meet together third time in LCM(5/60,3/60,5/60)
    =15/60 hour=15 min

    But options are: 5 ,7, 11and 13....


  • Being MBAtious!


    @sumit-agarwal
    Assumption that they travel 100 m together between every meetings is not correct. Like if A is 10 m from one of the end while B crossed him, then the next meeting would happen once B comes back from that end. They won't cover 100 m together in this scenario right?



  • @zabeer
    Yes sir, that's right.
    But there is a similar question in gyan room-arithmetic shortcuts ,where the scenario is same.. the meetings are given...and one has to find speed....
    The same logic of combined distance is applied there.
    So,what else could one apply here, instead of tabulation as used in the original answer....


  • Being MBAtious!


    @sumit-agarwal

    Question in Gyan room is not the same.
    Will share some thoughts on the question in discussion here.

    We know, A would take 50/3 = 16.66 seconds and B would take 10 seconds to complete one lap.Let X and Y be the two sides of the pool, X being the start point. So,

    (0 - 10 seconds) : A : X --> Y & B : X --> Y (ignore this slot as both are traveling in same direction)
    (11 - 20 seconds) : A : (X --> Y)(Y --> X) & B : (Y --> X) (there is one meeting point here as they are traveling in opposite directions and B completes one full lap in this interval)
    (21 - 30 seconds) : A : (Y --> X) & B : (X --> Y) (second meeting here)
    (31 - 40 seconds) : A : (Y --> X)(X --> Y) & B : (Y --> X) (third meeting here)
    (41 - 50 seconds) : A : (X --> Y) & B : (X --> Y) (ignore as both are in same direction)
    (51 - 60 seconds) : A : (Y --> X) & B : (Y --> X) (again, same direction. So ignore. Race ends here and B wins)

    So together 3 meeting points.

    Note : We took interval of 10 seconds as B (fastest) would complete a full lap during this time.

    Sure, it took some time to write all this, but it won't take more than a minute to solve with a rough sketch.



  • @zabeer
    Thank u. Got it :)





  • @zabeer
    0_1517479460706_DOUBT.png

    sir i got RS =6sqrt 3
    dont know how to proceed further.......................................



  • @vinaycat2017
    RS=6Sqrt3
    Angle R =90
    PR =12 (by similarity)
    Apply pythagores in tri PRS
    PS=sqrt(144+108)=6sqrt7

    Hope it helps..



  • 3 individuals john wright, greg chappell and gary kristen are in the race for the appointment of new
    coach of team india. The probabilities of their appointment are 0.5, 0.3 and 0.2 respectively. If john
    wright is appointed then probability of ganguly appointed as a captain will be 0.7 and corresponding
    probability if greg chappell or gary kristen is appointed are 0.6 and 0.5 respectively. find the overall
    probability that ganguly will appointed as a captain


  • Being MBAtious!


    @rahul-jodavula
    In general,
    If you have OR, ADD the probabilities.
    If you have AND, multiply the probabilities.

    Here, Ganguly can be the captain in the following manner
    (John AND Ganguly) OR (Greg AND Ganguly) OR (Gary AND Ganguly)
    (0.5 * 0.7) + (0.3 * 0.6) + (0.2 * 0.5) = 0.63 should be our answer.
    What's the OA ?


 

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