Friday, December 17, 2010

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TCS Openseesame aptitude questions with solutions part2 TCS Open seesame questions tcs open seesame

  • Friday, December 17, 2010
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  • 11.A hare and a tortoise have a race along a circle of 100 yards diameter. The tortoise goes in one direction and the hare in the other. The hare starts after the tortoise has covered 1/5 of its distance and that too leisurely. The hare and tortoise meet when the hare has covered only 1/8 of the distance. By what factor should the hare increase its speed so as to tie the race?

    a. 8

    b. 37.80

    c. 5

    d. 40

    ANS :b

    hint:in the first case time taken by hare to reach 1/8 of distance and tortoise to cover 1-1/5-1/8=27/40 of distance is same.in the second case time taken by tortoise to reach 1/8 of distance and hare to cover 7/8 of distance is equal.so equate these 2 eqns to get the increase in speed by hare!
    let x be total distance
    in first case
    (27/40)x/speed of tortoise=(1/8)x/speed of hare
    in 2nd case
    (1/8)x/speed of tortoise=(7/8)x/new speed of hare
    divide eqn 1 by 2 to get increase in speed by hare!




    12.Anoop managed to draw 7 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the radius of the circles to the side of the square.

    a. (2+ 7√2):1

    b. 1:(2+ 6√2)

    c. 1:(4+ 7√3)

    d. 1:(2+ 7c2)

    ANS: b

    hint 12r+2√2r=l√2

    where r is the radius and l is length of side of square.here is a pic of how to solve this question

    13.On the planet Oz, there are 8 days in a week- Sunday to Saturday and another day called Oz day. There are 36 hours in a day and each hour has 90 min while each minute has 60 sec. As on earth, the hour hand covers the dial twice every day.
    Find the approximate angle between the hands of a clock on Oz when the time is 12:40 am.
    a. 251
    b.111
    c.71
    d.89
    ANS: d




    14.Alok and Bhanu play the following min-max game. Given the expression N = 9 +X + Y– Z where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu
    would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal
    strategies, the value of N at the end of the game would be
    a. 27
    b. 18
    c. 20
    d. 0.0

    ANS:c
    hint:just trial and error,the values taken by alok are 7,9 and 9(check yourself!)in our aptitude exam(cusat) the same question was asked with 9 changed as 7.in that case just remember the formula x+y-z=11! and apply!


    15.A person drives with constant speed and after some time he sees a milestone with 2 digits. Then travels for 1 hours and sees the same 2 digits in reverse order. 1 hours later he sees that the milestone has the same 2 digits with a 0 between them. What is the speed of the car?
    a.54.00 mph
    b.45.00 mph
    c.27.00 mph
    d.36.00 mph
    ANS: b
    s=speed
    "Then travels for 1 hour and sees the same 2 digits in reverse order."
    10x + y + 1s = 10y + x
    10x - x + s = 10y - y
    9x + s = 9y

    1 hours later he sees that the milestone has the same 2 digits with a 0 between them."
    10y + x + s = 100x + y
    10y - y + s = 100x - x
    9y + s = 99x

    Rearrange the above two equations for elimination
    9x - 9y + s = 0
    -99x+9y + s = 0
    ----------------adding eliminates y
    -90x + 2s = 0
    2s = 90x
    s = 45x
    x has to equal 1, then s = 45 mph (If x=2 and s=90 mph, when added to
    a two digit milestone, could not be at another two digit milestone.)


    16.

    Fermat’s Last Theorem is a statement in number theory which states that it is impossible to separate any power higher than the second into two like powers, or, more precisely- If an integer n is greater than 2, then the equation a^n b^n = c^n has no solutions in non-zero integers a, b, and c. Now, if the difference of any two numbers is 9 and their product is 17, what is the sum of their squares?
    a.43
    b.45
    c.98
    d.115
    ANS: d



    17 .India with a burgeoning population and a plethora of vehicles (at last count there were more than 20 million of them) has witnessed big traffic jams at all major cities. Children often hone their counting skills by adding the wheels of vehicles in schoolyards or bus depots and guessing the number of vehicles.

    Alok, one such child, finds only bicycles and 4 wheeled wagons in his schoolyard. He counts the totalnumber of wheels to be 46. What could be the possible number of bicycles?
    a.25
    b.5
    c.4 d.8
    ANS: b


    18.

    Given a collection of points P in the plane, a 1-set is a point in P that can be separated from the rest by a line; i.e. the point lies on one side of the line while the others lie on the other side. The number of 1-sets of P is denoted by n1(P). The maximum value of n1(P) over all configurations P of 19 points in the plane is
    a.10
    b.9
    c.3
    d.5

    ans:a the maximum value is 19,as it is not given it is 10


    19.Both A and B Alice and Bob play the following chip-off-the-table game. Given a pile of 58 chips, Alice first picks at least one chip but not all the chips. In subsequent turns, a player picks at least one chip but no more than the number picked on the previous turn by the opponent. The player to pick the last chip wins. Which of the following is true?
    a.In order to win, Alice should pick 14 chips on her first turn.
    b.In order to win, Alice should pick two chips on her first turn.
    c.In order to win, Alice should pick one chip on her first turn.

    ans:b


    20.30 teams enter a hockey tournament. A team is out of the tournament if it loses 2 games. What is the maximum number of games to be played to decide one winner?
    a.60
    b.59
    c.61
    d.30
    e.34
    ANS: b
    29 teams should have to fail two games, so 29*2=58, the winner loses one game.so 59(58+1)

    you must also read
    tcs open seesame questions part3
    tcs open seesame questions part4

    2 Responses to “TCS Openseesame aptitude questions with solutions part2 TCS Open seesame questions tcs open seesame”

    Unknown said...
    December 29, 2010 at 4:01 AM

    gud yar mind blowing


    Unknown said...
    December 30, 2010 at 6:48 PM

    thanks:)


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