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  • A jewellery shop owner conducts his business in the following manner.
    Every once in a while he raises his prices by the same percentage. After one such up down cycle of increasing and then decreasing his price by x% the price of a jewel decreases by rs100. In the next cycle he increases and decreases his price by (x/2% and then sells the jewel for rs2346. What is the initial price of the jewel (in rs) .... ?
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  • Sagar07

    Member • Oct 3, 2011

    m getting some fractional errors..... will that do??????
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  • Radhika Deshpande

    Member • Oct 3, 2011

    Try it. It wont give you any fractional error.
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  • Sagar07

    Member • Oct 4, 2011

    ok, i will try to solve it...................
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  • Radhika Deshpande

    Member • Oct 4, 2011

    No other takers ?

    Should I post the answer ?😕
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  • Ramani Aswath

    Member • Oct 5, 2011

    Frooty
    No other takers ? Should I post the answer ?😕
    The problem as stated has an answer. However it is not simple.
    Let P be the initial price. For interest rate r = x/100, (working with a fraction is easier than using % as you avoid dividing by 100. There is no difference to the calculations),

    P - P(1+r)(1-r) =100
    Solving for r,
    r = 10/sqrt(P)........(1)

    For interest rate r/2,

    (P -100)(1 + r/2 )(1-r/2) = 2346 (given)
    Substituting for r from (1), collecting like terms and simplifying we get a quadratic,
    P[SUP]2[/SUP] - 2471P + 2500 = 0

    Solving the quadratic we get,

    P = 2470 (rounded of to the nearest rupee).
    r=.2012 (rounded to the 4th place)
    x = 20.12%
    P(1 - x[SUP]2[/SUP]) = 2370 (rounded off value) This is INR 100 less than the initial price, which is given.

    The solution is self consistent and so correct.

    I think that there must be some error in the problem statement.

    On the other hand, if the price after the second round of x/2 rate is 2376 instead of 2346 (which is given), then the answer is really simple.

    The initial price then becomes INR 2500, x = 20%,the price after first round becomes INR 2400 (which matches the INR 100 specified) and the final price after x/2 (10%) raising and lowering is INR 2376.
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  • Sagar07

    Member • Oct 5, 2011

    Initial price=2470rs.... x=20.12%..... then after first cycle of increasing and decreasing by x%... price=2470*(1.2012)*(0.7988)=2370 rs............

    after second cycle of increasing and decreasing by x/2%..... price=2370*(1.1006)*(0.8994)=2346 rs..............
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  • Radhika Deshpande

    Member • Oct 9, 2011

    Sorry for not posting the answer ..


    Well from the given condition what i have is that the answer goes like this -

    2346/2 = 1173

    1173 - 100 = 1073
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  • Ramani Aswath

    Member • Oct 10, 2011

    The initial price has to be always more than the final price if the cycle involves initially raising by x% and then decreasing by the same x%.
    For each cycle if the initial price is P and the % is x then the final price = P(10000 - x[SUP]2[/SUP])/10000.

    Then the answer given by the other two posters has to be correct.

    1073 does not match the stated conditions, unless we misread the boundary conditions and the statement means something else.

    I shall be very much interested in knowing the detailed working behind the given answer.
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  • Sagar07

    Member • Oct 11, 2011

    even i didn't get the answer given by frooty.... will u plz elaborate?????
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  • Radhika Deshpande

    Member • Oct 11, 2011

    Yes you both are correct .. Even I am thinking there is some mistakes in given conditions. .. I will recheck it and post the answer as soon as I get it .
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  • Ramani Aswath

    Member • Oct 11, 2011

    Frooty
    Yes you both are correct .. Even I am thinking there is some mistakes in given conditions. .. I will recheck it and post the answer as soon as I get it .
    The simplest explanation is tht there is a typo in the condition of the second cycle value of 2346. It is possible that instead of 7, 4 was typed in. 4 minus the horizontal stroke looks a lot like 7.

    If the second value is indeed 2376 we do get a simple, elegant and satisfying solution.
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