steve reiny proportion help maths reiny!!!

The illumination of a small object by a lamp varies
directly as the candle-power of the lamp and inversely
as the square of the distance between the lamp and
the object.If a light-bulb of 8 candle power,fixed
150cm above a table,is replaced by a 5 candle power
bulb,how far must the new light be lowered to give the object thesame illumination as before

plz show step

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  1. i = k cp /d^2

    k*8/150^5 = k*5/d^2

    8 d^2 = 5*150^2

    d^2 = (5/8)150^2
    d = 150 sqrt(5/8)
    d = 150(.79)
    d = 118.5cm
    lower it 150-118.5 = 31.5 cm

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  2. If
    y is the illumination
    c is the candlepower
    d is the distance, then

    y = kc/d^2

    so, if y is constant,

    c/d^2 = y/k, which is constant

    So, you want d such that

    8/150^2 = 5/d^2
    d^2 = 5*150^2/8
    d = 118.56 cm

    or, using a proportion, you use
    c2/c1 = d2^2/d1^2
    d2/d1 = √(c2/c1)

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  3. Let illumination be I and C.P be the candle-power and distance be d. I=k(C.P /dxd .Let I=1
    dxd/C.P=k
    Where d=150
    C.P=8
    150x150/8 =8
    K=2812.5
    I=2812.5(C.P/dxd)
    Where C.P=5
    I=2812.5(5/dxd)
    2812.5x5/dxd=1
    d=118.59m(by finding the square root )
    d=150m_118.59m
    d=31.41m.

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  4. Nice work

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