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    I've been struggling on this question for a while now:

    A bakery sells two sizes of doughnuts.
    The small size has a mass of 50g.
    The large size has a mass of 168.75g.
    If the larger doughnut has a diameter of 15cm, find the diameter of the small size.

    (Side note: Both the doughnuts are hollow.)
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    4.44cm. If the large doughnut has a diameter of 15cm for 168.75 grams, then one gram of doughnut has a diameter of 15/168.75 = 0.08888... cm per gram. 0.08888 x 4 = 4.44cm (2dp)
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    If it helps for questions like these, put the numbers into a ratio so you can see the proportions clearly.
    15:168.75 divide my 15 to find cm per gram
    1:0.08888...9 then multiply this to the 50g (0.8888...9) as you want to find the small one.
    50: 4.4444
    4.44cm is the answer.
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    (Original post by skillzaismine)
    I've been struggling on this question for a while now:

    A bakery sells two sizes of doughnuts.
    The small size has a mass of 50g.
    The large size has a mass of 168.75g.
    If the larger doughnut has a diameter of 15cm, find the diameter of the small size.

    (Side note: Both the doughnuts are hollow.)
    Is this is a question about mathematically similar shapes.
    If so.....
    Since the two doughnuts are made from the same material the scale factor for mass is the same as the scale factor for volume.
    With mathematically similar shapes the scale factor for volume is the cube of the scale factor for length - conversely the scale factor for length is the cube root of the scale factor for volume.


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    (Original post by gdunne42)
    Is this is a question about mathematically similar shapes.
    If so.....
    Since the two doughnuts are made from the same material the scale factor for mass is the same as the scale factor for volume.
    With mathematically similar shapes the scale factor for volume is the cube of the scale factor for length - conversely the scale factor for length is the cube root of the scale factor for volume.


    Posted from TSR Mobile
    Thank you! I managed to answer this question before you replied, but thanks for the help anyway
 
 
 
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