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Help with a math/percentage question please!?

I'm terrible at math. Could someone please show me how to work this question correctly? Thank you!

In 2000, 490 women became pregnant in City X. This represents an 11% increase from 1999. How many women became pregnant in 1999?

3 Answers

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  • David
    Lv 7
    10 years ago
    Favorite Answer

    x + .11x = 490

    x(1 + .11) = 490

    1.11x = 490

    x = 490/1.11

    x = 441.44144144144144144144144144144

    about 441 women were pregnant in 1999

  • Anonymous
    10 years ago

    Take the total 490 x 10% = 49 - 441 x11% = 48.51 + 441 = 486.51 round up to next number equals 490 there for 441 people were pregnant in 1999

  • 10 years ago

    It's easy to do 10% & 1%...

    1/10 is 49

    & 1/100 is 4.9

    ...they add up to 53.9 [11%]

    ...but this doesn't help because it's 11% of the unknown number that has been added on to make 490.

    So the sum is for 100% (called 'n') + 11% expressed as 'x 1.11',

    n x 1.11 = 490

    ...to find n, we turn the sum around;

    490 / 1.11 = n

    and this gives the value 441.4414414414414

    which is 441 to the nearest 'whole person'

    ...to check this, we can go back to the top of the answer and use the simple 10 + 1 %s

    10 % of 441 is 44.1

    1% is 4.41

    they add up to 48.51 (11%)

    so, 441 + 48.51 = 489.51

    ...which to the nearest whole person, is 490 :)

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