H asked in Science & MathematicsMathematics · 9 years ago

Maths Problem help needed? Urgent, please help!?

Hi,

I am having some difficulty with these problems below. No calculators are allowed so I cannot get the answer that way. Any ideas please help & please show working.

Write as a decimal:

1) 4/27

2) 11/111

Write as a fraction in its simplest form:

1) 0.015 (both the 0 & 5 have a recurring symbol above it- not the 1 therefore, written out long hand it is 0.01505050505 etc.)

2) 0.015 (this time both the 1 & 5 have a recurring symbol above it not the 0 therefore, written out long hand it is 0.015151515 etc.)

3) 0.015 (this time only the 5 has a recurring symbol above it but not the 0 & 1 therefore, written out long hand it is 0.015555555 etc.)

Any ideas would be much appreciated & please show working with it.

2 Answers

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  • Anonymous
    9 years ago
    Best Answer

    Write as decimal: -- using known sequences; this method might not be the best for complicated numbers

    1) 4/27

    27 is divisible by 9 and 3

    You know 4/9 equals .444...

    4/9 divided by 3 equals 4/27, so .444... divided by 3 equals 4/27; alternately do 4/3 divided by 9

    2) 11/111

    111 is divisible by 3 and 37

    do the same thing as a above with 11/3 or 11/37

    Write as a fraction in its simplest form:

    1) 0.015(05)... (that's less ambiguous notation)

    0.015 alone is 15/1000 or 3/200

    let variable d equal 0.000(05)...

    Understand that 0.015(05)... = d + (3/200) because 0.015 + 0.000(05)... = 0.015(05)...

    100,000d = 00005.0505... (this is equal to 100,000 times the original number)

    5.0505... - 0.0505... = 5

    0.0505... is equal to 1000d so this equation can be

    rewritten 100,000d - 1000d = 5, which simplifies to 99,000d = 5; divide both sides of the equation by 99,000 and you get 5/99,000 = d

    Since 0.015(05)... = d + (3/200) it can be represented as 5/99,000 + 3/200

    The least common denominator is 495, so multiply 3 and 200 by that to get 1,485/99,000

    The sum is thus 1,490/99,000 or: 149/9,900 (149 is prime)

    Try to apply this to the other problems. I'm done.

  • ?
    Lv 6
    9 years ago

    I can't be bothered writing out the workings for these problems, but the first recurring decimal should be written in longhand as 0.015015015 etc, as the recurring part is that between and including the recurring symbols.

    Good luck.

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