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Re: A is the hundredths digit in the decimal 0.1A and B is the thousandths [#permalink]
1
For A And B can be 0 to 9
A=0, B=0
0.10/0.020=5 Which is less than 10
A=9, B=9
0.19/0.029=6.55 which is less than 10
So, B is always greater, Ans is B

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A is the hundredths digit in the decimal 0.1A and B is the thousandths [#permalink]
Since A and B are non-zero digits, the range of values for A and B are {1-9}

Essentially, I want to minimize the denominator and maximize the numerator here.

Removing the decimals will make this clearer.

Take \(\frac{A}{B}\).

If I want to maximize this fraction, I would want to make B as small as possible and A as large as possible.

Notice how when I increase the denominator, the fraction gets smaller:

\(\frac{10}{1} = 10\)

\(\frac{10}{2} = 5\)

\(\frac{10}{3} = 3.3333...\)
.
.
.

And when I decrease the denominator, the fraction gets bigger:

\(\frac{10}{10} = 1\)

\(\frac{10}{9} = 1.1111.....\)

\(\frac{10}{8} = 1.25\)
.
.
.

So when we look at:

\(\frac{0.1A}{0.02B}\)

Let's make A as big as possible (A = 9) and B as small as possible (B = 1)

So we get (would recommend the calculator):

\(\frac{0.19}{0.021} < 10\)

Therefore the answer is B

______

There is a more algebraic approach, however I believe simple intuition is good enough here.
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Re: A is the hundredths digit in the decimal 0.1A and B is the thousandths [#permalink]
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