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If X is the hundredths digit in the decimal 0.1X and if Y is the thous [#permalink]

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16 Jun 2017, 08:04

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If X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal 0.02Y ,where X and Y are nonzero digits,which of the following is closest to the greatest possible value of \(\frac{0.1X}{0.02Y}\) ?

If X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal 0.02Y ,where X and Y are nonzero digits,which of the following is closest to the greatest possible value of \(\frac{0.1X}{0.02Y}\) ?

A. 4 B. 5 C. 6 D. 9 E. 10

To MAXIMIZE the value of 0.1X/0.02Y, we must MAXIMIZE the numerator (0.1X) and MINIMIZE the denominator (0.02Y)

So, the numerator is maximized when X = 9. So, the numerator is 0.19 The denominator is minimized when Y = 1. So, the denominator is 0.021

So, we must determine the value of 0.19/0.021

IMPORTANT: We need not calculate the value of 0.19/0.021 Instead, just recognize that 0.19/0.02 = 9.5, which is halfway between 9 and 10 Since 0.021 is a bit bigger than 0.02, we know that 0.19/0.021 is a bit LESS THAN 9.5 So, 0.19/0.021 must be closest to 9

Re: If X is the hundredths digit in the decimal 0.1X and if Y is the thous [#permalink]

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07 Sep 2017, 22:12

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AbdurRakib wrote:

If X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal 0.02Y ,where X and Y are nonzero digits,which of the following is closest to the greatest possible value of \(\frac{0.1X}{0.02Y}\) ? A. 4 B. 5 C. 6 D. 9 E. 10

The way I happened to solve it is: Naturally for the largest fraction we need the largest possible numerator and the smallest denominator. So we assume x = 9, and y = 1. Therefore, 0.19/0.021 = Answer if we assume that the denominator was 0.020 instead of 0.021 we realise that the number we'll get will be smaller than the actual answer.

Therefore, 0.19/0.02 = 19/2 = (Little less than) 9.5 Therefore, the closest answer is 9 Answer D

If X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal 0.02Y ,where X and Y are nonzero digits,which of the following is closest to the greatest possible value of \(\frac{0.1X}{0.02Y}\) ?

A. 4 B. 5 C. 6 D. 9 E. 10

We are given two decimals:

0.1X and 0.02Y

To make 0.1X the greatest, we can let X = 9 and we have:

0.19

To make 0.02Y the smallest, we can make Y = 1 (since Y = 0 is not allowed) and we have:

0.021

Thus:

0.19/0.021 = 190/21 = 9 1/21

So, the greatest possible value is about 9.

Answer: D
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We're told that X is the hundredths digit in the decimal 0.1X and if Y is the thousandths digit in the decimal 0.02Y and that X and Y are NON-ZERO digits. We're asked which of the following is closest to the GREATEST possible value of 0.1X/0.02Y

To make the value of this positive fraction as large as possible, we need to make the numerator as large as possible AND make the denominator as small as possible... so we should make X = 9 and Y = 1.

.19/.021

To 'clean up' this fraction, let's multiply both the numerator and denominator by 1000....

190/21

At this point, you might find it useful to compare 'multiples' of 21... (21)(9) = 189 (21)(10) = 210

189 is considerably closer to 190 than 210 is, so that fraction is closest to 9.