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

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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, 09:04
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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 $$\frac{0.1X}{0.02Y}$$ ?

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

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

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27 Jun 2017, 15:36
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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

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

Answer:

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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, 10:40
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$$\frac{0.1X}{0.02Y}$$ = $$\frac{0.1+0.0X}{0.02+0.00Y}$$

For the largest value to this expression, x must be maximum and y minimum
If x=9,y=1

Expression become $$\frac{0.1+0.09}{0.02+0.001}$$ = $$\frac{0.19}{0.021}$$ = approximately 9 (Option D)
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Re: 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, 10:50
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For the largest value of a fraction, numerator must be max and denominator must be min.

Numerator = 0.1X.. Its largest value = 0.19
Denominator = 0.02Y.. Its smallest value = 0.021

Hence required fraction = 0.19/0.021 = 190/21 .. This is approx 9. Hence D answer
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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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Updated on: 16 Jun 2017, 11:28
Maximize X and minimize Y

--> $$\frac{0.19}{.021} \approx 9.4xx$$, First decimal place is <5, thus rounding down.

Answer is D

Originally posted by roadrunner on 16 Jun 2017, 10:57.
Last edited by Bunuel on 16 Jun 2017, 11:28, edited 1 time in total.
Formatted.
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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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18 Jun 2017, 23:32
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Excellent question.

Here is what I did on this one =>

For any fraction P/Q to be maximum => P should be maximum and Q should be minimum.
Hence => Max value => at x= 9 and y = 1(as both are non zero)

=> 0.19/0.021 => 190/21 => 9(approx)

Hence D.

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

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27 Jun 2017, 15:29
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pushpitkc wrote:
$$\frac{0.1X}{0.02Y}$$ = $$\frac{0.1+0.0X}{0.02+0.00Y}$$

For the largest value to this expression, x must be maximum and y minimum
If x=8,y=0

Expression become $$\frac{0.1+0.08}{0.02+0.000}$$ = $$\frac{0.18}{0.02}$$ = 9(Option D)

Be careful; the numerator is maximized with x = 9, and we're told that y cannot equal 0

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

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04 Jul 2017, 14:23
pushpitkc wrote:
$$\frac{0.1X}{0.02Y}$$ = $$\frac{0.1+0.0X}{0.02+0.00Y}$$

For the largest value to this expression, x must be maximum and y minimum
If x=8,y=0

Expression become $$\frac{0.1+0.08}{0.02+0.000}$$ = $$\frac{0.18}{0.02}$$ = 9(Option D)

Pushpit,
y needs to be non-zero as per question stem. so X=9 and y =1 gives the correct value.
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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, 23:12
1
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
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Re: If X is the hundredths digit in the decimal 0.1X and if Y is the thous  [#permalink]

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15 Nov 2017, 17:20
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

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

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18 Nov 2017, 17:00
1
Hi AbdurRakib,

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.

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

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18 Mar 2018, 20:05
Quote:
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 $$\frac{0.1X}{0.02Y}$$ ?

pushpitkc niks18 Hatakekakashi
amanvermagmat Bunuel chetan2u

Is below approach correct?

For greatest value of fraction, numerator needs to be greatest and denominator needs be smallest.

Using round off procedure

$$\frac{0.1X}{0.02Y}$$ = $$\frac{0.19}{0.021}$$

approximates to $$\frac{19}{2.1}$$

approximates to 9 (I approximated 2.1 to 2 and 19 to 18 and 18/2 = 9)

Here answer options D and E are too close for me to correctly approximate.
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Re: If X is the hundredths digit in the decimal 0.1X and if Y is the thous  [#permalink]

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18 Mar 2018, 21:08
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adkikani wrote:
Quote:
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 $$\frac{0.1X}{0.02Y}$$ ?

pushpitkc niks18 Hatakekakashi
amanvermagmat Bunuel chetan2u

Is below approach correct?

For greatest value of fraction, numerator needs to be greatest and denominator needs be smallest.

Using round off procedure

$$\frac{0.1X}{0.02Y}$$ = $$\frac{0.19}{0.021}$$

approximates to $$\frac{19}{2.1}$$

approximates to 9 (I approximated 2.1 to 2 and 19 to 18 and 18/2 = 9)

Here answer options D and E are too close for me to correctly approximate.

I would not approximate the way you did because the options are not widespread. You should do the way proposed in the solutions above:

$$\frac{0.19}{0.021}=\frac{190}{21}=9\frac{1}{21}\approx 9$$.

Or $$(\frac{190}{20} = 9.5) > \frac{190}{21}$$, so 190/21 is closer to 9 than it is to 10.
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Re: If X is the hundredths digit in the decimal 0.1X and if Y is the thous  [#permalink]

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18 Mar 2018, 23:02
adkikani wrote:
Quote:
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 $$\frac{0.1X}{0.02Y}$$ ?

pushpitkc niks18 Hatakekakashi
amanvermagmat Bunuel chetan2u

Is below approach correct?

For greatest value of fraction, numerator needs to be greatest and denominator needs be smallest.

Using round off procedure

$$\frac{0.1X}{0.02Y}$$ = $$\frac{0.19}{0.021}$$

approximates to $$\frac{19}{2.1}$$

approximates to 9 (I approximated 2.1 to 2 and 19 to 18 and 18/2 = 9)

Here answer options D and E are too close for me to correctly approximate.

Agree with what bunuel has to say.. Don't try to approximate

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

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04 May 2019, 01:13
As many have already stated: we maximise the numerator, minimise the denominator

0.19 / 0.021

BUT I prefer to translate this to a fraction as fractions are easier to work with:

19/100 / 21/1000
= 19*1000/ 21*100
= 190/21
= 9.1
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Re: If X is the hundredths digit in the decimal 0.1X and if Y is the thous  [#permalink]

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06 May 2019, 01:59
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

.1x/.02y

After maximization
.19/.021

Now, we see that.19/.02 =9.5

So .19/.021 must be a bit less than 9.5.

Thus, maximum possible value is 9.

Answer: D

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Re: If X is the hundredths digit in the decimal 0.1X and if Y is the thous   [#permalink] 06 May 2019, 01:59
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