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

Question

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

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

BrentGMATPrepNow · Accepted Answer

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:  D

RELATED VIDEO
 https://www.youtube.com/watch?v=718lE2Hvlnk

pushpitkc · Answer

\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)

avigutman · Answer

Video solution from Quant Reasoning:
 https://www.youtube.com/watch?v=NaOGGV_9O_0

amanvermagmat · Answer

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

stonecold · Answer

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.

BrentGMATPrepNow · Answer

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

Cheers,
Brent

aashaybaindurgmat · Answer

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

ScottTargetTestPrep · Answer

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

EMPOWERgmatRichC · Answer

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.

Final Answer:  D

GMAT assassins aren't born, they're made, 
Rich

adkikani · Answer

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.

Bunuel · Answer

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.

MathRevolution · Answer

0.1X / 0.02Y

=>  \frac{{1000 * 1X} }{ {100 * 2Y}}

=>  \frac{10 * 1X }{ 2Y}

For a fraction to be maximum, Numerator has to be maximum and the Denominator should be minimum.

For X = 9 and Y = 1, fraction is maximum

=>  \frac{10 * 19 }{ 21}

=> 9

Answer D

GMATinsight · Answer

Answer: Option D

Video solution by  GMATinsight

https://www.youtube.com/watch?v=itc1WFZUep4

babayega4975 · Answer

why x cannot be 2 and y = 4 that way 120/24= 5 answer.

BrentGMATPrepNow · Answer

If x = 2 and y = 4, then  ONE possible value  of 0.1X/0.02Y is 5.

However, the question asks us to find the  greatest  possible value of 0.1X/0.02Y

Kimberly77 · Answer

Hi Brent  BrentGMATPrepNow , to clarify since question state If X is the hundredths digit in the decimal 0.1X, so we are looking for 1 digit value therefore answer choice D. 10 is out of scope here? Thanks Brent

BrentGMATPrepNow · Answer

Correct. If X and Y are non-zero digits, then 0.1X/0.02Y cannot equal 10.

Kimberly77 · Answer

Great thanks BrentGMATPrepNow for confirmation

ctorres4358 · Answer

How do we know to choose 9 as the maximum value of X? Why can we not choose 90 or 900? Wouldn't that maximize the value?

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

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