If x is an integer, can the number (5/28)(3.02)(90%)(x) be

Question

If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits?

(1) x is greater than 100
(2) x is divisible by 21

Source: Jamboree

Bunuel · Accepted Answer

THEORY: Reduced fraction \frac{a}{b} (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only b (denominator) is of the form 2^n5^m , where m and n are non-negative integers. For example: \frac{7}{250} is a terminating decimal 0.028 , as 250 (denominator) equals to 2*5^3 . Fraction \frac{3}{30} is also a terminating decimal, as \frac{3}{30}=\frac{1}{10} and denominator 10=2*5 . Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not. For example \frac{x}{2^n5^m} , (where x, n and m are integers) will always be terminating decimal. (We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \frac{6}{15} has 3 as prime in denominator and we need to know if it can be reduced.) Questions testing this concept: does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html 700-question-94641.html is-r-s2-is-a-terminating-decimal-91360.html pl-explain-89566.html which-of-the-following-fractions-88937.html if-r-and-s-are-positive-integers-141000.html BACK TO THE ORIGINAL QUESTION: If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits? First of all \frac{5}{28}*3.02*\frac{9}{10}*x=\frac{5*302*9*x}{28*100*10}=\frac{5*302*9*x}{7*(4*100*10)}=\frac{5*302*9*x}{7*(2^2*2^2*5^2*2*5)} . Now, according to the theory above, in order this number to be termination decimal, 7 must be reduced by a factor of x (no other number in the numerator has 7 as a factor and all other numbers in the denominator have only 2's and 5's), so it'll be a terminating decimal if x is a multiple of 7. (1) x is greater than 100. Not sufficient. (2) x is divisible by 21. Sufficient. Answer: B. Hope it's clear.

BangOn · Answer

If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero decimal digits?
(1) x is greater than 100
(2) x is divisible by 21

Ok. WHat do we have at the denominator in fraction. 100 and 28.
To be non recurring or finite decimal, Denominator should be made of multiple 2 or 5 or both.
100 is not a problem. 28 has 7 which is an issue.

B says X is divisible by 21. 7 goes in denominator. Hence Sufficient.

sambam · Answer

If x is an integer, can the number (5/28)(3.02)(90%)(x) be represented by a finite number of non-zero
decimal digits?

(1) x is greater than 100 
(2) x is divisible by 21

Bunuel · Answer

Merging similar topics. Please refer to the solutions above. Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to the rule #3: the name of a topic (subject field) MUST be the first 40 characters (~the first two sentences) of the question.

WarriorGmat · Answer

"Brilliant Bunuel"

shilpabhagwat · Answer

What happens if x has many 000s in the end so that there is no decimal left e.g. if x= 21000000 - there will no finite number of non-zero decimals. I think (b) is also not sufficient. Please help!!

Bunuel · Answer

It's VERY easy to test that. Plug x = 21,000,000 and see what you get.

dqi2016 · Answer

So when the question says CAN the number be a terminating decimal, it's actually asking is it ALWAYS a terminating decimal?

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Harley1980 · Answer

Hello dqi.
It can't be  always  or  not always  it's or can be represent or cannot be represent.
If X doesn't contain 7 than this number can't be represent as terminate decimal. (always)
If X contain 7 than this number can be represent as terminate decimal. (always)

dqi2016 · Answer

But according to statement 1, if x>100, then if x is 280, that number would be a terminating decimal. But if x=101, then it's non terminating. So technically, abusing by statement 1, x "can" be a number to make it terminating, but isn't necessarily?

Harley1980 · Answer

I think now I correctly understand your first question about "always" 
Yes we should find statement (or both statements) that give us condition when X always make this number terminating. If statement give us variants - this is wrong statement.

In this task in second statement X always make this number terminating. And in first statement sometimes yes (x =280) - this number will be terminating and sometimes no (x = 101). 
This is general rule for any DS task. We should find condition (or combination) which always will give us needed result without any execeptions.

Divyadisha · Answer

A number can be represented as infinite decimal if the denominator is in the form 2^n 5^m

Looking at the expression, (5/28)(3.02)(90%)(x), we are fine with all numbers except 28 as it is a multiple of 7. If any statement can tell us that either the expression gets rid of this 7 or it doesn't, we can conclude if the expression is terminating or not. (Basically look for statement that tells us if x is or not a multiple of 7)

(1) x is greater than 100. x may or may not be multiple of 7. Not sufficient
(2) x is divisible by 21. That means x is a factor of 7; and hence it will leave 4 as the decimal. we will get a terminating decimal. Sufficient

B is the answer

Salsanousi · Answer

\frac{5}{28} * \frac{302}{100} * \frac{90}{100} * x

for this to be a terminating decimal the denominator must be in the form of  \frac{x}{2^a * 5^b}

Re-writing the question stem numbers to be  \frac{5}{7*2^2} * \frac{302}{2^2 * 5^2} * \frac{90}{2^2 * 2^5} * x

from statement 1) if we x > 100 it might include a multiple of 7 and it might not. so insufficient.

2) x is divisible by 21, meaning it could be 21,42,63 etc.. it is a multiple of 21 and 21 = 7 * 3

This will cancel the 7 in the denominator and will lead to the terminating decimal form, sufficient.

Answer choice B

energetics · Answer

\frac{5}{28} *\frac{302}{100} * \frac{90}{100} * x

\frac{x*5*2*151*9*10}{2^{4}* 7 * 10^{4}}

\frac{x * 9 * 151}{2^{4} * 7 * 10^{2}}

Reworded Question:  is x=7 or a multiple of 7?

(1) x is greater than 100 
if x = 700 then YES
if x = 101 then NO
Insufficient.

(2) x is divisible by 21 
If x is divisible by 21, then it's divisible by 7.
Sufficient.

bumpbot · Answer

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If x is an integer, can the number (5/28)(3.02)(90%)(x) be

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