OG 11 DS problem 143 (help with an alternate solution)

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This isssue refers to OG 11 DS problem 143

# 143

If m>0 and n>0, is (m+x)/(n+x) > m/n

(1) m<n
(2) x>0



Please suggest whether i can cross multiply the inequality as following

(m+x)/(n+x) > m/n =mn+nx/mn+mx>1

= mn+nx>mn+mx =nx>mx= n>m

so 1) is sufficient and i marked a)
but the answer is C)

Please suggest what is wrong and any efficient method of solving the problem as i find the method in OG very complex
Source: — Data Sufficiency |

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ashish1354 wrote:This isssue refers to OG 11 DS problem 143

# 143

If m>0 and n>0, is (m+x)/(n+x) > m/n

(1) m<n
(2) x>0



Please suggest whether i can cross multiply the inequality as following

(m+x)/(n+x) > m/n =mn+nx/mn+mx>1

= mn+nx>mn+mx =nx>mx= n>m

so 1) is sufficient and i marked a)
but the answer is C)

Please suggest what is wrong and any efficient method of solving the problem as i find the method in OG very complex
No, you cannot cross-multiply here, unless you know that x is positive. If x is negative, then n+x could be negative, and when you multiply both sides of the inequality by n+x, you would need to reverse the inequality.

One can see why both statements together are sufficient without doing any algebra, incidentally. Imagine you are in the middle of taking a test, and you've answered m questions correctly out of a total of n, where m < n. What percent have you answered correctly so far? Well, m/n. If you then answer the next x questions correctly, your percentage must improve- you've gotten x answers right in a row. And what would your new percentage be? (m+x)/(n+x).

So if m < n and x > 0, m/n must be less than (m+x)/(n+x).
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solving further!!

by ashish1354 » Sat Aug 30, 2008 12:38 am
1) is not sufficient since we do not know value of x
2) is not sufficient as it does not tell value of m & n

problem is combining 1) and 2)

Q) should we pick numbers for m n and x plug in and figure out (it takes too long to solve the problem) or is there a faster way ??

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Re: solving further!!

by lunarpower » Sat Aug 30, 2008 10:01 pm
ashish1354 wrote:1) is not sufficient since we do not know value of x
2) is not sufficient as it does not tell value of m & n

problem is combining 1) and 2)

Q) should we pick numbers for m n and x plug in and figure out (it takes too long to solve the problem) or is there a faster way ??
ok, so the first thing you have to know here - and that you should know, for all future inequality problems - is that there's really no such thing as "cross multiply".
that's right - there is no magic operation that allows you to multiply across an equals sign. in fact, what you know as "cross multiply" is really just multiplication by both denominators.
try it for yourself:
take a/b = c/d and multiply both sides by bd (the product of the denominators). you'll get ad = bc, the very result of so-called "cross multiplication".

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therefore, we have the following result:
you cannot "cross multiply" an INEQUALITY unless you know the sign of the product of the denominators.
in the equation above, a/b = c/d, this means you can't cross multiply unless you know the sign of the product bd.
notice that this doesn't mean you have to have the signs of b and d themselves; all you have to know is whether those signs are the same (which would make bd positive) or opposite (which would make bd negative). if the latter is the case, then, when you cross multiply, you'll have to reverse the sign of the inequality, just as with any other multiplication/division by a negative number.

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as long as statement (2) is true - either by itself or in concert with statement (1) - m, n, and x are all positive, and therefore both denominators are positive.
this means that, as long as statement (2) is true, you can "cross multiply" the inequality to give "is mn + nx > mn + mx?", from which mn can be subtracted to give "is nx > mx?"
once you have that, you can divide both sides by x (which is known to be positive) to give the final rephrase: "is n > m?"
statement (2) isn't sufficient to answer this question; statement (1), which does indeed declare that n > m, isn't good enough by itself because you can't cross multiply the inequality if you only have that statement.

if you have both statements together, then, as described, the question can be rephrased as "is n > m?", and statement (1) tells you that n > m. that's sufficient.
Ron has been teaching various standardized tests for 20 years.

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by lunarpower » Sat Aug 30, 2008 10:03 pm
you can also plug in numbers, of course. when you use statement (1) or (2) by itself, just take care to plug in numbers that represent the full range of available numbers (note that this is the object of DS number plugging in general).

this means that, for statement (1), you must plug in numbers featuring both positive and negative values of x, and, for statement (2), you must plug in numbers featuring m < n and m > n (and perhaps m = n).
in other words, when you're trying the individual statements, you should make sure to plug in numbers that are disallowed by the other statement!
otherwise, you may be accidentally testing the statements together when you're trying to test one of them by itself.
Ron has been teaching various standardized tests for 20 years.

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