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C is not the right answer.cubicle_bound_misfit wrote:IMO ans is C.
if,
a+y/b+y <a> ab + by <ab> by<ay> (a-b)y>0
So, question is is (a-b)y>0
stmt 1:
a<b>0, b>0 hence a-b <0 but Y not known INSUFF
stmt 2: y<0, a-b not known INSUFF
together
a-b<0,y<0>0 SUFF
ans C
let me know if my approach is wrong
regards,
Where is the question from? It's an exact copy of q143 from the Official Guide (orange 1th edition, DS section), except that a couple of the inequalities are flipped around. The difference, however, is that there's a 10-second solution to the OG question (if you think of the fractions as ratios), and we can't use that solution here.beeparoo wrote:Searched, but could not find this question in the forum...
If a > 0, b > 0, and (b + y) does not = 0, is
(a + y)/(b + y) < a/b ?
1) a < b
2) y < 0
Oooh, yes! I did not pick numbers that made b + y negative, which misled me to think there was sufficiency. The fact that the OA is E really stumped me until now.Ian Stewart wrote:Where is the question from? It's an exact copy of q143 from the Official Guide (orange 1th edition, DS section), except that a couple of the inequalities are flipped around. The difference, however, is that there's a 10-second solution to the OG question (if you think of the fractions as ratios), and we can't use that solution here.
...
Note that this illustrates the potential danger of choosing numbers here (which, however, you should definitely try if you get stuck with the algebra). If you never pick numbers b and y that make b+y negative, you won't get the right answer here.
with a problem like this, it's probably best to go with strategic number picking, UNLESS you are ABSOLUTELY AMAZING at algebra - meaning that you can manipulate variables flawlessly, and you are possessed of an equally flawless understanding of signs, inequalities, and the like (and the way they transform under different operations).beeparoo wrote:Searched, but could not find this question in the forum...
If a > 0, b > 0, and (b + y) does not = 0, is
(a + y)/(b + y) < a/b ?
1) a < b
2) y < 0
IMO Abeeparoo wrote:Searched, but could not find this question in the forum...
If a > 0, b > 0, and (b + y) does not = 0, is
(a + y)/(b + y) < a/b ?
1) a < b
2) y < 0
nope. incorrect.fruti_yum wrote:From the question stem itself
we are given a > 0 and b>0
therefore it is safe to cross multiply
Let us rephrase the questionIf a > 0, b > 0, and (b + y)!= 0, is
(a + y)/(b + y) < a/b ?
1) a < b
2) y < 0
a > 0, b > 0, and (b + y)!= 0,
if (b+y)<0 Is by > ay ? or
if (b+y)>0 Is by < ay ?
We don't know if (b+y)<0 or if (b+y)>0 and if the value of y is > or < 01) a < b
The value of b+y < 0 (b = 4 y = -5) or > 0 (b = 4 y = -1),2) y < 0
Now that we don't know the value of b+y, let us check the answer for the questions in both the conditions!From 1 and 2
That is *only* true if your numerator and denominator do not become negative after you subtract. If you start, say, with the fraction 5/9, and you subtract 4 from the numerator and denominator, then our numerator and denominator will still be positive, and the value of the fraction will certainly go down - you get 1/5. However, if you subtract 10 from the numerator and denominator, the value of the fraction will go up - you'll get (-5)/(-1) = 5.kumadil2011 wrote: In number property as we know for a proper fractions adding same numbers, the value increases, and subtracting decrease..
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