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,
Inequality with Multiple Variables and Fractions
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cubicle_bound_misfit
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beeparoo
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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,
Also, cubi, your solution seems to have been garbled in a couple of areas.
For instance:
"a+y/b+y <a> ab + by <ab> by<ay> (a-b)y>0" and
"a0, b > 0" and
"a-b<0,y<0>0 SUFF"
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durgesh79
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looking at the queastion and statements, its between C and E
my approcah to this was a bit different. i actully tried to prove that it is E with values,
a=10, b=20 and y=-1
a=10, b=20 and y=-100
i agree that this approcah is not recommended in case of C.
my approcah to this was a bit different. i actully tried to prove that it is E with values,
a=10, b=20 and y=-1
a=10, b=20 and y=-100
i agree that this approcah is not recommended in case of C.
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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
I'm sure we can agree that the answer must be C or E. Now, we know:
y < 0
a < b
ay > by [we multiplied by y, which is negative, so reverse the inequality]
ab + ay > ab + by [just adding ab to both sides]
a(b+y) > b(a+y) [factoring on both sides]
a(b+y)/b > (a+y) [we know b>0, so no need to reverse the inequality]
Now, if we divide both sides by b+y, we get a/b on one side, and (a+y)/(b+y) on the other. The problem is, we don't know whether b+y is positive or negative, so we don't know if we need to reverse the inequality. The answer to the original question might be yes, if b+y>0, or no, if b+y < 0. E.
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.
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beeparoo
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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.
Still, your solution around this problem, a backwards approach, if you will, is a trick I would never think of during the actual test. *Nail biting!* Makes for a good, tough question, I guess!
To answer your question, this Q is from a Kaplan diagnostic test and I am surprised that you referenced a similar, albeit different, question from the OG 11th ed. ...
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lunarpower
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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
if you're going to pick numbers here, then, as with other number-picking operations, you should pick numbers that are relevant to the operations at the heart of the problem. in other words, you should NOT pick numbers at random.
here, the problem is centered around an inequality involving two ratios. such inequalities differ from other inequalities principally in that their behavior changes radically according to the signs of the numerators and denominators.
therefore, you want to pick numbers that will mess with the signs of those numerators/denominators in different ways.
to wit:
taking both conditions together,
let a = 2, b = 4
(a) take y = -1. then a+y and b+y are still positive.
this yields 1/3, which IS less than 2/4.
(b) take y = -3. then a+y is negative, but b+y is still positive.
this yields -1/1, which IS less than 2/4 (because it's negative).
(c) take y = -5. then both a+y and b+y are negative.
this yields -3/-1 = 3, which is NOT less than 2/4.
insufficient
answer = e
there it is, folks: even number picking is decidedly nonrandom. it depends almost entirely on the context of the problem; the more you understand about the underlying number properties, the more likely you are to nail the number picking.
Ron has been teaching various standardized tests for 20 years.
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fruti_yum
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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
From the question stem itself
we are given a > 0 and b>0
therefore it is safe to cross multiply
we get (a+y)b <a(b+y)
by<ay
b<a
we are given a<b in statement 1
hence stufficient
anyone agree/disagree?
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lunarpower
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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
the question stem doesn't give any information about y, so we can't tell with any certainty whether (b + y) is positive or negative. that information would be required for "cross multiplication".
remember, by the way, the following important takeaway:
there's actually no such thing as "cross multiplication".
"cross multiplication" is really just MULTIPLICATION BY BOTH DENOMINATORS.
therefore, you don't actually need the signs of both denominators - all you really need is to know whether those signs are the same (in which case you can cross multiply and keep the ">" or "<"), or opposite (in which case you can cross multiply and switch it).
problem is, we don't know either of these. so no, you can't cross multiply.
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vikram_k51
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Rephrase: Is (b-a)y<0?
1) a < b
2) y < 0
Now if we take both A and B the above inequality holds.Hence,the answer is C
Sorry it should be E.a and b can be negative as well.
1) a < b
2) y < 0
Now if we take both A and B the above inequality holds.Hence,the answer is C
Sorry it should be E.a and b can be negative as well.
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Why not C...
In number property as we know for a proper fractions adding same numbers, the value increases, and subtracting decrease..
its clearly given here that its a proper fraction (a LT b) and y is -ve and hence the value should decrease..
RON/IAN/Mitch/Sturat can some one correct me..
In number property as we know for a proper fractions adding same numbers, the value increases, and subtracting decrease..
its clearly given here that its a proper fraction (a LT b) and y is -ve and hence the value should decrease..
RON/IAN/Mitch/Sturat can some one correct me..
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kumadil,
Let me solve the problem in my way and let me see if I can help you !
(a + y)/(b + y) < a/b ?
If b+y > 0,
The question (a + y)/(b + y) < a/b ? can be rephrased as shown below.
(a + y)*b < (b + y)*a (multiply both sides with (b+y)*b. The inequality doesn't change because (b+y)*b is a positive number)
ab + by < ab + ay ?
by < ay ?
If b+y < 0
(a + y)/(b + y) < a/b ?
The question (a + y)/(b + y) < a/b ? can be rephrased as shown below.
(a + y)*b > (b + y)*a (multiply both sides with (b+y)*b. The inequality reverses because (b+y)*b is a negative number)
ab + by > ab + ay ?
by > ay ?
The question can now be rephrased to
Insufficient!
If the value of b+y < 0
The question can be rephrased to if (b+y)<0 Is b < a ?
Answer : Donno !
--------------------------------------------------------
If the value of b+y > 0 (b = 4 y = -1)
The question can be rephrased to if (b+y)>0 Is b > a ?
Answer : Donno ! Insufficient!
If the value of b+y < 0 Is b < a ? - No
If the value of b+y > 0 Is b > a ? - Yes
Two different answers, hence insufficient!
Option E
Phew ! - Read it a couple of times, step by step, if you are one of the victims of glazed eyes !
Correct me if I am wrong!
Let me solve the problem in my way and let me see if I can help you !
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 + y)/(b + y) < a/b ?
If b+y > 0,
The question (a + y)/(b + y) < a/b ? can be rephrased as shown below.
(a + y)*b < (b + y)*a (multiply both sides with (b+y)*b. The inequality doesn't change because (b+y)*b is a positive number)
ab + by < ab + ay ?
by < ay ?
If b+y < 0
(a + y)/(b + y) < a/b ?
The question (a + y)/(b + y) < a/b ? can be rephrased as shown below.
(a + y)*b > (b + y)*a (multiply both sides with (b+y)*b. The inequality reverses because (b+y)*b is a negative number)
ab + by > ab + ay ?
by > ay ?
The question can now be rephrased to
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
Insufficient!
The value of b+y < 0 (b = 4 y = -5) or > 0 (b = 4 y = -1),2) y < 0
If the value of b+y < 0
The question can be rephrased to if (b+y)<0 Is b < a ?
Answer : Donno !
--------------------------------------------------------
If the value of b+y > 0 (b = 4 y = -1)
The question can be rephrased to if (b+y)>0 Is b > a ?
Answer : Donno ! Insufficient!
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
If the value of b+y < 0 Is b < a ? - No
If the value of b+y > 0 Is b > a ? - Yes
Two different answers, hence insufficient!
Option E
Phew ! - Read it a couple of times, step by step, if you are one of the victims of glazed eyes !
Correct me if I am wrong!
Anil Gandham
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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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