If a and b are integers, and |a| > |b|, is a · |b| < a – b?

[maggiely07]

If a and b are integers, and |a| > |b|, is a · |b| < a - b?

(1) a < 0

(2) ab 0

The answer is E.

The explanation provided is very long, is there a way to solve this within 2 min?

[RonPurewal]

If a and b are integers, and |a| > |b|, is a · |b| < a - b?

(1) a < 0

(2) ab 0

The answer is E.

The explanation provided is very long, is there a way to solve this within 2 min?

for a problem this unconventional and obnoxious, i would just start PLUGGING IN NUMBERS.
the given condition, |a| > |b|, is easy to satisfy, so this problem is pretty optimal for number-picking. basically, "|a| > |b|" just means that 'a' is a BIGGER number than 'b'; we just don't know whether either of them is positive or negative. (also, b could be 0.)

so, here's a fairly complete list of plug-ins: (a first, then b)
1, 0
-1, 0
2, 1
2, -1
-2, 1
-2, -1
if we go through all of these, we can be fairly confident of our answer.

statement (1)
1, 0 doesn't apply; skip it
-1, 0 --> is 0 < -1? NO
-2, 1 --> is -2 < -3? NO
-2, -1 --> is -2 < -1? YES
insufficient

i can't tell what statement (2) is supposed to say; it currently says "ab 0". but, if you keep using these plug-ins, you should find that it's also insufficient (and that the combination of statements is also insufficient).

[rajkapoor]

If a and b are integers, and |a| > |b|, is a · |b| < a - b?

(1) a < 0

(2) ab 0

The answer is E.

The explanation provided is very long, is there a way to solve this within 2 min?

for a problem this unconventional and obnoxious, i would just start PLUGGING IN NUMBERS.
the given condition, |a| > |b|, is easy to satisfy, so this problem is pretty optimal for number-picking. basically, "|a| > |b|" just means that 'a' is a BIGGER number than 'b'; we just don't know whether either of them is positive or negative. (also, b could be 0.)

so, here's a fairly complete list of plug-ins: (a first, then b)
1, 0
-1, 0
2, 1
2, -1
-2, 1
-2, -1
if we go through all of these, we can be fairly confident of our answer.

statement (1)
1, 0 doesn't apply; skip it
-1, 0 --> is 0 < -1? NO
-2, 1 --> is -2 < -3? NO
-2, -1 --> is -2 < -1? YES
insufficient

i can't tell what statement (2) is supposed to say; it currently says "ab 0". but, if you keep using these plug-ins, you should find that it's also insufficient (and that the combination of statements is also insufficient).

thanks Ron.I should have jumped on plugging numbers than trying to solve it by algebra.
btw ,the second statement is ab >-= 0

[RonPurewal]

thanks Ron.I should have jumped on plugging numbers than trying to solve it by algebra.
btw ,the second statement is ab >-= 0

hi -

(by the way, you can write ">". just type a regular ">" sign, but use an underline.)

if you have ab > 0, then, using the same plug-ins,
1, 0 --> is 0 < 1? YES
-1, 0 --> is 0 < -1? NO
insufficient

if you have the two statements together:
1, 0 --> doesn't satisfy statement 1
-1, 0 --> is 0 < -1? NO
2, 1 --> doesn't satisfy statement 1
2, -1 --> doesn't satisfy either statement
-2, 1 --> doesn't satisfy statement 2
-2, -1 --> is -2 < -1? YES
still insufficient

ans (e)

--

what does the answer key do?

[arturocb86]

Ron,

What is a good strategy for picking numbers? I miss one of the possible combination of values!

I ask you because in this case a cannot be 0... but imagine a could be 0.

you will have 2^3 = 8 different combinations... should we try them all in a DS prob.? Is this is likely to occur?

Thank you!

I have another question:

Quoting from your process,

-1, 0 --> is 0 < -1? NO Can you say from here it is insufficient?
2, 1 --> doesn't satisfy statement 1
2, -1 --> doesn't satisfy either statement
-2, 1 --> doesn't satisfy statement 2
-2, -1 --> is -2 < -1? YES
still insufficient

[RonPurewal]

Ron,

What is a good strategy for picking numbers? I miss one of the possible combination of values!

it's going to vary from problem to problem.

essentially, your task is to notice which types of problems demand the use of which types of numbers, and get takeaways to use on future problems.

for instance, positives, negatives, and zero are important on lots of problems, but not all; other problems will turn on such things as fractions, perfect squares, odds and evens, or other such things.

the only way you can really get good at this is to go through lots of problems, and make connections between the appearance of the problem and the types of numbers that matter.

I ask you because in this case a cannot be 0... but imagine a could be 0.

you will have 2^3 = 8 different combinations... should we try them all in a DS prob.? Is this is likely to occur?

actually, you would have 3^2 = 9, not 2^3 = 8 combinations:
+ +
+ -
+ 0
- +
- -
- 0
0 +
0 -
0 0

it might be the case that you'd actually have to try all nine of these combinations. most likely you wouldn't, but it's not unreasonable to expect that you could.

the key, as usual, is to start using plug-in methods as soon as possible -- basically, the instant you realize that you don't know the "textbook" way to solve the problem, and not a bit later.

-1, 0 --> is 0 < -1? NO Can you say from here it is insufficient?

whoa, no.
this is a misunderstanding of the fundamental nature of data sufficiency -- make sure you clear this one up right away.

if you have a yes/no question:
* DEFINITE YES is SUFFICIENT
* DEFINITE NO is SUFFICIENT
* "MAYBE" (sometimes yes, sometimes no) is insufficient

it's not good enough just to get a "no" answer to the question prompt; you haven't established insufficiency until you get BOTH a "yes" and a "no".

[jcartano]

Is this a realistic question? It seems like quite a bit of work.

[RonPurewal]

Is this a realistic question? It seems like quite a bit of work.

it's a little bit on the labor-intensive side, but it's definitely not totally unreasonable.
in particular, it's quite useful in terms of encouraging a quick transition to number plugging -- most people spend too much time staring unsuccessfully at algebra before deciding to try backup methods.

[arjunagarwala]

I timed myself number plugging after knowing how to do the answer, lets say that a 750+ student knows exactly what to do immediately. Writing things down took me a surprising amount of time - even though the calculations are really easy.

(1) 1:17 min:seconds
(2) 1:04 min:seconds
(1) + (2) 1:04 min:seconds

Not only did it take 3:25 min:seconds. Only in the last four seconds did I realize that the answer was (E) and not (C) -- after you test the special case where b=0.

Take away: I could have completed 95% of the problem, wasted a ton of time and got it wrong by picking (C). I think for almost everyone this should be a question that you throw away.

[jlucero]

If picking numbers is laborious for you, then absolutely, this is a question you might want to throw away on test day. But I'll echo Ron's point here- this is a good problem to show students when Algebra just won't cut it and to quickly go to picking numbers (or guessing and using your time elsewhere).