-
DanaJ
- Site Admin
- Posts: 2567
- Joined: Thu Jan 01, 2009 10:05 am
- Thanked: 712 times
- Followed by:550 members
- GMAT Score:770
I've decided to start this thread hoping that it might help some community members with their quant performance. It's not designed as something 100% solid, it's just going to be a collection of my thoughts on various topics or problems that we face when trying to Beat The GMAT. I'm not a GMAT expert, but I am considering teaching/tutoring some day, so any feedback is more than welcomed.
Thought I'd start with something I hate, so here goes...
NUMBER PICKING STRATEGY
I personally don't like this one. There's a good reason for this: most of the time it's a risky strategy. If you're not paying attention, the set of numbers you've selected might turn the whole thing around! I honestly prefer ye old algebraic method of solving things, since it's 100% safe (if employed correctly, of course).
However, there are a few instances when number picking is the recommended strategy.
COUNTEREXAMPLES
This is actually the only instance when I fully support number picking. Counterexamples are extremely useful in two instances:
- DS questions, when you're not really interested in solving a particular problem, but in establishing whether the info provided is or is not sufficient
- PS questions when an algebraic approach would take up too much time OR you're not really sure of how to solve a question by using the classical method
I'll try to illustrate each situation with a few official questions.
I. DS-ing
If n is an integer, is n + 2 a prime number?
1. n is a prime number
2. n + 1 is not a prime number
We analyze the stmt 1 first. Here comes your first number picking tip: try to pick the smallest numbers that fit the description. If you go for bigger numbers, your calculations might suffer. In this case, let's pick 2 and 3. 2 + 2 = 4 and 4 is not a prime number. 3 + 2 = 5 and 5 is indeed a prime number. So 1 is insufficient. If I were to pick 29 and 31, for instance, I'd get similar results, but there's no need to move too far away from zero.
Stmt 2 is up next: in this case, let's pick 3 and 7. 3 + 1 = 4 is not a prime number, but 3 + 2 = 5 is a prime number. 7 + 1 = 8 is not a prime number and 7 + 2 = 9 is not a prime number. So 2 is insufficient as well.
Put the two stmts together to get the same thing: use counterexamples 3 and 7.
If n is a positive integer, is 150/n an integer?
1. n < 7
2. n is a prime number
Search for a counterexample for stmt 1. Start from 0: 150 is divisible by 2, 150 is divisible by 3, but 150 is not divisible by 4. So 1 is out.
For stmt 2, we stray a bit further from zero with say 7: 20*7 = 140, so 150 will not be divisible by this prime number. But pick either 2 or 3 and you get a divisibility.
Since we've eliminated choices A, B and D, time to go for choice C: if we take the two stmts together, then we get three prime numbers that are smaller than 7: 2, 3 and 5. All divide 150 evenly, so here's your answer.
II. PS-ing
Which of the following describes all values of x for which 1 - x^2 ≥ 0?
A. x ≥ 1
B. x ≤ -1
C. 0 ≤ x ≤1
D. x ≤ -1 or x ≥ 1
E. -1 ≤ x ≤ 1
This is an easy one. Most test-takers will not hesitate in solving it algebraically, but here goes: for A, pick x = 2: x^2 = 4 and 1 - 4 = -3, which is definitely smaller than 0. For B, pick -2 with the same results. Since D can be eliminated on the same examples, you're basically left with two choices: C and E. Here's where your real skills kick in: you know that the square of an integer is also the square of the integer's opposite. This means that, if a certain a is in the interval that you're looking for, -a will also be in there. So that means that choice E is indeed your answer.
If x is an integer and y = 3x + 2, which of the following CANNOT be a divisor of y?
A. 4
B. 5
C. 6
D. 7
E. 8
This question is most easily solved when noticing that y is not a multiple of 3. This means that it's also not a multiple of 6, so C is your answer. But if you don't notice this, then start picking numbers. For 4, so for 8 = 3*2 + 2. For 5, go for 18 = 3*6 + 2. For 7, pick 14 = 3*4 + 2. For 8, pick 8 itself = 3*2 +2. As you can see, all numbers are pretty close to zero, so as a general rule don't stray to far from it!
The examples above bring us to the second number picking tip: counterexamples work best with divisibility and intervals. If you see this type of problem and you can't solve it algebraically, then try number picking.
FRACTIONS/PERCENTAGES
Another relatively safe bet for number picking is percentages and NOT TOO TANGLED fractions. Percentages are OK as far as this strategy is concerned because a percentage is basically a fraction with 100 as the denominator, so picking that one is a pretty solid approach.
The organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance?
A. 45%
B. 56%
C. 64%
D. 75%
E. 80%
Say that last year, 100 people attended the fair. This means that this year, we were expecting 125 people, but only 80 showed up. How much is 80 out of 125? Well, it's (80/125)*100 = 1600/25 = 64, with C the correct answer.
In a certain city, 60 percent of the registered voters are Democrats and the rest are Republicans. In a mayoral race, if 75 of the registered voters who are Democrats and 20 percent of the registered voters who are Republicans are expected to vote for Candidate A, what percent of the registered voters are expected to vote for Candidate A?
A. 50%
B. 53%
C. 54%
D. 55%
E. 57%
We'll assume that our city has 100 voters, so 60 are Democrats and 40 are Republicans. 75% of Democrats or 45 of them vote for A, while 8 Republicans also vote for A. This makes 45 + 8 = 53 votes for A, or 53% of registered voters.
Picking numbers when you have slightly different fractions is tricky. Even though this strategy might produce the correct result, it should be used with caution. Even experts might not follow the exact proportions required by the problem!
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
A. 1/5
B. 2/5
C. 3/5
D. 4/5
E. 5/8
So the night workers load three quarters of the number of boxes that the day workers load. This means that you need to pick a number of boxes loaded by day workers in such a way that it's divisible by 4. Go for the smallest: say each day worker loads 4 boxes. This means that night workers load 3 boxes. Then comes your next tricky choice: the number of workers. Again, you're interested in picking a number of day workers that's divisible by 5, so pick 5. You have 5 day workers and 4 night workers. Now, this means that your day workers will load 5*4 = 20 boxes, while your night workers will load 3*4 = 12boxes. This brings us to 32 boxes in total, out of which 20 were loaded by the day crew. 20/32 = 5/8, so the answer's E.
I admit that these are not the most challenging examples one can think of, but unfortunately I guess I need to work on finding the harder ones in the OG books... Will try to do better next time!
Thought I'd start with something I hate, so here goes...
NUMBER PICKING STRATEGY
I personally don't like this one. There's a good reason for this: most of the time it's a risky strategy. If you're not paying attention, the set of numbers you've selected might turn the whole thing around! I honestly prefer ye old algebraic method of solving things, since it's 100% safe (if employed correctly, of course).
However, there are a few instances when number picking is the recommended strategy.
COUNTEREXAMPLES
This is actually the only instance when I fully support number picking. Counterexamples are extremely useful in two instances:
- DS questions, when you're not really interested in solving a particular problem, but in establishing whether the info provided is or is not sufficient
- PS questions when an algebraic approach would take up too much time OR you're not really sure of how to solve a question by using the classical method
I'll try to illustrate each situation with a few official questions.
I. DS-ing
If n is an integer, is n + 2 a prime number?
1. n is a prime number
2. n + 1 is not a prime number
We analyze the stmt 1 first. Here comes your first number picking tip: try to pick the smallest numbers that fit the description. If you go for bigger numbers, your calculations might suffer. In this case, let's pick 2 and 3. 2 + 2 = 4 and 4 is not a prime number. 3 + 2 = 5 and 5 is indeed a prime number. So 1 is insufficient. If I were to pick 29 and 31, for instance, I'd get similar results, but there's no need to move too far away from zero.
Stmt 2 is up next: in this case, let's pick 3 and 7. 3 + 1 = 4 is not a prime number, but 3 + 2 = 5 is a prime number. 7 + 1 = 8 is not a prime number and 7 + 2 = 9 is not a prime number. So 2 is insufficient as well.
Put the two stmts together to get the same thing: use counterexamples 3 and 7.
If n is a positive integer, is 150/n an integer?
1. n < 7
2. n is a prime number
Search for a counterexample for stmt 1. Start from 0: 150 is divisible by 2, 150 is divisible by 3, but 150 is not divisible by 4. So 1 is out.
For stmt 2, we stray a bit further from zero with say 7: 20*7 = 140, so 150 will not be divisible by this prime number. But pick either 2 or 3 and you get a divisibility.
Since we've eliminated choices A, B and D, time to go for choice C: if we take the two stmts together, then we get three prime numbers that are smaller than 7: 2, 3 and 5. All divide 150 evenly, so here's your answer.
II. PS-ing
Which of the following describes all values of x for which 1 - x^2 ≥ 0?
A. x ≥ 1
B. x ≤ -1
C. 0 ≤ x ≤1
D. x ≤ -1 or x ≥ 1
E. -1 ≤ x ≤ 1
This is an easy one. Most test-takers will not hesitate in solving it algebraically, but here goes: for A, pick x = 2: x^2 = 4 and 1 - 4 = -3, which is definitely smaller than 0. For B, pick -2 with the same results. Since D can be eliminated on the same examples, you're basically left with two choices: C and E. Here's where your real skills kick in: you know that the square of an integer is also the square of the integer's opposite. This means that, if a certain a is in the interval that you're looking for, -a will also be in there. So that means that choice E is indeed your answer.
If x is an integer and y = 3x + 2, which of the following CANNOT be a divisor of y?
A. 4
B. 5
C. 6
D. 7
E. 8
This question is most easily solved when noticing that y is not a multiple of 3. This means that it's also not a multiple of 6, so C is your answer. But if you don't notice this, then start picking numbers. For 4, so for 8 = 3*2 + 2. For 5, go for 18 = 3*6 + 2. For 7, pick 14 = 3*4 + 2. For 8, pick 8 itself = 3*2 +2. As you can see, all numbers are pretty close to zero, so as a general rule don't stray to far from it!
The examples above bring us to the second number picking tip: counterexamples work best with divisibility and intervals. If you see this type of problem and you can't solve it algebraically, then try number picking.
FRACTIONS/PERCENTAGES
Another relatively safe bet for number picking is percentages and NOT TOO TANGLED fractions. Percentages are OK as far as this strategy is concerned because a percentage is basically a fraction with 100 as the denominator, so picking that one is a pretty solid approach.
The organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance?
A. 45%
B. 56%
C. 64%
D. 75%
E. 80%
Say that last year, 100 people attended the fair. This means that this year, we were expecting 125 people, but only 80 showed up. How much is 80 out of 125? Well, it's (80/125)*100 = 1600/25 = 64, with C the correct answer.
In a certain city, 60 percent of the registered voters are Democrats and the rest are Republicans. In a mayoral race, if 75 of the registered voters who are Democrats and 20 percent of the registered voters who are Republicans are expected to vote for Candidate A, what percent of the registered voters are expected to vote for Candidate A?
A. 50%
B. 53%
C. 54%
D. 55%
E. 57%
We'll assume that our city has 100 voters, so 60 are Democrats and 40 are Republicans. 75% of Democrats or 45 of them vote for A, while 8 Republicans also vote for A. This makes 45 + 8 = 53 votes for A, or 53% of registered voters.
Picking numbers when you have slightly different fractions is tricky. Even though this strategy might produce the correct result, it should be used with caution. Even experts might not follow the exact proportions required by the problem!
At a loading dock, each worker on the night crew loaded 3/4 as many boxes as each worker on the day crew. If the night crew has 4/5 as many workers as the day crew, what fraction of all the boxes loaded by the two crews did the day crew load?
A. 1/5
B. 2/5
C. 3/5
D. 4/5
E. 5/8
So the night workers load three quarters of the number of boxes that the day workers load. This means that you need to pick a number of boxes loaded by day workers in such a way that it's divisible by 4. Go for the smallest: say each day worker loads 4 boxes. This means that night workers load 3 boxes. Then comes your next tricky choice: the number of workers. Again, you're interested in picking a number of day workers that's divisible by 5, so pick 5. You have 5 day workers and 4 night workers. Now, this means that your day workers will load 5*4 = 20 boxes, while your night workers will load 3*4 = 12boxes. This brings us to 32 boxes in total, out of which 20 were loaded by the day crew. 20/32 = 5/8, so the answer's E.
I admit that these are not the most challenging examples one can think of, but unfortunately I guess I need to work on finding the harder ones in the OG books... Will try to do better next time!
Last edited by DanaJ on Tue Sep 07, 2010 8:58 am, edited 2 times in total.
