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89 Cards in this Set

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Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour.
Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?

P= Favorable/Total
Total = 6C2 = 6*5/2 = 15

Favorable:
Pick 1 sour candy out of 2 sour candy
AND
Pick 1 sweet candy out of 4 sweet candy

2C1*4C1 = 2*4=8

P=8/15

When you use combination method, it is picking all possible cases and the order doesn't matter. Whereas, upon choosing probability method to solve, order matters.

Thus,
Total Probability:
Probability of choosing sour candy first AND Probability of choosing sweet candy
OR
Probability of choosing sweet candy first AND Probability of choosing sour candy

2/6* 4/5 + 4/6 * 2/5
= 8/30 + 8/30
= 16/30
= 8/15

Why: 2/6* 4/5 + 4/6 * 2/5
P=2/6: total sour candies= 2, total candies=6
1 Sour candy Picked.
AND means "*"
Now, total candies=5(because one picked). Sweet candies=4(First picked was sour so 4 sweet candies still left)
P=4/5
2/6 * 4/5

OR Means "+"
4/6: total sweet candies= 4, total candies=6
1 Sweet candy Picked.
AND means "*"
Now, total candies=5(because one picked). Sour candies=2(First picked was sweet so 2 sour candies still left)
P=2/5
4/6 * 2/5

Total Cost ($)
Unit Price ($ per unit) x Quantity purchased (units)
Total Sales or Revenue
Unit Price x Quantity sold
Profit
Revenue - Cost (all in $)
Unit Profit
Sale Price - Unit Cost
Total Earnings/Salary ($)
Wage Rate ($ per hour) x Hours worked
Miles (distance)
Miles per hour x Hours
Miles (gas)
Miles per gallon x Gallons
Complicated age problems can be effectively solved with an ..., which puts people in rows and times in columns. Such a chart helps you keep track of one person's age at different times (look at a row), as well as several ages at one time (look at a column).
Age Chart
8 years ago, George was half as old as Sarah. Sarah is now 20 years older than George. How old will George be 10 years from now?
2G-16=G+12
G=28
In general, if you have a whole number constraint on a Data Sufficiency problem, you should suspect that you can answer the question with .... This pattern is
not a hard-and-fast rule, but it can guide you in a pinch.
Very little information
If Kelly received 1/3 more votes than Mike in a student election, which of the following could have been the total number of votes cast for the two candidates?

(A) 12 (B) 13 (C) 14 (0) 15 (E) 16
Let M be the number of votes cast for Mike. Then Kelly received M + (1/3)M, or (4/3)M votes. The total number of votes cast was therefore "votes for Mike" plus "votes for Kelly" or M + (4/3)M. This quantity equals (7/3)M, or 7M/3.
Because M is a number of votes, it cannot be a fraction-specifically, not a fraction with a 7 in the denominator. Therefore, the 7 in the expression 7M/13 cannot be canceled out. As a result, the total number of votes cast must be a multiple of 7. Among the answer choices, the only multiple of 7 is 14, so the correct answer is (C).
A store sells erasers for $0.23 each and pencils for $0.11 each. If Jessica buys both erasers and pencils from the store for a total of $1.70, what total number of erasers and pencils did she buy?
Let the number of erasers bought be 'e' and the number of pencils bought be 'p'
Then

0.23e + 0.11p =1.7

(Remember that bought e and p have to be integers)
Multiplying both sides of the equation by 100 so that it looks better.

23e + 11p = 170

Lets plug in a few values for 'e' starting from 1.
when we put e = 1 or 2 or 3 or 4 in the equation above, we don't get integral values for p. But when we put e = 5, we get an integral value for p which is 5.

e + p = 10

And, thats the answer.
RATE RELATIONS
Twice / half/ n times as fast as

Train A is traveling at twice the speed of Train B.
(Do not reverse these expressions!)
(Do not reverse these expressions!)
RATE RELATIONS
Relative rates

Car A and Car B are driving directly toward each other.
For example, if Car A is going 30 miles per hour and Car B is going 40 miles per hour, then the distance between them is shrinking at a rate of 70 miles per hour. If the cars are driving away from each other, then the distance grows at a rate of (...
For example, if Car A is going 30 miles per hour and Car B is going 40 miles per hour, then the distance between them is shrinking at a rate of 70 miles per hour. If the cars are driving away from each other, then the distance grows at a rate of (a + b) miles per hour.
Either way, the rates add up.
RATE RELATIONS
Relative rates

Car A is chasing Car B and catching up.
For example, if Car A is going 55 miles per hour, but Car B is going only 40 miles per hour, then Car A is catching up at 15 miles per hour-that is, the gap shrinks at that rate.
For example, if Car A is going 55 miles per hour, but Car B is going only 40 miles per hour, then Car A is catching up at 15 miles per hour-that is, the gap shrinks at that rate.
RATE RELATIONS
Relative rates

Car A is chasing Car B and falling behind.
TIME RELATIONS
Slower / Faster

Joey runs a race 30 seconds faster than Tommy.
These signs are the opposites of the ones for the "slower I faster" rate relations. If Joey runs a race faster than Tommy, then Joey's speed is higher, but his time is lower.
These signs are the opposites of the ones for the "slower I faster" rate relations. If Joey runs a race faster than Tommy, then Joey's speed is higher, but his time is lower.
TIME RELATIONS
Left . .. and met / arrived

"Sue left her office at the same time as Tara left hers. They met some time later.
Sue and Tara traveled for the same amount time
Sue and Tara traveled for the same amount time
TIME RELATIONS
Left . .. and met / arrived
TIME RELATIONS
Left . .. and met / arrived

Sue and Tara left at the same time, but Sue arrived home 1 hour before Tara did."
Sue traveled for 1 hour less than Tara.
Sue traveled for 1 hour less than Tara.
TIME RELATIONS
Left . .. and met / arrived

Sue left the office 1 hour after Tara, but they met on the road.
Again, Sue traveled for 1 hour less than Tara.
Again, Sue traveled for 1 hour less than Tara.
The Kiss (or Crash)

Car A and Car B start driving toward each other at the same time. Eventually they crash into each other.
The Chase

Car A is chasing Car B. How long does it take for Car A to catch up to Car B?
The Round Trip

Jan drives from home to work in the morning, then takes the same route home in the evening.
Following footsteps:

Jan drives from home to the store along the same route as Bill.
Often, you can make up a convenient value for the distance. Pick a Smart Number-a value that is a multiple of all the given rates or times.
Often, you can make up a convenient value for the distance. Pick a Smart Number-a value that is a multiple of all the given rates or times.
Second guessing

Jan drove home from work. If she had driven home along the same route 10 miles per hour faster ...
No matter what situation exists in a problem, you will often have a choice as you name variables. When in doubt, use variables to stand for either Rate or Time, rather than Distance. This strategy will leave you with easier and faster calculations...
No matter what situation exists in a problem, you will often have a choice as you name variables. When in doubt, use variables to stand for either Rate or Time, rather than Distance. This strategy will leave you with easier and faster calculations (products rather than ratios).
Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy,how far from her original destination has Heather walked when the two meet?
(A) 7 miles (B) 8 miles (C) 9 miles (0) 10 miles (E) 12 miles
Because Stacy started walking earlier than Heather,
you should not simply add the rates in this scenario. You can only add the rates for the period during which the women are both walking
Because Stacy started walking earlier than Heather,
you should not simply add the rates in this scenario. You can only add the rates for the period during which the women are both walking
Stacy and Heather are 20 miles apart and walk towards each other along the same route. Stacy walks at a constant rate that is 1 mile per hour faster than Heather's constant rate of 5 miles/hour. If Heather starts her journey 24 minutes after Stacy,how far from her original destination has Heather walked when the two meet?
(A) 7 miles (B) 8 miles (C) 9 miles (0) 10 miles (E) 12 miles
If Lucy walks to work at a rate of 4 miles per hour, but she walks home by the same route at a rate of 6 miles per hour, what is Lucy's average walking rate for the round trip?
If an object moves the same distance twice, but at different rates, then the average rate will NEVER be the average of the two rates given for the two legs of the journey. In fact, because the object spends more time traveling at the slower rate, ...
RT=D
1"(5) = 24
r = 4.8 miles per hour
Work rate =?
Almost every work problem with multiple people or machines doing work has those people or machines working together. Thus, we can almost always follow this rule: If two or more agents are performing simultaneous work, ... the work rates.
add
The population of a certain type of bacterium triples every 10 minutes. If the population of a colony 20 minutes ago was 100, in approximately how many minutes from now will the bacteria population reach 24,000?
Notice that if there are only ... parts in the whole, you can derive a part-whole ratio from a part-part ratio, and vice versa.
Two
Remember that ratios only express a relationship between two or more items; they do not
provide enough information, on their own, to determine the exact ... for each item.
Quantity
To save time, you should cancel factors out of proportions before cross-multiplying. You can cancel factors either ... within a fraction or across an equals sign. Never cancel factors diagonally across an equals sign. Always cross-multiply.
Vertically, horizontally
When can you use the Unknown Multiplier in ratio problems?
You can use it ONCE per problem. Every other ratio in the problem must be set up with a proportion. You should never have two Unknown Multipliers in the same problem.
When should you use the Unknown Multiplier?
You should use it when neither quantity in the ratio is already equal to a number or a variable expression. For example, in this question no need to use it:

The ratio of girls to boys in the class is 4 to 7. If there are 35 boys in the class, how many girls are there?

We can just set up a simple proportion to solve the problem.
In a box containing action figures of the three Fates from Greek mythology, there are three figures of Clotho for every two figures of Atropos, and five figures of Clotho for every four figures of Lachesis.
(a) What is the least number of action figures that could be in the box?
(b) What is the ratio of Lachesis figures to Atropos figures?
(a) In symbols, this problem tells you that C: A = 3 : 2 and C: L = 5 : 4. You cannot instantly combine these ratios into a single ratio of all three quantities, because the terms for C are different. However, you can fix that  problem by multiply...
(a) In symbols, this problem tells you that C: A = 3 : 2 and C: L = 5 : 4. You cannot instantly combine these ratios into a single ratio of all three quantities, because the terms for C are different. However, you can fix that problem by multiplying each ratio by the right number, making both C's into the least common multiple of the current values.
(b) L : A = 12 : 10, which reduces to 6 : 5.
Sale Price
Unit Cost + Markup
If an object moves the same distance twice, but at different rates, then the average rate will ... be the average of the two rates given for the two legs of the journey.
Never. In fact, because the object spends more time traveling at the slower rate, the average rate will be closer to the slower of the two rates than to the faster.
In order to find the average rate, you must first find the TOTAL combined time for the trips and the TOTAL combined distance for the trips.
In combinatorics, we are often counting ...
The number of possibilities: how many different ways you can arrange things.
Fundamental Counting Principle: If you must make a number of separate decisions, then ... the numbers of ways to make each individual decision to find the number of ways to make all the decisions.
MULTIPLY. To grasp this principle intuitively, imagine that you are making a simple sandwich. You will choose ONE type of bread out of 2 types (Rye or Whole wheat) and ONE type of filling out of 3 types (Chicken salad, Peanut butter, or Tuna fish). How many different kinds of sandwich can you make?

2 breads x 3 fillings = 6 possible sandwiches.
An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?
The number of ways of putting n distinct objects in order, if there are no restrictions, is ...
n! (n factorial).
In staging a house, a real-estate agent must place six different books on a bookshelf. In how many different orders can she arrange the books?
Using the Fundamental Counting Principle, we see that we have 6 choices for the book that goes first, 5 choices for the book that goes next, and so forth. Ultimately, we have this total:
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720 different orders.
An ... is a rearrangement of the letters in a word or phrase.
Anagram
How many different anagrams (meaningful or nonsense) are possible for the word ATLANTA?
The number of anagrams of a word is ...
The factorial of the total number of letters, divided by the factorial(s) corresponding to each set of repeated letters.
If seven people board an airport shuttle with only three available seats, how many different seating arrangements are possible?
You can also solve this problem using the Slot Method previously introduced for the lock-code problem. You do not need slots for the people left standing; you only need three slots for the actual seats. Fill in the slots, noting that 1 person is "...
You can also solve this problem using the Slot Method previously introduced for the lock-code problem. You do not need slots for the people left standing; you only need three slots for the actual seats. Fill in the slots, noting that 1 person is "used up" each time. The total number of seating arrangements is therefore 7*6*5 = 210.
If three of seven standby passengers are selected for a flight, how many different combinations of standby passengers can be selected?
At first, this problem may seem identical to the previous one, because it also involves selecting 3 elements out of a set of 7. However, there is a crucial difference. This time, the three "chosen ones" are also indistinguishable, whereas in the e...
At first, this problem may seem identical to the previous one, because it also involves selecting 3 elements out of a set of 7. However, there is a crucial difference. This time, the three "chosen ones" are also indistinguishable, whereas in the earlier problem, the three seats on the shuttle were considered different. As a result, you designate all three flying passengers as F's. The four non-flying passengers are still designated as N's. The problem is then equivalent to finding anagrams of the "word" FFFNNNN.
To calculate the number of possibilities, we follow the same rule-factorial of the total, divided by the factorials of the repeated letters on the bottom. But notice that this grid is different from the previous one, in which we had 123NNNN in the bottom row. Here, we divide by two factorials, 3! for the F's and 4! for the N's, yielding a much smaller number:
The I Eta Pi fraternity must choose a delegation of three senior members and two junior members for an annual interfraternity conference. If I Eta Pi has 12 senior members and 11 junior members, how many different delegations are possible?
This problem involves two genuinely different arrangements: three seniors chosen from a pool of 12 seniors, and two juniors chosen from a separate pool of 11 juniors. These arrangements should be calculated separately.
Because the three spots in the delegation are not distinguishable, choosing the seniors is equivalent to choosing an anagram of three Y's and nine N's, which can be accomplished in 12! / (9!*3!) = 220 different ways. Similarly, choosing the juniors is equivalent to choosing an anagram of two Y's and nine N's, which can be done in 11! / (9!*2!) = 55 different ways.

Since the choices are successive and independent, multiply the numbers: 220 x 55 = 12,100 different delegations are possible.
If a GMAT problem requires you to choose two or more sets of items from separate pools, count the arrangements ... - perhaps using a different anagram grid each time. Then multiply the numbers of possibilities for each step.
Separately
For problems in which items or people must be next to each other, pretend that the items ... are actually one larger item.
"stuck together" (glue method)
Greg, Marcia, Peter, Jan, Bobby, and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each other, in how many different arrangements can the six people sit?
This is a simple arrangement with one unusual constraint: Marcia and Jan will not sit next to each other. To solve the problem, ignore the constraint for now. Just find the number of ways in which six people can sit in 6 chairs.
6! = 6*5*4*3*2*1 = 720
Because of the constraint on Jan and Marcia, though, not all of those 720 seating arrangements are available. So you should count the arrangements in which Jan is sitting next to Marcia (the undesirable seating arrangements), and subtract them from the total of 720.
We imagine that Jan and Marcia are "stuck together" into one person. There are now effectively 5 "people": JM (stuck together), G, P, B, and C. The arrangements can now be counted. These 5 "people" can be arranged in 5! = 120 different ways.
Each of those 120 different ways, though, represents two different possibilities, because the "stuck together" moviegoers could be in order either as J-M or as M-J. Therefore, the total number of seating arrangements with Jan next to Marcia is 2*120 = 240.
Finally, do not forget that those 240 possibilities are the ones to be excluded from consideration. The number of allowed seating arrangements is therefore 720 - 240, or 480.
Probability
Number of desired or successful outcomes / Total number of possible outcomes
Assume that X and Y are independent events. To determine the probability that event X AND event Y will both occur, ... the two probabilities together
MULTIPLY
Assume that X and Y are mutually exclusive events (meaning that the two events cannot both occur). To determine the probability that event X OR event Y will occur, ... the two probabilities together
ADD
What is the probability that a fair die rolled once will land on either 4 or 5?
1/6+1/6=2/6
Molly is playing a game that requires her to roll a fair die repeatedly until she first rolls a 1, at which point she must stop rolling the die. What is the probability that Molly will roll the die less than four times before stopping?
1/6+5/6*1/6+5/6*5/6*1/6=91/216
If an "OR" problem features events that cannot occur together, then you can find the "OR" probability by adding the probabilities of the individual events.

If an "OR" problem features events that can occur together, then use the following formula to find the "OR" probability:
P(A or B) = P(A) + P(B) - P(A and B)
What is the probability that, on three rolls of a single fair die, AT LEAST ONE of the rolls will be a six?
We could list all the possible outcomes of three rolls of a die 0-1-1, 1-1-2, 1-1-3, etc.), and then determine how many of them have at least one six, but this would be very time-consuming. Instead, it is easier to think of this problem in reverse before solving.

Failure: What is the probability that NONE of the rolls will yield a 6?

The probability that on all 3 rolls the die will not yield a 6 is 5/6*5/6*5/6=125/216.

1-125/216=91/216.
If on a GMAT problem, "success" contains multiple possibilities - especially if the wording contains phrases such as at least and at most - then consider ...
Finding the probability that success does not happen. If you can find this "failure" probability more easily (call it x), then the probability you really want to find will be 1- x
Binomial Distribution
In each individual test, the probability of A happening is p and not happening is 1-p. What is the probability of A happening exactly k times in n repeated tests?
Formula:
C(n,k) * p^k * (1-p) ^ n-k
Example:
When a coin has tossed, it will show a head at 50% probability and a tail at 50% probability. What is the probability of getting two heads among four tosses?
C(4, 2) * (1/2) ^2 * (1/2) ^2 = 6/16 = 3/8
C(4, 2) = 4с2
You can compute the effect of a new data point on the existing average of a set with the following formula, which can be derived from the average law
Change in mean = (New term - Old mean) / New number of terms
A charitable association sold an average of 66 raffle tickets per member. Among the female members, the average was 70 raffle tickets. The male to female ratio of the association is 1 : 2. What was the average number of raffle tickets sold by the male members of the association?
Assumed new number of
1) females = 3;
2) males = 1

(66-70)=(x-70)/3
-12=x-70
x=58
Matt gets a $1,000 commission on a big sale. This commission alone raises his average commission by $150. If Matt's new average commission is $400, how many sales has Matt made?
150=(1000-250)/x
150x=750
x=5
On a particular exam, the boys in a history class averaged 86 points and the girls in the class averaged 80 points. If the overall class average was 82 points, what was the ratio of boys to girls in the class?
Let b stand for the number of boys, and let g stand for the number of girls. If the average score for all the boys is 86, then the sum of all the boys' scores is 86b.
Likewise, the sum of all the girls' scores is 80g. So the sum of all the scores...
Let b stand for the number of boys, and let g stand for the number of girls. If the average score for all the boys is 86, then the sum of all the boys' scores is 86b.
Likewise, the sum of all the girls' scores is 80g. So the sum of all the scores is 86b + 80g.
Since the average of all the scores is 82, and the overall number of scores is b + g, we can set up an equation using the relationship Sum = Average * Number of terms.
Sam earned a $2,000 commission on a big sale, raising his average commission by $100. If Sam's new average commission is $900, how many sales has
he made?
Change in mean = (New term - Old mean) / New number of terms

100 = (2000-800)/x
x=12
Change in mean = (New term - Old mean) / New number of terms

100 = (2000-800)/x
x=12
Weighted average = ?
If the "weights" are fractions or percentages, then the denominator of the weighted-average expression (sum of weights) will already be ...
1
Having just the ... of the weights will allow you to find a weighted average. Simply write the ... as a fraction, and use the numerator and the denominator as weights.
Ratios
A mixture of "lean" ground beef (10% fat) and "super-lean" ground beef (4% fat) contains twice as much lean beef as super-lean beef. What is the percentage of fat in the mixture?
Given the set {x, x, y, y, y, y}, where x < y, is the median greater than the arithmetic mean?
Standard Deviation (SD) indicates ...
How far from the average (mean) the data points typically fall
A small SD indicates that a set is clustered ... around the average (mean) value.
Closely
A large SD indicates that the set is spread out ..., with some points appearing far from the mean.
Widely
You should also know the term "variance," which is just the ... of the standard deviation.
Square
For GMAT problems involving only two categorizations or decisions, the most efficient tool is
Double-Set Matrix
Double-Set Matrix
A Venn Diagram should be used ONLY for problems that involve ... sets. Stick to the double-set matrix for two-set problems.
Three
Venn Diagrams are easy to work with, if you remember one simple rule
Work from the Inside Out
Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from
Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?
The first step is to find out how many groups you can make with the people from each division separately, ignoring the other divisions. There are enough Division A people for 20 groups, but only enough Division B people for 15 groups (= 30 people -7- 2 people per group). As for Division C, there are only enough people for 13 groups, since 40 people -7- 3
people per group = 13 groups, plus one person left over. So the limiting factor is Division C: only 13 complete groups can be formed. These 13 groups will take up 13 Division A people (leaving 20 - 13 = 7 left over) and 26 Division B people (leaving 30 - 26 = 4 left
over). Together with the 1 Division C person left over, 1 + 4 + 7 = 12 people will be left over in total.
How many days after the purchase of Product X does its standard warranty expire? (1997 is not a leap year.)
(1) When Mark purchased Product X in January 1997, the warranty did not expire until March 1997.
(2) When Santos purchased Product X in May 1997, the warranty expired in May 1997.
Rephrase the two statements in terms of extreme possibilities:
(1) Shortest possible warranty period: Jan. 31 to Mar. 1 (29 days later)
Longest possible warranty period: Jan. 1 to Mar. 31 (89 days later)
Note that 1997 was not a leap year.
(2) Shortest possible warranty period: May 1 to May 2, or similar (1 day later)
Longest possible warranty period: May 1 to May 31 (30 days later)
Even taking both statements together, there are still two possibilities-29 days and 30 days -so both statements together are still insufficient. Note that, had the given year been a leap year, the two statements together would have become sufficient! Moral of the story: Read the problem very, very carefully.
Five identical pieces of wire are soldered together to form a longer wire, with the pieces overlapping by 4 cm at each join, If the wire thus made is exactly 1 meter long, how long is each of the identical pieces? (1 meter = 100 cm)
Exponential formula to represent population growth?
The number of combinations of r items, chosen from a pool of n items, is
n! / (n-r)!*r!
The number of permutations of r items, chosen from a pool of n items, is
n! / (n-r)!
If the mean of the set (97, 100, 85, 90, 94, 80, 92, x) is 91, what is the value of x?
Residual = Data point - Mean

You could certainly solve this problem with the traditional formula for averages, but you would waste a lot of time on arithmetic. Instead, just use the given mean of 91 to compute the residuals for all the terms except x: +6, +9, -6, -1, +3, -11, +1. These residuals sum to +1.Therefore, x must leave a residual of -1, since all the residuals sum to zero. As a result, x is one less than the mean, or 90.
You only need to know one thing about residuals: for any set, the residuals sum to ...
Zero
if the population is cut in half in the interval, the formula is
S*0.5^(t/I)

I, or interval, is the amount of time given for the quantity to decay.
S, or starting value, is the size of the population at time zero or "now."
t, or time, is the variable.