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(1) Rate × Time = Distance

Rates and work

Interest Formulas

Compound interest = P * (1+r/n)^nt

P = principal

r = rate (in decimal form)

n = number of times per year

t = number of years

BUT you don't need to know this formula!

Treat these problems like successive percent problems.

e.g. after 2 years of 8% annual compound interest, the new amt is Principal*1.08*1.08

** Be careful of when the principal compounds versus the rate. Everytime the principal compounds, the amount increases. e.g. if you have $100 that earns 8% annual interest and compounds quarterly, then by Q3 it has compounded 6%

Percent Change = Change in value / Original Value

Smart numbers

Sometimes I will have to plug in numbers to solve a problem, and I'll need to choose a number

The best number (aka smart number) is going to be one which is a multiple of the denominators of all fractions in the problem.

Problem: If tank #1 is 4/5 full and tank #2 is 1/2 full, and if all liquid in tank #2 is moved to tank #1, then how full is tank#1?

The smart number needed to solve this problem is 10!!

**You can ONLY pick ONE smart number per problem - NEVER PICK MORE THAN ONE SMART NUMBER!!!! You can't do this because the ratio of the two smart numbers can vary

Comparing fractions using cross-multiplication

Is 3/13 bigger or smaller than 2/9?

Sometimes I can quickly determine this (because I'm the shit...). Other times, I waste time figuring stupid stuff like this out.

So, to quickly solve, use cross-multiplication!!

3/13 ??? 2/9

9x3 ??? 13x2

27 ??? 26

3/13 > 2/9!!!!!!

Notice how important it is to avoid mixing up sides during the calculations!

Mixture charts

When I am adding/subtracting ingredients from two different items, a mixture chart is a great way to figure out in an organized fashion what the new %'s are for each ingredient in the overall mixture.

Think of mixture charts like a simple, straightforward table. See p. 114-115 of the FDP book for examples

+ The columns show the original, change, and new (original + change = new)

+ The rows show the ingredients - e.g. alcohol, water, total solution

+ The individual cells are the volumes for each ingredient

*** Fill out what you know, then solve for unknowns!

Decimals to memorize

This will help me (a) identify solutions faster, (b) easily convert decimals to fractions so I can do math easily, (c) help me avoid careless errors. The decimals I need to know are:

+ Multiples of 1/6, Multiples of 1/7, Multiples of 1/8, Multiples of 1/9, and Multiples of 1/11

+ 1/6 --> 0.16, 0.33, 0.50, 0.66, 0.83

+ 1/7 --> 0.14, 0.29, 0.43, 0.57, 0.71, 0.86

+ 1/8 --> 0.125, 0.250, 0.375, 0.50, 0.625, 0.75, 0.875

+ 1/9 --> 0.11, 0.22, 0.33, 0.44, 0.55, 0.66, 0.77, 0.88

+1/11 --> 0.09, 0.18, 0.27, 0.36, 0.45, 0.54, 0.63, 0.72, 0.81, 0.90

Converting decimals to fractions

This is a step-by-step bulletproof way to convert any decimal to a fraction, which is VERY useful if I can't figure out the fraction

Step 1: Count the #'s to the right of the decimal

Step 2: Get rid of the decimal - you now have the numerator

Step 3: The denominator starts with a 1, and then has as many 0's as you counted numbers in step 1

Example: 0.1234

1) there are 4 #'s

2) 1234 is the numerator

3) 10000 is the denominator

--> I can then reduce the fraction as needed

Terminating decimals

Terminating decimals can all be written as:

Some integer ÷ some power of 10

+ If the denominator, after being fully reduced, only has factors of 2 and/or 5, then the decimal will terminate.

+ If there are any other prime factors, then the decimal will not terminate.

Rates and Work

(2) Rate × Time = Work

Rates and Work - Relative rates

If you are dealing with 2 separate and distinct bodies that are moving towards each other, away from each other, or in the same direction at diff rates, then the calculations could take a lot of time to work through.

***For all 3 of the above scenarios, one strategy to save time is to create a third RT = D equation for the rate at which the distance between the 2 bodies is changing

e.g. if two people move towards each other at 5 mph and 6 mph, then they are decreasing the distance between themselves at a rate of 5+6=11 mph.

e.g. if two cars are going opposite directions at a rate of 30 mph and 45 mph, then they are increasing the distance between them at a rate of 75 mph

Rates and Work - Relative rates

e.g. Two people are both going in the same direction but one is moving at 8 mph and the other is moving at 5 mph. Thus, they are decreasing the distance between them at a rate of 3 mph.

e.g. Two people are 14 miles apart but walk towards each other. One walks 3 mph, the other 4 mph. How long till they meet each other?

--> 3+4 = 7 mph. thus, 7 mph × t = 14 miles. --> t = 2 hours.

Rates and Work

If you are dealing with two workers who are both doing the same work, then you can add the rates at which they're doing the work.

e.g. Lucas finishes something in 1/3 of an hour, and Martha finishes something in 1/2 an hour, so combined they can complete some task in 5/6 of an hour.

But if one worker is negating the effect of the other, you'd subtract the rates.

e.g. Hose A is filling a pool at 3 gallons a minute, Hose B is withdrawing water from the pool at a rate of 1 gallon per minute. Combined, the pool is being filled at a rate of 2 gallons a minute.

When working distance and work problems, ALWAYS write rates so that the time appears on the bottom.

It may sound obvious, but if I take a bad shortcut and forget to do this, I could easily miss the problem (as practice problems have shown).

Averages

Sometimes you have a problem which involves an old average, a new average, and an additional entry in the data set that has causes the old average to shift (to the new average).

How do you solve these problems?

The best way to solve these problems is to make a table (similar to the d=rt or w=rt problems) in which you make all the data laid out in a nice and orderly fashion.

Only then can you see the connections between the numbers.

Though it's obvious, this is a good formula to keep in mind: Average * number of entries = sum.

Weighted averages (1/3)

There's a quick and easy way to solve problems that involve (a) an overall average, and (b) >= 2 groups which each have their own averages

Take this problem: A mixture of "lean" ground beef (10% fat) and "super-lean" ground beef (3% fat) has a total fat content of 8%. What is the ratio of "lean" to "super-lean"?

Step 1: Recognize that the "lean" has a +2 differential from the avg., and that "super-lean" has a -5 differential

Step 2: Make these differentials cancel out by multiplying each differential by a weighted number.

x(+2) + y(-5) = 0

Step 3: PICK NUMBERS in the above equation.

If x=5 and y=2, then the numbers cancel out. Thus, you have 5 parts "lean" for every 2 parts of "super-lean"

Weighted Averages (2/3)

Another way to solve the problem is to work it out algebraically

10x + 3y = 8 (x+y)

Eventually you realize that 2x=5y, and x=5 when y=2

So with these problems you can solve it either through the algebraic approach above, or using the differential approach described in the previous flash card.

Weighted averages (3/3)

ALWAYS ALWAYS solve these weighted average problems through using differentials, and picking #'s!!

Let's say you have a charity that sold an average of 66 tickets to each member. The female members averaged 70 tickets per member. The male:female ratio is 1:2.

What is the average number of tickets sold by the male members?

To solve this problem, you MUST think about differentials (and weighted averages).

(1) Women sold an average of 70 tickets, 4 higher than the total average. Women have a +4 differential.

(2) You also know that the ratio of men to women is 1:2. So, you multiply the women's differential by 2.

(3) The men's differential multiplied by 1 must cancel out the women's differential multiplied by 2.

(4) 1×m + 2×4 = 0 --> m = -8 --> Men sold avg. of 58 tickets

Standard Deviation (SD)

SD's tell you how far from the mean the data points typically fall.

Variance is the square of SD (Variance = SD^2)

+ Small SD means data is clustered near the mean

+ Large SD means data is spread out widely

You do not need to know how to calculate SD! You only need to know (a) how SD changes when you change the set, or (b) how SD's compare between two sets of #'s.

Standard Deviation

Set 1 = {5, 5, 5, 5} --> avg spread = 0, SD = 0

Set 2 = {2, 4, 6, 8} --> avg spread = 2, SD = sqrt(5)

Set 3 = {0, 0, 10, 10} --> avg spread = 5, SD = 5

Question: If each data point in a set is increased by a factor of 7, does the set's SD increase, decrease, or stay the same?

Answer: It depends!

+ If the set consists of different numbers, then the SD will go up since the gap btwn each number will go up.

+ BUT if all the #'s in the set are the same, then the SD is 0 since all numbers will still be the same after being multiplied by 7.

Consecutive integers - Evenly spaced sets 101

"Evenly spaced sets" are numbers that increase/decrease by a fixed amt (aka the increment)

The following properties apply to all evenly spaced sets:

(1) The mean and median are equal to each other

e.g. {4, 8, 12, 16, 20, 24} --> mean=14, median=14

(2) The mean and median of the set are equal to the average of the first and last terms --> see above example

Consecutive integers - Evenly spaced sets - # terms

Strategies for dealing with these sets:

If you have to count the number of items in a set, use this formula:

(Last - First) ÷ Increment + 1

**You add 1 because of the "off-by-1 bug"

You divide by the increment so that you don't overcount if the set has an increment of 3

e.g. {3, 6, 9}

(9-3) ÷ 3 + 1 = 3

--> There are 3 items in the set

Consecutive integers - Summing consecutive integers

Let's say you have to sum the integers from 20 to 100 inclusive. What's a quick way to do this?

Doing this sum takes way too much time on the GMAT. YOU HAVE TO ELIMINATE THE TIME!

Therefore, use the rules for evenly spaced sets in order to create shortcuts!

How do you sum all #'s from 20 to 100, inclusive???

Step 1: Find the average of the set

--> 100 + 20 = 120, and 120 / 2 = 60

Step 2: Count the # of terms

--> (100 - 20) / 1 + 1 = 81 items

Step 3: Multiply the average by the number of terms to find the sum

60 x 81 = 4860

Consecutive integers - General rules about averages of these sets

+ The average of an odd number of consecutive integers is always an integer

{1, 2, 3, 4, 5} --> 3

+ The average of an even number of consecutive integers is never an integer

{1, 2, 3, 4} --> 2.5

Consider this question:

Question: The sum of k consecutive integers is divisible by k. Is k^2 odd?

Answer: The sum divided by k is an integer

Overlapping sets - How to properly create a table

Remember, overlapping sets are all about describing two pieces of independent information for a single group of people.

Creating tables is very useful for these problems (e.g. of 30 students, 15 play golf and 20 play tennis and golf, so how many play both sports?)

To solve these problems, create a table - make sure the rows correspond to one decision (to play golf) and the columns correspond to another decision (to play tennis). The last row and last column show the totals.

Overlapping sets - Table tips and tricks (1/2)

Tip #1: Be careful when a problem says something like "20% of all students love dogs" vs. "20% of those students who have money love dogs". These two statements describe two different numbers! One relates to the "grand total" i.e. total number of students; while the other relates only to a subtotal, i.e. the students who have money.

--> Don't write "20" in my table, write (1/5 of subtotal)

Tip #2: There's a right way and wrong way to make these tables. The right way is intuitive, but still need to be aware of how the right way should look.

+ Right way - The rows correspond to the mutually exclusive options for one decision. The columns correspond to the mutually exclusive options for the other decision. Remember - NO OVERLAP!

+ Wrong way - the rows/columns are not mutually exclusive; this results in overlap, e.g. rows are "right/wrong", and columns are "prob 1, prob 2".

Overlapping sets - Table tips and tricks (2/2)

Tip #3: Be prepared to use variables to relate 2 distinct squares to one another

Tip #4: Be careful when the problem says "25% of students are males who took math" vs. "25% of the males had taken math." These 2 statements are different enough that they lead to 2 different tables!

Tip #4 is all about making/not making an assumption! Thus, Tip #4 (and the one below) has big implications in DS!!

Tip #5: Be mindful that sometimes you will be given 2 groups that are setup so that you can't have a value in one of the squares.

For example, this happens if you have a group that plays instruments, and so the box that covers people who don't play violin or don't play cello is 0!

Tables vs. Venn diagrams

+ Tables are used for problems that have two sets. We've seen this described briefly in the previous flash card.

+ Venn diagrams are used for problems that have three sets. These problems usually involve 3 teams/clubs, and there are people that are either ON or NOT ON these teams/clubs.

When making a Venn diagram -- first draw the venn diagram, then fill in the numbers FROM THE INSIDE OUT! This is the easiest and fastest way to work them!

Remember to remove the people you've already accounted for in each team/club as you work your way from the inside-out (i.e. avoid double counting!)

--> Not removing people I've already accounted for (i.e. double counting) is a major source of errors on the GMAT!

Venn diagram example / tricks (1/2)

Take this problem as an example: There's students taking 3 electives - gym, music, and art. 20 in music, 23 in gym, 24 in art.

How many kids are in 2 electives when 5 take all 3 and the total # kids is 52?

+ Our # kids sums up to 67. We need to start assigning kids to groups, from the inside-->out.

+ If we put 5 in the middle then we must subtract 5 from all 3 groups (to avoid double counting). Thus our new total # kids comes to 57. Still too high, so there must be people taking 2 courses.

See next card for rest of solution...

Venn diagram example / tricks (2/2)

+ To find out how many kids are in 2 courses, you must realize that your only goal is to bring the total # kids down to 52.

+ When accounting for the kids in the middle, you went from 20 + 23 + 24 to (20-5) + (23-5) + (24-5) + 5 = 15 + 18 + 19 + 5. Thus, to find the # of kids in both groups, run a similar calculation.

+ This calculation looks like: (15-n) + (18-n) + (19) + n = 52-5. The (52-5) comes from the fact that you've accounted for the 5 kids in all 3 electives. You only subtract n from two groups because (a) it doesn't matter which 2 electives the kids are in, and (b) the math works out the same anyways.

+ GMAT trick: when n students are added to the 3-way intersection, 2n students come out of the total # kids. But, when n students are added to the 2-way intersection, n students are subtracted from the total.
Backsolving technique

Sometimes you'll have to plug in answer choices to find the right answer (rather than doing math that is tedious). There's a right way and wrong way to do this.

+ The right way involves choose an answer choice to begin with that lets you eliminate others (not always possible, but works great when it is possible).

+ The wrong way involves plugging in the first answer choice you see without thinking ahead.

Grouping problems

Grouping problems involve drawing items out of a large pool and grouping them together somehow. They look like this:

e.g. There's a conference where groups are being formed. You need 1 person from Division A, 2 from Division B, and 3 from C. There are 20 people from A, 30 from B, and 40 from C. What is the smallest # of people who will not be able to be assigned to a group?

To solve these types of problems, 1 way to go is to determine the limiting factor on the number of complete groups. That is, if you need diff types of items for a complete group, just find out how many groups you can make with each item, while ignoring the other types (as if you had an unlimited number of those other items). Then, simply compare results!

----> e.g. in the above problem, you'll see that for A, you can form 20 groups, and with B you can form 15 groups, and with C you can form 13 groups.

Consecutive integers - advanced

(1) The product of k consecutive integers is always divisible by k!

(2) For any set of consecutive integers with an ODD # of items, the sum of all the integers is ALWAYS a multiple of the # of items.

--> This works b/c the sum equals the average times the # of items. For odd-type sets, the average is an integer, so the sum is a multiple of the # of items.

(3) For any set of consecutive integers with an EVEN # of items, the sum of all the items is NEVER a multiple of the # of items.

--> This is b/c the sum equals avg. times the # of items. For even-type sets, the avg. is never an integer. Thus, the sum can't be a multiple of the # of items.

e.g. Is the sum of integers from 54 to 153, inclusive, divisible by 100?! HELL NO

About this deck

Author: Kunal K.

Created: 2014-04-12

Updated: 2014-07-05

Size: 32 flashcards

Views: 33

Created: 2014-04-12

Updated: 2014-07-05

Size: 32 flashcards

Views: 33

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