GMDSS GOC RENEWAL VIDEO TUTORIAL

GMDSS GOC RENEWAL VIDEO TUTORIAL

INCLUDING FAQS


https://youtu.be/npb7BuPmu-0

Viral Content- 6 secret ingredients to a make a viral content.

IS CONTENT MARKETING A NEW STRATEGY?

Today’s buzz word is content. Every business is talking about right content strategy. But, content is not a new thing at all. It was always there, just the medium has changed. The world is shifting to digital age on a very fast pace and the result of this is that today’s customer behavior and expectations are also shifting. Earlier most of the customers used to buy things directly from the shops but now trend of online shopping is increasing including in the developing countries. For example, if you want to buy a condo what will you be doing? For sure, you will go online and search on google “3 bedroom condo in brooklyn MA”. You will go through few links on the first few pages, then look for a realty agency. These are only few initial steps, you will go further, and you will ask your friends on facebook, you will check reviews given about those realty agencies. After getting convinced then you might like to see the condo, first online and then in real. This is what we all are doing now for everything, may it be reviews about a movie, a book, a restaurant, a tutor or anything to everything.

This shift in customer behavior has made the business to think about content in a different perspective. Companies understood that their customers are spending more time on internet than on any other medium of information. Therefore, Content has become the buzz word. Any business cannot be a sustainable business if they only focus on traditional marketing, they have to start thinking about the content and digital marketing for long term sustainability. Today’s customers know exactly what they are looking for and where to look for, so only good content can hold their attention. Therefore, companies should understand their customers’ need and provide the content to help them and not to promote their own brand. If customers like the content they will promote the company, as customers are the face of a company.

Are Likes Always Good?

If you are an entrepreneur or a marketing manager then it is very important for you to identify the true ROI of social media advertisement expenditure. You need to understand that yes, it is important to get more LIKES and VIEWS but organic and not fake. Try to experiment with your marketing strategy to increase the engagement, for example, as the survey shows that people interact more if you ask a question or conduct a contest, so make your post asking some questions or conduct some contest. So be wise and enjoy the fruit of smart digital marketing.

Skills you must have in the Digital Age?

In today's fast developing world, learning and adapting with the new technology is very important for the ongoing success. The biggest challenge company's are facing is the gap between the technical skills and the business skills. Human resource are good at either of these but for the long term sustainability a person should have the technical skills and the understanding that how that technical skills can be used to earn profit. Read more about these skills in the Capgemini article ...
http://ebooks.capgemini-consulting.com/The-Digital-Talent-Gap/#/5/zoomed 

What is Inbound marketing?

Inbound marketing is all about sharing content with the world. For successful inbound marketing, content sharing is the key. One must share the content they think their customer needs, it can be any thing from introduction to complete training manual for the best usage of the product. To ensure that the customers are reading your content you have to be systematic in publishing your content. For example, you want to publish all questions & answers content on every Wednesday and tips to use the product on every Friday. By this the customers will understand the schedule and read the content of their interest, just like as we watch our favorite TV show without missing.

BSc. Nautical Science Degree Equivalent Certificate from Mumbai University on behalf of Master(F.G.) CoC

Procedure for obtaining BSc. Nautical Science Degree Equivalent Certificate from Mumbai University on behalf of Master(F.G.) CoC

Mariners holding a valid Class 1 Deck officer CoC i.e Master(F.G.) CoC are eligible for a Equivalent Certificate (Letter of Equivalency) of BSc. Nautical science Degree from Mumbai University.
The procedure is as follows:
Visit Mumbai University Kalina Campus.
Address: University of Mumbai
Vidyanagari
Kalina
Santacruz (E)
Mumbai 400 098
Tel Nos. : +91-22-26543215/ 26543216


Go to Dr. Baba Saheb Ambedkar Bhavan, a Building within the Campus.After entering the building, turn left on the ground Floor and enter the first room on the right side.
Ask for BSc. Nautical Science Degree Certificate and you will be directed to the person in-charge.
You will be given a form to fill up.

Application Form                                                                                                                            Cash Receipt                                                                                                                  Letter of Equivalence

Fill up the form and submit to the same person along with copies of following Documents:
Passport

CDC
Master (F.G.) CoC (UK CoC also considered)
(10+2)- Marksheet
₹5000/- in cash

Carry the originals of above Certificates / Documents for verification at the time of application.


Deposit fees before 1430 Hrs as cash collection counter closes at 1430 Hrs.

University Office is open Monday to Friday and also on 1st & 3rd Saturdays.

Its closed on all Sundays, Public Holidays and 2nd & 4th Saturdays.



Also you have to visit the university in person to get this letter. To my knowledge, there is no other procedure available as of now.

You will get the Letter of Equivalence on the same day within 2 to 3 hours from the time of Application.

Please note this information was correct as of 19th May 2013.

Relevance and Importance of this Certificate:
Though this Certificate is only a Letter of Equivalence and not a true Graduation Degree but this may fulfill the requirements for exams/ admissions that require graduation in most of the institutions/ universities worldwide.

This Letter has already been successfully used by fellow seafarers for obtaining admission in Post Graduation Courses in various UK universities and also for appearing in UPSC examinations in India that require Graduation.

                         Learn anything-Free-sharmasknowledgeschool.com

Instruction for filling the application form for GMDSS Endorsement in INDIA

INSTRUCTIONS & GUIDELINES
(Instruction for filling the application form for GMDSS Endorsement)

 

 

 

 

 

 

1.         Application should be accompanied by the photocopies of the following certificates:

            a) GMDSS (GOC) Provisional Certificate OR GMDSS (GOC) Certificate.

              b) Certificate of Competency OR College Attendance Certificate

              c) PSSR

              d) Proficiency in Survival Craft and Rescue Boat

              e) Advance Fire Fighting

              f) Medical First Aid

             g) C.D.C. OR Passport

             h) Medical Certificate

             i) 2 recent passport size color photographs

 

2.         For renewal of GMDSS Endorsement

               a) All the above mentioned certificates

             b) Approved seagoing service, performing functions appropriate to the certificate held , for a period of at least one year in total during the preceding 5 years.

           

3.         Applicant should apply and collect certificate in person

           

4.         Medical certificate submitted, should be valid.

           

5.         Candidates have to produce all the above mentioned certificate in original at the time of collection of GMDSS Endorsement.

           

6.         Record of Sea service

             a) Record of Sea service required is to be filled starting from latest.

             b) For foreign flag vessels the sea service has to be duly verified by the local representative (in India) of the Owner/Manager on letterhead.

           

7.         Candidate should ensure that form is correctly filled up and all required documents are attached otherwise candidate will be declared ineligible for endorsement.

           

8.         Any person who makes any false representation or knowingly gives any false information is liable for punishment for cheating under section182 and 420 of the Indian Penal Code.

           

9.         For payment of Fees, Challan form to be collected along with application form and requisite amount to be deposited in bank.

 

CHECKLIST FOR GMDSS ENDORSEMENT

FOLLOWING DOCUMENTS ARE REQUIRED FOR GMDSS ALONG WITH APPLICATION FOR ENDORSEMENT.

XEROX COPY OF EACH DOCUMENT MUST BE ARRANGED IN THE FOLLOWING SEQUENCE.

NAME: _____________________________ RANK: _____________________

(If Name and Place of birth is different in CDC and Passport, Notarized Affidavit to be submitted in original).

SR. NO.	DESCRIPTION OF DOCUMENT	YES	NO
1	Application Form in original duly filled up & signed
2	Seafarer Registration Print Out on E-Governance of MMD, Mumbai (Copy attach with application no.)
3	Certificate of Competency (For Officers) / College Attendance (For Cadets)./ Revalidation Course.
4	For Radio Officer – Copy of RTG / COP License.
5	GOC Certificate / Provisional Certificate
6	OLD Original GMDSS Endorsement Certificate (For Renewal).
7	Original Company Letter for 1 year sea time in last 5 years.
8	Letter issued by the owner of the vessel / Agent in original regarding sea service with RPSL Number & E-mail Address.
9	PSSR
10	Proficiency of Survival Craft & Rescue Boat
11	Advance Fire Fighting.
12	Medical First Aid.
13	Indos Number
14	Copy of CDC (Relevant Sea Service Pages)
15	Copy of Passport (First & last page)
16	Copy of Medical Fitness Certificate Latest within 6 moths
17	Two recent colour photographs passport size(White shirt)
18	Copy of Challan of Rs. 1000/- for each endorsement
19	Above all documents to be kept in one green envelope size 10 x 14 inches with name, rank, contact no./e-mail, sr. no. of register entry & type of endorsement.

Please note : Seafarers Registration Compulsory for all Candidates at the time for

applying GMDSS Endorsement.

Signature : ___________________________

Item 19 not required for noida mmd.

Instruction to fill US Visa -INDIA

Resources for filling DS-160, paying SEVIS Fee, paying DS-160 fee and Visa interview question (India only)

Information to fill – DS-160 -
https://docs.google.com/file/d/0B2AI_Q6bf5gXZ3o0NEtSczgxbTg/edit?usp=sharing

... US Visa interview questions for F1 visa only
https://docs.google.com/file/d/0B2AI_Q6bf5gXZ3hpTnZCNnVZVUk/edit?usp=sharing

Complete process of filling DS-160 to visa interview
https://docs.google.com/file/d/0B2AI_Q6bf5gXRlpBQkE3ZDN3Vnc/edit?usp=sharing

Please note that while paying DS-160 fees online through net-banking you will asked to write the type of account of the beneficiary, in which you can write Domestic and current account, as this worked out for me

 

1)A U.S. visa in his/her passport gives a foreign citizen permission to apply to enter the U.S. A visa by itself doesn’t authorize entry to the U.S. A visa simply indicates that your application has been reviewed by a consular officer at an American embassy or consulate, and that the officer determined you’re eligible to travel to a U.S. port-of-entry for a specific purpose. The port of entry can be an international airport, a seaport or a land border crossing.
　
2)At the port-of-entry, a U.S. immigration officer of the Department of Homeland Security (DHS) decides whether to allow you to enter and how long you can stay for any particular visit, as part of the Admission process. Only the U.S. immigration officer has the authority to permit you to enter the U.S.--I 94

Visa Expiration:
The visa expiration date is shown on the visa along with the visa issuance date. The time between visa issuance and expiration dates is called your visa validity. The visa validity is the length of time you are permitted to travel to a port-of-entry in the U.S.

Depending on your nationality, visas can be issued from a single entry (application) up to multiple/unlimited entries.

A visa issued for a single entry (denoted on the visa under "Entries" with the number 1) is valid, or can be used from the date it is issued until the date it expires to travel to a U.S. port-of-entry one time.

A visa issued for multiple entries (denoted under "entries" with a certain number (2, 3, etc.) or "M" for multiple/unlimited entries) is valid, or can be used from the date it is issued until the date it expires to travel to the U.S. port-of-entry as many times as your visa states, provided that:

Applying for a new visa is not necessary if your visa has not expired and you have not exceeded the number of entries permitted on your visa.

Multiple uses of a visa must be for the same purpose of travel allowable on the type of visa you have.

Please be aware, a visa does not guarantee entry to the U.S. Additionally, the visa expiration date shown on your visa does not reflect how long you are authorized to stay within the U.S. Entry and the length of authorized stay within the U.S. are determined by the Customs and Border Protection (CBP) Officer at the port-of-entry each time you travel--I 94.

RENEWAL OF EXPIRED GMDSS IN INDIA

NEW POST
GMDSS GOC RENEWAL VIDEO TUTORIAL

INCLUDING FAQS

https://youtu.be/npb7BuPmu-0



For GMDSS renewal

Fill one application form avaible on the Ministry of Communication (MoComm) website. Make a draft of Rupees 500/- (other details from the website). Do the online registration and save the code generated over there. Make photocopies of CDC and GoC. Now have the attestation done from gazettered officerSend the photocopies , + the application form ,+ bank draft to the MoCC. it takes 2~3 months for the renewal leaf to arrive, (it is reportedly similar to our GOC Endorsement page).

 http://www.wpc.dot.gov.in/, go to licence, renewal of licence , creat your userid and password, again go to same renwal section log in with userid , selct GMDSS renewal , enter your licence no , you will get the details and can apply for it.

Go to http://www.wpc.dot.gov.in/

Make log in id. and account

Fill details and register.

Log in with newly created id and password.

                                      
                            On left hand side,Renewal lic link is there,click on it.

From drop down menu select gmdss and enter your present GMDSS LIC no.

GMDSS LIC renewal link will come and first you have to enter details of draft and then it will produce receipt.You have to take a print out of that  receipt and send along with all other documents to Ministry of Telecommunication office.

Required documents for appearing in examinations and Issuance / Renewal of COP Licences :

Copies of alll required  documents should be attested by gazetted officers of State governments or Central Governments service

Demand Draft /Banker’s Cheque should be in favour of PAO (HQ ), DOT, New Delhi

GMDSS(GOC) - Renewal

a)      Application Form – COP 23

      b)      Attested copies of CDC or Company Experience Certificate (6 months of last 5 years)

        or renewal by test as per order in force.

c)      Demand Draft of rupees Five Hundreds (500/-)

d)      On line registration acknowledgement slip

e)      Late fees as per rules.

Download COP 23 form:   www.wpc.dot.gov.in/DocFiles/cop-exam-app.doc

Free education for all: www.sharmasknowledgeschool.com

Educate a woman; you educate a generation

You educate a man; you educate a man. You educate a woman; you educate a generation.”
― Brigham Young
                                  www.sharmasknowledgeschool.com

WHAT YOUNG CHILDREN SHOULD BE LEARNING

                 WHAT YOUNG CHILDREN SHOULD BE LEARNING
                    sharmasknowledgeschool.com

The question of what should be learned must be addressed by all teachers at every level. In terms of broad goals, most teachers and parents readily agree that children should learn whatever will ultimately enable them to become healthy, competent, productive, and contributing members of their communities. But when it comes to the specifics of what should be learned next month, next week, or on any particular day, agreement is not so easily achieved.

The answers will depend partly on the ages of the learners. In other words, the question of what should be learned to some extent depends upon when it is to be learned. Although the what question deals with the goals and objectives of education, the when question involves considerations of what we know about the nature of development and how it relates to learning.

What should be learned takes on new importance as states begin to establish standards for student performance, and as new concern is voiced about "social promotion." The interest in standards, competencies, and promotion policies is likely to have a renewed "push-down" effect on prekindergarten education. It is interesting to note that the recent legislation reappropriating funds for Head Start establishes performance standards and stipulates that all Head Start graduates must learn 10 letters of the alphabet (National Head Start Association, 1998, p. 5). What the letters are expected to mean to the children has not been addressed; these new requirements are apparently intended to address the issue of readiness for formal instruction in literacy and numeracy.

This Digest first defines the concept of development and then outlines some ways to approach both the "what" and "when" questions in terms of what we are learning from research about the effects of various curriculum approaches.

THE NATURE OF DEVELOPMENT

The concept of development includes two major dimensions:

1. The normative dimension concerns the typical or normal capabilities as well as limitations of most children of a given age within a given cultural milieu.
   
   
2. The dynamic dimension concerns the sequence and changes that occur in all aspects of the child's functioning with the passage of time and increasing experience, and how these changes interact dynamically (Saarni, Mumme, & Campos, 1998).

Although the normative dimension indicates a probable range of what children typically can and cannot be expected to do and to learn at a given age, the dynamic dimension raises questions about what children should or should not do at a particular time in their development in light of possible long-term dynamic consequences of early experience. In many preschool programs and kindergartens, for example, young children are given instruction in phonics and are expected to complete worksheets and recite number facts in rote fashion. But just because young children can do those things, in a normative sense, is not sufficient justification for requiring them to do so. Most young children willingly do most things adults ask of them. But their willingness is not a reliable indicator of the value of an activity. The developmental question is not only, "What can children do?," rather it is also, "What should children do that best serves their development and learning in the long term?"

FOUR CATEGORIES OF LEARNING GOALSThe four categories of learning outlined below are relevant to all levels of education-especially to the education of young children:

1.KNOWLEDGE
In early childhood, knowledge consists of facts, concepts, ideas vocabulary, stories, and many other aspects of children's culture. Children acquire such knowledge from someone's answers to their questions, explanations, descriptions, and accounts of events, as well as through active and constructive processes of making the best sense they can of their own direct observations.

2.SKILLS
Skills are small units of action that occur in a relatively short period of time and are easily observed or inferred. Physical, social, verbal, counting, and drawing skills are among a few of the almost endless number of skills learned in the early years. Skills can be learned from direct instruction or imitated based on observation, and they are improved with guidance, practice, repetition, drill, and actual application or use.

3.DISPOSITIONS
Dispositions can be thought of as habits of mind or tendencies to respond to certain situations in certain ways. Curiosity, friendliness or unfriendliness, bossiness, generosity, meanness, and creativity are examples of dispositions or sets of dispositions, rather than of skills or items of knowledge. Accordingly, it is useful to keep in mind the difference between having writing skills and having the disposition to be a writer, or having reading skills and having the disposition to be a reader (Katz, 1995).

Dispositions are not learned through formal instruction or exhortation. Many important dispositions, including the dispositions to learn and to make sense of experience, are in-born in all children-wherever they are born and are growing up. Many dispositions that most adults want children to acquire or to strengthen-for example, curiosity, creativity, cooperation, openness, friendliness-are learned primarily from being around people who exhibit them; they are strengthened by being used effectively and by being appreciated rather than rewarded (Kohn, 1993).

To acquire or strengthen a particular disposition, a child must have the opportunity to express the disposition in behavior. When manifestations of the dispositions occur, they can be strengthened as the child observes their effectiveness and the responses to them and experiences satisfaction from them. Teachers can strengthen certain dispositions by setting learning goals rather than performance goals. A teacher who says, "See how much you can find out about something," rather than, "I want to see how well you can do," encourages children to focus on what they are learning rather than on an external evaluation of their performance (Dweck, 1991).

4.FEELINGS
Feelings are subjective emotional states. Some feelings are innate (e.g., fear), while others are learned. Among feelings that are learned are those of competence, confidence, belonging, and security. Feelings about school, teachers, learning, and other children are also learned in the early years.
LEARNING THROUGH INTERACTIONContemporary research confirms that young children learn most effectively when they are engaged in interaction rather than in merely receptive or passive activities (Bruner, 1999; Wood & Bennett, 1999). Young children therefore are most likely to be strengthening their natural dispositions to learn when they are interacting with adults, peers, materials, and their surroundings in ways that help them make better and deeper sense of their own experience and environment. They should be investigating and purposefully observing aspects of their environment worth learning about, and recording and representing their findings and observations through activities such as talk, paintings, drawings, construction, writing, and graphing. Interaction that arises in the course of such activities provides contexts for much social and cognitive learning.

RISKS OF EARLY ACADEMIC INSTRUCTIONResearch on the long-term effects of various curriculum models suggests that the introduction of academic work into the early childhood curriculum yields fairly good results on standardized tests in the short term but may be counterproductive in the long term (Schweinhart & Weikart, 1997; Marcon, 1995). For example, the risk of early instruction in beginning reading skills is that the amount of drill and practice required for success at an early age seems to undermine children's disposition to be readers. It is clearly not useful for a child to learn skills if, in the process of acquiring them, the disposition to use them is lost. In the case of reading in particular, comprehension is most likely to be dependent on actual reading and not just on skill-based reading instruction (Snow, Burns, & Griffin, 1998). On the other hand, acquiring the disposition to be a reader without the requisite skills is also not desirable. Results from longitudinal studies suggest that curricula and teaching should be designed to optimize the simultaneous acquisition of knowledge, skills, desirable dispositions, and feelings (Marcon, 1995). Another risk of introducing young children to formal academic work prematurely is that those who cannot relate to the tasks required are likely to feel incompetent. Students who repeatedly experience difficulties leading to feelings of incompetence may come to consider themselves stupid and bring their behavior into line accordingly (Bandura et al., 1999).

VARIETY OF TEACHING METHODSAcademically focused curricula for preschool, kindergarten, and primary programs typically adopt a single pedagogical method dominated by workbooks and drill and practice of discrete skills. It is reasonable to assume that when a single teaching method is used for a diverse group of children, many of these children are likely to fail. The younger the children are, the greater the variety of teaching methods there should be, because the younger the children, the less likely they are to have been socialized into a standard way of responding to their social environment.

In this way, it is more likely that children's readiness to learn school tasks is influenced by background experiences that are idiosyncratic and unique. For practical reasons, there are limits to how varied teaching methods can be. It should be noted, however, that while approaches dominated by workbooks often claim to individualize instruction, individualization rarely consists of more than the day on which a child completes a particular page or other routine task. As suggested by several follow-up studies, such programs may undermine children's in-born disposition to learn-or at least to learn what the schools want them to learn (Schweinhart & Weikart, 1997; Marcon, 1995).

THE LEARNING ENVIRONMENTAs for the learning environment, the younger the children are, the more informal it should be. Informal learning environments encourage spontaneous play in which children engage in the available activities that interest them, such as a variety of types of play and construction. However, spontaneous play is not the only alternative to early academic instruction. The data on children's learning suggest that preschool and kindergarten experiences require an intellectually oriented approach in which children interact in small groups as they work together on projects that help them make increasing sense of their own experience. Thus, the curriculum should include group projects that are investigations of worthwhile topics. These projects should strengthen children's dispositions to observe, experiment, inquire, and examine more closely the worthwhile aspects of their environment. They usually include constructions and dramatic play as well as a variety of early literacy and numeracy activities that emerge from the work of the investigation and the tasks of summarizing findings and sharing the experiences of the work accomplished.

This Digest is a revision of the 1987 Digest WHAT SHOULD YOUNG CHILDREN BE LEARNING? by Lilian Katz.

      GO TO THEM , TEACH THEM ---MAKE TOMMORROW'S WORLD

 

ANSWERS TO LAST POSTED 100 GMAT Quant – Numbers / Inequalities / Absolute Values (Modulus) / General Algebra

ANSWERS TO LAST POSTED 100 GMAT Quant – Numbers / Inequalities / Absolute Values (Modulus) / General Algebra

 

 

1.

OA: A

the word "parity", as used in the following discussion, means "whether something is even or

odd", in the same way in which "sign" means whether a number is positive or negative.

the best way to approach things like this is to break down the compound statements into

statements about the parity of the individual variables.

to do that, you'll almost certainly have to split the statements into cases.

"xy + z is odd"

two numbers can add to give an odd sum only if they have opposite parity. hence:

case 1: xy is odd, z is even

there's only one way this can happen:

x = odd, y = odd, z = even. (1)

case 2: xy is even, z is odd

there are 3 ways in which this can happen:

x = even, y = even, z = odd (2a)

x = odd, y = even, z = odd (2b)

x = even, y = odd, z = odd (2c)

this is a bit awkward, but, once you've divided the question prompt up into cases, all you have to

do is look at your results, check the cases, and you'll have an answer.

––

statement (1)

the easiest way to handle expressions like this is to factor out common terms.

you can handle

the statement without doing so, but it's more work that way.

pull out x:

x(y + z) is even.

this means that at least one of x and (y + z) is even.

* if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes"

and we're done.

* the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it

doesn't satisfy any of the cases above.

therefore, the answer must be "yes".

sufficient.

––

statement (2)

this means that y and xz have opposite parity.

* y = even, xz = odd ––> this means x = odd, y = even, z = odd. that's case (2b), which gives

"no" to the question.

at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se

you know that "yes" MUST be a possibility with this statement (as statement #1 gives

exclusively "yes" answers).

if you use this statement first, you'll have to keep going through the cases.

insufficient.

ans = a

2.

this problem involves two fractions that are added together. for no other reason than that 'it's the

normal thing to do with two fractions added together', let's find the common denominator:

w/x + y/z = wz/xz + xy/xz = (wz + xy)/xz

therefore

the question can be rearranged to:

is (wz + xy)/xz – which is the same thing as w/x + y/z – odd?

–– (2) alone ––

if wz + xy is an odd integer, then all of its factors are odd. this means that (wz + xy)/xz, which is

guaranteed to be an integer**, must also be odd – because it's a factor of an odd number.

sufficient

**we know this is an integer because it's equal to w/x + y/z, which, according to the information

given in the problem statement, is integer + integer.

–– (1) alone ––

try to come up with contradictory examples**:

w=2, x=1, y=3, z=1 (so that wx + yz = 5 = odd, per the requirement):

w/x + y/z = 2 + 3 = 5 = odd

w=2, x=2, y=3, z=1 (so that wx + yz = 7 = odd, per the requirement):

w/x + y/z = 1 + 3 = 4 = even

insufficient

**of course, if you're at a loss for the theory, you should try this for statement (1) too ... but

you'll find that all the examples you get are odd.

––

answer = b

3.

JUST PLUG IN NUMBERS.

statement (1)

let's just PICK A WHOLE BUNCH OF NUMBERS WHOSE GCF IS 2 and watch what

happens. let's try to make the numbers diverse.

say,

4 and 6

6 and 8

8 and 10

10 and 12

...

4 and 10

6 and 14

6 and 16

8 and 18

8 and 22

...

in all nine of these examples, the remainders are greater than 1. in fact, there is an obvious

pattern, which is that they're all even, since the numbers in question must be even.

in fact, i just thought of this, which is a much nicer, more ground-level approach to statement

one:

in statement 1, both m and p are even. therefore, the remainder is even, so it's greater than

1.

done.

sufficient.

--

statement (2)

just pick various numbers whose lcm is 30.

notice the numbers selected above:

5 and 6 --> remainder = 1

10 and 15 --> remainder = 5 > 1

insufficient.

ans (a)

4.

Answer: B

(1)

this is a disguised way of saying 'n is prime'

therefore, insufficient

(2)

this says any two factors. that means any two factors – i.e., ALL pairs of factors have an odd

difference.

there's only one way to do this: one odd factor and one even factor. (as soon as you get 2 odd

factors or 2 even factors, you get an even difference by subtracting them.)

2 is the only # with only 1 odd factor and only 1 even factor.

therefore, sufficient

5.

Take a prime number and figure out a specific soln for that prime number.

Let p = 5. So, excluding 1, the other numbers that have no factors common with 5 are 2,3,4.

Let p = 7. So, excluding 1, the other numbers that have no factors common with 7 are 2,3,4,5,6

Do you see the pattern? For any prime number, all the numbers less than it will have no factors

in common with it except 1.

So f(p) = p – 2 Answer is B. NOOOOOOOOOO

We need to include 1 and hence the correct answer is p–1 (A).

Pick a prime number for p. Let's say p=5.

The positive integers less than 5 are 4, 3, 2, and 1.

5 and 4 share only 1 as a factor

5 and 3 share only 1 as a factor

5 and 2 share only 1 as a factor

5 and 1 share only 1 as a factor

There are four positive integers, therefore, that are both less than 5 and share only 1 as a factor.

In other words, we include 1 in this set of integers.

6.

Let's first consider the prime factors of h(100). According to the given function,

h(100) = 2*4*6*8*...*100

By factoring a 2 from each term of our function, h(100) can be rewritten as

2^50*(1*2*3*...*50).

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by

any of these prime numbers will yield a remainder of 1.

Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The

correct answer is E.

7.

n is an integer

is n odd?

yes/no question, so I will try to prove it wrong (that is, get a yes and a no based upon the

statements)

(1) n/3

n could be 6 (that is divisible by 3). Is n odd? No

n could be 9 (that is divisible by 3). Is n odd? Yes

Elim A and D

(2) 2n has twice as many factors as n

n could be 1, which has one factor; 2n would be 2, which has two factors; is n odd? Yes

n could be 2, which has two factors; 2n would be 4, which has three factors. Oops, can't use this

combo of numbers (has to make statement 2 true, and this combo doesn't)

What's going on here?

general rule: 2n will be divisible by 2 and also by whatever number 2n is.

If I make n an even number, even numbers are already divisible by 2. So 2n will only be

divisible by one new number, equal to 2n. That is, I add only one new factor for 2n. [editor:

there's a mistake in this explanation - see below for a correction]

Any even number, by definition, has at least two factors - 1 and 2. So I would need to add at least

two more factors to double the number of factors. But I can't - the setup of statement 2 only

allows me to add one new factor if n is even. So I can never make statement 2 true using an even

number for n.

Sufficient. Answer is B.

let's say a number has "n" different factors.

when you multiply this number by 2, you POTENTIALLY create "n" MORE factors - by

doubling each factor.

HOWEVER,

the only way that ALL of these factors can be NEW (i.e., not already listed in the

original n factors) is if they are ALL ODD.

if there are ANY even factors to start with, then those factors will be repeated in the original list.

(for instance, note that 2, 4, 26, and 52 all appear in both lists above.) therefore, if the number is

even, then the number of factors will be less than doubled

because of the repeat factors.

thus if statement (2) is true, then the number must be odd.

8.

Start with statement 2. This doesn't tell us one value of d, so elim. B and D.

Statement 1: 10^d is a factor of f. This isn't going to be sufficient. If you're not sure why try the

easiest possible positive integers. Is 10^1 = 10 a factor of f? Yes, so 1 is a possible value for d. Is

10^2 = 100 a possible factor of f? Yes, so 2 is a possible value for d. I just found 2 possible

values for d. Elim. A. Only C and E are left.

d is a pos int (given in stem) and is greater than 6 (statement 2). Smallest possibility, then, is 7. If

d is anything greater than 7, then 7 will work too (eg, if d actually is 8, then 7 would also satisfy

both statements and we wouldn't be able to tell, just from the statements, whether d is 7 or 8). So

it's either 7, exactly, which is sufficient, or it's something greater than 7, which is not sufficient.

So how many 10's are in f?

write down the numbers that contain 2s and 5s (only those)

30*28*26*25*24*22*20*18*16*15*14*12*10*8*6*5*4*2

Now ask yourself Is my limiting factor going to be 5 or is it going to be 2?

It's going to be 5 because there are many more 2's up there. So circle the numbers that contain

5's:

30, 25, 20, 15, 10, 5

How many 5's do you have? Seven 5's (don't forget – 25 has two 5's!), so you can make seven

10's. That's it. Answer is C.

"limiting factor" means "which is least common or likely." Think of it this way: there are many

more multiples of 2 than there are multiples of 5. In probability terms, a number is more likely to

be even than to be a multiple of 5. In divisibility terms, take some large number that is divisible

by both 2 and 5, and it is likely to have more factors of 2 than 5.

For example: 400 = 4*10*10 = (2*2)(2*5)(2*5) = (2^4)(5^2).

I know, numbers with more factors of 5 than factors of 2 exist...this is just a bet we make to ease

the computation.

In general, the larger the factor, the less likely it is to divide evenly into a number. The larger the

factor, the more of a "limiting factor" it is.

here's all you have to do:

forget entirely about 10, 20, and 30, and ONLY THINK ABOUT PRIME

FACTORIZATIONS.

(TAKEAWAY: this is the way to go in general – when you break something down into primes,

you should not think in hybrid terms like this. instead, just translate everything into the language

of primes.)

each PAIR OF A '5' AND A '2' in the prime factorization translates into a '10'.

there are seven 5's: one each from 5, 10, 15, 20, and 30, and two from 25.

there are waaaaaaayyyyy more than seven 2's.

therefore, 30! can accommodate as many as seven 10's before you run out of fives.

––

statement 2 is clearly insufficient.

statement 1, by itself, means that d can be anything from 1 to 7 inclusive.

together, d must be 7.

ans (c)

9.

96 =

6*8*2 or 2*8*6 or 826 or 628.....

3*8*4 or 483 or 843 or ....

According to 1: the number is odd; We have only one odd digit: 3. – correct

while II says: hundreds digit of m is 8; There are many combination: 682 or 483 or.... incorrect.

96 = 2*2*2*2*2*3,

statement 1: m is odd, so unit's digit could be 1,3,5,7,9.

But we have only one odd factor in 96(product of digits of m) i.e. 3. Therefore, unit's digit of m

is 3. – sufficient

statement 2: hundred's digit is 8, so we are left with 2*2*3. Therefore, m could be 826, 843, 834,

862. So no unique unit's digit. Insufficient

10.

the correct answer: B

if we are told that four different prime numbers are factors of 2n then can't i further assume that

one of those four prime numbers is 2 (since it's 2n)

it's possible that 2 is already a factor of n to start with, in which case n itself would still have 4

different prime factors (because, in that case, the additional 2 would not change the total number

of prime factors).

for instance, if n = 3x5x7 = 105 (which has three prime factors), then 2n = 2x3x5x7 = 210 has

four prime factors.

if n = 2x3x5x7 = 210, which has four prime factors, then 2n = 2x2x3x5x7 = 420, which still has

two prime factors.

therefore, #1 is not sufficient.

11.

the question is asking whether k has a factor that is greater than 1, but less than itself.

if you're good at these number property rephrasings, then you can realize that this question is

equivalent to "is k non–prime?", which, in turn, because it's a data sufficiency problem (and

therefore we don't care whether the answer is "yes" or "no", as long as there's an answer), is

equivalent to "is k prime?".

but let's stick to the first question – "does k have a factor that's between 1 and k itself?" – because

that's easier to interpret, and, ironically, is easier to think about (on this particular problem) than

the prime issue.

––

key realization:

every one of the numbers 2, 3, 4, 5, ..., 12, 13 is a factor of 13!.

this should be clear when you think about the definition of a factorial: it's just the product of all

the integers from 1 through 13. because all of those numbers are in the product, they're all factors

(some of them several times over).

––

consider the lowest number allowed by statement 2: 13! + 2.

note that 2 goes into 13! (as shown above), and 2 also goes into 2. therefore, 2 is a factor of this

sum (answer to question prompt = "yes").

consider the next number allowed by statement 2: 13! + 3.

note that 3 goes into 13! (as shown above), and 3 also goes into 3. therefore, 3 is a factor of this

sum (answer to question prompt = "yes").

etc.

all the way to 13! + 13.

works the same way each time.

so the answer is "yes" every time ––> sufficient.

––

in this problem, the prompt asks, "Is there a factor p such that...?"

this means that, if you can show that there is even one such factor, then it's "sufficient" and you

are DONE.

we have ascertained that every one of the "k"s in that range has at least one such factor.

to wit, 13! + 2 has the factor 2; 13! + 3 has the factor 3; ...; 13! + 13 has the factor 13.

that's all we need to know.

sufficient.

you are right that it's difficult to ascertain whether numbers greater than 13 are factors of these

"k"s. luckily, we don't have to care about that.

12.

OA: D

Since it has only 2 prime factors but 6 factors (4 of which are 1, 3, 7, k) this means that the prime

factors must be combined to generate the other 2 factors – the other 2 can only be either 3 which

means 3x3=9 and 3.7=21 is a factor OR 7 which means the other 2 factors are 21 and 49.

SHORTCUT METHOD:

if you know the following useful fact, then you can solve this problem much more quickly.

USEFUL FACT: if a, b, ... are the EXPONENTS in the prime factorization of a number,

then the total number of factors of that number is the product of (a + 1), (b + 1), ...

example:

540 = (2^2)(3^3)(5^1), in which the exponents are 2, 3, and 1. therefore, 540 has (2 + 1)(3 + 1)(1

+ 1) = 3 x 4 x 2 = 24 different factors.

with this shortcut method, realize that 6 (the total number of factors) is 3 x 2. therefore, the

exponents in the prime factorization must be 2 and 1, in some order.

therefore, there are only two possibilities: k = (3^2)(7^1) = 63, or k = (3^1)(7^2) = 147.

statement (1) includes 63 but rules out 149, so, sufficient.

statement (2) includes 63 but rules out 149, so, sufficient.

answer = (d).

––

IF YOU DON'T KNOW THE SHORTCUT:

statement (1)

if 3^2 is a factor of k, then so is 3^1.

therefore, we already have four factors: 1, 3^1, 3^2, and 7.

but we also know that (3^1)(7) and (3^2)(7) must be factors, since 3^2 and 7 are both part of the

prime factorization of k.

that's already six factors, so we're done: k must be (3^2)(7). if it were any bigger, then there

would be more than these six factors.

sufficient.

statement (2)

if 7 is a factor of k, but 7^2 isn't, then the prime factorization of k contains EXACTLY one 7.

therefore, we need to find out how many 3's will produce six factors when paired with exactly

one 7.

in fact, it's data sufficiency, so we don't even have to find this number; all we have to do is

realize that adding more 3's will always increase the number of factors, so, there must be exactly

one number of 3's that will produce the correct number of factors. (as already noted above, that's

two 3's, or 3^2.)

sufficient.

13.

when you take the product of two numbers, all you're doing, in terms of primes, is

throwing all the prime factors of both numbers together into one big pool.

therefore, the original question – 'what's the greatest prime factor of the product?' – can be

rephrased as,

what's the greatest prime that's a factor of

either t or n?

(1)

because the gcf only tells us which primes are in BOTH t and n. there could be great big fat

primes that are factors of only one of them, and they wouldn't show up in the gcf.

insufficient.

(2)

the lcm of two numbers contains EVERY prime that appears in either one of the two numbers

(because it's a multiple of both numbers). therefore, whatever is the largest prime factor of the

lcm is also the largest prime that goes evenly into either t or n.

sufficient.

––

if you don't realize why the relationships between lcm/gcf and primes, stated above, are what

they are, you can just try a few cases and watch the results for yourself. for instance, consider the

two numbers 30 (= 2 x 3 x 5) and 70 (= 2 x 5 x 7).

the gcf of these 2 numbers is 10 (= 2 x 5), which doesn't show anything about the presence of the

prime factor 7 in one of the numbers.

the lcm of these 2 numbers is 210 (= 2 x 3 x 5 x 7), which contains all of the primes found in

either number.

Ans. B

OR

(1) The first statement does not tell us about factors that are not common to n and t. One of those

factors might be greater or less than 5. Tis statement alone is NOT SUFFICIENT

(2) The LCM is 105 which can be factored as 3*5*7. Since LCM incorporates all the factors of n

and t we know the greatest factor for the two numbers is 7. Hence this statement is sufficient.

OR

consider the 'prime box' approach (= an imaginary box that contains all the numbers in the prime

factorization of a number, for those of you who are uninitiated into our curriculum).

you're looking for the greatest prime # that would be in the 'box' obtained from dumping all the

factors of n and all the factors of t, including all repetitions, into a bigger box. (this is what

multiplication does: it multiplies the complete factorization of one number by that of another. for

instance, 12 = 2x2x3 and 20 = 2x2x5, so 12 x 20 = 2x2x2x2x3x5.) therefore, the question can be

rephrased as follows: what is the greatest prime # that is a factor of either t or n ?

(1) this only tells is that the greatest number that is in both factorizations – those of n and t – is

5. but there could be a larger factor that is part of only one of the factorizations. for instance:

– it's possible that n = t = 5. then the greatest prime factor of nt is 5.

– it's possible that n = 5 and t = 35. then the greatest prime factor of nt is 7.

insufficient.

(2) the least common multiple contains every factor of t or n at least once. (it has to; if, say, t had

a factor that wasn't contained in it, then it would fail to be a multiple of t.) so, the biggest prime

factor of this # will also be the biggest prime factor of the product nt.

sufficient.

try a few combinations of n and t if you aren't convinced.

answer = b

14.

Answer is A.

From (1), I could figure (t+3)(t+2) will always have a remainder 2, hence SUFFICIENT

I had trouble with (2) as I could not come up with an algebraic approach. I understand I can plug

numbers to see that t^2 = 36 and t^2 = 64 fit the criterion BUT yield different remainders when

the corresponding values of t are plugged into t^2 + 5t +6 and hence INSUFFICIENT.

OR

one fact that's pretty cool, and which happens to apply to this problem, is that you can do normal

arithmetic with remainders, as long as all the remainders come from division by the same

number.

the only difference is that, if/when you get numbers that are too big to be authentic

remainders (i.e., they're equal to or greater than the number you're dividing by), you have to take

out as many multiples of the divisor as necessary to convert them back into "legitimate"

remainders again. you can think of the remainders as on an odometer that rolls back to 0

whenever you reach the number you're dividing by.

so with statement (1), all the remainders are upon division by 7, so we can do normal arithmetic

with them:

if t gives a remainder of 6, then t^2 = t x t gives a remainder of 6 x 6 = 36 ––> this is more than

7, so we take out as many 7's as possible: 36 – 35 = 1.

if t gives a remainder of 6, then 5t gives a remainder of 5(6) = 30 ––> this is more than 7, so we

take out as many 7's as possible: 30 – 28 = 2.

and finally, 6 itself gives a remainder of 6.

therefore, the grand remainder when t^2 + 5t + 6 is divided by 7 should be 1 + 2 + 6 = 9 ––> take

out one more seven ––> remainder will be 2.

sufficient.

by the way, much more generally (and therefore perhaps more importantly), the patterns in

remainder problems will always emerge fairly early when you plug in numbers.

therefore, if

you don't IMMEDIATELY realize a good theoretical way to do a remainder problem, you

should get on the number plugging RIGHT AWAY.

with statement (1), generate the first 3 numbers for which the statement is true: 6, 13, 20.

try 6: 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.

try 13: 169 + 65 + 6 = 240, which yields a remainder of 2 upon division by 7.

try 20: 400 + 100 + 6 = 506, which yields a remainder of 2 upon division by 7.

i'm convinced. (again, remember that PATTERNS EMERGE EARLY in remainder problems. 3

examples may not be enough for other types of pattern recognition, but that's usually pretty good

in a remainder problem.)

with statement (2), as a poster has already mentioned above, find the first two t^2's that actually

do this, which are 1^2 = 1 and 6^2 = 36.

if t = 1, then 1 + 5 + 6 = 12, which yields a remainder of 5 upon division by 7.

if t = 6, then 36 + 30 + 6 = 72, which yields a remainder of 2 upon division by 7.

insufficient.

OR

remainder problems usually show patterns after a very, very small number of plug–ins.

statement (1):

it's easy to generate t's that do this: 6, 13, 20, 27, ... (note that 6 is a member of this list, and an

awfully valuable one at that; it's quite easy to plug in)

try 6: 36 + 30 + 6 = 72; divide by 7 ––> remainder 2

try 13: 169 + 65 + 6 = 240; divide by 7 ––> remainder 2

try 20: 400 + 100 + 6 = 506; divide by 7 ––> remainder 2

by this point i'd be convinced.

note that 3 plug–ins is NOT good enough for a great many problems, esp. number properties

problems. however, as i said above, remainder problems don't keep secrets for long.

sufficient.

statement (2):

it's harder to find t's that do this. however, the gmat is nice to you. if examples are harder to

find, then the results will usually come VERY quickly once you find those examples.

just take perfect squares, examine them, and see whether they give the requisite remainder upon

division by 7.

the first two perfect squares that do so are 1^2 = 1 and 6^2 = 36.

if you don't recognize that 1 ÷ 7 gives remainder 1, then you'll have to dig up 6^2 = 36 and 8^2 =

64. that's not that much more work.

in any case, you'll have

1 + 5 + 6 = 12 ––> divide by 7; remainder = 5

36 + 30 + 6 = 72 ––> divide by 7, remainder = 2 (the work for this was already done above; you

should NOT do it twice. i'm reproducing it here only for the sake of quick understanding.)

or

36 + 30 + 6 = 72 ––> divide by 7, remainder = 2 (the work for this was already done above; you

should NOT do it twice. i'm reproducing it here only for the sake of quick understanding.)

64 +40 + 6 = 110 ––> divide by 7, remainder = 5

either way, insufficient within the first two plug–ins!

answer (a)

15.

A) (8*n + 5)/4 –> sufficient

B) p is odd. So in the sum one is odd and one is even.

p = e^2 + o^2

e^2 will be divisible by 4.

o^2 = (2n+1)(2n+1) / 4 = (4n^2 + 4n + 1)/4 –> sufficient

you can always plug in a bunch of numbers until you've satisfied yourself that the statements are

sufficient.

for (1), just find the first few numbers that give remainder 5 upon division by 8: 5, 13, 21, 29, 37,

etc. all of these give remainders of 1 upon division by 4, so that's convincing enough. sufficient.

(note: the gmat WILL NOT give problems on which a spurious pattern appears, only to be

broken after the 40th or 50th number; if you see a pattern persist for 4–5 cases, you can take it on

faith that the pattern persists indefinitely.)

for (2), you should make the same realization you made above: one of the numbers has to be odd

and the other even. then just try a bunch of possibilities:

1^2 + 2^2 = 5

2^2 + 3^2 = 13

3^2 + 4^2 = 25, etc

1^2 + 4^2 = 17

2^2 + 5^2 = 29

3^2 + 6^2 = 45, etc

all these give a remainder of 1 upon division by 4. sufficient.

16.

1. p can be represented as p = 8n+5

so p can take values , 13, 21, 29,37, 45.....just plugging in various values of integer n.

Now the next task is to represent all these odd numbers as sum of 2 perfect squares.

13 = 4+9 = 2^2+3^2 this implies x=2, y=3 . As question already told us that y is odd.

21 = cant represnt as sum of two +ve integers

29 = 4+ 25 = 2^2+ 5^2 ....implies x=2, y=5.

37 = 1+36 ....implies y=1 , x=6

45 = 9+36....implies y=3, x=6

So it tells us that x = 2, 6...not divisble by 4.

Hence SUFFICIENT

2. Condition B tells us

x–y= 3 and y is odd

so y=1 , x=4 Div by 4

y= 3 , x= 6 Not Div by 4

y= 5, x= 8 Div by 4

y= 7, x=10, Not DIv by 4

So INSUFFICIENT

17.

TAKEAWAY:

in REMAINDER PROBLEMS:

if you don't INSTANTLY see the algebraic solution, then IMMEDIATELY start

LOOKING FOR A PATTERN.

there's also a fact that you should know concerning this problem statement:

fact:

REMAINDERS UPON DIVISION BY 10 are simply UNITS DIGITS.

for instance, when 352 is divided by 10, the remainder is 2.

since remainders are fundamentally based on stuff repeating over and over and over again, it

shouldn't be a surprise that patterns emerge early and often among remainders.

this solution isn't necessarily "easier" – that judgment depends upon how comfortable you are

with the algebra and theory – but it can be quite efficient.

using (1)

9*3^4n + m becomes 9*3^8 + m

considering only units digit , 9*1 + m

INSUFFICENT

instead, you can just realize that, since m can be anything at all, you can have any units digit you

want.

using (2)

9*3^4n + 1 , as shown above, for all values of n, units digit 3^4n remains the same. ( UD of

3^4=1,UD of 3^8 =1)

Now , considering only units digit

9*1 + 1 = 10 ,Hence B SUFFICIENT

yeah.

technically, you should also throw away the "1" in your sum of 10, reducing to a final units digit

of 0.

The answer is 'B', but I don't get it!!! if

m is one, you still don't know what 3^(4n+2) is... right?

all we know is it's a power of 3... so it's units digit could be any number between 0–9.... thus, we

still don't know what the remainder would be if divided by 10.... please help!!!!

Rephrase the expression: –

3^(4n+2) +m = (9)*3^(4n) + m

statement (1) n =2 so the expression = (9)*3^8 + m but we do not know what m is – so cannot

predict the value of the expression. INSUFFICIENT

statemtn (2) m = 1 which makes the expression:

(9)*3^(4n) + 1 Since we know n is +ve integer, now it gets tricky: –

for n = 1,2,3,4 the exponential component of the expression will be

3^4, 3^8, 3^12 or

9^2, 9^4, 9^6 or

81, 81^2, 81^3 ans so on... the unit digit of all these values will be 1, now this value will be

multiplied by 9 and '1' will be added to the result. It will make the unit digit of the result – 0. It

means the result will be prefectly divided by 10. So the remainder will be 0

SUFFICIENT, So the answer is (B)

18.

(1)

if n = 3, then (n – 1)(n + 1) = 8, so the remainder is 8

if n = 5, then (n – 1)(n + 1) = 24, so the remainder is 0

insufficient

(2)

if n = 2, then (n – 1)(n + 1) = 3, so the remainder is 3

if n = 5, then (n – 1)(n + 1) = 24, so the remainder is 0

insufficient

(together)

the best approach, unless you're really good at number properties, is to try the first few numbers

that satisfy both statements, and watch what happens.

if n = 1, then (n – 1)(n + 1) = 0, so the remainder is 0

if n = 5, then (n – 1)(n + 1) = 24, so the remainder is 0

if n = 7, then (n – 1)(n + 1) = 48, so the remainder is 0

if n = 11, then (n – 1)(n + 1) = 120, so the remainder is 0

...you can see where this is headed.

here's the theory:

– if n is not divisible by 2, then n is odd, so both (n – 1) and (n + 1) are even. moreover, since

every other even number is a multiple of 4, one of those two factors is a multiple of 4. so the

product (n – 1)(n + 1) contains one multiple of 2 and one multiple of 4, so it contains at least 2 x

2 x 2 = three 2's in its prime factorization.

– if n is not divisible by 3, then exactly one of (n – 1) and (n + 1) is divisible by 3, because every

third integer is divisible by 3. therefore, the product (n – 1)(n + 1) contains a 3 in its prime

factorization.

– thus, the overall prime factorization of (n – 1)(n + 1) contains three 2's and a 3.

– therefore, it is a multiple of 24.

– sufficient

answer = c

takeaway:

once you're established "insufficient", do not bother testing additional cases!

the fact that n = 2 and n = 5 are both of the form (3k + 2) is random coincidence.

two:

if you look at the treatment of the 2 statements together, i have included both (3k + 1) and (3k +

2)–type cases in that treatment. unlike statement (2) alone, the combination of the 2 statements

turns out to be sufficient, so this time i must consider all of the possibilities.

therefore, i do.

if n is not divisible by 3, then exactly one of (n – 1) and (n + 1) is divisible by 3

if n – 1 is divisible by 3, then n has the form 3k + 1.

if n + 1 is divisible by 3, then n has the form 3k + 2.

both have been considered.

19.

We can rephrase the statement as such:

Is: n(n^2 – 1) divisible by 4?

Is N(N–1)(N+1) divisible by 4?

Is the product of three consecutive integers divisible by 4?

Final rephrasing:

Is N an odd integer or is N a multiple of 4?

Evaluate the statements:

1) n = 2k + 1, where K is an integer.

2K + 1 will give us an odd integer for N. (YES)

The problem I had was with plugging in 0 for K.

2(0) + 1 = 1 0x1x2 = 0 (OA: A 0 is divisible by every positive integer.)

note the following:

the only way you will encounter this sort of query is if you plug in your own numbers. in other

words, the official problems WILL NOT require you to decide the issue of whether 0 is divisible

by n (for whatever n); they restrict the scope of divisibility problems strictly to positive divisors

and positive dividends.

however, you should still know this fact, because, as you have seen here, you will often

encounter "extra" questions like this as artifacts of plugging in your own numbers. therefore,

even though the gmat won't test the concept directly, you may still have to rely on it to solve the

problem because of your number plugging.

––

as long as we're at it, if you encounter "negative multiples" in your number plugging adventures,

then yes, those are divisible too. for instance, –4 is divisible by 4, as are –8, –12, and the whole

lot.

20.

Method 1: Visual/Number Line approach.

(1) r is 3 times farther away from 0 than m is. But we have no "distances" given, nor any info

about sign (i.e. is m left or right of 0?)

(2) On a number line, put a dot at 12. Put two dots on either side of it for m and r. What can

vary? The distance between m and r--they can be very close to 12, or both very far away. Also,

we don't know whether m is the dot to the left or to the right of 12.

(1)&(2) together: We still don't know distances (from 12 or 0), or whether m is left or right of r.

We can either have (case A) r = 18 and m = 6 or (case B) r = 36 and m = -12.

Method 2: Algebra approach

(1) r = +/-3m

(2) r-12 = 12-m, or r+m = 24.

(1)&(2) together: r+m= (+/-3m)+m = 24. Either 4m = 24 (i.e. m=6) or -2m = 24 (i.e. m = -12).

since the natural instinct is to try only positive values for m and r, this is a very tricky problem.

Statement (1) tells us that r = 3m or r = –3m (as either case would result in an r with an absolute

value that is three times that of m). Insufficient. Eliminate AD from AD/BCE Grid.

Statement (2) tells us that (r+m)/2 = 12. Insufficient. Eliminate B from remaining BCE Grid.

By substituting each equation from Statement (1) into the equation from Statement (2), the

statements together tell us that 3m + m = 24, so m = 6 and r = 18, or that –3m + m = 24, so m = –

12 and r = 36. As there are still two possible values for r, the correct answer is E.

OR

if you'd rather conceptualize it (which is always a good idea for number–line problems like this

one), you can think of it this way:

r is 3 times as far away from 0 as is m, but we don't know in which direction.

that's the big thing.

since 12 is halfway between m and r, imagine m and r both starting out at 12, and 'sliding'

equally in opposite directions, with r moving to the right and m moving to the left. (you can't

slide r to the left and m to the right, because, if you do so, then r will be closer to 0 than is m.)

when the numbers have 'slid' a certain distance – specifically, 6 units each, so that m = 6 and r =

18 – they'll arrive at a point where the distance between m and 0 is 1/3 of the distance between r

and 0. that's the first point that satisfies both criteria.

now keep sliding the points away from 12.

eventually, m will pass through 0 itself, and will come out on the negative side. if you keep

sliding, you'll reachanother point at which the distance from 0 to m is 1/3 of the distance from 0

to r, only this time m is negative. (specifically, this will happen when m = –12 and r = 36.)

21.

the OA is C

Consider Data 2 independently

We have two possibilities:

1) Keep S on the right side of zero and satisfy the condition

2)Keep S on the left side and again satisfy the condition

In both ways the data is sufficient, but our quest is whether zero is halfway ...this is where u

seem to have miss out...

If u dont consider the Data 1 then u may or may not get zero halfway...

Thats y the answer is C

<––––––––––––––––––R––––––––––S–––––T––––––––––––>

OR

Choice (B) does not eliminate the possibility that R & S are zero. Combining the two statements

eliminates zero as an answer and gives us a definite "yes" as an answer.

watch those assumptions.

the distance between t and (–s) must be a positive number, but the problem is that we don't know

which way to subtract to get that positive number. if t > –s, then the distance is t – (–s), as you've

written here. however, if –s > t, then the distance is actually (–s – t) instead.

if s is to the left of zero, then –s will be to the right of zero – which could well place –s to the

right of t. if that happens, then the distance will become (–s – t), rendering your calculation

inaccurate. try drawing out this possibility – put zero WAY to the right of both s and t on the

number line, then find –s, and watch what happens).

if s lies to the right of zero, then –s must lie even further to the left than does s itself. since s is

already to the left of t, it then follows that –s is also to the left of t. therefore, in that case, you

can definitively write the distance as t – (–s), and your calculation is valid. therefore, (c).

––

ironically, the presence of statement (1) should make it easier to see that statement (2) is

insufficient. specifically, statement (1) calls your attention to the fact that s could lie to the left of

zero, in which case you could get the alternative outcome referenced above. that's something you

might not think about if statement (1) weren't there.

plug in numbers to the number line here:

Statement 1)

if the line reads: r=–1, zero, s=1, t=3, then zero is halfway between r and s.

if the line reads: zero, r=1, s=2, t=3, then zero is not between r and s.

Insufficient.

Statement 2)

by definition zero is halfway between s and –s.

a) if the line reads: –s=r=–2, zero, s=2, and t=4, then (t to r)=(t to –s)=6.

zero is halfway in between r and s.

b) if the line reads: r=–4, s=–2, t=–1, zero, and –s=2, then (t to r) = (t to –s) = 3.

zero is between t and –s.

Insufficient.

Together)

Forces the case 2a). Sufficient.

OR

the particular trap you may have fallen into in your interpretation of (2) is that of assuming "–s"

is to the LEFT of "t". there is no good reason whatsoever to make this assumption, and, what's

more, at least one good reason (viz., "the gmat loves to test exactly these sorts of

assumptions) not to make it.

of course, you don't need reasons to be very careful about your assumptions; that should be your

default state.

if "–s" is to the right of "t", then you have

<––r–––––––s–––t–––––––––––(–s)––>

in which case 0 is in no–man's–land between "t" and "–s".

in this case, note that "s" is negative. also note that (–s) is positive in this case, a situation that is

difficult to digest for most students.

taking statements (1) and (2) together eliminates the above possibility, leaving only the case that

you have outlined.

––

incidentally, the fault in the algebraic approach lies in writing the distance between t and (–s) as t

– (–s). this writing is correct only if t is greater than (–s), an assumption that, as we've seen, is

unjustified.

the correct way to write the distance is |t – (–s)| = |t + s|, an expression that is thoroughly

unhelpful in solving this problem.

22.

The answer is A.

S and t are different numbers on the line segment, Is s+t=0?

We need to know where s and t are in the line segment

Using BDACE Grid ,

2 says 0 is between s and t

In a line segment s and t are two points and 0 is between them. Let says s at –7 in the coordinate,

t could be in 3 and 0 is between them. It does not give a statement that s+T=0. Insuff

1 says distance between s and o is = d( betwen t and o)

Clearly, 0 is between S and t because distance from s to 0 is equal to distance from t to 0.

This gives a way to solve for s+t=0. Hence A is sufficient

Statement 2 is insufficient because 0 is between s and t. But that means s can equal –5 and t can

equal +3. In such a case, 0 is still between s and t but that does not make them equidistant from

0. Or, s and t can be –4 and +4 respectively in which case they are equidistant from 0. Therefore,

this statement doesn't necessarily answer the question because it can have different results.

The question states that s and t are different numbers, so they cannot both be –5. Therefore, they

must be opposites of each other.

just as in normal parlance, "between" only means "between", and carries no connotations of

equidistance from the two points.

for instance, it's quite true that 1 is between 0 and 100, but obviously false that 1 is the midpoint

between 0 and 100.

same thing with statement two. if 0 is between s and t, then all this means is that one of s and t is

positive and the other is negative. that is all; there's nothing barring possibilities such as s = –

1,000,000 and t = 1.

23.

well, first, think about the qualitative aspects of the sequence: if the sequence consisted entirely

of 7's, then there would be fifty terms in the sequence. these answer choices are reasonably close

to fifty, so it stands to reason that by far the majority of the terms will be 7's. therefore, try as few

77's as possible.

try only one 77:

remaining terms = 350 - 77 = 273

this would be 273 / 7 = 39 sevens

so ... you'd have one '77' and thirty-nine '7's

this works!

answer = c

OR

Since the units digit of 350 is zero, you know that the number of terms in the equation must be

such that:

n*7 = number with units digit of zero

The only time this is true is if n is 10 or a multiple thereof, and 40 is the only answer that

satisfies that.

24.

after the first two terms i.e (2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8) is a GP series with first

term as 2^2 and ratio as 2.

Using the formula for sum of GP series, for this part, the original equation becomes

2^2 + 2^2(1–2^7)/(1–2)

= 2^2 + 2^2(127)

= 2^2(1 + 127)

= 2^2 * 2^7

=2^9

OR

there are several ways.

(1) PATTERN RECOGNITION

it should be clear that there's nothing special about 2^8 as an ending point; in other words, they

just cut the sequence off at a random point. therefore, if we investigate smaller "versions" of

the sequence, we should be able to detect a pattern.

let's look:

first term = 2

sum of first 2 terms = 4

sum of first 3 terms = 8

sum of first 4 terms = 16

ok, it's clear what's going on: each new term doubles the sum. if you see a pattern this clear, it

doesn't matter whether you understand WHY the pattern exists; just continue it.

so, i want the sum of nine terms, so i'll just double the sum five more times:

32, 64, 128, 256, 512.

this is choice (a).

this is a general rule, by the way: IF SOMETHING CONTAINS MORE THAN 4–5

IDENTICAL STEPS, YOU SHOULD BE ABLE TO EXTRACT A PATTERN FROM

LOOKING AT SIMILAR EXAMPLES WITH FEWER STEPS.

(2) ALGEBRA WITH EXPONENTS ("textbook method")

the first two terms are 2 + 2. this is 2(2), or 2^2.

now, using this combined term as the "first term", the first two terms are 2^2 + 2^2. this is

2(2^2), or (2^1)(2^2), or 2^3.

now, using this combined term as the "first term", the first two terms are 2^3 + 2^3. this is

2(2^3), or (2^1)(2^3), or 2^4.

you can see that this will keep happening, so it will continue all the way up to 2^8 + 2^8, which

is 2(2^8) = (2^1)(2^8) = 2^9.

(3) ESTIMATE

these answer choices are ridiculously far apart, so you should be able to estimate the answer.

memorize some select powers of 2. notably, 2^10 = 1024, which is "about 1000". 2^9 = 512,

which is "about 500". and of course you should know all the smaller ones (2^6 and below)

by heart.

thus we have 2^8 is about 250, and the other terms are 128, 64, 32, 16, 8, 4, 2, 2.

looking at these numbers, i'd make a ROUGH ESTIMATE WITHIN A FEW SECONDS:

250 is 250.

128 is ~130.

64 and 32 together are ~100.

the others look like thirty or so together.

so, 250 + 130 + 100 + 30 = 510.

the only answer choice within shouting range is (a); the others are absurdly huge.

––

even if you have no idea how to do anything else, you should still be able to do out the

arithmetic within the two–minute time limit.

it won't be fun, but you should be able to do it. if you can't, then the reason is probably "you

stared at the problem for too long, and didn't get started when you should have".

yes ,the shortest method on this planet to solve the above question.

This formula may be of use:2^1+2^2+........+2^n =[2^(n+1)] – 2, where n equals to number of

terms.

question is: 2+2+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

this can be written as : 2+(2^1)+(2^2)+(2^3)+(2^4)+(2^5)+(2^6)+(2^7)+(2^8)

Therefore, 2+[2^(8+1)] – 2 = 2^9 (answer)

25.

OA is C

statement (1) means that the smallest and largest elements of the list have the same sign, i.e., are

both positive or both negative.

but, since those are the smallest and largest elements of the list, that means that all the

elements between have to have that same sign, too.

or:

you can't have 0 between two positive numbers, or between two negative numbers.

either everything in the list is positive, or everything in the list is negative.

From statement (1), we know the product of the highest and the lowest integer is + ve, it means

either both of them are +ve or –ve

For ex: –2,–1,1,2 in this list the product of –2&2 is –4. It proves both the highest and the lowest

terms have to be of same sign.

(+ve or –ve) the other factor that needs to be considered is the no. of terms in the list,

If the no. of terms is odd and all the integers are –ve, the product of all the integers will be –ve

this information is given by statement (2)

no. of terms in the list are even, hence the product of all the integers in the list will always be

+ve.

So if you combine (1) & (2), they are sufficient.

You have to multiply the smallest and the largest to satisfy case I. For exmple {+,–,–,+} does not

satisfy case I. You have taken both negative numbers in the middle. But the smallest number will

be one of the negative numbers and the largest one of the positive ones, giving a negative

product of as opposed to positive. Same holds for the third example you have used.

if you have numbers arranged from least to greatest, then any '–' numbers must show up to the

left of all '+' numbers. otherwise, you've created an impossible situation in which a negative

number is somehow bigger than a positive number.

26.

(1) imagine 0, 0, 0, 0 OR 0, 0, 0, 2 NS

(2) in order for the sum of "any 2 numbers" to = 0, all the numbers must equal 0

SUFFICIENT

statement 2 gives: the sum of ANY TWO...

So, since there are "more than 2" numbers in the set, the set contains (in your example): (–2, 2,

x). because the set contains at least that x, the sum of "any 2" numbers, for instance 2 + x, does

not have to equal zero.

so, INSUFFICIENT

you need to have more than 2 numbers in the set. the problem is that ANY two numbers have to

sum to zero – which means that if you pair the mystery third number with EITHER of the

existing two numbers, you must get a sum of zero.

if your first two numbers are 2 and –2, that's impossible: there's no number that will add to 2 to

give zero, and will ALSO add to –2 to give zero.

in general, if your first two numbers are –x and x, then your third number must be x (so that it

adds to –x to give zero), but it must ALSO be –x (so that it adds to x to give zero). the only way

that x can equal –x is if x is zero – which means that all three numbers are zero.

therefore, everything must be zero.

27.

The answer is A

With statement 1:

this function can only be addition or multiplication

with either of these two operations the left side does indeed equal the right...sufficient

With statement 2

this function can be either multiplication or division

with multiplication the left and right side equal one another

with division it doesn't...

hence 2 is insuffcient.

note the general takeaway here:

if you have a problem like this, in which a mystery symbol stands for

one or more of a

collection of operations, then your #1 goal is to figure out ANY AND ALL operations for

which that symbol can stand.

28.

Let's consider (1) N+1 >0. This clearly tells you nothing about p, so is insuffienct by itself, ruling

out A&D.

Let's look at (2). np >0. This tells us that n,p and either both positive or both negative. Therefore

it is insufficient to answer whether p >0, so we can eliminate B.

Now let's consider (1) and (2) together. (1) combined with the fact that N and P are integers tells

us that N>= 0. (2) tells us that N and P are either both positive or both negative and that neither

are equal to 0. Combined with (1) we therefore know what N is positive, and from (2) P must be

positive too. So (1) and (2) together are sufficient and the answer is C.

29.

(2)

Takeaway #1: when you plug numbers on a DS problem, YOUR GOAL IS TO PROVE

THAT THE STATEMENT IS INSUFFICIENT.

Therefore, as soon as you get a 'yes' answer, you should be TRYING to get a 'no' answer to go

along with it; and, as soon as you get a 'no' answer, you should be TRYING to get a 'yes' answer

to go along with it.

Statement (2)

you need to pick numbers such that x + y > z, per this statement.

First, pick a completely random set of numbers that does this: how about x = 1, y = 1, z =

0. These numbers give a YES answer to the prompt question, since 1^4 + 1^4 is indeed greater

than 0^4. Now remember: your goal is to prove that the statement is INSUFFICIENT. This

means that we have to try for a 'no' answer. This means that we have to make z^4 as big as

possible, while still obeying the criterion x + y > z.

Fortunately, this is somewhat simple to do:

just make z a big negative number. Try x = 1, y = 1, z = -100. In this case, x + y > z (satisfying

statement two), but x^4 + y^4 is clearly less than z^4, so, NO to the prompt

question. Insufficient.

Statement (2)

you need to pick numbers such that x^2 + y^2 > z^2, per this statement. First, pick a completely

random set of numbers that does this: how about x = 1, y = 1, z = 0 (the same set of numbers we

picked last time). These numbers give a YES answer to the prompt question, since 1^4 + 1^4 is

indeed greater than 0^4. Now remember: your goal is to prove that the statement is

INSUFFICIENT.

This means that we have to try for a 'no' answer. This means that we have to

make z^4 as big as possible, while still obeying the criterion x^2 + y^2 > z^2.

Unfortunately, this

isn't as easy to do as it was last time; we can't just make z a huge negative number, because z^2

would then still be a giant positive number (thwarting our efforts at obeying the criterion). So,

we have to finesse this one a bit, but the deal is still to make z as big as possible while still

obeying the criterion.

Let's let x and y randomly be 3 and 3. Then x^2 + y^2 = 18; we need z^2

to be less than this, but still as big as possible. So let's let z = 4 (so that z^2 = 16, which is pretty

close). With these numbers, x^4 + y^4 = 162, which is much less than z^4 = 256. Therefore, NO

to the prompt question, so, insufficient. Answer = e.

Takeaway #2: if a statement is sufficient, then you WILL be able to PROVE that it is,

algebraically or with some other form of theory.

In other words, you'll never get a statement that's sufficient, but for which you can only

figure that out by number plugging.

It's obvious that you can get a YES answer to the question; all you have to do is take ridiculously

big numbers for x and y, and a small number for z. for instance, x = y = 100, z = 0, satisfy both

statements, and clearly give a YES answer. So, you're trying for a NO answer. Try to make Z as

big as possible while still satisfying the criteria (i.e., less than x^2 + y^2). Let's let x = y = 3

then to satisfy both statements, we need z^2 less than 18, and z less than 6. We'll take z = 4,

which is pushing the limit of the first one. In this case, then, x^4 + y^4 = 81 + 81 = 162, but z^4

= 256, giving a NO answer. Insufficient Answer = e

30.

Ans. A

(1).. add 1 to both sides… you get r > w.

(2) gives contradictory answers… NS

31.

We're told x and y are positive but not whether they are greater than 1, so I have to consider

fractional possibilities. How do I know what to try?

When I take a square root:

Anything greater than 1 will get smaller (but remain larger than 1)

1 will stay the same

Anything between 0 and 1 will get bigger (but remain a fraction between 0 and 1)

When I take a reciprocal in each of the above cases:

1/something larger than 1 = something smaller than 1 (but still positive)

1/1 = 1

1/something smaller than 1 = something larger than 1

If I want to try numbers now, then I know I need to try a number from each set. Or I can continue

with logic and the algebraic representations. Do whichever you are most comfortable with.

For trying numbers, first try something greater than 1:

x=2, y=2 (I'm trying the same numbers b/c I'm trying to see if I can prove things false and funny

things happen when you use the same number for different variables). 1/(4)^.5 = 1/2.

Roman Numeral 1 (RN1): (4)^.5 / 2(2) = 2/4 = 1/2. Same, not greater, so elim RN1.

RN2: (2^.5 + 2^.5) / (4) = 2(2^.5) / 4. Well, 2^.5 is about 1.7. 2*1.7 = 3.4 / 4 = more than 1/2. So

RN2 is okay, at least with this instance.

RN3: (2^.5 – 2^.5) / 4 = 0/4 = 0. Elim RN3.

At this point, I don't know whether I have to try more numbers b/c the answer choices haven't

been listed. If I have both "none" and "II only" as options, then I have to try more numbers. If

"none" is not an option, then I'm done.

You will notice that equation 2 will always be more .

THE FASTEST WAY IS TO EXPRESS EACH EQUATION AS A FUNCTION OF

1/(X+Y)^0.5 . This aproach takes less than a minute.

there's little sense in dealing with #3 algebraically: because of the subtraction, it can clearly equal

0 (if x and y are the same number). since 1/v(x + y) is a positive number, the possibility of 0

rules out roman numeral III. (in fact, that expression can even be negative, as nothing

prohibits x from being smaller than y.)

if you want to compare two fractions, you can use the technique of cross products to perform

the comparison.

to use this technique, you take the two 'cross products' (one of the numerators, times the

denominator of the other fraction), and associate each of the cross products with whichever

fraction donated the numerator.

for instance, if you're comparing 2/3 vs. 11/17, then the cross products are 2 x 17 = 34

(associated with 2/3) and 3 x 11 = 33 (associated with 11/17). because 34 is greater than 33, it

follows that 2/3 is greater than 11/17.

notice that this technique only applies to positive fractions... but that's all you really need: if the

fractions have opposite signs, then the comparison is trivial (the positive one is bigger!), and if

the fractions are both negative, then the comparison is the opposite of whatever it would be if

they were positive.

find cross products in #(i):

v(

x + y)/2x vs. 1/v(x + y)

cross products are (x + y) vs. 2x

subtract one x from both sides ––> this comparison is the same as y vs. x

we don't know which is bigger.

find cross products in #(ii):

(vx + vy)/(x + y) vs. 1/v(x + y)

cross products are (vx + vy)v(x + y) vs. (x + y)

divide both sides by v(x + y) to give (vx + vy) vs. v(x + y) ––– remember that (quantity) divided

by v(quantity) is v(quantity) –– that's the definition of what a square root is.

since both of these quantities are positive, we can square them and compare the squares:

(vx + vy)^2 vs. (v(x + y))^2

x + 2vxy + y vs. x + y

left hand side is bigger

so the original fraction is bigger than 1/v(x + y)

ans = ii only

32.

OA: B

1. |x – 3| > y

Taking numbers:

x: –2, 1 both can satisfy the above equation. Insufficient.

2. |x – 3| < – y

Since |x–3| is an absolute value, the smallest it can go is 0.

And since y is given to be >0, thus – y will give a negative value which will cause the equation

to fall apart unless it is 0.

so |x–3| = 0.

x = 3.

33.

OA: C

we can rephrase this to x = z – y.

there are thus 3 possibilities for the absolute value |x| :

(a) if z – y is positive, then |x| = z – y, and will NOT equal y – z (which is a negative quantity).

(b) if z – y is negative, then |x| = y – z (the opposite of z – y).

(c) if z – y = 0, then |x| equals both y – z and z – y, since each is equal to 0.

TAKEAWAY: when you consider absolute value equations, you'll often do well by

considering the different CASES that result from different combinations of signs.

notice that (a) and (b), or (a) and (c), taken together prove that statement 1 is insufficient.

statement 2:

we don't know anything about y or z, so this statement is insufficient.**

if you must, find cases: say y = 2 and z = 1. if x = –1, then the answer is YES; if x is any

negative number other than –1, then the answer is NO.

together:

if x < 0, then this is case (b) listed above under statement 1.

therefore, the answer to the prompt question is YES.

sufficient.

––

WE could craft a statement that doesn't mention all three of x, y, z and yet IS STILL

SUFFICIENT.

here's one way we could do that:

(2) y < z

in this case, y – z is negative and therefore CAN'T equal |x| –– no matter what x is –– since |x|

must be nonnegative.

so, this statement is a definitive NO, and is thus sufficient even though it doesn't mention x at

all.

34.

Ans. C… standard property.

35.

If x = 0.1, then x^2 < 2x < 1/x (so 1 is possible)

If x = 0.9, then x^2 < 1/x < 2x (so 2 is possible)

(1) x^2<2x<1/x

This means that x^2<2x so divide by x to get x<2. The second one tells you that 2x<1/x which

simplifies to x < 1/sqrt(2). These can obviously both be satisfied at the same time, so (1) works.

(2) x^2<1/x<2x

This means that x^2<1/x which gives x^3<1, or x<1. The second half gives you 1/x<2x or

1<2(x^2) or x>1/sqrt(2). So any number that satisfies 1/sqrt(2)<x<1 will work.

(3) 2x<x^2<1/x. The first part gives 2x<x^2 or x>2. The second half gives x^2<1/x or x^3<1 or

x<1. Since the regions x>2 and x<1 do not overlap, (3) can not be satisfied.

The Answer choice is (4), 1 and 2 only

if x = 1/2, then:

x^2 = 1/4

1/x = 2

2x = 1

so the order would be x^2 < 2x < 1/x. So (a) is possible. Eliminate i and iii. (Looks like

you got this far)

if x = 3/4, then:

x^2 = 9/16

1/x = 4/3

2x = 3/2

so the order would be x^2 < 1/x < 2x. So (b) is possible. Eliminate ii. (Looks like this is

where you had trouble.)

36.

Notice that we're dealing with a fractional inequality, which, worse yet, CAN'T be multiplied by

the common denominator (since we don't know the sign of that denominator).

Therefore, pick numbers.

Just be careful to pick APPROPRIATE numbers. i.e., this problem contains sums and

differences,

as well as sign considerations, so i would pick:

* POSITIVES AND NEGATIVES (as allowed by the statements)

* DIFFERENT RELATIVE SIZES (i.e., "bigger" and "smaller" numbers) –– this is important

because of addition and subtraction.

so, for statement (1), i would pick:

1, 2

2, 1

1, –2

2, –1

1, 0

for statement (2), i would pick:

1, –2

2, –1

–1, –2

–2, –1

0, –1

for "together" i would look at the two common elements, which are (1, –2) and (2, –1).

note that this is a lot of plug–ins, but you don't wind up trying them all – you STOP as soon as

you get "insufficient".

GMAT's answer is (E)

this is a difficult problem, because it resists simple algebra. you CANNOT multiply through by

the denominator (x + y), because the sign of that denominator is unknown.

therefore, you have to leave the problem as written (ugly as it may be).

since there's no simple algebraic solution, a fallback is to recognize the types of numbers that

are important in the problem, and try numbers across those categories.

there are two things that matter in this problem (as may be deduced from an inspection of the

problem + experience with these sorts of things):

1. positive vs. negative

2. the relative magnitudes of x and y

let's try numbers across both of these categories.

statement (1)

x must be positive, but y could be positive or negative, and smaller or bigger (or the same) in

magnitude.

if x = 1 and y = 2 ––> answer = NO

if x = 2 and y = 1 ––> answer = NO

if x = 1 and y = –2 ––> answer = NO

if x = 2 and y = –1 ––> answer = YES

insufficient

[at this point you could notice that the last two examples also satisfy statement 2, and

therefore satisfy statements 1 and 2 together. this fact proves that the answer is E right

now, and you're done. if you don't notice this (most students won't), then go on.]

statement (2)

y must be negative, but x could be positive or negative, and smaller or bigger (or the same) in

magnitude.

if x = –1 and y = –2 ––> answer = NO

if x = –2 and y = –1 ––> answer = NO

if x = 1 and y = –2 ––> answer = NO

if x = 2 and y = –1 ––> answer = YES

insufficient

together

x must be positive and y must be negative, but the relative magnitudes can go any way

(bigger/smaller/same)

if x = 1 and y = –2 ––> answer = NO

if x = 2 and y = –1 ––> answer = YES

insufficient

ans (e)

37.

Let us take statement I – In this, it is given that 1/(k–1) >0. This implies that k must be postive

and k must be greater than 1. Hence, 1/k is definitely greater than zero. For example, k's value is

2, then 1/(2–1) = 1 which is > 0. This implies that 1=2, t=3, then zero is not between r and s.

34.

Ans. C… standard property.

But if k's value is –0.5, it will satisfy the second equation but 1/k will be –2 which is <0 and

hence, INSUFFICIENT.

Hence, A alone is sufficient to answer this question.

38.

The sign of a variable has nothing to do with addition and subtraction.

It is only MULTIPLICATION AND DIVISION that are affected by the sign of the

quantity.

There's a much worse problem here: you CAN'T SUBTRACT INEQUALITIES that face

the same way. You can add them, but you can't subtract them.

The answer is B

if you don't have an approach, then you should immediately start plugging in. you should do

ANYTHING to ensure that you're not just sitting there staring at a problem.

––

Statement (2):

All you need is y < w, which is EXACTLY equivalent to w – y > 0. There are two ways you

could figure this out:

So #2 is sufficient. they're just including x in there to try to get you to waste your time.

You cannot subtract two inequalities that face the same way.

Think about this:

x < 10

y < 10

if you try to subtract these, then you'll get x – y (?) 0.

but that clearly doesn't work, since you could create possibilities for "<" (e.g. x = 7, y = 8); "="

(e.g., x = y = 8); or ">" (e.g., x = 8, y = 7).

Incidentally, if two inequalities face in OPPOSITE ways, then you can subtract them. But if

that's the case, it's easier to just multiply one of them by –1 and then add them.

The question is asking is w–y>0

And After reading the the St.1 , things you know are

w+x <0 from Question Stem &

x+y <0 from St.1

Nothing more is given, don’t use the inequality used in the Question.

So the above statements can be true

w=1, y=2 and x= –10

w+x = –9<0

w+y = –8 <0

or

w=2, y=1, x=–10

w+x = –8<0

w+y = –9<0

And as you can see we cant say whether w>y or w<y and hence insufficient.

39.

OA: A

The quick way to approach will be pick a number x<0

Lets pick –5

so we know x=–5

sqrt (–x|x|) = sqrt (–(–5)|–5|)

= sqrt (5*5) = sqrt (25) = 5 = –(–5) = –x

So Answer A.

40.

If statement 2 contains statement 1, then ELIMINATE (b) and (c).

If statement 1 contains statement 2, then ELIMINATE (a) and (c).

Ans– D ( BOTH SUFFICIENT)

SQRT ( (x–5)^2 = |x–5|

squaring a quantity, and then square–rooting, is equivalent to taking the absolute value.

we can make this more clear:

|x – 5| can be either (x – 5), the actual quantity within the absolute–value bars, or (5 – x), the

opposite of that quantity.

if it's to be the original quantity (x – 5), then that quantity must be at least 0: x > 5.

if it's to be the opposite (5 – x), then that opposite quantity must be at least 0. for that to happen,

x < 5.

(notice that, if x is actually 5, then |x – 5| equals both (x – 5) and (5 – x), since both of them are

zero.)

therefore, we can rephrase the question:

is x < 5?

when you see this statement, it may bewilder you at first, but you should look at it and think:

"ok, just absolute–value bars and negative signs. no other numbers; no other operations; this

could only possibly have to do with the sign of x."

then just test it to see whether it works for PNZ (positive, negative, zero).

turns out that it only works for negative numbers.

therefore, rephrase:

1. –x|x| > 0

the above is possible only for x < 0, therefore (x–5) < 0

|x–5| = 5–x this becomes – – – > –(x–5) = 5–x and hence sufficient

2. Clearly states that (x–5) < 0 so this is sufficient and the answer is D

(1) x < 0

this is sufficient, since x is definitely less than 5 if it's negative.

41.

the absolute value will do one of two things to a quantity:

(a) LEAVE THE QUANTITY ALONE, if the quantity is POSITIVE;

(b) REVERSE THE SIGN of the quantity, if the quantity is NEGATIVE.

if the quantity is exactly 0, then both of these result in the same number, so it doesn't matter

which of them you call it.

therefore:

the expression |x – 3| will equal one of two expressions:

LEFT ALONE as (x – 3), if x – 3 is POSITIVE –– i.e., if x is greater than 3;

REVERSED to (3 – x) (which is the same as –x + 3), if x – 3 is NEGATIVE –– i.e., if x is less

than 3;

EITHER of these (since both equal 0) if x is exactly 3.

therefore, we now have a rephrase of the question.

REPHRASE:

is x

= 3 ?

so, the answer is (b).

you can also solve this problem, perhaps more easily, by PLUGGING IN NUMBERS.

statement (1):

since 3 is clearly a pivotal number in this problem, try numbers that are greater than 3 and

numbers that are less than 3.

try x = 0:

v((

x – 3)^2) = 3

3 – x = 3

answer to prompt question = YES

try x = 5:

v((

x – 3)^2) = 2

3 – x = –2

answer to prompt question = NO

insufficient.

––

statement (2)

this statement is an obnoxious way of stating that x is a negative number. (you still have to

figure that out –– no way around it)

if you try plugging in a vast array of negative numbers – big, small, even, odd, etc. – you'll find

that the equality in the prompt question holds for ALL of them.

sufficient.

42.

A

When multiplying or dividing an inequality by a negative, we have to switch the sign. We're not

told whether the quantity (a–b) is pos or neg, so we have to check this both ways.

IF a–b is pos, then 1 < (a–b)(b–a) (and you can simplify further from there)

IF a–b is neg, then 1 > (a–b)(b–a) (and again you can simplify further from here)

But the point is that I have to follow both possibilities through and any particular statement is

only sufficient if BOTH equations give me the SAME definitive answer, yes or no.

I'll stop there – try this again and come back if you have more questions.

Is 1/(a–b) < b – a?

Rephrase: 1< (a–b) (b–a)

1) a < b

If a, b are both +ve, Both –ve, or –ve & +ve

1 > (a–b) (b–a)

Statement (1) is sufficient.

2) 1 < |a–b|

This statement mentions |a–b| and nothing about (b–a)

(b–a) can be +ve or –ve.

Therefore, statement (2) is not sufficient.

43.

if you have

| QUANTITY 1 | = | QUANTITY 2 |

(with NOTHING ADDED to, or SUBTRACTED from, the abs. values)

then

just SOLVE TWO EQUATIONS:

* QUANTITY 1 = QUANTITY 2

* QUANTITY 1 = –(QUANTITY 2)

this is all you need. (there are also the possibilities with a negative sign in front of quantity 1, but

those will just be equivalent the two already written here.)

statement (1)

for EQUATIONS involving absolute value, like this one, the key realization is that the absolute

value of a quantity can signify either the quantity itself or the opposite of the quantity. therefore,

if you try each of the sign combinations (pos/neg) of the absolute values in the problem, you'll be

guaranteed to find all of the solutions.

(note: in what follows, "+" means leaving the expression within the absolute value bars alone; "–

" means reversing the sign of that expression)

in this equation, there are ostensibly four sign combinations, +/+, +/–, –/+, –/–, but it's only

necessary to try two of them:

** first, either +/+ or –/–, in which both or neither of the absolute value expressions are flipped.

may as well go with +/+ (i.e., leaving both of the absolute value expressions alone while

removing the bars): x + 1 = 2(x – 1), or x = 3. plugging this back into the original equation shows

that it works.

** second, either +/– or –/+, in which one of the absolute value expressions is flipped. let's go (at

random) with flipping the first one: –x – 1 = 2(x – 1), or x = 1/3. plugging this into the original

equation also shows that it works.

therefore, statement 1 means that x = 3 or x = 1/3.

statement (2)

two ways to interpret absolute value inequalities like this one:

** memorize the template of the solution (preferred for efficiency's sake): you should just know

that |expression| > a means "either expression > a or expression < –a".

** conceptualize absolute value as distance: in this case, |x – 3| means the distance

between x and 3. therefore, this statement means that the distance between x and 3 is greater than

0 (in either direction).

either of these interpretations means that x < 3 or x > 3, or, equivalently, x is not equal to 3.

statement 1 is insufficient, because 1/3 gives a "yes" and 3 gives a "no". statement 2 is also

insufficient, because every number except 3 is possible. taken together, though, the two

statements are sufficient because they yield a unique value, 1/3, for x.

notice that there's no reason even to figure out whether 1/3 gives "yes" or "no" at this point;

it's one value, meaning that it is guaranteed to be sufficient no matter what the answer.

answer = c

44.

I is definitely true.

For II. z > y > x. Consider, x=1/4, y=1/3, z=1/2. This satisfies given primary inequality. Hence,

II also holds true.

For III. Consider x=1, y=1/3,z=1/2. This satisfies given primary inequality. Hence, III also holds

true.

Hence, E.

basically, the idea is that fractions (i.e., numbers between 0 and 1) "act funky" when they're

raised to powers.

so do negatives.

therefore, when you pick numbers, you MUST consider these sorts of numbers!

you can take 3 cases,

1. When X,Y,Z are positive integers in that case:

I. X > Y> Z holds true

2. When X, Y, Z are fractions:

Z > Y > X

1/2 > 1/3 > 1/4

X>Y^2>Z^4

1/4 > 1/9 > 1/16

3. X > Z > Y (Negative, positive)

20 > –3 > –4

X>Y^2>Z^4

20 > 16 > 9

Therefore answer is E.

45.

Take –4,–3,–2–,–1 as the 4 consecutive integers. The answer is A

the numbers can be –4,–3,–2,–1 as well but do we know that those are the numbers?

–2,–1,0,1.. then 0 > –1, however, if the numbers were 1,2,3,4 then 3<8.

If p, q, r, and s are consecutive integers, with p < q < r < s and pq < rs;then r>0.5 and cannot be

0. In fact pr<qs making statement 1 sufficient.

Statement 2 give no information other than the universal truth 2>0.

The correct answer is indeed A.

46.

Statement 1 is sufficient.

Stmt 2 is y < 1 , since y is an integer , it can be 0 only if it is less than 1.

So Stmt 2 is sufficient too.

Both statements are sufficient enough to deduce if y = 0 .

Answer is D

47.

M – 3Z > 0

–M + 4Z > 0

if you add the two equations, M cancels off and you are left with only one Z which is practical :

M–M –3Z +4Z > 0

gives :

Z>0

NOTE : obviously you can do this only if the two inequality signs face in the same direction!

So we now Z is positive. Now consider "M – 3Z > 0". We can conclude M is positive

so M + Z ===> positive + positive so is positive

Thus C.

48.

OA is D.

believe it or not, it turns out that JUST THE PROBLEM STATEMENT IS ALREADY

SUFFICIENT

on this problem!

in other words, this problem is already "sufficient", even without EITHER of the two

statements!

yes, you read that correctly.

proof:

* "zy < xy < 0" means that z and y have opposite signs, and x and y have opposite signs.

therefore, x and z have the same sign.

furthermore, z must be farther away from zero than x (because the magnitude of zy is greater

than the magnitude of xy).

therefore, there are only 2 possibilities (shown on number line):

y–––––––0–––––––x––––––––z

or

z–––––––x–––––––0–––––––y

now let's turn to the problem statement.

|x – z| is the distance between x and z.

|x| is the distance between 0 and x.

|z| is the distance between 0 and z.

using these interpretations, it's plain that |x – z| + |x| = |z| is ALREADY true for both of these

statements.

neither of statements (1) and (2) is necessary.

technically, there's no answer choice that does this ("the problem statement is already

sufficient"), although it's clear that you should pick (d) should this situation ever arise on the real

exam.

There are different breeds of absolute value problems, so (as usual) there's no one neat, solid

answer to a question like that. however:

* if a problem contains the symbols "> 0" or "< 0" at any point, you can rest assured that the

crux of the problem involves the signs of quantities. (the problem in this thread is a perfect

example.)

if you encounter such a problem, you should immediately devote all of your energy to rephrasing

the question prompt and/or statements to equivalent formulations involving 'positive'/'negative'.

for instance, if you see

zy < xy < 0

you should think:

* z and x have the same sign

* y must have the opposite of whatever sign those two have

* therefore, (x y z) is either (+ – +) or (– + –)

that sort of reasoning will be an excellent start. from there, there's no telling which way the wind

will blow – just study your number properties, and you should be able to figure out the rest.

oh yeah, you should avoid 'solving' if at all possible: you should try to think in the abstract about

the signs of the numbers, and about the situation resulting from each possible combination of

signs. if that sort of reasoning gets you nowhere, then try plugging in numbers and solving as

plan b.

49.

Well Z could be –1 and n could be any even integer. Then the result is 1. So from (1) we can say

that Z is either 1 or –1. So the correct answer should be C.

For z^n = 1, and n being a non zero integer, there are 3 possible ways.

a. 1^1 = 1

b. 1^– 1 = 1

c. – 1^2 = 1

Statement 1 not conclusive. Z could be 1 or –1.

Statement 2: z > 0.

==> we need both statements to solve for z.

Answer: C.

50.

Answer is D

the first thing you should do here is rephrase the question.

big takeaway:

if you see the ABSOLUTE VALUE OF A DIFFERENCE, you should recast it as the

DISTANCE BETWEEN THE TWO THINGS on the number line.

therefore, |y – a| is the distance between y and a, and so on.

hence:

QUESTION: is y closer to a than to b ?

(1) z is closer to a than to b

(2) y is closer to a than z is to b

statement (1):

note that the distance y–a is less than the distance z–a, because y is placed between a and z.

also, note that the distance y–b is greater than the distance z–b, since z lies between y and b.

therefore:

distance y–a < distance z–a < distance z–b < distance y–b (note that these are color–coded to the

statements above)

so, distance y–a < distance y–b

"yes"

SUFFICIENT

statement (2):

same thing as statement (1), except for the second term of the inequality above isn't there

anymore.

i.e., distance y–a < distance z–b < distance y–b.

SUFFICIENT

ans = (d)

OR

First, let's try to clear the absolute values.

Because we know that a < y < z < b, we know

abs(y – a) = y – a

abs(y – b) = b – y (since y – b is negative)

We can rephrase the question:

Is y – a < b – y?

or

Is 2y < a + b?

Statement (1) can be rephrased: z – a < b – z, so 2z < a + b. We also know that since y < z, then

2y < 2z. So 2y < 2z < a + b. (1) is sufficient.

Statement (2) can be rephrased: y – a < b – z, so y + z < a + b. Since y < z, we can add y to both

sides: 2y < y + z. So 2y < y + z < a + b, so (2) is sufficient as well.

51.

Answer: E.

Specifically, rounding rounds ALL 5's up, ALL the time, even if they're followed by nothing at

all.

This is why the problem doesn't contain the word "round": according to traditional rounding, 4.5

rounds to 5

the wording in the actual problem, though, is completely unambiguous: "4 is the integer that is

closest to x + y".

this statement actually rules out BOTH 3.5 and 4.5, because each of those numbers

is equidistant from two integers: the former from 3 and 4, and the latter from 4 and 5.

therefore, here are the CORRECT rephrases:

(1) 3.5 < x + y < 4.5

(2) 0.5 < x – y < 1.5

all four of those signs are strict inequalities. there are no <'s or >'s in this problem.

you can add all 3 corresponding parts of the inequalities directly:

3.5 < x + y < 4.5

0.5 < x – y < 1.5

_____________________

4 < 2x < 6

therefore

2 < x < 3

notice that all this discussion of <'s, <'s, >'s, and >'s is immaterial in the final analysis, because

there are still numbers greater than 2.5 (which are closest to 3) and numbers less than 2.5 (which

are closest to 2). therefore, insufficient even if you misinterpret the question prompt as referring

to "rounding".

but they could, easily, write a problem that would turn on the inclusion/exclusion of a number

such as 4.5.

here's an example:

what number results if the number x is rounded to the nearest hundred?

(1) the multiple of 20 that is closest to x is 140.

(2) x is within ten units of 140.

here, statement (1) means that 130 < x < 150. that's a strict inequality, which doesn't apply to 130

and 150 themselves (since 130 is just as close to 120 as to 140, and 150 is just as close to 160 as

to 140).

all of these numbers give 100 when rounded to the nearest hundred, so this statement is

sufficient.

statement (2), on the other hand, means that 130 < x < 150. this inequality includes 130 and 150.

since 150 rounds to 200, this statement is insufficient.

in this problem, the inclusion vs. exclusion of 150 makes all the difference.

52.

you can ADD TWO INEQUALITIES TOGETHER if the inequalities are BOTH "<" OR

BOTH ">".

(you can't add them if one is "<" and the other is ">". if that's the situation, then you should

multiply one of the inequalities by –1, or just turn it around, so that both of them are either "<" or

">".)

–q>n–p

+

q>p

_________

0>n

therefore, C.

53.

statement (1)

all we know is that z^3 is AN INTEGER. in particular, we can't deduce that z^3 is a

perfect cube.

if z^3 is a PERFECT CUBE, such as 1, 8, or 27, then z will be an integer.

if z^3 is NOT a perfect cube, such as 2, 3, 4, etc., then z will NOT be an integer.

therefore, INSUFFICIENT.

(notice that you can easily find this by PLUGGING IN NUMBERS. in fact, the very first two

positive integers, 1 and 2, give "yes" and "no" respectively, so that's a clear "insufficient".)

if we assume that z^3 is a perfect cube, then we're assuming that z is an integer. if we make that

(totally unfounded) assumption, then we shouldn't be surprised when we find a specious answer

of "yes".

––

statement (2) is insufficient

––

together is actually SUFFICIENT.

* consider all the numbers that satisfy statement (2):

1/3, 2/3, 1, 4/3, 5/3, 2, etc.

* of these, the only ones that satisfy statement (1) as well are 1, 2, 3, ...

(all the fractional ones will still be fractions when you cube them)

* since these – the numbers that satisfy BOTH statements – are all integers, we have

TOGETHER = SUFFICIENT.

answer = (c)

54.

(2) tells us that the magnitude of x is more than the magnitude of y and x and y are either both

positive or both negative.

If x and y are both positive, x has to be more than y… if x and y are both negative, x has to be

less than y (-2 is less than -1).

(1) tells us that x – y = ½… so x has to be more than y…

Combining, x and y are both +ve. Ans. C

55.

if you know that x > y, then you know that x – y is positive, and vice versa.

if you know that x < y, then you know that x – y is negative, and vice versa.

it's not hard to manipulate to get these statements; for instance, merely subtracting y from both

sides of x > y will give x – y > 0.

but that's not the point; the point is to recognize, INSTANTLY, that knowing the status of

the inequalityinvolving x and y (i.e., whether x > y or x < y) is equivalent to knowing the sign of

x – y.

well, the question prompt is:

is (m – k)(x – y) > 0?

based on the considerations above, statement #1 gives us the sign of the expression (m – k), and

statement #2 gives us the sign of the expression (x – y).

if we have both of these signs, then we can figure out the sign of their product, so both

statements together are sufficient.

(note that we don't even have to figure out the actual signs; it's good enough to realize that we

can find them)

so, should be (c)

OR

mx + ky > kx + my

mx – kx > my – ky

(m–k)x > (m – k)y

(m–k)x – (m–k)y > 0

(m–k)(x–y) > 0

ie m > k and x > y

Answer C

56.

"500 is the multiple of 100 that is closest to X"

this means that, of all multiples of 100, 500 comes closest to x.

in other words, 500 is closer to x than is 100, 200, 300, 400, or 600, 700, 800, ...

if you think about this for a sec, you'll realize that it means x has to be strictly between 450 and

550.

Since the numbers don't have to be integers, you have

1. 450 < x < 550 (excluding BOTH endpoints) – note that x could be 450.00001 or 549.99999

2. 350 < y < 450 (again excluding both endpoints)

also, watch your <'s and <'s.

Range of X: 450<X<550

Range of Y: 350<X<450

By 1: Say X=499

Now if Y = 449 then nearest multiple of 100 to X+Y will be: 900

Say X=449 Now if Y = 350 then nearest multiple of 100 to X+Y will be: 800

Similarly you can prove that it is E

57.

Here are the TWO DEADLY ASSUMPTIONS:

1. NEVER assume that numbers are integers,

unless you're told, or can infer, that they are.

2. NEVER assume that numbers are positive,

unless you're told, or can infer, that they are.

There are numbers between –2 and –1, and those numbers are precisely the reason why the

answer to this problem is (e).

58.

IS 1 / p > [r / (r2 + 2)]

St 1. case 1 P=r=2

a=1/p = 0.5 and b=r/(r^2+2) = 1/3 = 0.33

a>b

case 2 p=r=-2

a=1/p = 1/-2 = -0.5 and b= r/(r^2+2) = -1/3 = -0.33

a < b

not suff

St 2 don't know anything about p not sufficient

combine both, as shown in statement 1 whenever p or r > 0, a > b. sufficient

answer C

59.

8/9 + 1/8 = 73/72, which is greater than 1.

therefore, knowing that x + y < 73/72 is insufficient to address the issue of whether x + y < 1,

because x + y could be, say, 1/2 ("yes") or any value between 1 and 73/72 ("no").

therefore, (e).

60.

statement (1):

(a + b)/(a – b) < 0

therefore, a + b and a – b have opposite signs. we can divide this statement into 2 cases.

CASE 1: a + b is positive and a – b is negative

a – b is negative ––> immediately know a < b

also, in this case, a + b > a – b, so therefore b > –b, so therefore b is positive.

that's all we know, though; we know nothing about the sign of a. (note that this case works for (a,

b) = (2, 4) but also (–2, 4))

CASE 2: a + b is negative and a – b is positive

a – b is positive ––> immediately know a > b

in this case, a + b < a – b, so therefore b < –b, so therefore b is negative.

again, that's all we know. (this case works for (a, b) = (2, –4) but also (–2, –4))

this is insufficient, because there's a case in which a < b and a case in which a > b.

statement 2:

obviously insufficient

together:

this has to be CASE 2 above, so therefore a > b.

sufficient.

61.

The first step is rephrase the question. IS my=rx?

Statement 1 states that m/y = x/r

This does not help us to calculate as to whether m/r is equal to x/y. So, MAYBE!

(INSUFFICIENT)

Statement 2 states m+x/r+y = x/y

You can cross multiply rewriting the equation as y(m+x) = x(r+y) –––––> my+yx = rx+yx

Subtracting yx from both sides, the equation then becomes my=rx, which is the rephrase of the

question itself. SUFFICIENT.

B is the answer.

* if you don't know what else to do with a proportion, cross–multiply it.

(1) the reason this is valuable is because there are all sorts of versions of the same proportion that

LOOK different as proportions, but which are shown to be the same when cross–multiplied. for

instance, ALL of the following proportions

a/b = c/d

a/c = b/d

d/b = c/a

d/c = b/a

are equivalent, as all of them multiply to give ad = bc (as do countless others, such as (a + c)/(b +

d) = c/d, after cancellation).

(2) this applies only to proportions with EQUALS SIGNS in them, NOT to inequalities. if

you have an inequality such as a/b < c/d, then you can't cross–multiply it unless you know the

sign of the product of the two denominators, bd (because that's all cross multiplication is:

multiplying by both denominators at once on both sides). if bd is positive, then the sign won't

flip; if bd is negative, then the sign must flip.

62.

(1) is surely enough.

if you have a simultaneous equation and inequality, then solve the equation and then

substitute it into the inequality.

Correct answer is D

2x+5y= 20

Or, y= (20–2x)/5

–2x>3y

Substiture for y

–10x>60–6x

–4x>60

The only way this would be possible is if x is negetive

Hence 2 is sufficient

63.

note that the expression c + d appears in the question prompt. therefore, solve for this expression

in statement 2: c + d = 300 – b.

now, substitute this into the question prompt, and also substitute a + b = 200:

is 200 > 300 – b ?

rephrase by solving ––> is b > 100?

thus, it still comes down to observation that b must be more than 100, because it's the larger one

of two numbers that add to 200 and therefore must be greater than half of 200.

but if you rephrase the question in this way, it's much more clear that you actually have

to think about whether b > 100.

TAKEAWAY:

takeaway: if 2 numbers add up to n, then the larger number is more than n/2, and the

smaller number is less than n/2.

it is given a+b = 200, a < b , is a+b > c+d

rephrase the question

we know a+b = 200, so

is 200 > c+d or is c+d < 200 ? This is a YES/NO question

1. c+d < 200 ( Sufficient )

2. b + c +d = 300 – eq 1

add a to both the sides

a + b + c + d = 300 + a

we know a + b = 200, so

200 + c + d = 300 + a

c + d = a + 100

Now we know that a < b , a was equal to b a would be 100 and b would be 100, thus since a < b,

a < 100

therefor a + 100 < 200 and c + d < 200 –– > Sufficient

Answer D

64.

when you consider a problem like this, in which you are GIVEN INEQUALITIES, you should

always CONSIDER THE EXTREMES of the given inequalities.

this technique is very simplistic, yet very powerful: consider the extremes to find the extremes.

therefore, it's sufficient to think about, say, 0.1 and 0.9 for r, and 1.1 and 1.9 for s.

statement (i): 0.1/1.1, 0.9/1.1, 0.1/1.9, and 0.9/1.9 are all less than 1, so you're good.

statement (ii): works for (0.1)(1.1), (0.9)(1.1), and (0.1)(1.9), but NOT (0.9)(1.9).

statement (iii): only works for 1.1 – 0.9, doesn't work for any of the other pairs.

notice that this method is systematic: you don't just generate numbers at random, you generate

numbers at the extremes of the intervals dictated by the inequality/ies.

I. Any number between 0 & 1 divided by any number between 1 & 2, will always be < 1

II. 2 cases: Consider r = 0.9 and s = 1.5, rs = 1.35. Consider r = 0.1 and s = 1.1, then rs = 0.11, so

not true

III. 2 cases: 1.9 – 0.1 = 1.8 (this is > 1), 1.1 – 0.9 = 0.2 (< 1)

hence only I

65.

statement (1):

there's a statement called the pigeonhole principle, which basically says the following two

things:

* if the AVERAGE of a set of integers is an INTEGER n, then at least one element of the set

is > n.

* if the AVERAGE of a set of integers is a NON–INTEGER n, then at least one element of the

set is > the next integer above n.

this principle is easy to prove: if you assume the contrary, then you get the absurd situation in

which every element of a set is below the average of the set. that is of course impossible.

specifically, statement (1) is a case of the first part of the principle: the average of the set is 6/3 =

2, so at least one element of the set must be 2 or more.

again, you can prove this by reductio ad absurdum: if no one had sold 2 or more tickets, then

you'd have a set in which everyone sold either 0 or 1 ticket, but the average is somehow still 2.

that's untenable.

––

statement (2):

there are only two ways not to sell at least 2 tickets: sell 0 tickets, and sell 1 ticket.

if everyone sells a different # of tickets, then you can't fit three people into these two categories.

therefore, someone must have sold at least 2 tickets.

(1) if they sold 6 together, the possibilities (2,2,2), (1,2,3), (0,3,3) (different variations of these).

In all cases, there is at least one with 2 or more.

(2). This I think is real cool.. if one of them is 0, the other is 1, the third one has to be 2 or more,

hence sufficient.

Hence the answer is D.

66.

we have the equation zt < –3, and it wants to know whether z < 4.

(1) alone: If z < 9, then we still don't know whether z < 4. (For instance, z could be 3 [yes] or 5

[no].)

(2) alone: If t < –4, then we know that z is a positive number (because the product is negative

and t is negative). you can't really divide two inequalities in any simple way, so just try plugging

numbers. let t be, say, –10, and try z's that are greater than 4 and less than 4 (remember, the point

here, as on all DS problems, is to prove 'MAYBE'). make sure that you don't violate the

condition zt < –3.

if t = –10 and z = 1, then the condition zt < –3 is satisfied, and z is not < 4.

if t = –10 and z = 5, then the condition zt < –3 is satisfied, and z < 4.

these two results show that (2) alone is insufficient.

in fact, the same two results show that statements (1) and (2) TOGETHER are insufficient (since

both of the z's selected here happen to be < 9).

answer = e

OR

The answer is E.

It is a YES/NO question whether z < 4

Looking at it 1st statement, for the 1st inequality to be true either z is –ve or t –ve, however both

cannot be negative.

1. z < 9 – Insufficient as z = 8 , t = –9 then zt < –3 , similarly z = 1 and t = –5, zt < –3 thus it is

insufficient to state that z <4

2. t < –4 Insufficient since this only provides us that z > 0, but does not given any indication

whether z > 4

for example t = –6, z = 2 , zt = –12

t = –5, z = 5 zt = –25

Now taking them together: z < 9 and t < –4

If one has to be negative, for the 1st statement to hold true than z is +ve between 1 and 9 and t –

ive –4 to –infinity.

z = 8, t = –5 ; zt = –40

z = 3, t = –5 ; zt = –15

Thus one cannot conclusively state whether z< 4, therefore the correct answer is E.

If you simplify the stem as below:

zt < –3

z < –3/t

and since t < –4 from Statement 2:

if t = –5 ––> z < –3/–5 = 0.6 ––> z < 0.6 is less than 4

if t = –10 ––> z < –3/–10 = 0.3 ––> z < 0.3 which is less than 4

WRONG…

you can't divide by t unless you've ascertained whether t is positive or negative. and moreover, if

t is negative, then you have to switch around the inequality sign ("<" becomes ">").

so if t < –4, then you actually have z > –3/t, not "<".

67.

remember that if c is positive, then the statement is guaranteed to be true (because b is already

greater than a, so adding something positive will keep it that way).

statement (1)

in this case, you know that each of b and c is greater than a, but that's all you know.

if b and c are positive, then this is good enough, because then b + c will be greater than either b

or c (both of which are already greater than a). so that's a Yes.

with negative values, though, you can get a No. if b = –2, c = –3, and a = –4, then b > a and c >

a, but b + c < a. that's a No.

insufficient.

(2) means one of the following: (a) all three are positive, (b) two are negative and one is

positive.

if all 3 are positive, then a fortiori c is positive, so b + c must be greater than a.

but

if b is positive, and a and c are negative, then it's possible that b + c is not greater than a. if you

don't like plugging actual numbers, then consider the idea that c could be a REALLY BIG

NEGATIVE that cancels out the positive–ness of b. for instance, if a = –1, b = 2, and c = –100,

then b > a, but b + c <<<< a.

insufficient.

(together)

in this case, a is the smallest of the 3 numbers.

3 cases:

* all positive: this is a yes, as established before

* a < b < 0 < c: this is also a yes, because c is positive

* a < c < 0 < b:

rearrange the inequality to "is b > a – c ?"

in this case, notice that b is positive and a – c is negative, so this is still a yes.

always yes

sufficient

ans = c

68.

OA is B

the best way to solve this problem is to notice that its subject matter is POSITIVES AND

NEGATIVES. how do you know this? because it deals only with absolute values and inequality

signs – no other numbers or non–absolute values in sight to mess things up.

there is no way to 'quickly solve' this one algebraically, unfortunately. in fact, even at the highest

levels of mathematics, the best (and really the only) way to solve problems like these is case–

wise, considering the different possibilities for + and – one case at a time.

A) y < x

a case for >

x = 2, y = –3

LHS = 5 > RHS = –1

Case for = or <

x = anything, y = 0

LHS = RHS

Insufficient

B ) xy < 0. x and y are on opposite sides of 0 on the number line

| x – y | – distance of x from y

|x| – distance of x from 0

|y| – distance of y from 0

If you imagine a number line

like this

x––––––––0–––––––––––y

or

y––––––––0–––––––––––x

you can conclude that the distance between x and y is greater than the difference between x,0 and

y,0.

OR

Lets take (1)

y<x

y 2 –4

x 4 2

Substitute in the equation

|x–y|>|x|–|y|

Using the above values,

|2|>|4|–|2| First Column Values

Not true

|2–(–4)|>|2|–|–4| Second Column Values

|6|>2–4

6>–2 ––True

Hence A is Insuff

(2)

xy<0 which means either x or y should be negative

x 7 –7

y –3 3

Substitute the values in the eq

|x–y|>|x|–|y|

|10|>|7|–|–3|

10>4

|–7–3|>|–7|–|3|

|–10|>–7–3

10>–10

Hence Suff

Hence B.

69.

REPHRASE

is y <1

statement 1) doesnt tell us anything about Y

statement 2) if y<0 then y must be <1

B is the answer

70.

Ans: E

1) 7x–2y>0

7(8)–2(3) = positive

7(8)–2(–3) = positive

y can be neg or pos. NOT sufficient.

2) –y<x

x = 8 , y can equal 3

x = 8, –3<8, so y can be positive

x = 8, y can equal –3

x=8, –(–3)<8, 6<8, y can be negative too. Not sufficient.

c) combined: Look at the breakdown from statement (1) with numbers that also satisfy statement

(2)'s criteria. y can be neg OR positive.

Answer is Choice E.

71.

1. If that is that case.

i) is insufficient. there is not enough information on the values of x,y,z.

ii) is sufficient. there are 3 possibilities: x<y=z, x=y=z, y=z<x. in any case z is the

median.

Hence the answer is B

If you need convincing about answer a, then remember that your goal on these types of problems

is to try to prove 'maybe' (i.e., insufficient). so try to find 2 groups of numbers, one of which

gives a 'yes' answer and one of which gives a 'no' answer:

(x, y, z) = (1, 3, 5): z is not the median

(x, y, z) = (1, 5, 3): z is the median

insufficient.

important note:

make sure that you understand that this is the direction of the logic in all data sufficiency

questions. you are always trying to prove/disprove the prompt question, based on the evidence

given in the 2 statements. you have written the question backwards – your 'if' and 'then' construct

a logic that runs in precisely the opposite direction – which will make it essentially impossible

for you to answer questions correctly.

––

treatment of the question:

rephrase of initial question:

this is the same as asking: is z equal to the middle number of the three numbers?

statement (1)

this statement tells nothing about the order of the three numbers. it could be true regardless of the

order of the 3 numbers, and, more to the point, regardless of the position of z in the ordered list.

examples:

x = 1, y = 2, z = 3: z is not the median

x = 1, y = 3, z = 2: z is the median

insufficient

statement (2)

if y and z are equal, there are three possibilities:

––– they are the two largest #s in the list. in this case, both of them equal the median of the list.

––– they are the two smallest #s in the list. in this case, both of them equal the median of the list.

––– all three numbers in the list are the same. in this case, all of them equal the median.

in any of these cases, z is the median.

sufficient

answer = b

72.

OA = E

try to draw out the cases. for example

(1) xyz < 0 and |xy| > |xz| . this implies the following are possible (with the Y–axis indicated as

the vertical bar)

yx | z

z | xy

xz | y

yz | x

(2) xy < 0 means that x & y are on opposite sides of the vertical axis. and |xy| > |xz|

y | zx

y | xz

x | zy

the cases illustrate the answer

1. xyz < 0 –– so either all three or one of the three is negative

2. xy < 0 –– either x or y is negative, BUT not both.

Scenario A ) where XYZ < 0.

2 positives 1 negative

3 negatives.

For three negatives on the number the order can be z, x, y, 0...or x, z, y, 0.

So A is insufficient.

Scenario B) hwere XY < 0.

One negative one positive. So the order can go. x 0 z y...or z x 0 y. So B is insufficient.

Scenario A+B.

If X and Y are both of oppositie signs (XY<0)..then Z has to be postive in order for XYZ < 0.

So One Negative and two positives.

Order can be.....x 0 z y...or y 0 x z.

Thus E.

73.

x + y = 1

If y>=0.15, then try some:

If y=0.15, x = 0.85 in which case, no, x is not less than 0.8.

If y=0.3, then x=0.7, in which case, yes, x is less than 0.8.

Contradictory answers = indufficient.

y = 1–x

If C >=7.30, then try some:

If C=7.30, then 7.3 = 6.5x + 8.5(1–x)

7.3 = 6.5x + 8.5 – 8.5x

–1.2 = –2x

1.2/2 = x

x = 0.6, so yes, x < 0.8.

If x+y=1, and I need to create a larger C, I have to make x smaller and y larger (because I

multiply y by 8.5 in the formula while only multiplying x by 6.5). So, x will decreas as C

increases. As a result, I can always say that x < 0.8. Sufficient.

74.

The correct answer is D

(1) if m<p, mv<pv<0 will only hold if V is +ve and M and P are –ve. For other two cases that

you mentioned mv>pv so they are not valid.

Hence, for mv<pv<0 V to hold, V has to +ve. So (1) is sufficient.

Luci, you were on the right track with your process, but you didn't actually work out the

numbers, so you didn't notice that some of them were inconsistent with the given condition

(mv<pv<0). I love trying numbers as a technique, but make sure you follow through just a bit

more on the calculations.

If you had, you would have seen:

ex. m=3, p=5, and let's make v =–2

mv = –6

pv = –10

–6<–10<0

Which is not true, so that combination is invalid – I can't use it to test the statement. The only

way to make it work would be to make p less than m, but statement one gives the condition m<p,

so I can't do it.

75.

E