MUST HAVE 15 GMAT full-length tests with video explanations, rigorous analytics, 200+ conceptual videos, and a set of 12 sentence correction e-books. $50!Know More
Dedicated Thread for the Experts' Students - Experts' Global
Announcement Announcement Module
Collapse
No announcement yet.
Dedicated Thread for the Experts' Students Page Title Module
Move Remove Collapse
This is a sticky topic.
X
X
Conversation Detail Module
Collapse
  • Filter
  • Time
  • Show
Clear All
new posts

  • #91
    ​​Sumer's Doubts with solutions...

    P1:

    g^2<1

    -1

    I know that the sign changes in inequalities when we multiply with –ve sign but if we solve the

    above problem as

    g^2-1<0

    (g+1)(g-1)<0

    g<-1 and g<1

    Please explain where i’m making a mistake.


    (g+1)(g-1)<0 means -1
    The concept has been explained in great detail in the Algebra video; please watch it again.


    P2:

    Q-13, Pg-34, Math-Stage I

    P is a multiple of 7 then it should give a different answer every time but answer is A

    You're right; we need to get this corrected in the material. We've made a note of this; thanks for bringing this to our notice.


    P3: from Number System I introduction question

    83^3743^742

    I think answer should be 9 because 3 has a cyclicity of 4 and 3743/4 = 3 and 3^3= 7 and 7^742

    gives 2 as remainder and answer should be 9. But right answer is 3 please explain.

    The mistake you're committing is in dividing just 3743 by 4 and not 3743^742 by 4. Think, 3%4 = 3 but 3^2%4=1; the power matters
    While finding the remainder, divide the entire power by 4
    (3743^742)%4
    =(3740+3)^742%4
    =3^742%4
    = 9^371%4
    =1^171%4 = 1
    Now, 3^1 = 3 remainder.
    Don't worry much if you don;t get it; this just a teaser, beyond the GMAT levels.


    P4:

    Pg-61,62, math stage I

    Q- Maximum/Minimum

    When we try to solve the below question with similar approach it does not work

    Q-In an office where working in at least one department is mandatory, 78% of the employees are

    in operations, 69% are in finance and 87% are in HR. What are the maximum percentage of

    employees that could have been working in all three departments?

    If I take 69% as the max then the total becomes more than 100% please explain.

    Please watch the video carefully; the maximum is simply the smallest of all values; the answer to this question is simple 69% as that's the lowest of the three values.









    MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

    GMAT Preparation: The Most 'Complete' Program Ever

    GMAT Classroom Program: Maxximus Teaches in Noida!

    Comment


    • #92
      Roseena's Doubt...


      1. Is x>0?
      a) 1/x<1
      b) IxI is not greater than x

      Statement a is clear. Not sufficient.

      I did not understand statement 2, the explanation given in the book is not clear.

      Regards

      ================================================== =========================================


      Explanation:

      In all such questions, try to get a 'yes' as well as 'no' to prove each statement insufficient.

      Is x > o?
      |x| is not greater than x
      When x=1, |x| is not greater than x and we get 'Yes', x>0
      When x=0, |x| is not greater than x and we get 'No' for x>0

      Yes and No, both are possible and hence, Statement 2 is insufficient.

      On combining, the exception of x=0 can be eliminated and we shall get a consistent 'Yes'.

      Answer: C
      MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

      GMAT Preparation: The Most 'Complete' Program Ever

      GMAT Classroom Program: Maxximus Teaches in Noida!

      Comment


      • #93
        Roseena's Doubt...

        Question:
        There are 10 stations on a certain railway line. How
        many different kinds of tickets of class llnd must be
        printed in order that a passenger may go from one
        station to any other by purchasing ticket?

        Solution:


        Of the 10 stations, if we find how many 'pairs' can be made, we will have a route (or tickets) between all two stations.

        Ans: 10C2 = 45

        MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

        GMAT Preparation: The Most 'Complete' Program Ever

        GMAT Classroom Program: Maxximus Teaches in Noida!

        Comment


        • #94
          Rachita's Query...

          Please help me with this sentence:

          Sentence

          The consultant is looking for a cafe that has comfortable chairs and that provides
          free internet access.

          Query

          Is "that provides free internet access" an independent clause?

          If yes, then why don't we have a comma before and?

          Thanks





          Solution:



          Hi Rachita,

          Avoiding the grammatical jargon in my explanation, I suggest that the sentence says "...a cafe that has X and provides Y"; hence, the two phrases "cafe that has X" and "cafe that provides Y" are parallel and all is fine. The sentence could have done without the second 'that' and the message would have been "...a cafe that...has X and provides Y"- even better.

          Please try looking at sentences through the logical approach (as the one suggested above) rather than a hardcore, technical/rule-book way as you shall always be missing some nuances. The finer detail that you are missing here it will be incorrect to call either clause independent as both have equal weight- the Consultants needs X and Y; both X and Y become important.

          Such concepts have been elucidated in a very logical, easy to understand way in the Mofieirs video; I recommend that you watch the video once more- the concepts do take a while to settle and multiple look at the conceptual matter is often needed.

          All the best!
          MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

          GMAT Preparation: The Most 'Complete' Program Ever

          GMAT Classroom Program: Maxximus Teaches in Noida!

          Comment


          • #95
            Atul's doubts with explanations in line...

            In how many arrangements can a teacher seat 3 girls and 3 boys
            in a row of 6 seats if no two children of the same gender are to
            be adjacent?
            (A) 6
            (B) 12
            (C) 36
            (D) 72
            (E) 144
            Solution according to me :
            Let’s begin with seating 3 boys
            BBB : Now there are four places for girls like this : 1 B 2 B 3 B 4 so total no is 4P3 x 3! = 144 – so choice E
            Jeff solution :
            There are 6 slots to fill so one way :
            1. GBGBGB : - 3x3x2x2x1x1 = 36
            2. BGBGBG:- = 36
            3. Total ways :- 72 so choice D

            Jeff’s solution is correct.
            Problem with your solution: If girls take positions 1,3,4, first two boys are together; many such extra cases have been counted and thus, your answer is going to be much bigger than the correct answer. This concept has been very well explained in the P&C video; please watch the video once more.



            Problem 53 :
            A certain television show has 15 sponsors, including Company
            X, each of which has produced a 30-second advertisement to
            be televised during the show. The first commercial break will
            consist of 4 of these 30-second advertisements, each of which
            will represent a different sponsor. What is the probability that
            Company X.s advertisement will be one of the first two shown
            during the first commercial break?
            (A) 1/225
            (B) 1/30
            (C) 1/15
            (D) 2/15
            (E) 4/15

            My Solution :
            Probability that X gets picked up in first slot = 1/15
            Probability that X gets picked up in second slot = 1/14
            Probability that either one of these happen = 1/15+1/14= 29/210 – No choice matches this
            Jeff Solution :
            Probability that X gets picked up in either of 2 slots = 2/15


            Though the answer shall be 2/15, I am not very happy with Jeff’s simplistic solution; it’s not that straightforward.
            Probability that X gets first slot but not the second slot: 1/15 x 14/14= 1/15
            Probability that X gets second slot- Probability that X doesn’t get first slot x gets second slot=(1-1/15) x 1/14 = 14/15 * 1/14 = 1/15
            The mistake you committed is that, in the second case, you didn’t take the probability of X not getting first slot as only then it can get the second slot.


            Problem no 68 : (DS)
            Each person who attended a certain concert paid the same
            price for a ticket. How many people attended the concert?

            (1) If the ticket price had been $6 more and 100 fewer
            people had attended the concert, the total revenues
            from ticket sales would have been the same.

            (2) If the ticket price had been $6 less and 100 more people
            had attended the concert, the total revenues from ticket
            sales would have been $1200 less.

            My solution :
            Let people / tickets = x , price per ticket = p : looking for the value of x ,
            Statement 1 : (p+6)(x-100)= px after simplifying 100p-6x + 600 = 0 ……INSUFFICIENT
            Statement 2 : (p-6)(x+100)= px-1200 after simplifying 100p-6x+600 = 0 – INSUFFICENT
            Since both the statements are leading to identical equations so together also INSUFFICEINT
            Thus – choice E
            Jeff solution : ( I am copying it verbatim )
            C

            Explanation: Given the actual number of tickets and the ticket price, the
            total revenues are np, where n is the number of tickets and p is the price of each
            ticket.

            Statement (1) tells us that np is equal to (n - 100)(p + 6), which gives us a
            single equation with two variables. That’s not enough to solve for the value of
            n.
            Statement (2) is similar: it tells us that np is $1200 more than (p-6)(n+100),
            another two-variable equation that doesn’t allow us to solve for the value of n.

            Taken together, the statements are sufficient. Given two equations with the
            same two variables, you can solve for each of the variables. Don.t spend the
            time to do it in this case: it would be time-consuming, and that’s not what the
            GMAT is after.


            That’s an inadvertent mistake by Jeff; he assumed that the two equations won’t simplify down to give same equation. Good observation. The answer should be E.



            Problem no 12 – DS challenge

            If p is an integer, then p is divisible by how many positive integers?
            (1) p = 2x, where x is a prime number.
            (2) p = x2, where x is a prime number.
            My Solution
            Statement 1 : if x is a prime number ; 2x will have exactly (x+1) factors including 1 , thus – SUFFICIENT
            Statement 2 : if x is a prime number ; x2 will have exactly three factor ( 1 , x and x2) , thus – SUFFICIENT
            Therefore choice D


            That’s a howler you are committing!! The answer cannot be in form of ‘x’, it has to be a unique, arithmetic answer. Something clearly explained multiple times in the initial Math videos.



            Jeff’s solution ( verbatim )
            B
            Explanation: Statement (1) is insufficient: 2x could be a wide range of
            numbers with very different numbers of factors. To take just two examples: if
            x = 2, 2x = 4, which has three factors (1, 2, and 4). If x = 3, 2x = 8, which has
            four factors (1, 2, 4, and 8). Statement (2) is sufficient: while p could be a very
            wide range of numbers, the square of a prime always has the same number of
            factors. three. Those factors are one, the prime number, and the square of the
            prime. That fact isn’t extensively tested, but it’s very handy to know should it
            come up.
            Problem No 16 – DS
            If y is an integer such that 2 < y < 100 and if y is also the square of
            an integer, what is the value of y?
            (1) y has exactly two prime factors.
            (2) y is even.
            My Solution :
            Statement 1 : y can be 4 , 9 , 25 etc …in fact the square of any prime number from 2 – 9 because
            4 : 2<4<100
            4=22
            4=2x2 ( exactly two primes )


            That’s again a howler; 2 and 2 don’t form two prime factors it’s still only one prime factor; 2 and some other prime factor- say 3, will mean two prime factors.



            9 : 2<9<100
            9=32
            9 = 3 x 3 ( exactly two primes ) INSUFFICENT
            Statement 2 : y can be all the “even square number “ like 4 , 16 , 36 and so on ….INSUFFICIENT

            Statement 1&2 : Only 4 satisfies all the conditions
            4: 2<4<100
            4 22
            4 2x2 ( exactly two primes)
            4 is even – SUFFICIENT
            Thus choice - D
            Jeff’s solution ( verbatim)
            A
            Explanation: Given the parameters of the question, there are a limited num-
            ber of possible values for y: any square of an integer less than 10. For instance,
            y could be 1( ??? how Y can be 1 when y is between 2 and 100 ) , 4, 9, 16, etc. Statement (1) limits the possibilities even further: if y has exactly two prime factors, that eliminates 1, 4, 9, 16, 25, 49, 64, and 81,
            leaving only 36. Recognize that if a square has exactly two prime factors, its
            square root has the same two prime factors: it may be faster to check numbers
            1 through 9 than squares 1 through 81. Either way, you.re left with only one
            possible value, and (1) is sufficient
            On its own, Statement (2) is insufficient : y could be any even square, such
            as 4, 16, or 36.
            MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

            GMAT Preparation: The Most 'Complete' Program Ever

            GMAT Classroom Program: Maxximus Teaches in Noida!

            Comment


            • #96
              Khushbu's doubt in Geometry - Quant..
              Doubt: See the attachment.
              =====================
              Solution:
              Dear Khushbu,

              We don't really need AE to be perpendicular to BD. All we need is the height of the triangle AEF. See the entire solution the following way.
              We know that BD = 5, Pyragorus theorem.
              Let's now see the triangle ADB. This is half the area of ABCD. So, area of ADB = 1/2 * 3 * 4 = 6.

              In both triangles - ADB and AEF, the height is the same.... so let's say that height is h.
              ADB ==> 1/2 * h * BD = 6 ==> h = 12/5....Note that this may or may not be AE. And it doesn't matter

              So, AEF ==> 1/2 * h * EF = 12/ * 12/5 * 2 = 12/5.

              So, the method is the same and your doubt is correct, but we don't need AE to be a perpendicular to BD.
              =====================
              Attached Files
              Last edited by GMATMentor; 05-30-2016, 09:55 AM.
              MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

              GMAT Preparation: The Most 'Complete' GMAT Program Ever!

              ISB Admissions Consulting: 90% Interview Invitations since 2010!

              Comment


              • #97
                An Interesting Probability Q
                ===============

                A dog breeder currently has 9 breeding dogs. 6 of the dogs have exactly 1 littermate, and 3 of the dogs have exactly 2 littermates. If 2 dogs are selected at random, what is the probability that both selected dogs are NOT littermates?

                =============== The simplest way to think of it is as follows.

                9 breeding dongs - 6 have one littermate and 3 have 2 littermates.
                6 have one littermate - Means there are groups of 2 littermates. There are 6/2 = 3 such groups. i.e. (1,2), (3,4), (5,6). Verify that these are 6 dogs with one littermate each.
                3 have two littermate - Means there are groups of 3 littermates. There are 3/3 = 1 such groups. i.e. (7,8,9) . Verify that these are 3 dogs with two littermates each.
                So, the 9 dogs are split into - (1,2), (3,4),(5,6),(7,8,9). Let's call them Group 1 through Group 4.

                P(Not littermates) = P(We select two different groups).

                Group 1 and Group 2 = 2*2 = 4 selections
                Group 1 and Group 3 = 2*2 = 4 selections
                Group 1 and Group 4 = 2*3 = 6 selections
                Group 2 and Group 3 = 2*2 = 4 selections
                Group 2 and Group 4 = 2*3 = 6 selections
                Group 3 and Group 4 = 2*3 = 6 selections
                Total = 4+4+6+4+6+6 = 30


                Total number of ways in which 2 dogs are selected from 9 dogs = 9C2 = 36

                Required probability = 30/36 = 5/6.


                There are other ways of solving the same. We could do 1-P(both dogs belong to the same group).
                No of ways in which both dogs belong to the same group = Both belong to Group 1 + Both belong to Group 2 + Both belong to Group 3 + Both belong to Group 4 = 1 + 1 + 1 + 3 = 6
                Total number of ways in which 2 dogs are selected from 9 dogs = 9C2 = 36
                P(both dogs belong to the same group) = 6/36 = 1/6.
                Required probability = 1 - 1/6 = 5/6.

                ===============

                MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                ISB Admissions Consulting: 90% Interview Invitations since 2010!

                Comment


                • #98
                  Sonali's doubt
                  ================================================== ====================
                  I got the below doubt in Alligation:

                  Two ores, A and B, contained Iron and Copper in ratio 3:1 and 1:4. When the two were mixed, resultant ore had Iron and Copper in ratio 1:1. If the resultant ore had 110 kg Copper, what quantity of ore A was used?
                  ================================================== ====================
                  Solution:

                  There are two methods to solve this problem. Let's see the Algebra way first.

                  Method 1: Algebra
                  Ore A: Iron to Copper in 3:1 ratio. Let's say there is 3x Iron and x Copper, in a mass of 4x Ore A.
                  Ore B: Iron to Copper in 1:4 ratio. Let's say there is y Iron and 4y Copper, in a mass of 5y Ore B.
                  Mixing, we have (4x+5y) total mass with (3x+y) Iron and (x+4y) Copper.
                  In the mixture, the Iron:Copper ratio is 1:1.
                  So, 3x+y = x+ 4y ==> x:y=3:2
                  In the mixture, there was (x+4y) Copper. Converting everything to x, there is x+4(2x/3) = 11x/3 Copper. This is given to be 110 kg. So x = 110*(3/11) = 30 kg. So y = 2x/3=20 kg.


                  Let's verify it.
                  Ore A: Iron = 3x = 90 kg and Copper = x = 30 kg
                  Ore B: Iron = y = 20 kg and Copper = 4y = 80 kg
                  Mixing, we have total mass of 220 kg with 110 Iron kg and 110 kg Copper. This ratio is 1:1.


                  So, quantity of ore A is 30+30=120 kg.

                  Method 2: Alligation

                  Quality of copper in the A, B and resultant ore are as follows:

                  1/4 and 4/5

                  1/2

                  The differences will be as follows

                  4/5-1/2 = 3/10 and 1/2-1/4= 1/4

                  (3:10) : (1:4) = 6 : 5 (A : B in resultant ore)


                  Resultant ore contains 110 Kg copper and had Iron:Copper as 1:1. So, total quantity of resultant ore is 220 kg

                  Hence, quantity of A is 220 x 6/11 = 120 kg

                  Let's verify this.
                  We mix A = 120 kg (Iron A = 90 kg, Copper A = 30 kg) and B = 100 kg (Iron B = 20 kg, Copper B = 80 kg)

                  Total Final Quantity = 120 + 100 = 220 kg
                  Total Final Iron = 90 + 20 = 110
                  Total Final Copper = 30 + 80 = 110
                  Iron:Copper Ratio in the Final Quantity = 1:1.


                  ================================================== ====================
                  MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                  GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                  ISB Admissions Consulting: 90% Interview Invitations since 2010!

                  Comment


                  • #99
                    Probability DS and PS Qs by Sonali
                    ======================================
                    Probability DS Questions:

                    Q1)
                    Does the year XXY5 have 53 sundays?

                    * The year XXY4 has 52 Sundays.

                    * The year XXY4 is a leap year.

                    Ans: Is the answer E)? Please explain.



                    -->
                    For a regular year, there are 365 days. 364 of them accommodate in 52 whole weeks i.e. 52 Sundays. It is the last day that determines if there are 53 Sundays.
                    For a leap year, there are 366 days. 364 of them accommodate in 52 whole weeks i.e. 52 Sundays. It is the last 2 days that determine if there are 53 Sundays.

                    For XXY5 to have 53 Sundays, we first need to know whether it is a leap year. It cannot be a leap year as it will not be divisible by 4.
                    So, XXY5 has 365 days and all we need to know is whether its last day fell on a Sunday. (or whether its first day fell on a Sunday).

                    A) XXY4 has 52 Sundays. The means that XXY4's last day was NOT a Sunday (and second last day too was NOT a Sunday, if XXY4 was a leap year). but XXY5's first day could have been or could not have been a Sunday. Insufficient.

                    B) XXY4 is a leap year. This in itself does not tell where the year started or ended. Insufficient.

                    Combining both, XXY4's last two days did NOT have a Sunday but XXY5's first day could have been or could not have been a Sunday. Insufficient.

                    Hence, the answer is E.

                    Q2)
                    Does the year XXY5 have 53 sundays?

                    * The year XXY4 has 53 Sundays.

                    * The year XXY3 has 53 Sundays.


                    Ans: Is the answer B)? Please explain.



                    --> Similar logic.
                    For a regular year, there are 365 days. 364 of them accommodate in 52 whole weeks i.e. 52 Sundays. It is the last day that determines if there are 53 Sundays.
                    For a leap year, there are 366 days. 364 of them accommodate in 52 whole weeks i.e. 52 Sundays. It is the last 2 days that determine if there are 53 Sundays.

                    For XXY5 to have 53 Sundays, we first need to know whether it is a leap year. It cannot be a leap year as it will not be divisible by 4.
                    So, XXY5 has 365 days and all we need to know is whether its last day fell on a Sunday. (or whether its first day fell on a Sunday).

                    A) XXY4 has 53 Sundays. The means that XXY4's last day was a Sunday or second last day was a Sunday, if XXY4 was a leap year. In either case, XXY5's first day could not have been a Sunday. Sufficient to say that XXY5 does NOT have 53 Sundays.

                    B) XXY3 has 53 Sundays. The means that XXY3's last day was a Sunday. XXY3 cannot be a leap year so we will ignore the possibility of second last day being a Sunday.
                    i.,e, XXY4's first day was NOT a Sunday. It was a Monday.
                    Case 1: If XXY4 was a leap year, XXY4's last day was Tuesday. XXY5's first day was Wednesday. XXY5 does NOT have 53 Sundays.
                    Case 2: If XXY4 was not a leap year, XXY4's last day was Monday. XXY5's first day was Tuesday. XXY5 does NOT have 53 Sundays.

                    Sufficient to say that XXY5 does NOT have 53 Sundays.

                    Hence, the answer is D.

                    Probability PS Questions:

                    Q3)

                    3 cards are drawn from a pack of 52 cards; what is the probability

                    i) that all honor cards(J,Q,K,A) if it is known that all cards drawn are red cards?
                    Ans: Is the answer 8C3/52C3?



                    --> How many Red honor cards exists? 8
                    P = 8C3/52C3


                    ii) all honor cards if it is known that no card drawn is a club?
                    Ans: Is the answer 12C3/52C3?



                    --> How many non-Club honor cards exists? 12
                    P = 12C3/52C3
                    MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                    GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                    ISB Admissions Consulting: 90% Interview Invitations since 2010!

                    Comment


                    • An SC doubt from Sneh...
                      ==========================
                      Supercell, a continuously rotating updraft deep within a thunderstorm, is not visible in all thunderstorms because they require very high wind velocity and moisture.
                      A) because they require
                      B) because they will require
                      C) because it will require
                      D) because it requires
                      E) because of requiring

                      The answer given in the book (SC Grail) is- In this sentence the options A & B get eliminated because the plural ‘they’ cannot refer to the singular ‘Supercell’. Option C unnecessary adds the future tense will. Between Options D & E avoid the continuous tense ‘requiring’ in E because it does not specifically refer to an ongoing action and go with the simple present tense in D, the correct answer

                      I eliminated all the other choices correctly but chose E over D because the question says"continuously rotating".I'm not satisfied with the ans given in the book.can you pls explain?

                      ========================== Dear Sneh,
                      "because of requiring" is awkward, hence Option E is to be avoided. For example,
                      I do not watch TV shows because I require silence after a day's work. - This is simple and shows the meaning.
                      I do not watch TV shows because of requiring silence after a day's work. - Who is "requiring?" TV shows require silence or I require silence? This is not clear. Also, is there a need to show continuous tense? Is the "requiring" action happening right now? That is not needed.

                      So, D is the better choice.
                      ==========================
                      MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                      GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                      ISB Admissions Consulting: 90% Interview Invitations since 2010!

                      Comment


                      • Probability Doubt by Rahul
                        ===================================

                        The question says - A bag has 5 Red balls,3 White Balls; if three balls are randomly chosen, what is the probability that all are red ?
                        Ans. According to me it should be 1/8*1/7*1/6 but as per study material it is 5c3/8c3. Request you to please help me understand where is my thinking process going wrong.

                        I am assuming possibility of a red ball from 1st draw = 1/8, possibility of a red ball from 2nd draw = 1/7,possibility of a red ball from 3rd draw = 1/6.

                        ===================================

                        Dear Rahul, Your logic is fine, but the numbers are off. So you have a one-ball-at-a-time approach. Perfect! Let's see that....
                        First ball - 5 Red balls available from 8 total balls = P(first ball is red) = 5/8
                        Second ball - 4 Red balls available from 7 total balls = P(second ball is red) = 4/7
                        Third ball - 3 Red balls available from 6 total balls = P(third ball is red) = 3/6
                        Overall probability = 5/8 * 4/7 * 3/6.
                        This is the same as 5C3/8C3 again

                        Where you went wrong is in thinking that the probability for each ball was 1/8,1/7,1/6 etc.

                        ===================================

                        MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                        GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                        ISB Admissions Consulting: 90% Interview Invitations since 2010!

                        Comment


                        • DS doubt by Udyan
                          =========================
                          I have a small doubt regarding one Practice question (DS) in Numbers - II. I don't know whether I am thinking right in DS.. I am confused whether we need to get same answer everytime in DS so that the option can be termed as sufficient or whether getting any answer is sufficient to.. Please help.
                          =========================

                          You don't "have" to find the answer. It is still recommended that you solve all the way to the end, without getting into calculations.

                          Also, for any statement to be sufficient, we would have to get the "same" answer.
                          e,g. a Q could be "Is X positive?" In this Q,
                          - if we get ONLY negative values all times, we can say that it is sufficient. The answer to "Is X positive?" will be NO. It doesn't matter though.
                          - if we get ONLY positive values all times, we can say that it is sufficient. The answer to "Is X positive?" will be YES. It doesn't matter though.
                          - if we get one negative value of X and another positive value for X, the statement is not sufficient because we cannot be sure whether X is positive or not.

                          =========================
                          MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                          GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                          ISB Admissions Consulting: 90% Interview Invitations since 2010!

                          Comment


                          • A Grammar Doubt by Haresh
                            =====================================
                            Can we always consider " due to" as a wrong choice on SC Questions and prefer "Because of"?


                            Are there any specific scenarios where "Due to" is correct?
                            =====================================
                            Dear Haresh,

                            Basically, there are the following rules.

                            1. Generally, "Because of" is used for verbs while "due to" is used for nouns.
                            2. "Due to" is used in the meaning of "caused by". It is used with "to be" verb ONLY.
                            3. To know the best usage, reverse the sentences and they should fit into the following formats.
                            "Because of" - WHY did something happen? Note that "happen" is a form of action/verb. i.e. "Why" part makes "because of" modify a verb.
                            "Due to" - WHAT was something caused by? Note that "was" is a form of "to be".i.e. "What" part makes "because of" modify a noun.


                            E.g.
                            1.
                            (Because of/Due to) my arrogance, I lost my friendship.
                            What is the cause? My arrogance
                            What is the result? I lost my friendship. = Action.
                            We need to explain an action and there no "to be" form. i.e. we need to use "Why" part. So we will go ahead with "because of".

                            Because of my arrogance, I lost my friendship.
                            WHY did I lose my friendship? Because of my arrogance. CORRECT.

                            2. My loss of friendship is (due to/because of) to my arrogance.
                            What is the cause? My arrogance
                            What is the result? My loss of friendship = Noun
                            We need to explain a noun with a "to be" form. i.e. we need to use "What" part. So we will go ahead with "due to".

                            My loss of friendship is due to my arrogance.
                            WHAT was my loss of friendship caused by? Caused by (Due to) my arrogance. CORRECT.



                            =====================================
                            MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                            GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                            ISB Admissions Consulting: 90% Interview Invitations since 2010!

                            Comment


                            • Regarding DS Option choice D - A query by Nishant
                              ============================
                              What is the side of a square?
                              (I) Diagonal of the square is 10 cm
                              (II) Area of square is 85 cm²

                              My Answer : E
                              Actual Answer : D

                              I feel the answer is wrong as in case of data sufficiency both St (I) and St(II) should give same answer when option D is selected. But here as per St (I) and St(II) sides of square are coming different. Please resolve.

                              ============================
                              Dear Nishant,
                              Your understanding is not correct. Option D does NOT expect Statement I and Statement II to give the same answer. For example, let's see the following simple Q.

                              Question
                              Is x positive?
                              Statement I: x=5
                              Statement II: x=-5

                              Solution
                              Statement I: x=5. Is x positive? Yes. We can answer the Q. Sufficient.
                              Statement I: x=-5. Is x positive? No. We can answer the Q. Sufficient.

                              Since both Statements are sufficient, we go with Option D. Note that both Statement give us different answers, but they both are "Sufficient" which is what the DS question type asks us.
                              ============================
                              MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                              GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                              ISB Admissions Consulting: 90% Interview Invitations since 2010!

                              Comment


                              • A DS Problem by Rahul
                                ================================================== =====================================
                                The average of 5 different numbers is 14. What is the average (arithmetic mean) of the 3 largest numbers?

                                (1) The average (arithmetic mean) of the two smallest numbers is 5.

                                (2) The average (arithmetic mean) of the two smallest numbers is 1/4 of the average of the 3 largest numbers.


                                I feel he ans should be A & not D, because option B maximum takes us to the relation of Avg. b/w 3 large no. & 2 small no. & not to final answer of avg. of 3 largest no.
                                ================================================== =====================================
                                Dear Rahul,
                                The way to solve this Q is as follows.
                                Let's say the numbers are a,b,c,d,e, in increasing order. We know that their average is 14. i.e. (a+b+c+d+e) = 70. We are asked the average of c,d,e. i.e. (c+d+e+)/3
                                Statement I: Average of a,b is 5. So, a+b = 10. i.e (10+c+d+e) = 70. We can find (c+d+e)/3. Sufficient.
                                Statement II: Average (a,b) = 1/4 * average (c,d,e). So, (a+b)/2 = 1/4 * (c+d+e)/3. So.=, 6(a+b) = (c+d+e).
                                Let's say, (a+b) = x. So (c+d+e) = 6x
                                Now, (a+b+c+d+e) = 70 i.e x+6x=70. So x=10. i.e. 6x= 60. i.e. c+d+e=10. We can find (c+d+e)/3. Sufficient.


                                Both statements are sufficient individually. So the answer is D. Please note the part in BLUE, which you missed.
                                ================================================== =====================================
                                MBA Admissions Consulting: Every 10th Indian in US top 50 is our Student!

                                GMAT Preparation: The Most 'Complete' GMAT Program Ever!

                                ISB Admissions Consulting: 90% Interview Invitations since 2010!

                                Comment

                                Working...
                                X

                                Covered by...