Abhishek R

Forum Replies Created

Viewing 15 posts - 1 through 15 (of 56 total)
  • Author
    Posts
  • in reply to: Hello everyone #17625
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Welcome to the community John.

    Feel free to post your questions in this community as you work your way through the book.

    in reply to: Management Consulting Case Study Group #17539
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Hi C,

    I prepared for case study interviews myself. Originally I was trying to get into Management Consulting (McKinsey, Bain), then somehow I got into analytics which also required the use of case studies.

    I would suggest you post this question in Management Consulting Facebook groups. You will surely find someone there.

    in reply to: Hello #17538
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Hi Mathew,

    Did your ISP list any reason for the block? I will try to resolve it from my end.

    in reply to: Rounding up 5492 to 8000 “FAILURE” #17537
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Sorry for the late reply. I was on vacation but looks like you figured it out.

    The method in the book is just to round up to the nearest multiple of 10, 100 or 1000.

    In your question 5492 = 8000 – ____, the nearest multiple of 5492 is actually 6000 not 8000.

    So 5492 = 6000 – 508.

    If you want to round up to the next multiple of 1000 which is 7000 then you add 1000 to 508 to get 1508.

    If you want to round up to 8000 then you need to add 2 x 1000 = 2000.

    You can subtract the first digit of both numbers and then subtract the difference by one.

    In 5492 = 8000 – _____

    8 – 5 – 1 = 2

    9 – 4 = 5

    9 – 9 = 0

    10 – 2 = 8

    So the rounded up number is 2508

    5492 = 8000 – 2508

    Hope this provides more clarity.

    in reply to: Getting Through 2020 and beyond… #16411
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Welcome to the community.

    Hope you and your family are safe during these difficult times.

    Are you transitioning to online teaching? You will find that you can make a bigger impact for the same effort when you harness technology and teach online.

    When are you getting yourself a golden retriever? ?

    in reply to: Hello everyone #15657
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Just be careful about how you approach that one Yassine.

    Waking up early not really a habit itself. It is actually the outcome of other good habits like going to bed at the same time everyday and getting enough sleep.

    in reply to: Maths calculation #15387
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    What you are looking for is the XIRR formula. It’s easiest to calculate this in excel.

    You can also check in this online calculator below:
    https://www.calculator.net/investment-calculator.html?ctype=endamount&ctargetamountv=1000000&cstartingprinciplev=200&cyearsv=25&cinterestratev=10&ccontributeamountv=200&ciadditionat1=monthly&printit=0&x=63&y=25

     

    in reply to: Hello everyone #15196
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Welcome to the community Yassine.

    What habits are you looking to develop?

    We are here to help you. ?

    in reply to: Hello everyone #15195
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Welcome to the community Yassine.

    What habits are you looking to develop?

    We are here to help you if you need any help.

    in reply to: Hello Everybody #14983
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Welcome to the community.

    I hope you enjoy reading it. Feel free to ask any questions you have in the community.

    in reply to: Two Digit Multiplier in U T method 6671 x 386 #14390
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    I would recommend you use the Bridge or the Vitruvian man method for 3 digit multipliers to reduce the cognitive load.

    UT Method is recommended for two digit multipliers. However it is still possible to do the UT method with the a 3 digit multiplier.

    Here is the illustration of the Method

    If you want to multiply 6671 × 386 here are the steps.

    Step 1: Add the same number of zeros as the number of digits in the multiplier.

    So the multiplier 386 has 3 digits. So we add three zeroes in front of 6671 to get 000671

    0 0 0 6 6 7 1 x 386

    Step 2: Sum of two pair products is a single digit of the answer.

    Use your middle and forefinger to keep track of whether to calculate U or the T. Place each finger in the middle of two pairs.

    Step 2.1

    Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
    Using 8 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U8T8 = 00 + 00).
    Using 3 in 386, the pair product of 0 and 6 is 0 + 1 = 1 (U3T3 = 00 + 18).
    Adding the three number together we get 0 + 0 + 1 = 1.

    U6 T6
    =   U8 T8
    =          U3 T3
    0    0     0    6     6     7    1 x 386 = 1 _ _ _ _ _ _

    Step 2.2

    Moving to the next set of digits

    Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
    Using 8 in 386, the pair product of 0 and 6 is 0 + 4 = 4 (U8T8 = 00 + 48).
    Using 3 in 386, the pair product of 6 and 6 is 8 + 1 = 9 (U3T3 = 18 + 18).
    Adding the three number together we get 0 + 4 + 9 = 13.

    We attach 3 in 13 as the next digit. But 13 has two digits so we carry over the 1 in 13. So the last digit from the previous step which is 1 becomes 2 (got by carrying over 1+1).

    =   U6 T6
    =         U8 T8
    =                 U3 T3
    0    0     0    6     6     7    1 x 386 = 2 3 _ _ _ _ _

    Step 2.3

    Moving to the next set of digits

    Using 6 in 386, the pair product of 0 and 6 is 0 + 3 = 3 (U6T6 = 00 + 36).
    Using 8 in 386, the pair product of 6 and 6 is 8 + 4 = 12 (U8T8 = 48 + 48).
    Using 3 in 386, the pair product of 6 and 7 is 8 + 2 = 10 (U3T3 = 18 + 21).
    Adding the three number together we get 3 + 12 +10 = 25.

    We attach 5 in 25 as the next digit. But 25 has two digits, so we carry over the 2 in 25. So the last digit from the previous step which is 3 becomes 5 (got by carrying over 3 + 2).

    =          U6 T6
    =                 U8 T8
    =                        U3 T3
    0    0     0    6     6     7    1 x 386 = 2 5 5 _ _ _ _

    Step 2.4

    Moving to the next set of digits

    Using 6 in 386, the pair product of 6 and 6 is 6 + 3 = 9 (U6T6 = 36 + 36).
    Using 8 in 386, the pair product of 6 and 7 is 8 + 5 = 13 (U8T8 = 48 + 56).
    Using 3 in 386, the pair product of 7 and 1 is 1 + 0 = 1 (U3T3 = 21 + 03).
    Adding the three number together we get 9 + 13 +1 = 23.

    We attach 3 in 23 as the next digit. But 23 has two digits, so we carry over the 2 in 23. So the last digit from the previous step which is 5 becomes 7 (got by carrying over 5 + 2).

    =                 U6 T6
    =                       U8 T8
    =                              U3 T3
    0    0     0    6     6     7    1 x 386 = 2 5 7 3 _ _ _

    Step 2.5

    Moving to the next set of digits

    Using 6 in 386, the pair product of 6 and 7 is 6 + 4 = 10 (U6T6 = 36 + 42).
    Using 8 in 386, the pair product of 7 and 1 is 6 + 0 = 6 (U8T8 = 56 + 08).
    Using 3 in 386, the pair product of 1 and _ is 3 + _ = 3 (U3T3 = 03 + –-).
    Adding the three number together we get 10 + 6 + 3 = 19.

    We attach 9 in 19 as the next digit. But 19 has two digits, so we carry over the 1 in 19. So the last digit from the previous step which is 3 becomes 4 (got by carrying over 3 + 1).

    =                      U6 T6
    =                            U8 T8
    =                                  U3 T3
    0    0     0    6     6     7    1           x     386 = 2 5 7 4 9 _ _

    Step 2.6

    Moving to the next set of digits

    Using 6 in 386, the pair product of 7 and 1 is 2 + 0 = 2 (U6T6 = 42 + 06).
    Using 8 in 386, the pair product of 1 and _ is 8 + _ = 8 (U8T8 = 08 + –-).
    Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –-).
    Adding the three number together we get 2 + 8 + _ = 10.

    We attach 0 in 10 as the next digit. But 10 has two digits, so we carry over the 1 in 10. So the last digit from the previous step which is 49 becomes 50 (got by carrying over 49 + 1).

    =                              U6 T6
    =                                   U8 T8
    =                                             U3 T3
    0    0     0    6     6     7    1                x     386 = 2 5 7 5 0 0 _

    Step 2.7

    Moving to the next set of digits

    Using 6 in 386, the pair product of 1 and _ is 6 + _ = 6 (U6T6 = 06 + -).
    Using 8 in 386, the pair product of _ and _ is _ + _ = _ (U8T8 = –– + –-).
    Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –-).
    Adding the three number together we get 6 + _ + _ = 6.

    6 becomes the last digit of the answer. .

    =                                  U6 T6
    =                                            U8 T8
    =                                                   U3 T3
    0    0     0    6     6     7    1                   x     386 = 2 5 7 5 0 0 6

    This is your final answer.

    Hope this clarifies.

    in reply to: DD Method vs DS Method #12796
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Both DS and DD methods are simple checks to quickly verify your answer without having to do the calculation again.

    The only way to be 100% sure that you have the right answer is to do the calculation again.

    The only time that the DS method will fail is if the Digit Sum of the incorrect answer matches the Digit Sum of the Right answer.

    7 x 6 = 42 (DS is 6).

    So 15 x 16 should have DS 6 on both sides.

    However if you calculate incorrectly and the incorrect answer also has then digit sum 6 then the DS method will fail.

    The probability of that happening is approximately 10% of the time. It is exactly around 11.11% of the time. The remaining 88.88% of the time (approximately 90%) it should correctly identify the mistake.

    How do we calculate this probability?

    The incorrect answers digit sum can be between 1 to 9. The digit sum will identify the error 8 out of 9 times. But 1 out of 9 times it will match with the correct answer digit sum.

    So 8/9 = 88.88% (approximately 90%) the DS method will identify the error. But 1/9 times it will fail.

    In the example 15 x 16, the DS method will fail if the incorrect answer has DS 6. But it won’t fail if the answer has DS between 1 to 5 and 7 to 9.

    DD method has a similar error rate. However since the methods use different computations usually what fails in one method should be caught in the other method. However there are going to be exceptions where both methods fail.

    Both these methods are meant to be simple quick checks and nothing is going to replace doing the calculation again which is the only way I know of to get 100% accuracy.

    Hope this clarifies.

    in reply to: I am Qian Hong, nice to meet you!!! #12753
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Ah I see. Most of this seem to be arithmetic operations which are covered in the material.

    Let me know how it goes.

    in reply to: I am Qian Hong, nice to meet you!!! #12739
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    What kind of mental math do you do in school?

    The reason I am asking is to know whether the book and the course has everything that you do in school.

    Will include additional content based on your feedback.

    in reply to: I am Qian Hong, nice to meet you!!! #12737
    Abhishek R
    Keymaster
    • Topics: 0
    • Replies: 56

    Wow QianHong.

    Where are you from?

    Would definitely love to hear more about the things that are tested in eagle flight team club.

    Please share your techniques here with the community.

Viewing 15 posts - 1 through 15 (of 56 total)