MathOratory - Quant Notes - Set 4


  • IIM Lucknow | MathOratory


    If 10 persons have to sit around a rectangular table such that there are 2 each on shorter side and three each on longer side . In how many ways they can sit?

    First try to understand why is the number of arrangements of 10 people in a circular table is 9! and not 10!
    Before anyone sits, each seat in the table is symmetrical to each other.
    So, the first person, wherever he/she sits, it's the same thing. Thus, the first person can be seated in only 1 way. Now, once this person is seated, all the seats become unsymmetrical w.r.t. this person.
    Hence, the remaining 9 can be seated in 9! ways.

    In this problem, there is a rectangular table having three seats one the longer sides and two on the shorter sides.
    First draw the rectangle and mark seat numbers from 1 to 10, clockwise or counter-clockwise. Once you're done, you'll find that seat number 1 and seat number 6 are symmetrical to each other. Seat number 2 and seat number 7 are symmetrical to each other, and so on till the seats pair 5,10. Just rotate the rectangle by 180 degrees and you'll find the symmetrical pair of seats exchanging places.
    Thus, there are 5 pairs of symmetrical seats. But, a seat of any given pair is unsymmetrical to a seat of any other pair. For example, 1&2 are unsymmetrical, 1&3 are, etc..
    Therefore, here the first person can be seated in 5 different ways (i.e. one of the 5 unsymmetrical seats). Once the first person is seated, now again all seats become unsymmetrical. So, the remaining 9'people can be now seated in 9! ways.

    So, total ways = 5*9!

    A test consists of 4 sections. Each section has a maximum of 45 marks. What is the number of ways in which a student can qualify in the test, if the qualifying mark is 90?

    It's a good problem which you can simplify in the following manner.
    Let's first give the max 45 marks to each subject, we have distributed 180 marks now the marks that we take back should not exceed 90 or the cutoff is not cleared.

    Hence a+b+c+d < = 90
    Convert it to a+b+c+d+k = 90, 'k' being a slack variable
    Number of solutions of the above is (90+5-1)c(5-1) = 94c4

    Now, if we take back more than 45 from any particular subject, we get a negative marks, which is not possible.
    So scenarios like a=46+x, where x>=0, are not possible
    Let's plug this in
    46 +x+b+c+d+k = 90
    Or, x+b+c+d+k = 44
    Number of solutions of the above is (44+5-1)c(5-1) = 48c4

    Taking 46 and above back from each of the four variables is not possible, but taking back even a full 90 back from 'k' is possible, because that's just a slack variable. So, there will be 4 cases as above.
    Now, when we are taking back 90, we can't take back more than 45 from two variables, so our final answer should be:

    If the product of four consecutive natural number increased by a natural number p, is a perfect square, then find the value of p?

    I always like to take values/variables which simplifies calculations.
    So let us take the numbers as

    a-1.5, a-0.5, a+0.5, a+1.5
    (Don't think the use of decimals is going to slow you down, we are at most going to add some decimals, not multiplying them)

    Now, Multiplying the four numbers we get

    (a^2 - 0.25)(a^2 - 2.25) = a^4 - 2.5 * a^2 + 2.25 * 0.25

    Trying to make a perfect square let's consider the middle term 2.5*a^2 which should be of the form 2xy where x is a^2

    Hence, y must be 2.5/2=1.25

    Now, we need to make 1.25^2 by adding something to the last term, in order to have a perfect square

    The last term is 2.25 * 0.25=(1.25+1) * (1.25-1)=1.25^2 - 1

    Hence, to make this product 1.25^2, we must add 1

    P=1

    The sum of two numbers is 144 and their LCM is 420. What are the numbers?

    There's a famous process in numbers, which I'll try to simplify in the following manner:
    Let us start with an example, 48 = 12 * 4, 36 = 12 * 3; notice here that 12 being their HCF, the 4 and 3 that are left out are co-prime to each other.

    Also, let us add these two numbers: we must get 12 * (3+4), which also has the HCF 12. Further, the extra 3+4 should be co-prime to both 3&4.
    So, the sum and LCM must also have the HCF as 12

    Now, let us take our example. The sum is 144 and the LCM is 420.
    HCF = HCF (144,420) = 12
    If we divide the LCM by the HCF, we should get the product of the co-prime parts
    420/12=35
    Which can be either 35 * 1 or 5 * 7

    Recall that the numbers will be HCF * (one co-prime part)
    So, they can be either 12 * 1 & 12 * 35
    Or, in 12 * 5 & 12 * 7

    In the second case, we have the sum as 144; so the numbers should be 60 & 84

    A = {3,4,5,13,15}
    B = {6,12,14,16,17,18,22}
    C = {1,2,7,8,9,10,11}
    D = {19,20,21,23,24,25}
    In how many ways can we select ONE integer from EACH set, such that their sum is a multiple of 7?

    If you observe the set B, then you'd find that it has all possible remainders under 7, occurring exactly once. So, basically the trick is:

    If you select any element from sets A, C, D; then you will have EXACTLY one element from set B to make the sum a multiple of 7

    so basically select any element from A, C, D in 5 * 7 * 6 ways and correspondingly 1 element for B

    Total of 5 * 7 * 6 * 1 = 210 cases

    Find the remainder when 25^82 is divided by 27

    25 divided by 27, we can say the remainder is (-2)
    So 25^82 is equivalent to (-2)^82 = 2^82
    Now, Euler of 27 is 18... So 2^18 when divided by 27 leaves a remainder of 1
    And as 2^18 = (2^9)²... Thus 2^9 when divided by 27 must leave a remainder of +1 or -1
    If u check 2^9 when divided by 27 leaves a remainder of -1
    Thus 2^82 = 2 * (2^9)^9 = 2 * (-1)^9 = -2... Hence 25

    A is twice faster than B and together they finish a piece of work in 36 days. If A,B,C working together take 12 days to finish a work, in how many days will be taken by C alone to finish the work?

    Method 1
    efficiency ratio of A and B is 2:1
    Also time taken by A,B and time taken by A, B, C are 36 and 12 respectively
    So efficiency ratio of
    (A and B ) : (A, B, C) = 1:3
    Thus efficiency ratio of
    (A, B ) : C = 1: 2
    Thus C alone must take half the time as taken by A and B together, i. e. 18 days

    Method 2
    Let us say,
    Efficiency of A and B = 2 and 1 respectively
    So when working together net efficiency = 3
    As they both complete the total work in 36 days, thus total work = 36 * 3
    This can be completed by A, B, C in 12 days thus their combined efficiency = (36 * 3)/12 = 9
    But, out of this 9, the combined eff of A, B is 3.. So the eff of C = 6
    Thus C alone can finish the work in 36 * 3/6 = 18 days

    I will select 4 integers at random from 1 to 50. I will also construct lists of 20 numbers. How many such lists do I need in order to ensure that my 4 random integers will be in one of my lists?

    The number of ways you can select a list of 20 numbers from 50 will be 50c20
    Out of these, the number of lists having the 4 chosen numbers will be 46c16
    Let us first remove all of these 50c20 - 46c16. If we select these many sets then we still might not get all 4.
    So, 1 more has to be selected to ensure that we have all 4 numbers
    Answer: 50c20 - 46c16 + 1

    Find the sum of all four digit numbers that can be obtain by using the digits 2,3,5 and 7?

    In these questions, work the sum out by each place value separately.
    If we take 2 in the units place, the remaining places can be filled in 4 * 4 * 4 = 64 ways. I'm considering that repetition of digits in the number is allowed, as nothing has been mentioned, exclusively.
    Similarly, if I take any specific digit in the units place, it can be done in 64 ways. Thus, when we all the numbers, there will be 64 2's, 64 3's, 64 5's and 64 7's in the units place which when added will give us 64*(2+3+5+7).

    Similarly, fixing the tens place with a particular digit, we'll again have 64 possibilities each. But as the place value of these digits are 10, when we add the numbers, these digits affect the sum by 64*(2+3+5+7)*10

    Extending the same logic and using place values we can say that the sum of all these numbers should be
    64 * (2+3+5+7) * (1+10+100+1000)
    = 64 * 17 * 1111= 1208768

    What is the summation of (n^2+1) n! Upto k terms?

    I'll tell you a small trick about factorial summations: try to represent the given identity in a form where the coefficients of as many factorial terms we have are all CONSTANTS (i.e. The coefficient is independent of n)

    Firstly, you must know the following conversion
    n * n! = [(n+1)-1] * n! = (n+1) * n! - 1 * n! = (n+1)! - n! .... See we converted a factorial identity which had a coefficient of 'n' to one which is independent of it.

    Now, (n^2 +1) * n! = (n^2 + 2n +1) * n! - 2n * n!
    = (n+1)^2 * n! - 2 * [(n+1)!-n!]
    = (n+1) * (n+1)! - 2(n+1)! + 2n! ..... Multiplying one of the (n+1) with n!
    = (n+2)! - (n+1)! - 2(n+1)! + 2n!
    = (n+2)! - 3(n+1)! + 2n!

    I'll share another trick with you. Check the coefficients of the consecutive factorials, the positive coefficients are equal in magnitude with the negative one. This ensures that there WILL be cancellations when we add the terms.


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