Quant Boosters - Soumya Chakraborty - Set 1

  • Number of Questions : 30
    Topic : Quant Mixed Bag
    Solved ? : Yes
    Source : MathOratory

  • Q1) The number of permutation by using all the letter of MONDAY which are not beginning with M and not ending with Y?

  • The constraints are:
    Cannot begin with M
    Cannot end with Y
    The part where one need to be careful is: the word can begin with Y or end with M

    Now, if we start from the first position, we need to take two cases:

    Words starting with Y
    Words not starting with Y
    Case 1. Number of arrangements = 1∗4!∗5=1201∗4!∗5=120

    Explanation: As the word is starting with Y then the first place has 1 choice. As Y has already been positioned in the first place, the last place can take any of the remaining 5 letters and the remaining 4 positions in between can be filled in 4! ways

    Case 2. Number of arrangements = 4 ∗ 4! ∗ 4 = 3844 ∗ 4! ∗ 4 = 384

    Explanation: As the first letter cannot be M and we are not placing Y in the first place, there are 4 options for the first place. We have 5 letters left, out of which one is Y, so the last place has 4 options. The remaining 4 positions can be filled in 4! ways

    Total number of arrangements = 120 + 384 = 504

  • Q2) Each of 25 flags is either Red ,Black or White & has a no.(1-10) painted on it. If 1 flag is picked at random, what is probability-it is either W or has even no. ?
    (1) P that flag is both W & has even number painted is 0
    (2) P that ball is W minus P that flag has even number is 0.2

    a) (1) alone is Sufficient
    b) (2) alone is Sufficient
    c) Both statements Together are Sufficient
    d) Each statements alone is Sufficient
    e) Neither of the statements is Sufficient

  • So we are looking for P(W U E) = P(W) + P(E) - P(W ∩ E), ‘W’ being the event of getting white and ‘E’ being the event of getting even number.
    The first statement tells us that P(W ∩ E) = 0, but as we don't know P(W) and P(E), we can't answer the question
    The second statement tells us that P(W) - P(E) is 0.2, but again that isn't sufficient
    Even after combining the two info, we have P(W) - P(E) and P(W ∩ E) but still we have multiple possibilities for P(W) and P(E) which will result in different values of P(W) and P(E), hence we still can't answer the question.
    So, we can conclude that the given information is not sufficient and additional information is required to answer the question.

  • Q3) How many 3 - digit numbers abc (in base 10) are there such that abc + ab + bc + ca + a + b + c = 29 ?

  • Assuming, that “abc is a 3 digit number” implies ‘a’, ‘b’, ‘c’ are the digits of the number, let us try to factorise the LHS.
    abc + ab + bc + ac + a + b + c can be factorised as soon as we add a ‘1’ to it
    It factorises as (a+1) ∗ (b+1) ∗ (c+1)
    So, we have (a+1) ∗ (b+1) ∗ (c+1) = 30
    As, ‘a’, ‘b’ and ‘c’ are digits
    Each of (a+1), (b+1) and (c+1) have a range [2,10]
    First let us distribute 30 = 2 ∗ 3 ∗ 5 without any constraint over the 3 variables.
    This can be done in 3 ∗ 3 ∗ 3 ways, as each prime can be distributed to the 3 variables.
    But now those cases where any of those liberals takes up 15 or 30, must be removed. That's 3∗3 cases each, as there are 3 ways of distributing the ‘15’ and then 3 corresponding ways of distributing the ‘2’.
    We must also take off those cases where any of the literal takes 1. But the (1,1,30) cases have already been taken care of. So the remaining cases are 2^3/2 = 4
    So, if we count the allowable cases, it's 27 − 9 − 4 = 14 possibilities

  • Q4) How many pairs of 2 numbers are there whose LCM is 400?

  • Let us generalise this process. First of all we will try to find the ordered pairs of numbers. Then we will remove the order from it.
    LCM (a,b) = 400 = 2^4 ∗ 5^2
    Let us think from the perspective of each prime.
    First of all, both ‘a’ and ‘b’ must be factors of 2^4
    So we have 5 possibilities of both ‘a’ and ‘b’. Hence 5^2 possibilities.
    But from these possibilities we must remove those cases where ‘a’ and ‘b’ are neither 2^4 . This will happen when both ‘a’ and ‘b’ are factors of 2^3
    So that is 4^2 cases that must be removed.
    So, we have (5^2−4^2) cases where the LCM will be 2^4
    Similarly, for 5^2, we will have (3^2−2^2) cases where the LCM of ‘a’ and ‘b’ will be 5^2
    So, we have a total of (5^2 − 4^2) ∗ (3^2−2^2) = 45 cases where LCM (a,b) = 400
    These are the ordered solutions.
    In order to convert to unordered solution, we must first understand that except for the case (400,400), all cases have been counted twice.
    So, the number of unordered solutions will be (45+1)/2 = 23

  • Q5) In how many ways can we write 300 as the product of 3 integers ?

  • There are three parts of this problem.

    1. Find the number of ordered triplets (positive integers)
    2. Convert to unordered triplets (positive integers)
    3. Consider the negative integer cases

    First part

    This part is pretty easy. We have a∗b∗c = 300 = 2^2 ∗ 3 ∗ 5^2
    Number of ways of doing this is (2+3–1)C(3–1) ∗ 3 ∗ (2+3−1)C(3−1)
    Basically distributing the powers of each prime over the 3 variables
    So we have a total of 6 ∗ 3 ∗ 6 = 108 ordered triplets

    Second part

    As, we are simply looking for three positive integers, all those scenarios which have been counted more than once must be taken care of.
    Case 1: All identical. But then a * a * a = 300, which is not possible as 300 is not a perfect cube and ‘a’ must be an integer

    Case 2: Two of the values identical.
    So, we have a * a * c = 300, or a^2 * c = 300.
    In other other words, we need to find the number of perfect squares which are factors of 300
    As 300 = 3 ∗ (2 * 5)^2
    Number of perfect square factors will be simply given by factors of 2 * 5, which is 4
    These 4 cases have been counted 3!/2! = 3 times each

    Case 3: All distinct
    We are not counting these cases separately. But we need to understand that each of these cases have been counted 6 times in the ordered triplets
    So, the 4 cases which have been counted 3 times each, let's first count them 3 more times each
    108 + 4 * 3 = 120
    Now, as everything has been counted 6 times, we get the unordered triplets by simply dividing all the cases by 6
    120/6 = 20 triplets

    Third part

    Now, we need to consider the negative integers as well.
    Let us take ‘a’, ‘b’, ‘c’ positive
    In the 4 cases where two are identical, we can have
    (a,a,b); (-a,-a,b); (-a, a, -b)
    In the rest of the cases we will have
    (a,b,c); (-a,-b,c); (-a, b, -c); (a, -b,-c)
    So first let us multiply everything by 4
    20 * 4 = 80
    Now, as we have counted the 4 cases (where two are identical) one extra time, let us subtract each of them once.
    Hence we have a total of 80–4 = 76 solutions (answer)

  • Q6) How do I find how many 8 digit combinations can be formed by selecting 3 numbers from the given set of 4 numbers (say 1,2,3,4)?

  • I think inclusion-exclusion principle will be the best process in this scenario.
    Let us first find all the possible numbers which uses at most 3 of the given digits. 4c3 * 3^8
    Now, let us find all possible numbers which uses at most 2 of the given digits. 4c2 * 2^8
    Finally the numbers which uses only 1 of the digits is 4c1
    Applying inclusion-exclusion principle, the number of possible combinations = 4c3 *3^8 - 4c2 *2^8 + 4c1

    PS: Inclusion-Exclusion principle comes from set theory.
    n(AUBUC) = n(A) + n(B) + n(C) - n(A∩B) - n(A∩C) - n(B∩C) + n(A∩B∩C)

  • Q7) How many selection of atleast one red ball can be made from 4 red balls and three green balls if the ball of the same colour are different?

  • In this case, first thing that we should always look for is how many balls need to be selected. As there are no restrictions there, our focus should shift on each ball. The balls being distinct, each ball has uniquely two choices (either being selected or not being selected)

    So, without any constraint, we should have 2^7 choices.

    But, no red ball being selected is not allowed. So, not selecting any red ball (which is 1 choice), the green balls can be selected in 2^3 ways

    Thus, the number of ways of selecting at least one red ball would be:
    2^7 − 2^3 = 120 options.

  • Q8) Using only the digits 2, 3 and 9, how many six digit numbers can be formed which are divisible by 6?

  • Being a multiple of 6, the number has to be multiple of 3 as well as even number.

    The only even digit allowed being 2, it has to be used in the last place. So, the last digit is fixed at 2

    Also, being a multiple of 3, the sum of digits must be div by 3: as the digits 3 and 9 are themselves div by 3, hence we should use either three 2s or six 2s (at least one 2 is compulsory, from the previous constraint)

    Using six 2s, there is only one number 222222

    Using three 2s, the number will be of the form: _ _ _ _ _ 2, where each blank space will be occupied by a digit. Firstly, there will be two more 2s, occupying two of the five blanks, select those in 5C2 = 10 ways. Once done, the remaining three positions (whichever those are) will have two options (3 or 9) each. So, another 2^3 = 8 options. So total of 10*8 = 80 numbers in this scenario

    So, we will have 81 possibilities.

  • Q9) E= [1/3 + 1/50] + [1/3 + 2/50] +...+ upto 50 terms, then the exponent of 2 in E! Is?

  • There are couple of concepts in this question:

    1. In order to get the highest power of a certain prime number 'p' in a factorial N, we need to successively divide N by p and collect each quotient and add the quotients. So, here we need to successively divide 17 by 2 and collect the quotients. 17/2 = 8, 8/2 = 4, 4/2 = 2, 2/2 = 1.

    So total powers = 8 + 4 + 2 + 1 = 17

  • Q10) How many 4 digit numbers greater than 3000 can be formed using digits from the given set only: {2,2,3,3,3,4,4,4,4}?

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