Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 1



  • In the first case, there is a 2 added to the existing number of twos and in the second case, this 2 is removed and a 3 is added to the existing number of threes. If a number has 28 factors, the powers of the individual primes could be (27), (1,13), (3,6), (1,1,6).

    Similarly, if a number has 30 factors, the powers of the individual primes could be (29), (1,14), (2,9), (4,5), (1,1,1). The second case has to have one 2 lesser and one 3 more than the first. Only (3,6) and (4,5) satisfy this case and so, the number is in the form of 2^5 * 3^3. 6n becomes 2^6 * 3^4 and so, number of factors is 7 * 5 = 35



  • Q25) Anshul and Nitish run between point A and point B which are 6 km apart. Anshul starts at 10 a.m.from A, reaches B, and returns to A. Nitish starts at 10:30 a.m. from B, reaches A, and comes back to B. Their speeds are constant with Nitish’s speed being twice that of Anshul’s. While returning to their starting points they meet at a point which is exactly midway between A and B. When do they meet for the first time?



  • Anshul covers 9 km in say x hours. So, Nitish will also cover 9 km in (x - 0.5) hours. As the ratios of their speeds are known, we can form the following equation:

    9/(x-0.5) = 2 * 9/x
    x = 2x-1
    x=1 hour.

    Anshul covers 9 km in 1 hour and so, his speed is 9 km/hr
    Nitish covers 9 km in 0.5 hours and so, his speed is 18 km/hr

    For the first half hour, Anshul covers 4.5 km. So, 1.5 km need to be covered at the effective speed of 27 km/hr and so, will take 1.5/27 = 1/18 more hours ie. 3 1/3 minutes. So, a total of 33 1/3 minutes post 10.



  • Q26) What is the number of non-negative integer solutions for the equation x^2 – xy + y^2 = x + y?
    (a) 3
    (b) 4
    (c) 1
    (d) None of these



  • 2x^2 - 2xy + y^2 = 2x + 2y
    (x-y)^2+x^2+y^2-2x-2y+2=2
    (x-y)^2+(x-1)^2+(y-1)^2=2
    (1,1,0) gives (2,1) or (0,1)
    (1,0,1) gives (1,0) or (1,2)
    (0,1,1) gives (0,0) or (2,2)



  • Q27) A is the set of the first 100 natural numbers. What is the minimum number of elements that should be picked from A to ensure that at least one pair of numbers whose difference is 10 is picked?
    (a) 51
    (b) 55
    (c) 20
    (d) 11



  • A lot of students get confused in the minimum number of elements to be picked so as to be sure and might end up listing elements as (1, 21, 23, 34... 100) to get 10+1 = 11 elements. There is no real method for such questions but if you can simply think a bit laterally, you should be able to come up with an answer quickly.



  • Q28) A game consisting of 50 rounds is played among P, Q and R as follows: Two players play in each round and the player who loses in that round is replaced by the third player in the next round. If the only rounds in which P played against Q are the 3rd, 14th, 25th and 36th, then what can be the maximum number of games won by R?
    (a) 40
    (b) 42
    (c) 41
    (d) 36



  • The easier way to do this is to understand that P and Q played each other in those 4 matches. Which means that R would have lost in the previous round at least. Other than that there is no condition. So, R loses 4 matches and does not participate in 4 matches. So, max number of wins is 42.



  • Q29) 5f(x) + 4f[ (4x + 5) / (x – 4) ] = 9(2x + 1), where x ∈ R and x # 4. What is the value of f(7) ?
    (a) -17
    (b) -8
    (c) -7
    (d) None of these



  • Simply substitute x=7 and x=11. You will get 2 linear equations with 2 variables. The fact that f(7) is asked should make you suspicious. As f(x) has to be f(7) for you to get an answer, substitute 7 and then you might realize that the next step would be to plug x=11



  • Q30) There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairs of parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg, 101 kg and 104 kg. Which of the following can be the weight of one of the parcels?
    (a) 40 kg
    (b) 45 kg
    (c) 48 kg
    (d) 53 kg



  • 45 is correct. Let's say all parcels are of different weights such that a>b>c>d. So, there could be 6 possible combinations ab, ac, ad, bc, bd, cd. Out of them, there should be 2 pairs such that in each pair, addition of two quantities gives us the same result. If we look at it carefully, ab, cd, ac, bd cannot be equal to anything else. That leaves us with ad being equal to bc which defeats the given condition. So, there are 2 weights that are equal to each other and so, yield 4 cases.

    Addition of these 2 weights would also give a result. As all the weights are integral in nature, the sum should be an even number. So, either the equal weights are 47 kg each or 52 kg each.

    47, 47 will give us the other weights to be 50 and 54
    52, 52 will give us the other wights to be 49 and 45



  • @shashank_prabhu whats the logic behind 13c7



  • @Shivaraman-R

    Number of non-negative integral solutions = (n + r - 1)C(r - 1)
    Here, a + b + c + d + e + f + g + h = 6
    so r = 8 and n = 6
    Number of non-negative integral solutions = (8 + 6 - 1) C (8 - 1) = 13c7

    Read the below article if you want to refer more on this concept.
    https://www.mbatious.com/topic/887/solving-combinatorics-problems-using-stars-bars-concept



  • @shashank_prabhu sir but here only case is a > b > c but u have taken 6 cases??



  • @Sairam-Akella

    Hi Sairam, what is the confusion here ?
    Total solution of a + b + c = 60 is 62C2 = 1891.
    Now we need to remove cases with repetition.
    When a = b or b = c or c = a case.
    when a = b,
    2a + c = 60 --> 31 cases
    Similarly when b = c,
    a + 2b = 60 --> 31 - 1 = 30 cases (as a = b = c = 20 case is already counted in the previous case)
    Also, when c = a
    2a + b = 30 --> 31 - 1 = 30 cases (as a = b = c = 20 case is already counted)
    Total = 31 + 30 + 30 = 91 cases
    So we have 1891 - 91 = 1800 cases where a, b and c are different.

    Now we need to find only a > b > c ≥ 0 cases, the best method is already explained in shashank's solution.

    Will give something traditional to find the cases satisfying a > b > c (reference - math.stackexchange)
    let say d = a - c and e = b - c
    as a > b > c, we can say d > e > 0
    d + e = a + b - 2c = a + b + c - 3c = 60 - 3c
    d + e = 60 - 3c
    So the number of cases possible are [(60 - 3c - 1)/2]
    c can be anything from 0 to 19 (19 = [60/3] - 1).
    (a, b, c) cases would be (31, 29, 0) to (21, 20, 19)
    So our required answer is Summation of [(60 - 3c - 1)/2] where c = 0 to 19
    => 29 + 28 + 26 + 25 + 23 + 22 + .... + 2 + 1
    which is basically 1 + 2 + 3 + 4 ... + 30 - (3 + 6 + 9 + .... + 30)
    = 30 x 31/2 - 5 x (3 + 30) = 15 x 31 - 15 x 11 = 15 x 20 = 300

    As you can see, traditional method will take away a good chunk of your time. Hence it's best to go with the method explained by shashank or else you can mug up the below formula.
    Number of possible solutions for (a, b, c) when a + b + c = N and a > b > c ≥ 0 is [(n^2 + 6)/12]
    So here N = 60, so possible cases for a > b > c ≥ 0 is [(3606/12)] = [300.5] = 300
    Also, if we need to find out for a ≥ b ≥ c ≥ 0, then just add [N/2] + 1 to the above equation.

    For example,
    Find the possible solutions for a + b + c = 100 if
    a) a > b > c ≥ 0
    solution : [(100^2 + 6)/12] = 833
    b) a ≥ b ≥ c ≥ 0
    solution : 833 + [100/2] + 1 = 884



  • @shashank_prabhu Sir how do we get this 3 * 3 = 9 * x



  • @Sonam-Priya it is called chord theorem.

    0_1511241902316_5aa1dea2-c274-4e0b-8e25-f4a88d04069d-image.png

    This theorem states that A × B is always equal to C × D no matter where the chords are.


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