Quant Capsules by Shashank Prabhu - Set 14


  • CAT 100%iler, 5 times AIR 1, Director - learningroots.in, Ex ITC, Pagalguy, TAS


    Consider a four digit number for which the first two digits are equal and the last two are also equal. How many such numbers are perfect squares.

    The last two digits of any square will show a cyclicity around multiples of 50 for eg: 23^2, 27^2, 73^2, 77^2, 123^2, 127^2... will all end in 29. So, if you look at the squares of the first 25 numbers, you can see that there are only 2 cases where the last two digits repeat (00 and 44). Now 00 is not possible because you cannot have a square in the form of aa. So, we are left with 44 only. Need to try for 38^2, 62^2 and 88^2. 38^2 will be between 1225 and 1600 so, 1144 is not possible, 62^2 will be between 3600 and 4225 so again 3344 or 4444 are not possible. 88^2 will be between 7225 and 8100 and so, 7744 is possible. You need to check for only one value this way. And if you know a few numbers and their properties, it would be of some help. The likes of 1729, 145, 1001, 10001, 40585, 7744 and so on.

    There are four machines in a factory. At exactly 8 pm, when the mechanic is about to leave the factory, he is informed that two of the four machines are not working properly. The mechanic is in a hurry, and decides that he will identify the two faulty machines before going home, and repair them next morning. It takes him twenty minutes to walk to the bus stop. The last bus leaves at 8:32 pm. If it takes six minutes to identify whether a machine is defective or not, and if he decides to check the machines at random, what is the probability that the mechanic will be able to catch the last bus?
    A. 0
    B. 1/6
    C. 1/4
    D. 1/3
    E. 1

    There are 2 defective machines and 2 good machines. He can check two machines before he leaves for home. So, there are 2 cases that are in his favour. If the first two machines are good, he can identify the defective ones as the last two and leave. Also, if he gets the first two machines as defective, he would have known what he needed to know. The problem lies if he gets only one defective out of the first two and so, will have to check the third one. The four machines can be arranged in 6 ways (arrangements of ddgg). So, 2/6=1/3.

    In a convex hexagon, two diagonals are drawn at random. What is the probability that the diagonals intersect inside the hexagon?
    (1) 5/12
    (2) 1/2
    (3) 7/12
    (4) 2/5

    Number of diagonals is n(n-3)/2 ie. 9
    Two diagonals can be chosen in 9c2=36 ways
    For two diagonals to intersect inside the hexagon, the vertices should form a quadrilateral. Possible ways are 6c4=15
    Probability is 15/36 = 5/12

    The product of 3 natural numbers(N1,N2,N3) is 12 times of their HCF. Find the number of ordered triplets (N1,N2,N3).

    ax, bx, cx where a, b, c are coprime to each other.
    abc * x^2 = 12 =2^2 * 3
    x=1, (2,2,3)(4,3,1)(6,2,1)(1,1,12)
    x=2, (1,1,3)
    Total of 3+6+6+3+3=21 cases

    The micro manometer in a certain factory can measure the pressure inside the gas chamber from 1 unit to 999999 units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit 5 and moves directly from 4 to 6. What is the actual pressure inside the gas chamber if the micro manometer displays 003016?

    Simply put, it is a base 9 system with a missing 5 instead of the conventional missing 9. While it would be exactly the same till 3014 for conventional base 9 and the base 9 in this case, 3015 would have been skipped. So, 3016 in this form is nothing but 3015 in base 9. We have to convert this into decimal notation which will be 3×729+1×9+5×1=2201.

    If a^2 * b^3= (540)(35^2), then find the minimum value of 5a + 7b

    There is a specific way to solve these questions.
    Let the numbers be 5a/2, 5a/2, 7b/3, 7b/3 and 7b/3. Using AM>=GM, we get
    (5a+7b)/5>=(25 * 343/108 * a^2 * b^3)^1/5
    (5a+7b)/5>=(25 * 343 * 125 * 49)^1/5
    (5a+7b)/5>=35
    (5a+7b)>=175

    Let a, b, c, d and e be integers such that a = 6b = 12c and 2b = 9d = 12 e. Then which of the following pairs contains a number that is not an integer?
    a. (a/27, b/e)
    b. (a/36, c/e)
    b. (a/12, bd/18)
    d. (a/6,c/d)

    Converting everything into a common base is probably the easiest way to do this.
    As b is common to both the equations, we can get all other numbers in terms of b.
    a=6b, c=b/2, d=2b/9, e=b/6
    So, b has to divisible by 2, 9 and 6 and so, in the form of 18x. Now we get the values of the other integers
    a=108x, c=9x, d=4x, e=3x. Plug these in the options and get the answer as option 4.

    If x is the smallest positive integer that is not prime and not a factor of 50! What is the sum of the factors of x ?

    50! contains all numbers from 1 to 50. Hence, to find a number that is not a factor of 50!, we need to start from 51. Let's check the numbers one by one.
    51 = 3 * 17. Both 3 and 17 are a part of 50!. Hence, 51 is a factor of 50!
    52 = 2 * 26. Both 2 and 26 are a part of 50!. Hence, 52 is a factor of 50!
    53 is a prime number. Hence, this is the first number that will not be a factor of 50!. However, you need a non-factor that is also not a prime. Hence, this cannot be the number.
    In fact, all the numbers here on till 53 * 2 will either be prime or be factors of 50!. The smallest numbers that satisfies both the conditions i.e. prime as well as non-factor will be 53 * 2 = 106.
    Factors of 106 are 1,2,53 and 106. Hence, sum of factors = 162.

    The infinite sum 1 + 4/7 + 9/(7^2) + 16/(7^3) + 25/(7^4) + …

    As is the case with any mixed progression, you have to find the level at which you can get it as an infinite geometric progression. If we see the numerators carefully, we get:
    1....4....9....16....25
    ....3....5....7....9....
    ........2....2....2....
    So, the third level gives us a 2 in the numerator.
    Let S = 1 + 4/7 + 9/(7^2) + 16/(7^3) + 25/(7^4) ...
    S/7 = 1/7 + 4/(7^2) + 9/(7^3) + 16/(7^4) ...
    6S/7 = 1 + 3/7 + 5/(7^2) + 7/(7^3) + 9/(7^4) ...
    6S/49 = 1/7 + 3/(7^2) + 5/(7^3) + 7/(7^4) ...
    36S/49 = 1 + 2/7 + 2/(7^2) + 2/(7^3) + 2/(7^4) ...
    36S/49 = 1 + 2(1/7)/(6/7)
    36S/49 = 4/3
    S = 49/27

    The number of roots common between the two equations x^3 + 3x^2 + 4x + 5 = 0 and x^3 + 2x^2 + 7x + 3 = 0 is

    You have to simply equate the two and find out a solution.
    You will get x^2 - 3x + 2 = 0 and so, x as (1, 2). But these don't satisfy the parent equations.
    So, there are no solutions.


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