Quant Boosters - Quant Mixed Bag - Set 1


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    [Credits : Saquib Hasnain]
    9(a-b) = 63
    a - b = 7
    a + b = 9
    a =8 , b = 1
    8x+1 = 5(x+8 )
    3x = 39
    x = 13


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    Q8) 817 ^ 673 mod 100 = ?


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    [Credits : Saquib Hasnain]
    817^673 mod 4 = 1
    817^673 mod 25 = 12
    4k+1 = 25q +12
    k = 9..
    so remainder = 37


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    Q9) If the equation x^4 - 4x^3 + ax^2 + bx + 1 = 0 has 4 positive roots, then a + b = ?


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    [Credits : Saquib Hasnain]
    Product is 1..sum is 4..all roots are positive..so all of them have to be 1..
    a = 4C2 = 24/4 = 6
    b = -4C3 = -4
    so a+b = 6-4 = 2


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    Q10) A train after covering 1200 km meets with an accident and due to this it proceeds at 2/3 of its original speed and arrives at the station in 5 hours. Had the accident taken place after covering a distance of 300 km further, it would have arrived 2.5 hours earlier. What is the speed of the train ?


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    [Credits : Saquib Hasnain]
    So say the speed is v and 2/3v.. the guy saves 2.5hours..if the accident happens 300 further..so he saving that 2.5 hours in that 300m..
    Time taken to cover 300m at speed v is 300/v
    Time taken to cover 300m at speed 2/3v is 300/(2v/3)
    Difference is of 2.5hours..
    So 300 (3/2v - 1/v) = 2.5
    300 (1/2v) = 2.5
    v = 60


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    Q11) After three successive raises, Aftab’s salary became equal to 378/125 of his initial salary. By what percentage was the salary raised the first time if the third rise was twice as high (in percentage) as the second rise and the second raise was twice as high (in percentage) as the first rise?
    (a) 10%
    (b) 15%
    (c) 20%
    (d) 25%


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    [Credits : Saquib Hasnain]
    378 = 6 x 63 = 6 x 7 x 9
    125 = 5 x 5 x 5
    so, (1 + 1/5) (1 + 2/5) (1 + 4/5)
    1/5 = 20%, 2/5 = 40% and 4/5 = 80%


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    Q12) Find the sum of all the three digit numbers that contain only 1, 2, 3 or 4. It can contain the digits more than once in the same number too.


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    [Credits : Sabyasachi Mishra]

    Say a 3 digit number is abc.

    Now there are 4 ways to choose a, 4 ways for b and 4 ways for c = > There are 4 *4 * 4 = 64 three-digit numbers that can be made using 1,2,3 and / or 4.

    Let’s write all these numbers one below another to add them up. Now each digit would appear 16 times in ones place.

    Their sum = 16 *(1 + 2+ 3 + 4) = 160

    Similarly the tens place would add up to 160 and the hundreds place too.

    So the actual sum = 160 (1 + 10 + 100) = 160 * 111 = 17760


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    Q13) What is the Probability of selecting 3 numbers from 1-50 such that no two are consecutive?


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    [Credits : Sabyasachi Mishra]

    Imagine you have three 0s and 45 1s.

    Every sequence you make would look somewhat like this.

    10111111101111 … 1011 or 10011111 … 1011

    Let’s say after every sequence, I put one 1 after the first zero and another 1 after the second zero.

    Now I have a sequence of 50 digits, all in the form of 1 or 0. And no two zeroes are consecutive.

    Now 1 represents not-chosen and 0 represents chosen. So, I have picked three numbers between 1 and 50 such that no two of them are consecutive. (each 1 or 0 representing the status of numbers from 1 to 50 in order)

    The question simplifies into finding the number of ways three 0s and 45 1s can be arranged = > (48! / 45! 3!) or (48 C 3)

    Now ways of selecting 3 numbers of the 50 = 50 C 3

    Probability = 48C3 / 50C3 = 48 * 47 * 46 / 50 * 49 * 48 = 47 * 46 / 50 * 49 = 0.88245


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    Q14) how many numbers are there from 1 to 1000 which are not divisible by the digits 2, 3 and 5?
    a) 166
    b) 266
    c) 375
    d) 366


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    [Credits : Sabyasachi Mishra]

    Yes, this can be solved using a Venn diagram and intersections. But, if you are going to solve it in an exam, I would suggest a better and faster approach.

    Here, in our case LCM of 2,3 and 5 is 30.

    So, we can say that, if we write from 1 to 30 and find out the pattern, the same would be repeated across sets of 30 numbers thereafter.

    Step 1

    Write all numbers from 1 to 30

    0_1513687605195_fcccf411-09d7-4c57-b393-172992b20126-sm01.png

    Step 2

    Start with 2, the smallest divisor and cross out the numbers that are divisible. Then move to the next smallest one, in this case 3, circle out the numbers divisible by 3. And then do it for 5, use squares. The sheet would look like this.

    0_1513687621473_fe578874-0f70-4248-b0af-9984f75859c9-sm02.png

    Step 3

    Now that you need the numbers that are not divisible by either of these factors, let’s list them out.

    0_1513687639330_15842fd0-da67-4a82-9d3b-7de454854267-SM03.png

    Step 4

    Finding the answer

    1000 = 30 * 33 + 10

    So, the numbers not divisible by 2, 3 or 5 would be 8 * 33 + 2 = 264 + 2 = 266

    Note:

    1. This method can also be used to find the number of numbers between ranges that are divisible by specific factors. We need to start the count from the smallest number in the range and then list down a series of consecutive numbers (total count = LCM of factors)
    2. This method can also be used to find the number of numbers divisible by 2,3 and 5. In that case, instead of listing down the left ones, list down the crossed / highlighted ones.
    3. This method works faster when you have to do it on a small subset and extend the results. If the LCM is more than 100 or the factors are large numbers, Venn diagram method would be the way to go ahead.

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    Q15) How many AP can be formed with common difference 1 and sum is 840?


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    [Credits : Sabyasachi Mishra]

    Sum of an AP can be written as (n/2) * (2a + (n-1)d)
    Given:

    1. In our case, d = 1
    2. n is a natural number
    3. a is an integer

    Analysis:

    So, our equation now reduces to 2na+n(n-1) = 840 * 2 = 1680
    = > n^2 + (2a - 1)n - 1680 = 0
    Using the quadratic formula we can say that
    n = [ – (2a – 1) +/- root( ( 2a – 1)^2 – 4 (1) ( -1680)] / 2

    So, this is clear that for an integral value of a, there would be one positive n and one negative n = > For one integral value of a, there would be one possible n.
    (This makes sense as when we define the start point of the AP, the length inevitably gets fixed as the sum is constant)

    So, our question now changes to find the integral solutions of a.
    From simple observation, we can say a can both be odd or even.
    (2a-1)^2 + 4*1680 has to be a perfect square.
    Lets say 2a -1 = x
    now, x^2 + 4 * 1680 = y^2
    (Keep in mind that y is positive all the time because when we take the square root while solving a quadratic equation we only use the absolute value)
    y^2 - x^2 = 4 * 1680
    (y-x)(y+x) = 2a * 2b where ab = 1680

    If we find all solutions of ab = 1680 where a > 0, b > 0 then y would always be positive and x can be positive or negative.
    1680 = 2^4 * 3 * 5 * 7
    So, (4+1) * 2 * 2 * 2 = 40 factorsSo, a can be any of these 40 and b would be 1680 / a.
    There are 40 cases.
    (Keep in mind that we are already counting the cases where x is negative as when b < a, x is negative)

    So, x can have 40 values = > 2a -1 has 40 possible values and hence a has 40 possible integral values = > n can have 40 values

    = > There are 40 possible cases of writing an AP with common difference 1 where the sum would be 840.


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    Q16) How many numbers below 100 can be written as a difference of squares of natural numbers in only one way?


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    [Credits : Sabyasachi Mishra]

    Let’s say x= a^2 - b^2 = > x = (a+b)(a-b)

    If x has to be written as the difference of squares of two natural numbers in only one way, then there must be only one way to write x as (a+b) * (a-b) where (a+b) > (a-b).
    means that b can’t be zero and hence, (a+b) != (a-b)

    Now, (a+b) and (a-b) will either both be odd or both be even.

    1. Case : Both (a+b) and (a-b) are odd
      x = odd * odd = > odd
      If x is an odd number, it can always be represented as 1 * x. If x has more than one factor(excluding one and x itself), say two factors then x= c * d can always be written. So, we have to find x where x is either an odd prime or x is the square of an odd prime.
      x= k^2 (k being an odd prime) can be factorized as 1* k^2 or k * k.
      k*k would mean x = k^2 - 0^2 (not allowed as 0 is not a natural number)
      We have 25 prime numbers less than 100, hence 24 cases(excluding 2) and 9,25,49 as squares of odd primes = > 27 cases in all
    2. Case : Both (a+b) and (a-b) are even
      x = even * even = > even
      x = 2a * 2b = > x = 4 * (ab) & x < = 100
      = > ab < = 25 where a!=b & a and b are both natural numbers
      Lets say K=ab, K should be represented as ab in only one way. Similar to the above logic (except that numbers can be even too). K can be prime numbers or prime squares. So, k = 2,3,5,7,11,13,17,19,23 or 4,9
      = > 11 cases in all

    So there are 27 + 11 = 38 numbers between 1 and 100 which can be represented as the difference of natural squares in only one way.


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    Q17) How many three digit numbers are there whose sum of digits is divisible by 5?


 

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