Quant Boosters - HandaKaFunda - Set 4


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    A and B started at a time towards each other. After crossing each other, they took 16hrs, 25hrs respectively to reach their destinations. If A had 40kmph speed, what is the speed of B?

    Let us say that A and B started from point P and point Q respectively and met at point R in the middle after time 't'.

    Speed of A is given to us as 40 kmph.

    Let us assume the speed of B to be 's'.

    If we consider the distance PR, it was covered by A in 't' hours and B in 25 hours.
    => PR = 40 * t = s * 25

    If we consider the distance RQ, it was covered by B in 't' hours and A in '16' hours.
    => RQ = s * t = 40 * 16

    Let us divide the two equations. We will get

    40t/st = 25s/640
    => 40/s = 5s/128
    => s^2 = 40 * 128/5 = 8 * 128 = 1024
    => Speed of B = s = 32 kmph

    What degree occurs at a time 3:30 in the clock?

    At 3:30, the hour hand would be exactly between 3 and 4
    => 15 deg from both 3 and 4
    At 3:30, the minute hand would be exactly at 6
    The gap between the hour hand and the minute hand will be
    30 deg (between 5 and 6) + 30 deg (between 4 and 5) + 15 deg (between hour hand and 4) = 75 degrees

    You were a quarter of my age when I was twice your present age" said A to B. In five years from now, our ages will add up to 45. How old was I when you were born?

    Let us say my current age is a and your current age is b
    When I was 2b when you were 2b/4 or 0.5b
    => a - 2b = b - 0.5b
    => a = 2.5b

    In 5 years, I will be a + 5 and you be b + 5
    Sum of our ages will be 45
    => a + 5 + b + 5 = 45
    => a + b = 35
    Substitute the value of b
    => 2.5b + b = 35
    => 3.5b = 35
    => b = 10
    => a = 25

    So, when you were born i.e. 10 years ago; I was 25 - 10 = 15 years old

    What is the largest three-digit number that when divided by 6 leaves a remainder of 5 and when divided by 5 leaves a remainder of 3?

    Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of
    LCM(x,y,z)*n + constant

    The key in these questions is finding out the value of 'constant'. If all of them leave the same remainder 'r', constant = r. It can also be looked at as the smallest number satisfying the given property.

    In this question, we are given
    Remainder from 6 is 5
    Remainder from 5 is 3

    So, the number N = LCM(6,5)*n + constant = 30n + constant
    To figure out the constant, look at the numbers which give a remainder of 5 from 6.
    They are 5, 11, 17, 23, 29....
    Among these, find the one which leaves a remainder of 3 from 5. It is 23.

    So, our number N should be of the format of 30n + 23

    Biggest three digit number will occur when n is 32 = 30*32 + 23 = 983

    How many prime numbers less than 75 will leave the odd reminder when divided by 5?

    The number should leave a remainder of 1 or 3 from 5
    => It is of the form of 5k + 1 or 5k + 3
    => The number ends in 1 or 3
    I will just count out such numbers under 75, which are 16
    1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71 and 73
    From these, I would remove the non prime ones which are 1, 21, 33, 51 and 63
    So, I will be left with 11 numbers which fit the bill.
    3, 11, 13, 23, 31, 41, 43, 53, 61, 71 and 73

    What is the unit digit of LCM of (13^501 – 1) and (13^501 + 1)?

    13^501 – 1 and 13^501 + 1 are two consecutive even numbers.
    One of them will be of the format 4k and the other one will be of the format 4k + 2.
    They will only have 2 as a common factor.
    => HCF (13^501 – 1, 13^501 + 1) = 2

    We also know that HCF * LCM = Product of two numbers
    => LCM = (13^501 – 1)(13^501 + 1)/2 = (13^1002 – 1)/2

    Now, let’s try and find out the last digit of the LCM
    Last digit of 13^1002
    = Last digit of 3^1002
    = Last digit of 3^(4k + 2)
    = 9

    Last digit of 13^1002 – 1 = 9 – 1 = 8
    Last digit of (13^1002 – 1)/2 = 8/2 = 4

    What is the sum of all the integers less than 100 which leave a remainder 1 when divided by 3 and a remainder of 2 when divided by 4?

    Suppose we are given that a number when divided by x, y, and z, leaves a remainder of a, b, and c; then the number will be of the format of LCM(x,y,z) * n + constant

    The key in these questions is finding out the value of ‘constant’. If all of them leave the same remainder ‘r’, constant = r. It can also be looked at as the smallest number satisfying the given property.

    In this question, we are given
    Remainder from 3 is 1
    Remainder from 4 is 2

    If you look at the negative remainders
    Remainder from 3 is -2
    Remainder from 4 is -2

    So, the number N = LCM(3,4) * n – 2 = 12n – 2
    So, any number which is of the format of 12n – 2 will satisfy the given conditions.
    Positive Integers less than 100 which satisfy the above condition are
    10, 22, 34… 94
    Sum of these integers
    = (No. of terms/2) * (First term + Last Term)
    = (8/2) * (10 + 94) = 416

    Find the number of positive integral solutions of |x| + |y| = 10.

    Let |x| = a and |y| = b. First find the positive integral solution of a + b = 10.
    Number of positive integral solutions= (10-1)C(2-1) = 9.
    Now for each solution (a1, b1), the values of (x,y)= (a1, b1), (-a1, b1), (a1, -b1) and (-a1, -b1).
    So total number of positive integral solutions= 4 × 9 = 36.

    Find the total number of integral solutions of IxI + IyI + IzI = 15.

    First, let a = |x|, b = |y|, c = |z|.
    Now, we need to find the number of positive integral solutions of a + b + c = 15. The number of solutions are 14C2 = 91. Now for each value of a,b and c we will have two values of x, y and z each. Therefore, the total number of solutions = 91 x 2 x 2 x 2= 728.

    Now let one of the variables be equal to 0. For example, let x = 0 and |y| and |z| be at least equal to 1. Therefore, we need the positive integral solution of b + c = 15, where b = |y| and c = |z|. The number of solutions is 14C1 = 14. Each of these solutions will give two values of y and z and there are 3 ways in which we can keep one of the variables equal to 0. Therefore, total number of ways are 14 x 2 x 2 x 3 = 168.

    Now let two of the variables be equal to 0. In this case, the total number of solutions is equal to 6.

    Therefore, the total number of integral solutions = 728 + 168 + 6 = 902.

    Let g(x) = max (5 − x, x + 2). The smallest possible value of g(x) is
    a] 4.0
    b] 4.5
    c] 1.5
    d] None of these.

    To solve such kind of questions, in most cases, all you need to do is to equate the two values inside the function
    => 5 – x = x + 2
    => x = 3/2 = 1.5
    Please note that 1.5 is not the answer to the question.
    To find out the answer to the question, we need to find out the value of g(1.5)
    When we put x = 1.5, we get g(x) = max (5 – 1.5, 1.5 + 2) = max (3.5, 3.5) = 3.5
    So, our answer is 3.5 Option D – None of these

    I would strongly recommend that you watch the below video for better understanding of the solution of the question


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