Quant Boosters - Learn Quest - Set 4


  • Learn Quest


    N is a positive integer not more than 100. If 7^N + N^3 has a unit digit of 0, how many values can N take ?

    7^n ends with 7, 9, 3, 1
    7^(4k+1) --> 7
    7^(4k+2) --> 9
    7^(4k+3) --> 3
    7^(4k) ---> 1

    X^3 must end with 3, 1, 7 or 9

    Case (i) 7^(4k+1) =>7
    So we need
    7 + 3 = 0
    I.e x^3 ends with 3
    And it is possible only when last digit is 7
    here X should be of the form (4k+1)
    So 17,37,57,77,97
    5 possibilities

    Case (ii)
    9 + 1
    => 7^(4k+2) --> 9
    X is even
    And (even)^3 never ends with 1

    Case(iii)
    7^(4k+3) --> 3
    And we need x^3 --> 7
    Which is possible when last digit is 3
    As 3^3 => 27
    So we need to take x of the form (4k+3) whose last digit is 3
    So, 3,23,43,63,83
    5 possibilities

    Case (iv)
    7^(4k)--> 1
    So we need x^3 --> 9
    So X should end with 9
    But it says X is of the form 4k
    So can't be possible

    So overall 10 possibilities

    In triangle PQR, PS and QT are medians to sides PR and QR respectively. If PS = 16 cm and QT = 24 cm and PS is perpendicular to QT then find the area of triangle PQR.

    0_1508227972708_c7ed0cee-863d-4aed-9bac-f285c24e4a78-image.png

    We know PS and QT are two Medians and they are passing through a point and we know that point is called Centroid
    Now best part of centroid : It divides Triangle into 6 equal parts
    Now if some how I can calculate the area of PGT, I can calculate Area of whole triangle, just multiplying it by 6
    For that additional information is given to us
    that is PS is perpendicular to QT
    I.e PGT is a right angled Triangle
    So we just need base and height
    I.e length of GT and PG
    For that we know the property
    centroid divides the median in the ratio 2:1
    So GT => 1/3 of QT => 8
    And PG => 2/3 of PS => 32/3
    So area of Triangle PGT =>
    1/2 * (32/3) * 8
    => 128/3
    And area of whole Triangle
    (128/3 ) * 6 = 256

    A student was asked to find the sum of first n natural numbers. By mistake he missed one of the numbers in between and he got the average as 30 (7/29). Find the number he forgot to add

    Avg now = 30.xx

    Step 1: number of numbers will always be approx 2 * avg.
    So number of numbers = 2 * 30 = near to 60
    Avg= 877/29

    Step 2: number of numbers(after forgetting) will always be multiple of denominator.
    So 29 * 2 = 58 after forgetting.
    And he forgot 1, means earlier there were 58 + 1=59 numbers.

    Step 3:
    Now sum of 59 numbers would be 1 + 2 + ... + 59 = 59 * 60/2 = 59 * 30 = 1770
    Here sum for 58 numbers(after removing 1) 877 * 2= 1754

    Final step:
    One number removed = 1770 - 1754 = 16

    In how many ways can 18 identical balls be distributed among 3 identical boxes ?

    A + B + C = 18
    Total solution --> 20c2 => 190
    Now remove cases when two of them are equal
    2a + b = 18
    (0,..9)
    10 cases
    Which include one case (6, 6 * 6)
    So overall cases 9
    And these cases can be arranged in 3!/2! --> 3 ways
    9 * 3 = 27 ways
    Now (6,6,6) --> 1 case
    So
    total number of cases
    (190 - 27 - 1 )/6
    => 27 + 9 ( cases when two are equal ) + 1 ( 6,6,6)
    => 37

    The product of the number of diagonals of two distinct polygons with sides N1 and N2 is 40. What is the sum of N1 and N2.

    Product of diagonals = 40
    So possibile cases
    1 * 40
    2 * 20
    4 * 10
    5 * 8
    Now case (I) can't be possible
    I.e 1 diagonal can't be possible
    Now we know
    Number of diagonals is given by n(n-3)/2
    n(n-3)/2 = 2
    n = 4
    n(n-3)/2 = 20
    n(n-3) = 40
    n = 8
    Case (ii) is possible
    Now case (iii)
    n(n-3)/2 = 4
    Not possible
    Case (IV)
    n(n-3)/2 = 5
    n = 5
    n(n-3)/2 = 8
    Not possible
    So only possibility
    Case (ii)
    I.e polygons of side 4 and 8
    So 12

    Ram bought a few mangoes and apples spending an amount of at most Rs.2000. If each mango costs Rs.4 and each apple costs Rs.6, and Ram bought at least one fruit of each type, how many different possible amounts could he have spent in purchasing the fruits?

    4x + 6y < = 2000
    2x + 3y < = 1000
    atleast 1, so give 1,1 value
    2x + 3y < = 1000 - 2 - 3
    2x + 3y < = 995
    Rhs represents amount.
    Question is asking different possible values of amount. That can be anything 0 to 995 now.
    x, y= 0. Amount = 0
    x=1,y=0. Amount = 2
    x=0,y=1. Amount = 3
    x=2,y=0. Amount = 4
    x=1,y=1. Amount = 5 and so on.
    all amounts except amount 1 is possible
    hence o to 995 => 996 values
    Excluding 1, total 995 possible amount values.

    Beaker A and beaker B contain methanol, ethanol and phenyl in the ratio 1:3:2 and 2:1:5 respectively. Some parts of the solutions from beaker A and beaker B are thoroughly mixed and put into another beaker C.
    Which of the following cannot be the ratio of methanol, phenyl and ethanol in beaker C?
    (a) 10 : 23 : 15
    (b) 7 : 15 : 16
    (c) 6 : 13 : 13
    (d) 9 : 20 : 18

    Methanol --> (x/6 + 2y/8) => (4x +6y)/24
    Phenyl --> (2x/6 + 5y/8) =>
    (8x +15y)/24
    Ethanol --> (3x/6 + y/8)
    => (12x + 3y)/24
    Required ratio -->
    (4x +6y) : (8x +15y) : (12x +3y)
    Now let's check for the possibilities
    Case (I)
    X = Y = 1
    10: 23: 15
    Option (1) possible
    Similarly
    Case (ii)
    X = (1/4) , Y = (1/15)
    Option (2) is possible
    Case (iii)
    X = 1/2 and y = 1/6
    Option (3) is possible
    So option D

    If abcde is a 5 digit number divisible by 6 and a > b > c > d > e, then how many such 5 digit numbers exist ?

    a + b + c + d + e = 3k and e = even and since a, b, c, d are smaller than e hence e can be 6/8

    Case: e=6
    We need a + b + c + d = 3k form
    (1, 2, 4, 5) is a possibility.

    Case: e = 8
    Now a + b + c + d = 3k + 1 form
    we need to choose 4 numbers from (1, 2, 3, 4, 5, 6, 7) which is of the form 3k + 1
    Now 1, 4, 7 = 3k + 1
    2, 5 => 3k + 2
    3, 6 => 3k

    3k + 1 can be made by two 3k + two 3k+2 => 1 way
    one 3k + one 3k+2 + two 3k+1 => 2c1 * 2c1 * 3c2 = 12 ways

    Total=14 ways



  • number of numbers will always be approx 2 * avg.

    This 2 * avg is valid everywhere or only when [1,n] is given?


  • Being MBAtious!


    @gautam-arya
    sum of first n natural numbers = n(n+1) / 2
    Average = n(n+1) / 2n = (n + 1)/2
    2 * Average = (n + 1) ~ n
    So this should be in [1, n]


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