Quant Boosters - Sagar Gupta - Set 5


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    Author : Sagar Gupta - MBA student at Symbiosis Institute of Operations Management

    If a^3 + b^3 =10 and a^2 + b^2 = 60 then find a + b?

    (a+b)^3 = 10 + 3ab (a+b)
    (a+b)^2 = 60 + 2ab
    a+b = p
    (p^3 - 10 ) / 3p = ab
    p^2 = 60 + 2 [ (p^3 - 10 ) / 3p ]
    3p^3=180p + 2p^3 - 20
    p^3 +20 = 180p
    p=13.36

    A and B can build a wall in 4 days and 6 days respectively working alone, while C and D can destroy the wall in 8 days and 12 days respectively working alone. If three of them start working together and the wall was built in 24 days. Which of the four persons did not work?

    Total work =36 units
    A -> +9 units/day
    B -> +6 units/day
    C- > -4.5 units/day
    D- > -3 units/day
    24 (x+y+z) = 36
    x+y+z =1.5
    9-4.5-3 =1.5
    Hence B did not work

    Three persons A, B and C rent the grazing of a park for Rs. 570. A put 126 oxen in the park for 3 months, B
    puts in 162 oxen for 5 months and C puts in 216 oxen for 4 months. What part of the rent should each person
    pay?
    a. 105, 220, 245
    b. 125, 205, 245
    c. 105, 225, 240
    d. data inadequate
    d. None of these

    126 * x * 3 + 162 * 5 * x + 216 * 4 * x = 570
    x = 5/18
    A= 126 * 3 * 5/18 =105
    B=162 * 5 * 5/18 =225
    C=216 * 4 * 5/18=240
    Hence option C

    a,b and c are three positive real numbers. The minimum value that the expression [(a+b)/2 * (b+c)/2 * (c+a)/2] can take when the product of the three numbers is 3/2 is?

    abc = 3/2

    a+b+c > 3 [ 1.5 ]^1/3
    (a+b)/2 > 1.5[1.5]^1/3 - c/2 = 1.5^4/3 - c/2
    (b+c)/2> 1.5 [1.5]^1/3 - a/2 = 1.5^4/3 - a/2
    (a+c)/2 >1.5[1.5)^1/3 -b/2 = 1.5^4/3 - b/2
    a = b = c = (1.5)^1/3
    1.5^4/3 - 0.5* 1.5^1/3
    1.5^1/3 [ 1.5 - 0.5 ] =1*1.5^1/3 =1.5^1/3
    (1.5^1/3) ^3 = 1.5 = 3/2

    The number 'm' and 'n' are reciprocals of each other. Both m and n are positive real numbers. If both m^3 + n^3 = 65/8, determine (m+n).

    m = 1/n
    mn = 1
    m^3 + n^3 +3mn [ m+n ] = (m+n)^3
    65/8 + 3 x = x^3
    5/2 = x

    The median of five positive integers is 7. If the only mode is greater than the median and the mean is greater than 9, what is the lowest possible value of the mode of these 5 integers?

    a < b < c < d < e
    c = 7
    d and e are same
    a + b + c + d + e > 45
    5 = a
    b = 6
    c = 7
    18 + 2d > 45
    2d > 27
    d > 13.5
    d = 14
    e = 14
    Mode =14

    The number of factors of the number N = 4^6 + 6^8 is
    a. 18
    b. 36
    c. 54
    d. 72

    2^12 + 2^8 * 3^8
    2^8 [ 16 + 3^8 ]
    2^8 [ 16 + 81 * 81 ]
    2^8 [ 16 + 6561 ]
    2^8 [6577]
    9 * 2 = 18 factors

    When a natural number, N is divided by D, the remainder is 35. When 50N is divided by D, the remainder is 11. Find D
    A. 1739
    B. 43
    C. 47
    D. Cannot be determined

    n=ad +35
    50n=bd +11
    50n=50ad+1750
    50n=bd+11
    50ad+1750 = bd +11
    1739= (b-50a)d
    d can be 1739 or 47
    CBD

    In a row at a bus stop, A is 9th from the right and B is 7th from the left. They both interchange their positions. If there are 20 people in the row, what will be the new position of B from the left'?
    (1) 11th
    (2) 12th
    (3) 13th
    (4) 10th

    1,2,3,4,5,6,B,8,9,10,11,A,13,14,15,16,17,18,19,20
    So, 12'th from left

    Find the sum of all positive two-digit integers that are divisible by each of their digits

    {11,22...99}
    10a+b mod a=0 -> b mod a=0
    10a+b mod b=0 ->10a mod b=0
    when a=b , the above 2 conditions will be satisfied -> {11,22...99}
    also : {12,15,24,36,48 } will satisfy the above conditions
    sum =630


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