Quant Boosters - Hemant Yadav - Set 1



  • Method 1 :

    Say the sides are A B C and the heights are a,b,c respectively one those sides.
    So Aa = Bb = Cc
    Now IF B/A should be a/b and also their difference should be same B - A = a - b
    So it's only possible when A = B = C.

    Method 2:

    Since, Aa = Bb = Cc = k
    So, a = k/A, b = k/B and c = k/C
    From this we can say that A, B, C are in AP as well as HP. That means A = B = C. Hence an equilateral triangle.

    Method 3 :

    Let the sides be (a – x), a, (a + x) and corresponding altitudes be (h – y), h, (h + y)

    Since, (a – x)(h – y) = ah
    ay + hy = xy ..........(1)

    Also, (a + x)(h + y) = ah
    ay + hy = -xy .........(2)

    From (1) and (2), we can say that xy= 0

    Hence atleast one of x and y is zero.
    Say, x = 0, then from eq (1) we will get y = 0

    Hence, x = y = 0 and the triangle is equilateral triangle



  • Q24) A pyramid-shaped box has internal volume of 256 cubic cms and a square base of 8 cm * 8 cm. How many small solid pyramids, each with a volume of 4 cubic cms and a 2cm * 2cm square base ,can we pack wholly inside the box ?

    [OA : 44]



  • On the top, we can place 1 pyramid
    then in the next layer 2 * 2 + 1 Pyramids
    in the 3rd layer we can place 3 * 3 + 4 Pyramids
    and in the last layer we can place 4 * 4 + 9 Pyramids
    So Total = 1 + 5 + 13 + 25 = 44 Pyramids



  • Q25) ∆PQR has median lengths as 10, 24 and 26. What will be the area of ∆PQR?

    [OA : 160]



  • If medians are PX, QY and RZ and centroid be O.
    Extend OX to W such that OX = XW

    In triangle OWR, we can see that OW = (2/3)PX, OR = (2/3)RZ and RW = (2/3)QY
    So, triangle OWR will have 4/9th area of a triangle having sides equal to PX, QY and RZ.

    So, triangle OXR will have 2/9th area of triangle having sides equal to PX, QY and RZ.
    But triangle OXR has 1/6th area of triangle PQR

    Hence, triangle PQR has 4/3rd area of triangle sides equal to PX, QY and RZ.

    As, PX = 10, QY = 24 and RZ = 26
    Hence, area of triangle PQR = (4/3) * 10 * 24/2 = 160



  • Q26) What is the largest positive integer 'n' for which there is a unique integer 'k' such that 8/15 < n/(n + k) < 7/13
    a) 49
    b) 56
    c) 98
    d) 112
    e) 168

    [OA : 112]



  • 8/15 < n/(n + k) < 7/13
    => 13/7 < (n + k)/n < 15/8
    => 6/7 < k/n < 7/8

    Now, LCM of 7 and 8 is 56

    => 48/56 < k/n < 49/56
    => 96/112 < k/n < 98/112

    Here we can see that k can be only 97, now of we increase n then k will have 2 values.

    So, 112 is maximum possible value of n for which k will have a unique value.



  • Q27) Find the number of non-negative integral solutions for the equation 3a + 4b + 12c = 432

    [OA : 703]



  • Method 1 :

    Clearly a is a multiple of 4 and b is a multiple of 3. So, let’s say, a = 4k and b = 3n

    Hence, 12k + 12b + 12c = 432
    k + b + c = 36
    So, C(38, 36) = 703 solutions

    Method 2 :

    3a + 4b + 12c = 432 (1)
    RHS is multiple of 12 so must be LHS, that means
    3a + 4b = 12k (2)
    So, 12k + 12c = 432 = 12*36
    or, k + c = 36 (3)

    (3) has 37 solutions where k varies from 0 to 36. So we are to find the sum of number of solutions of (2) for this range of k.
    Now (2) has k + 1 number of solutions as a is a multiple of 4 varies from 0 to 4k.
    So basically we need to find the sum of an AP of 37 terms whose kth term is k + 1 and k varies from 0 to 36
    i.e. 1 + 2 + 3 + .... + 37 = 37 * 19 = 703



  • Q28) Find the number of ordered triplets (a,b,c) of positive integers such that
    LCM (a,b) = 1000
    LCM (b,c) = 2000 and
    LCM (c,a) = 2000

    [OA : 70]



  • LCM(a, b) = 1000 = 2^3 * 5^3
    LCM(b, c) = 2000 = 2^4 * 5^3
    LCM(a, c) = 2000 = 2^4 * 5^3

    a = 2^x1 * 5^y1
    b = 2^x2 * 5^x2
    c = 2^x3 * 5^x3

    So, x3 = 4
    => At least one of x1, x2 has to be 3, else LCM(a, b) will not be 1000.
    SO, 4^2 - 3^2 = 7 ways

    Now, at least two of y1, y2, y3 should be 3, else one of the LCM's will not be as desired.
    So, 3 * 4 - 2 = 10 ways

    So, answer will be 70 ways.



  • Q29) H.C.F. of three positive integers – A, B and C – is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?

    [OA : 18]



  • a + b + c = 48
    6k + 6n + 6m = 48
    k + n + m = 8, where gcd(k, n, m) = 1
    for k + n + m = 8, gcd(k, n, m) can take only two values 1 or 2

    total solutions of the equation k + n + m = 8 is 7C2
    when gcd is 2, p + q + r = 4, so 3C2 = 3 ways

    so, 7C2 - 3 = 18 triplets



  • Q30) a + 1/b = 7/3
    b + 1/c = 4
    c + 1/a = 1
    Find abc

    [OA : 1]



  • Multiply them to get
    (a + 1/b)(b + 1/c)(c + 1/a) = 28/3
    => a + b + c + 1/a + 1/b + 1/c + 2abc = 28/3
    => 22/3 + 2abc = 28/3
    => abc = 1



  • If h1, h2, h3 are the heights corresponding to the sides a, b, c, then

    a * h1 = b * h2 = c * h3 = 2A (where A is area of the triangle)

    => a = 2A/h1, b = 2A/h2 and c = 2A/h3

    Now, a + b > c
    => 2A/h1 + 2A/h2 > 2A/h3
    => 1/h1 + 1/h2 > 1/h3
    => 1/2 + 1/3 > 1/h3
    => h3 > 6/5

    Also, b + c > a
    => 2A/h2 + 2A/h3 > 2A/h1
    => 1/h2 + 1/h3 > 1/h1
    => 1/3 + 1/h3 > 1/2
    => h3 < 6

    So, 6/5 < h3 < 6
    => Integral values that h3 can take are 2, 3, 4, 5



  • minimum of |x|+|x-17|...+|x-440|
    median of 0,17,40,85,173,440 : (40+85)/2 = 62.5
    put x=62.5..so 641
    [swetabh kumar]



  • for 1st meet: ad/(a+b)= 400
    distance by A til 2nd meet: a/(a+b) * 3d = d + 300
    so 1200 = d + 300, d = 900
    [swetabh kumar]



  • ab - a - b = 35
    => (a - 1)(b - 1) = 36
    (a, b) = (2, 36), (3, 19), (4, 13), (5, 10), (7, 7)
    In last two cases a and b are not coprime



  • case 1: when there is no power of 2
    so 3 * 3 * 3=27 cases
    case 2: when there is only 1(2) in it
    one of the dice will have 2 or 6... and other two dices will have odd numbers.. so
    2 * 3 * 3 * 3c1=54
    case 3:
    when there are exactly two (2's)
    two dices will have (2,6) and one dice will have (1,3,5)
    so 2 * 2 * 3 * 3c1=36
    part2:
    exactly one dice will have 4 in it.. other two will have (1,3,5)
    so 3 * 3 * 3c1=27

    adding all 144 cases which will not have a multiple of 8.. remaining 72..
    so 72/216=1/3

    [jasneet dua]


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