Quant Boosters - Hemant Yadav - Set 1



  • It is (n +2)! + 2[(n + 1)!] - 4 = (n + 4)[(n + 1)!] - 4.
    Key lies in simply finding the nth term of the sequence which is (n + 1)[(n + 1)!] + 2n[n!].



  • Q19) How many natural numbers N < 40000 such that cube of sum of digits of N gives back the same number N.
    For eg, 8³ = 512, sum of digits is 8, same as cube root of 512.

    [OA : 6]



  • Since N < 40000, N < 35³

    It is clear from the given condition that the sum of digits and cube of sum of digits should leave same remainder when divided by 9.

    Hence sum of digits should be of form 9k + 1 or 9k - 1 or 9k

    1³ = 1 (sum of digits is 1, so possible)
    8³ = 512 (sum of digits is 8, so possible)
    9³ = 729 (sum of digits is 18, so not possible)
    10³ = 1000 (not possible)
    19³ = 6859 (sum of digit is not 19, so not possible)
    18³ = 5832 (sum of digits is 18, so possible)
    17³ = 4913 (sum of digit is 17, so possible)
    26³ = 17576 (possible)
    27³ = 19683 (possible)
    28³ = 21952 (not possible)

    So, 1, 512, 4913, 5932, 17576 and 19683 are the only such numbers



  • Q20) How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form i, j, k where i, j and k are positive integers not exceeding four?
    a) 60
    b) 72
    c) 64
    d) 76

    [OA : 76]



  • We can see that there are 12 planes (4 paralle to x-y plane, 4 parallel to y-z plane and 4 parallel to z-x plane) In every plane there would be 2 diagonals, so 24 lines. 4 diagonals of the cube and 16 lines parallel to every axis. So, total 4 + 24 + 48 = 76



  • Q21) Circles A, B and C are externally tangent to each other, and internally tangent to a bigger circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the center of D. What is the radius of circle B?

    [OA : 8/9]



  • O is the centre of the bigger circle. and A,B,C are the centres of the remaining circles. Circles with centres B and C have the same radius. Radius of circle with centre A = 1.
    Obviously, the radius of the bigger circle is 2.
    Let OD = x (D is the mid-point of BC)
    Then we have: r^2 + x^2 = (2-r)^2
    and (x+1)^2 + r^2 = (1+r)^2
    Taking the diff of the above equations gives x = 3r - 2
    using which we get r = 8/9



  • Q22) For any positive integer n , f(n) is the highest power of 2 that divides n!
    Find f(1) + f(2) + f(3) + ..... + f(1023)

    [OA : 518656]



  • It is similar to finding the highest power of 2 in 1! * 2! * 3! * ...... * 1023!

    First we will count that how many numbers are divisible by 2 in every factorial, then by 4, then by 8, ...., and lastly by 512

    So, S = (0 + 1 + 1 + 2 + 2 + ..... + 511 + 511) + (0 + 0 + 0 + 1 + 1 + 1 + 1 + .... + 255) + ..... + (0 + 0 + .... + 1 + 1 + 1 (512 times))
    = 511 * 512 + 512 * 255 + 512 * 127 + 512 * 63 + 512 * 31 + 512 * 15 + 512 * 7 + 512 * 3 + 512
    = 512(511 + 255 + 127 + 63 + 31 + 15 + 7 + 3 + 1) = 512 * 1013 = 518656

    => f(1) + f(2) + f(3) + ..... + f(1023) = 518656



  • Q23) The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Then what can be said about the triangle
    a) It's a rt angle triangle
    b) It's an equilateral triangle
    c) It's an obtuse angle triangle
    d) Nothing can be said about it



  • 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]


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