Quant Marathon by Gaurav Sharma - Set 10


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    Compute 1/10^0 + 1/10^1 + 2/10^2 + 3/10^3 + 5/10^4 + 8/10^5 + 13/10^6 + …

    1. 100/89
    2. 100/81
    3. 100/99
    4. None of these

    S = 1/10^0 + 1/10^1 + 2/10^2 + 3/10^3 + 5/10^4 + 8/10^5 + 13/10^6 + …

    S/10 = 1/10^1 + 1/10^2 + 2/10^3 + 3/10^4 + 5/10^5 + 8/10^6 + 13/10^7 + …

    S – S/10 = 1/10^0 + 0 + 1/10^2 + 1/10^3 + 2/10^4 + 3/10^5 + 5/10^6

    S – S/10 = 1/10^0 + 1/10^2 (1/10^0 + 1/10^1 + 2/10^2 + 3/10^3 + 5/10^6)

    S – S/10 = 1 + S/10^2

    S (1 – 1/10 – 1/10^2) = 1

    S = 100/89

    In the given figure what is the radius of the inscribed circle

    1. 3/2
    2. 5/2
    3. 7/5
    4. None of these

    Ar(ACD) + Ar(CDB ) = Ar(ABC)

    5r/2 + 3r/2 = 12/2

    8r = 12

    r = 3/2

    How many numbers from 1 – 2000 are such that at least two of their digits are same?

    1. 800
    2. 757
    3. 758
    4. 750

    Single digit number with no digit repeated = 9

    Two digit numbers with no digit repeated = 9C1 x 9C1 = 81

    Three digit numbers with no digit repeated = 9C1 x 9C1 x 8C1 = 9 x 9 x 8 = 648

    Four digit numbers less than 2000, with no digit repeated = 9C1 x 8C1 x 7C1 x 1 = 9 x 8 x 7 = 504

    Total numbers with no digit repeated = 9 + 81 + 684 + 504 = 1242

    Numbers with at least two digits same = 2000 – Numbers with no repetition

    2000 – 1242 = 758

    A triangle has sides 6, 7 and 8. The line through it’s incenter parallel to the shortest side is drawn to meet other two sides at P and Q. Then find the length of the line segment PQ

    1. 4
    2. 30/7
    3. 15/7
    4. Cannot be determined

    Area of triangle = r x s

    21r/2 = 6h/2 = 3h

    r/h = 2/7

    APQ and ABC are similar thus,

    ( h – r)/h = PQ/6

    Or 1 – r/h = PQ/6 = > 1 – 2/7 = PQ/6

    PQ = 30/7

    N = 99^3 – 36^3 – 63^3. How many factors does N have?

    1. 48
    2. 84
    3. 96
    4. 134

    N = 99^3 – 36^3 – 63^3

    N = 99^3 + (– 36) ^3 + (– 63) ^3

    N = 3 x 99 x 36 x 63 [if a + b + c = 0, then a^3 + b^3 + c^3 = 3abc]

    N = 2^2 x 3^7 x 7 x 11

    Number of factors of N = (2 + 1) (7 + 1) (1 + 1) (1 + 1) = 96

    A person is travelling on the given grid and has to go from A to B through C – D. in each move he can take a step in either the north direction of the east direction. Following the given instructions, in how many different ways can he travel from A to B

    Number of ways to travel from A to C = 6! / (3! X 3!) = 20

    Number of ways to travel from C to D = 1

    Number of ways to travel from D to B = 6!/(4! X 2!) = 15

    Total ways = 20 x 15 = 300

    AC is the diameter of a circle with center O. OE and OF are perpendicular to AD and AB respectively such that A – F – B and A – E – D. If perimeter of ABCD is x, then what will be the perimeter of AEOF?

    1. x/2
    2. 2x/3
    3. x/3
    4. root(2)x/3

    Triangle AEO ~ Triangle ADC ( AA Similarity)

    EO/DC = AE/AD = AO/AC = 1/2

    Similarly, Triangle AFO ~ Triangle ABC (by AA Similarity)

    FO/BC = AF/AB = AO/AC = 1/2

    AD + DC + CB + AB = x

    1/2 (AD + DC + CB + AB ) = x/2

    AE + OE + OF + AF = x/2

    How many positive divisors does 7^12 + 7^13 + 7^14 + 7^15 have

    1. 200
    2. 199
    3. 195
    4. 197

    7^12 + 7^13 + 7^14 + 7^15 = 7^12 (1 + 7 + 49 + 343)

    = 7^12 x 400

    = 7^12 x 2^4 x 5^2

    Number of factors:

    (12 + 1) (4 + 1) (2 + 1) = 195

    There are 10 numbers and none of them are divisible by 3. What will be the remainder when sum of their squares is divided by 3?

    a) 0
    b) 1
    c) 2
    d) 1 or 2

    The Square of a natural number other than one is either a multiple of 3 or exceeds a multiple of 3 by 1. In other words, a perfect square leaves remainder 0 or 1 on division by 3.

    Here it is said that none of them is divisible by 3. Hence each square term will give a reminder of 1

    So for 10 terms, 1 + 1 + 1 … + 1 (10 times) = 10

    10 mod 3 = 1

    Find the value of [root (1)] x [root (2)] x [root (3)] x … x [root (100)]

    1. 2^87 x 3^59 x 5^13 x 7^14
    2. 2^88 x 3^58 x 5^12 x 7^15
    3. 2^81 x 3^59 x 5^12 x 7^17
    4. 2^79 x 3^61 x 5^21 x 7^12

    To simplify, find the number of 7’s in the product [root (1)] x [root (2)] x [root (3)] x … x [root (100)]

    7s will be from [root (49)] to [root (63)]

    i.e 63 – 48 = 15

    From the given options only b satisfies the condition.

     


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