Quant Boosters by Hemant Malhotra - Set 3


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Ramu chooses five distinct numbers. In how many different ways can he assign these numbers to the variables p, q, r, s, and t so that p < s, q < s, r < t, and s < t?
    (A) 4
    (B) 5
    (C) 6
    (D) 8
    (E) 15

    t is definitely the largest no..
    now there are 2 cases when r is > s then there are 2 possible cases
    t r s p q and trsqp
    when r < s then we have t s _ _ _
    we have 6 possible cases
    hence total 8

    How many positive integral solution does the following equation has:-
    14x + 21y + 42z = 840
    2x+3y+6z=120

    now x should be multiple of 3 and y should be multiple of 2
    so x=3k and y=2m
    so 6k+6m+6z=120
    so k+m+z=20
    so 19c2=19*9=171

    If x and y are integers with (y − 1)^(x + y) = 4^3, then the number of possible values for x is
    (A) 8
    (B) 3
    (C) 4
    (D) 5
    (E) 6

    (1) (y-1)^(x+y) = 64
    y = 65 and x = -64
    (2) (y-1)^(x+y) = 8^2
    y = 9 and x = -7
    y = -7 and x = 9
    (3) (y-1)^(x+y) = 4^3
    y = 5 and x = -2
    (4) (y-1)^(x+y) = 2^6
    y = 3 and x = 3
    y = -1 and x = 7
    so 6 values

    What is the largest power of 2 that divides the number K = 75! – 71!

    71! ((75 * 74 * 73 * 72 - 1))
    75 * 74 * 73 * 72 - 1 is odd number so no power of 2
    so we basically need power of 2 in 71!
    [71/2]+ [71/4] + [71/8] + [71/16] + [71/32] + [71/64]
    35+17+8+4+2+1=67

    The hypotenuse of a right triangle is 10 cm, and the radius of the inscribed circle is 1.5 cm. Find the perimeter of the triangle.

    x - 1.5 + y - 1.5 = 10
    x + y = 13,
    x + y + 10 = 23

    0_1486535326287_img_1.png

    0_1486535341377_img2.png

    Now mug this formula too
    min(sqrt(a^2+(b+c)^2),sqrt(b^2+(a+c)^2),sqrt(c^2+(a+b)^2))
    Here a=4, b=3, c=2 so min will be sqrt((4^2+(3+2))^2)=sqrt41

    Suppose a function f is defined by f(1)=1,f(2)=2
    f(n)=2f(n-1) + f(n-2) for n≥3.
    find f(100) - 2f(99) - 2f(97) - f(96)

    f(100) = 2f(99) + f(98)
    f(98) = 2f(97) +f(96). Adding them =0

    In which of the base is the operation 61 × 61 = 4541 valid?

    let base n
    (61)n*(61)n=(4541)n
    1+6n)(1+6n)=1+4n+5n^2+4n^3
    1+12n+36n^2=4n^3+5n^2+4n+1
    so n=8

    What is the remainder when 27^11+9^11+3^11+1 is divided by 3^10-1

    3^10=x
    so 27 * 3^30 + 3^2 * 3^20 + 3 * 3^11 + 1
    so 27 * x^3 + 9x^2 + 3x + 1 put x=1
    so 27 + 9 + 3 + 1 = 40

    What is the last digit of the LCM of 7^5678 - 1 and 7^5678 + 1 ?

    HCF = 2
    LCM * HCF = first number * 2nd number
    LCM * 2 = ( 7^5678 - 1) * (7^5678 + 1)
    unit digit of 7^(5678)) = 9
    so unit digit of 7^(5678) - 1 will be 8
    so unit digit of LCM will be 7^(5678) + 1 will be 0
    so * * * 0 * * * 8/2=0 (((but u have to check that 2nd digit is even or odd )) its even that's y OA+0

    2nd method to remove this thing
    LCM * 2=((7^5678 -1) * (7^(5678+1)
    = ((49^(5678)-1))
    (49-1)(1 + 49 + 49^2 + ... 49^5677)/2 = 24(1 + 49 + 49^2 + ... 49^5677
    49^k is mod 10=1 when k is even and is -1 when k is odd
    now check inside portion is zero


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