Algebra Concepts by Anurag Chauhan - Set 1


  • QA/DILR Mentor


    if a + b + c = 0 , find the value of a^2/bc + b^2/ca + c^2/ab ?

    first method – find lcm of denominator i.e ( a^3+b^3+c^3)/abc = 3abc/abc= 3
    Since a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

    Secod method- put value of a,b,c such that rhs should be zero as
    a = 1 , b= 2 and c= -3
    put these value in the expression u will get required ans .

    If a + 1/a =2 find a^37 + 1/a^100 ?

    Put a=1 , and rhs satisfied so u will get ans as 2

    If (a - 1)^2 + (b + 2)^2 + (c + 1)^2 = 0 , then the value of 2a – 3b + 7c ?

    Sum of perfect squares should be zero if every term individually becomes zero .so
    a-1 = 0 => a = 1
    b + 2 = 0 => b = -2
    c + 1 = 0 => c = -1
    so 2 * (1) – 3(-2) +7(-1) = 2 + 6 - 7 = 1

    Minimum value of x^2 + 1/(x^2 + 1) - 3 ?

    Minimum value of perfect square should be zero .. so put x =0
    You will get ans -2

    If x = 3 + 2√2 , Find √x + 1/√x ?

    As (a+b)^2=a^2 + b^2 + 2ab ,
    Now equate 2ab = 2√2
    ab = √2 x 1
    so a= 1 and b = √2 , now
    √x = (1+√2)
    1/√x = √2-1
    On adding it becomes 2√2

    If x^3 + y^3 = 35 , and x + y = 5 , find 1/x + 1/y .

    Given that x+y = 5 , cubing both sides
    x^3 + y^3 + 3xy (x + y) = 125
    35 + 15 xy = 125
    xy = 6
    now x+y/xy = 5/6
    1/x + 1/y = 5/6

    #ALTER
    Put x = 3 and y = 2 … equations satisfied so
    1/3 + 1/2 = 5/6

    If a= 70, b= 71 , c = -141 , find a^3 + b^3 + c^3 ?

    Here a + b + c = 0, using identity
    a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
    a^3 +b^3 +c^3 = 3abc ( since a+b+c=0)
    so ans -3 x 70 x 71 x 141

    If p =124 , find (p(p^2 + 3p + 3) + 1)^1/3 ?

    First of all we open a bracket and try to it a identity so ..
    p(p^2 + 3p + 3) + 1 = p^3 + 3p^2 + 3P + 1 = (p + 1)^3
    Its cube root will be = (p + 1)
    And p= 124
    Ans = 125

    #Alter
    For better approach you have to take p=1 and solve u will get 8 and cube root of 8 is 2
    It means if we take a value of p =1 , we get 2 , similarly if p = 125 we will get p = 125
    No need so solve further for identity

    The unit digit of 27^16 – 18^48 ?

    This question seems easier but it is little tricky, now solve
    Unit digit of any number can be find out by cyclicity.
    cycle of unit digit is four so expression reduces into 7^4k – 8^4k = …..1 - ….6 = (-5) but unit digit can not in negatives so we take carry as 10 -5=5

    If a/b = c/d = e/f = 3/1 , then (2a^2 + 4e^2 + 3c^2)/(2b^2 + 3d^2 + 4f^2) ?

    For smarter approach you have to take values in this manner
    Let a/b = 3/1 => a=3b
    c = 3d , e = 3f
    Put these values u will get 9 .

    NOW WE COME TO SOME MODERATE QUESTIONS OF BASIC ALGEBRA …

    Find all ordered pairs of real numbers which satisfy 17a^2 + 2b^2 - 10ab - 6a + 2 = 0

    Write equation in this way as 2b^2 – 10ab + (17a^2 - 6a +2) = 0
    Eqn is quadratic
    Find discriminant D = b^2 – 4ac ( ax^2 + b x +c = 0 )
    D = (-10a)^2 – 4*2 (17^2 -6a+2)= -4(3a-2)^2 , this is negative unless a= 2/3 , now solving for b , it is 5/3
    (a,b)= (2/3,5/3)

    Find the sum of digits in the product 199999997 * 200000003 ?

    Use identity ( a-b) (a+b)= a^2 –b^2 , take a= 2 * 10^8 , b = 3 , so the product is
    39999…(15 times)1
    Sum = 3 + 15 * 9 + 1 = 139

    A three digit natural number xyz is written side by side to form a six digit natural number n = xyzxyz . then it must be divisible by ?

    N = abc * (1001)
    It must be divisible by 7, 11,13 because 7 * 11 * 13= 1001
    Also it must be divisible by factor of abc

    If (x-7)^{ (x^2 - 29x + 154)/(x^2 - 12x + 32)} = 1 , how many real values of x satisfy the given eqn ?

    (x-70^{(x-7)(x-22)/(x-8)(x-4)} , now let we take f(x) = (x-7) and G(x) = (x-7)(x-22)/ (x-8)(x-4) , now we take three cases .
    i) If f(x) = 1 and g(x) can be anything
    so x = 8 , for this value g(x) can not be defined
    ii) if f(x)=-1 , g(x) is an even , then x = 6 , g(x) is defined
    iii) if f(x) does not zero and g(x) is zero ,
    (x-7)(x-22) = 0 => x=7,22 , but for x=7 the eqn is not defined , so eqn is defined for two reals x i.e 6,22

    If x and y are two natural numbers such that by x = (5y-4)(5y+1) if 1 ≤ y ≤ 100 , what is the harmonic mean of all possible values of x ?

    Sol. When x=1 , y = 1 * 6
    x=2, y= 6 * 11
    x= 200, y= 996 * 1001
    harmonic mean = 2 * 100/{1/(1 * 6)+(1/6 * 11) +………..+1/(996 * 1001)}
    denominator can be solved as ….
    1/(1 * 6) = 1/5{ 1- 1/6} , 1/6 * 11 = 1/5( 1/6-1/11) …..
    Now , H.M = 200/{ 1/5( 1- 1/1001) } = 1001.



  • @anurag_chauhan The last sum is some sort of miscalculation I guess, because according to the sum when y=1 then x = 1 * 6 and not vice versa. In that case the value of y extends upto 100 and x upto 496*501 for which the ans should be 501.


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