Quant Boosters by Hemant Malhotra - Set 7


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


    If x = ( 9 + 4 * root(5))^48 = [x]+f,
    where [x] is defined
    as integral part of x and f is a fraction,
    then x(1–f) equals –
    A. 1
    B. Less than 1
    C. More than 1
    D. Between 1 and 2
    E. None of the above

    Note -

    1. in R=(a+b * rtc)^k
      if k is odd then f=X
      fractional part will be f=(a-b * rtc)^n
    2. and if k is even then f+X=1
      f=1-(a-b * rtc)^k

    here x=(9+4sqrt5)^48
    so f=1-(9-4sqrt5)^48
    so x * (1-f)=(9+4sqrt5)^48 * (9-4sqrt5)^48 =1

    p, q and r are roots of the equation x^3 - 7x^2 - 6x + 5 = 0. Find the value of (p + 1)(q + 1)(r + 1).

    we need (p+1) (q+1) (r+1)
    so need equation whose roots are p+1 ,q+1 and r+1
    so apply transformation replace x by x-1
    (x-1)^3-7(x-1)^2-6(x-1)+5=0
    x^3-1-3x(x-1)-7x^2-7+14x-6x+11=0
    x^3-10x^2+11x+3=0
    so product of roots will be -3

    p, q, r are roots of the equation x^3 - 7x^2 - 6x + 5 = 0. Find the value of (p + q)(q + r)(r + p).

    p + q + r = 7 so p + q = 7 - r
    q + r = 7 - p
    and p + r = 7 - q
    so we need (7-p)(7-q)(7-r)
    means we need equation whose roots are 7-p,7-q,7-r or
    x^3-7x^2-6x+5=(x-p)((x-q))(x-r))
    put x=7
    so 7^3-7 * 7^2-6 * 7+5 =- 37 OA

    If three medians of the triangle are of the lengths 9, 12 and 15, find the area of the triangle.

    Method 1 – Direct formula
    Area of triangle using medians, A = 4/3 ( Sqrt(s(s-p)(s-q)(s-r))
    p, q and r are medians and s = (p+q+r)/3

    Method2- ABC is the triangle, AD, BE, CF are medians and G centroid
    extend GD to a point M such that GD = DM, then CM = BG
    so, triangle MCG; has sides 6, 8 and 10
    area = 24
    area of ABC = 24 * 3 = 72.

    Method 3
    4 * (a^2+b^2+c^2)=3 * (9^2+12^2+15^2)
    so a^2+b^2+c^2=3 * (81+225+144)/4
    now we have a^2+b^2+c^2
    apply a^2+b^2=2*(AD^2+(c^2/4))

    Given 3x^2+x=1, find the value of 6x^3-x^2-3x+2015.

    6x^3+2x^2 = (2x)(3x^2+x) = 2x
    and -(3x^2+x) = -1
    so 6x^3-x^2-3x+2015 = 2x-1-2x+2015 = 2014.

    Let a, b and c be the roots of x^3 - 2x^2 + 2x + 4 = 0.
    Find the value of 1/(a^2 + b^2) + 1/(b^2 + c^2) + 1/(c^2 + a^2) + 1/4.

    a + b + c = 2
    ab + bc + ac = 2
    abc = -4
    here a+b+c=ab+bc+ca so a^2+b^2+c^2=0
    question reduces to
    -(1/a^2 + 1/b^2 + 1/c^2) + 1/4
    now we can easily find the value by solving this

    Another method : x^3-2x^2+2x+4=0
    roots are a,b,c equation whose roots are a^2,b^2,c^2
    will be (x^3/2)-2x+2sqrtx+4=0
    x*sqrtx-2x+2sqrtx+4=
    x^3+4x+4x^2=4x^2+16-16x
    so x^3+20x-12=0
    now roots of this equation are a^2,b^2,c^2
    so equation whose roots are 1/a^2,1/b^2,1/c^2 will be reverse of this
    so 1/x^3+20/x-12=0
    -12x^3+20x^2+1=0
    so 12x^3-20x^2-1=0
    so 1/a^2+1/b^2+1/c^2=5/4
    so -5/4+1/4=-1

    From 100 litres of milk, 10 litres of water is added and then 20 litres of solution is removed. Next 30 litres of the water is added and 20 litres of solution is removed. Find the amount of milk, in litres, in the solution now.

    Approach- #MULTIPLYING FACTOR
    Method to tackle These Type Of Questions
    always take initial as those part which is not added,
    here milk as it is never added
    here in first step 10 litre of water is added so solution=100+10=110
    in the 1st time 20 is removed out off 110 so
    left-out part is 90/110 so our multiplying factor is 90/110
    no 30 is added to 90 so 120 solution
    in the 2nd case 20 is removed from 120 so left out is 100/120
    so amount of milk= 100 * 90/110 * 100/120=750/11
    PS- Always remember Our concentration should be on total solution and removed part .... Added part will only increase the value of solution ..

    A shopkeeper claims to sell his items at a loss of 10% but instead he secretly makes a profit of 20% by using a false weight. what is the effective profit percentage he gets on every 100gm he sells?

    Method1- Approach= He claims to sell 120 when he actually sells 100 (using false weights)
    Now, he claims to sell them at a loss of 10%
    He makes 120*0.9 = 108
    so 8% profit

    Method2- = Multiplying factor
    100 * 9/10 * 12/10=108 so 8%

    Method3- 20-10+(20*(-10)/100=8%

    A dishonest trader, at the time of selling and purchasing, weighs 10% less and 20% more per kilogram respectively. Find the percentage profit earned by treachery. (Assuming he sells at Cost Price)
    a) 30%
    b) 20%
    c) 25%
    d) 33.33%

    Method1- Let's say original weight is 100 now he is taking 20% extra so total 120 and selling 10% less so 90
    So (120-90/90) * 100
    = 30/90 * 100= 33.33%

    Method2-
    multiplying factor
    1 * 10/9 * 6/5=1.33 so 33%

    A beaker contains milk and water in the ratio 5 : 3. If a person removes 6 L of this mixture and replaces it with pure water, the ratio of milk to water becomes 1 : 1. What is the amount of milk present in beaker initially

    Always remember one thing
    Final =Initial * fraction left 1 * fraction left 2 * fraction left 3 *( so on)
    Answer = 75/4


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