Quant Boosters by Hemant Malhotra - Set 9


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


    Determine the number of positive integral solutions (x, y) to x^y = y^(90)

    Factors of 90=2 * 3^2*5=12
    1,2,3,5,6,9,10,15,18,30,45,90
    x=y^(90/y)
    now check for square
    2^2 will satisfy
    3^3
    and 6^2 will satisfy

    Determine the number of positive integral solutions (x, y) to x^y = y^{60}.

    x=y^(60/y)
    all factors of 60 will satisfy
    1,2,3,4,5,6,10,12,15,20,30,60
    now check for squares of factors
    y = 16
    x = 16^((60/16)=2^15
    now no cubes is satisfying
    so OA=13

    Determine the number of positive integral solutions (x, y) to x^y = y^{72}.

    x^y=y^72
    now we will look for factors of 72 and their perfect square
    factors 1,2,3,4,6,8,9,12,18,24,36,72
    now squares of them
    16,,144 will satisfy
    now cubes
    3^3 , 6^3 will satisfy so overall 16 solutions

    Determine the number of positive integral solutions (x, y) to x^y = y^{100}
    x=y^(((100/y))
    factors of 100
    1,2,4,5,10,20,25,50,100
    now check for squares of factors
    at y=16 u will get x integer
    rest no value satisfying
    OA= 10

    If x, y and z are real numbers such that x + y + z = 4 and x^2 + y^2 + z^2 = 6 then x, y and z lie in
    (1) [3/2,2]
    (2) [2/3,2]
    (3) [0,2/3]
    (4) None of these

    x+y+z=4 and x^2+y^2+z^2 =6
    apply Cauchy
    (y * 1 + z * 1)^2 < = (y^2+z^2)(1^2+1^2)
    (4-x)^2 < = ((6-x^2)*2
    16+x^2-8x < = 12-2x^2
    3x^2-8x+4 < = 0
    2/3 < = x < = 2
    so OA=B

    Between 2 junctions A and B there are 12 intermediate stations. The number of ways in which a train can be made to stop at 4 of these stations, so that no two halting stations are consecutive is?

    Method1- Direct formula when u want to choose r numbers out of n numbers such that no two values are consecutive = 12-4+1C4=9c4

    Method2-
    A... A1 .... A2 ... A3... A4 ..... B
    here A1,A2,A3 and A4 are 4 halting stations
    now A1,A2,A3,A4 should not be consecutive
    NOW let x1 be d number of stations before A1
    so x1>=0 (( its possible that after A , next station is A1 )
    A1 and A2 should not be consecutive so there must be atleast one stations
    so let number of stations x2
    so x2>=1
    same for A2 and A3
    x3>=1
    same for A3 and A4
    x4>=1
    now station After A4 could be zero also
    so x5>=0
    total station=12
    and 4 are halting and 8 are other stations
    so x1+x2+x3+x4+x5=8
    where x1>=0 letx1=a
    x2>=1 so x2=1+b
    x3=1+c
    x4=1+d
    x5=e
    so a+b+c+d+e=5
    so number of ways=5+5-1C5-1=9c4

    Find the number of ways of distributing 27 ladoos to Swetabh, Raman and Gaurav such that number of ladoos Swetab gets are more than the number of ladoos Raman gets which in turn is more than the number of ladoos Gaurav gets?

    method 1
    let G=x then R=x+y+1 and S=x+y+z+2
    so 3x+2y+z+3=27
    so 3x+2y+z=24
    now make cases and solve u will get ur ans

    method 2
    S+R+G=27
    now ordered=29c2
    now here find unordered
    case when 2 same one different
    =2S+G=27
    so G=27-2S
    so S will vary from 0 to 13 so 14 values but 1 value will be there when all same
    so 2 same one different case will be 13
    and when all same
    3S=27 so 1 case
    so 6 * a + 3 * 13 + 1=29C2
    so find a and that will be your answer

    If numbers 320ab and 298cd are divisible by 35 and 65 respectively then find the number of divisors of [398cd – 320ab + 2]^10 where 320ab is the largest possible number satisfying the given conditions

    320ab mod 35
    means div by 5 and div by 7
    so b could be o or 5
    0ab-32 mod 7=0
    ab-4 mod 7=0
    ab mod 7=4
    when b=5 then a = 9
    max value of ab=95
    now same process and u will find cd=45
    OA: 41261 ; numbers will be 32095 and 39845.

    A salesman sells two kinds of trousers: cotton and woollen. A pair of cotton trousers is sold at 30% profit and a pair of woollen trousers is sold at 50% profit. The salesman has calculated that if he sells 100% more woollen trousers than cotton trousers, his overall profit will be 45%. However he ends up selling 50% more cotton trousers than woollen trousers. What will be his overall profit?
    A. 37.5%
    B.40%
    C. 41%
    D. 42.33% approach

    Let Cotton CP=a
    Woolen CP=b
    now let Cotton = x then Woolen=2x
    So SP of cotton=1.3 * x * a
    and Wollen =1.5 * 2x * b=3 * x * b
    now overall profit =x*((1.3a+3b))-((ax+2x*b))/((ax+2bx))
    ((0.3a+b))/(a+2b))=45/100=9/20
    6a+20b=9a+18b
    so 3a=2b

    now let woollen =2x then cotton=3x
    so CP = 2x * a + 3x * b
    and SP = 2x * 1.3a + 3x * b * 1.5
    so profit=((2.6a+4.5b-2a-3b)/(2a+3b)
    ((0.6a+1.5b))/(2a+3b))
    here 3a=2b
    so 40% approx

    3 men and 5 women together can finish a job in 3 days. Working on the same job 3 women take 5 days more than the time required by 2 men . What is the ratio of efficiency of a man to a woman ?

    let total work=60 unit
    3 M + 5 W per day work=20 unit
    let M work done in one day=a
    then work work done=(20-3a)/5
    2 M work done in one day=2a
    and 3 work done in one day=((60-9a)/5
    now 60/2a = ((60/((60-9a)/5 -5
    so 30/a +5 =300/(60-9a)
    find a then ratio


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