Quant Boosters by Hemant Malhotra - Set 10


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


    Two workers A and B completed a job in 10 days, in which A did not work for last 2 days. In the first seven days they completed 80% of the work. In how many days would A alone complete the work.
    A 12
    B 10
    C 16
    D 14

    let total work=100 unit
    in first seven days they completed 80 unit work
    so work done in one day =80/7 unit
    now work done on 8th day =80/7
    so 80*8/7=640/7 unit
    so remaining work=60/7 which is completed by B in 2 days so one day work=30/7
    so one day work of A=50/7
    so days taken by A=100/(50/7)=14 days

    In a bag there are certain number of black socks, brown socks and grey socks - all in pair of different brands. The number of socks of different colours satisfy the following conditions:
    (i) The minimum number of socks to be drawn from the bag to ensure one pair of socks of the same brand (of any colour) is equal to 32.
    (ii) The minimum number of socks to be drawn from the bag to ensure one pair of grey socks of any particular brand is nine less than the total number of socks.
    (iii) The number of brands of brown socks is equal to the number of black socks in the bag.

    1. What is the total number of socks in the bag?
    2. What is the minimum number of pairs of socks to be drawn from the bag to ensure a pair of black socks from the bag?
    3. What is the minimum number of socks to be drawn from the bag to ensure a pair of black socks of any particular brand from the bag?

    LET TOTAL NUMBER OF BROWN BRANDS= 2b
    Total number of brown shoes=4b
    total number of black brands=b
    total number of black shoes=2b
    total number of grew brand=c
    total number of grew shoes=2c
    The minimum number of socks to be drawn from the bag to ensure one pair of socks of the same brand (of any colour) is equal to 32.
    so 2b+b+c+1 will give us one brand pair of shocks
    so 3b+c+1=32 so 3b+c=31
    The minimum number of socks to be drawn from the bag to ensure one pair of grey socks of any particular brand is nine less than the total number of socks.
    so 4b+2b+c+1=((4b+2b+2c-9)
    so c=10
    so 3b=21 so b=7
    so brands- brown=14, black=7 and green=10
    socks= brown=28 , black=14 and green=20

    Remainder when 6^22 + 8^12 is divided by 49

    6^22+8^22 mod 49
    (7-1)^22 +(7+1)^22 mod 7^2
    22Co * 7^22 - 22c1 * 7^21 * + 22c2 * 7^20+................ 22c21 * 7^1(-1)^22))+22c0 * 7^22+.........+...... 22c1 * 7 +22c22 * 1
    so everything will be div except
    -22c1 * 71+1+22c1 * 7+1=2 so remainder=2

    P and Q finish a work in 15 days, p and r take 2 days more to complete the same work than q and r .p q and r can complete the work in 8 days in how many days p can complete the work separately?

    let work =120
    so P+Q's 1 day work=8 unit
    let P's one day work=a
    Q's one day work=8-a
    and P+Q+R's one day work=15 unit
    so R's one day work=7 unit
    now P+R's one day work=a+7 so time taken to complete =120/(a+7)
    and Q+R's one day work=15-a so 120/(15-a)
    120/(a+7)=120/(15-a) +2
    120((22)/(a+7) * (15-a)=2
    so 120 * 11=(a+7)(15-a)
    solve and find answer

    A & B throws a pair of dice and winner is one who gets a sum of 9 first.if A starts throwing first Then what is the ratio of A winning over B?

    Total possibilities=36
    now to find sum=9 we have
    3+6
    4+5
    6+3
    5+4
    so 4 possibility so to get sum 9 prob=1/9
    to not get sum 9 prob=8/9
    if A won prob=1/9
    if A lose B won prb =8/9*1/9
    if A lose B Lose A won =8/9 * 8/9 * 1/9
    so prob of A winning =1/9+(8/9))^2 * 1/9+(8/9)^4 * 1/9+.....
    Infinite Gp sum=1/9/(1-((64/81))=9/17
    prob of B winning =8/17 so ratio=9:8

    A circular pizza is cut into 4 identical pieces with 2 diametrical cuts perpendicular to each other and topped any one of the 5 toppings available such that no 2 adjacent slices have the same topping. In how many ways can the 4 pieces be topped?

    all different = 5C4 * 3!=30
    2 same 2 different = 5C1 * 4C2 = 30
    2 same 2 other same = 5C2 = 10
    Total = 70

    x, y and z are positive real numbers such that y – x = z and xyz = 4. Find the least value of Y
    2y = x + z
    x y z = 4
    Apply AM > = GM
    (x+y+z)/3 >= (xyz)^1/3
    (3y/3)>=(4)^1/3
    So y>=2^(2/3)

    If x = (60^99 – 58^99)/ (60^98 – 58^98), which of the following holds true?
    a) 0 < x 2.0

    Method1- x^n - a^n=(x-a)(x^n-1+x^(n-2)a+(x^(n-2) * a^2......a^(n-1)
    so (60)^99-(58)^71
    = (60-58)(60^98+60^97 * 58+...58^98)
    2((60^98+60^97*58+...58^70)/60^98+58^98
    so >2
    Method2- - lower all the powers accordingly... because all these equations follow general trend... let 99--- 3 , 98 --- 2 ,97---1 ,96 --- 0 ... the equation becomes (60^3 - 58^3/(60^2 + 58^2) . solve this, it will give 2.99

    Let f(x) = x^4 + ax^3 + bx^2 + cx + d be a polynomial whose roots are all negative integers. If a + b + c + d = 2009, find d.

    let roots=p,q,r,s
    if u will try to find sum of roots , product of roots and other things , then it will be very lengthy .. so don't try to form linear equations in this type of questions where u have only one information given
    let roots -p,-q,-r and -s
    so x^4+ax^3+bx^2+cx+d=(x+p)(x+q)(x+r)(x+s)
    now we have given value of a+b+c+d ... so put x=1 that will reduce our equation into a+b+c+d+1
    put x=1
    1+a+b+c+d=(1+p)(1+q)(1+r)(1+s)
    1+2009=2010=(1+p)(1+q)(1+r)(1+s)
    2010=2 * 3 * 5 * 67
    so p=66, q=4 , r=2 and s=1
    so our roots are -66,-4,-2,-1
    now we want to find d .. .. d is actually product of roots
    so d=-66*-4*-2*-1=528

    Find the sum of real roots of the equation (2x^2 – x – 1) * (2x^2 – 9x + 9) = 40
    a) 5/2
    b) 5
    c) 15/2
    d) 10

    2x^2-x-1
    2x^2-2x+x-1
    2x(x-1)+1(x-1)
    (2x+1)(x-1)
    now (2x^2-6x-3x+9)=2x(x-3)-3(x-3)=(2x-3)(x-3)
    so (x-1)(x-3)(2x+1)(2x-3)=40
    (x-1)(2x-3)(x-3)(2x+1)=40
    (2x^2-5x+3)(2x^2-5x-3)=40
    let 2x^2-5x=y
    Solve and answer would be A


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