Quant Boosters - Hemant Malhotra - Set 2


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


    Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source - Elite's Grid Prep Forum


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


    Q1) Find the sum of 3/2 + 5/4 + 9/8 + 17/16 + ... to 99 terms
    a) 100 - 1/2^99
    b) 101 - 1/2^99
    c) 100 + 1/2^99
    d) 100 + 1/2^100


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


    We are asked for 99 terms (n = 99) and is Option A is of form (n+1) - 1/2^n
    now if this is true then we can represent other numbers also in this form
    3/2 = (2 - 1/2^1) = 3/2 ( n = 1)
    3/2 + 5/4 = (3 - 1/2^2) = 11/4 ( n = 2) and so on.
    so OA = 100 - 1/2^99


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


    Q2) A and B working together can build a wall, 221 m long in 11 1/9 days. If they work on alternate days with A starting the work it will take 22 1/4 days to build it. If A and B work together and build a wall of twice the length and earn a total of Rs 1800 for it, then B's share of earning will be


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


    Wall length = 221m ; Time taken in days 100/9
    so in one day A and B can complete 221 * 9/100 of the work.
    so when they work in alternate way they will do this work in 2 days
    so in 22 days they will do 221 * 99/100 unit work so remaining work = 1/100
    and days =89/4
    so remaining days = 1/4
    so in 1/4 days A wil do 1/100 work
    so in 1 day A will do 1/25 th work
    so work done by B = 9/100 - 1/25 = 5/100 = 1/20
    so ratio = 4 : 5
    so B's sharing will be 5/9 * 1800 = 1000 rs


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


    Q3) The nth term and the sum of the first n terms of a sequence are Tn and Sn respectively. If Tn = T(n-1) - T(n-2) and Tn ≠ 0, then which of the following is definitely true?
    a) S88 = S188
    b) S66 = S160
    c) S100 = S160
    d) S120 = S142
    [note:digits after S is in subscript]


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


    Tn = T(n-1) - T(n-2) has cycle of 6 & sum of any six consecutive terms of this function is zero
    so subscript should yield the same remainder when divided by 6. use this in options and only S100 = S160 satisfies. (Both 100 and 160 yield remainder as 4 when divided by 6)


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


    Q4) If f(x) = 1/(1 + x) then f(3x) when expressed in terms of f(x) will be
    a) 3f(x)
    b) f(x)/(3 + 2f(x))
    c) f(x)/(3 - 2f(x))
    d) f(x)/(2 - 3f(x))


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


    f(x)=1/(1+x)
    f(3x)=1/(1+3x)
    now Question is simple and date is not that much . if data is lengthy then we can apply Options approach
    take x=1
    so f(1)=1/2
    and f(3)=1/4
    now f(3)=f(1)/2
    so f(1)=2f(3)
    a) out of picture
    b) f(1)/(3 + 2f(1) = 1/2/(3 + 1)) = 1/6 not equal to f(3)
    c) f(1)/3 - 2f(1) = 1/2/((3 - 1)) = 1/4 so OA = C


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


    Q5) Five friends attended a party. At the end of the night, they each randomly grabbed a left shoe and a right shoe. Find the number of ways such that each person left with exactly one of their own shoes?
    a) 104
    b) 112
    c) 120
    d) 128


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


    Conventional method :
    Each person left with exactly one of their own shoes means either they will get rite one or left one so derangement of 5
    so D(5) = 44
    so 2 * 44 = 88
    now 4 of them will get right then obviously 5th one will get right
    now 3 of them correct then
    so 5c3 * 2 = 20
    and 2 of them correct = 2 * 5c3 = 20
    so 88 + 40 = 128

    Quicker method:
    Every friends has two options (left or right) so ans must be multiple of 2^5 and only 128 satisfies.


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


    Q6) A contractor hired a group of men and women to do a piece of work in 12 days. But everyday for the first 12 days, one worker, either a man or a woman, was absent, and hence the work was delayed by 3 days. The contractor made an interesting observation that if the number of days for which women and men remained absent were interchanged, then the work would take 16 days to complete. Approximately how many days would it take for one man and one woman to complete the work?
    a) 16
    b) 20
    c) 24
    d) Cannot be determined


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


    Since its a 12 day work, in first case they lost 1/4th job
    and in second case they lost 1/3rd job
    12 men and 12 woman can do (1/3) + (1/4) = 7/12 work in 1 day
    so They can finish the work in 12/7 days
    so 1 woman and 1 man can finish the work in 12 * 12/7 = 144/7 or 20.57 days


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


    Q7) Let S(n) denote the sum of first n terms of an Ap. If S(2n) = 3 * Sn then find ratio of S(3n)/S(n)
    a) 4
    b) 6
    c) 8
    d) 10


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


    let first term = 1 and n = 1
    so S(2) = 3 * S(1)
    S(1) is sum of first term means first term itself.
    S(2) = 3 and S(1) = 1
    So sum of 2 terms = 3
    Sum of first term = 1
    So 2nd term = 3 - 1 = 2
    First term = 1
    Second term = 2
    So d = 1
    Means third term = 3
    So S(3) = 1 + 2 + 3 = 6
    S(3)/S(1) = 6


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


    Q8) There are 30 cards. Each card has a single number written on it, such that all the integers from 1 to 30 are represented in the cards. Two cards (which have numbers a and b written on them) are arbitrarily chosen, and are removed from the pack. The above two cards are replaced with a single card, on which the number (ab + a + b) is written. This constitutes one operation. After 29 such operations, what will be the number written on the sole surviving card?
    a) 31! − 1
    b) 15 × 31
    c) 30! − 1
    d) 31! + (15 × 31)
    e) 30! + (15 × 31) – 1


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


    if we take out 1 and 2, then inserted card number 2 + 1 + 2 = 5 = 3! - 1
    so here in this case 31! - 1


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


    Q9) Six jars of cookies contain 18 , 19 , 21 , 23 , 25 and 34 cookies respectively. One jar contains almond cookies only. The other 5 jars contain no almond cookies. A takes three of the jars and B takes two of the others. Only the jars of almond cookies remains. If A gets twice as many cookies as B , how many almond cookies are there ?


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


    Sum of all the jars is 140
    it is of the form 3k + 2
    to make it multiple of 3 (A got 2x and B got x = > Total 3x) we should subtract a number of the form 3k + 2.
    23 is present here in form of 3k + 2

    Another approach:
    A has even number of cookies
    even = even + even + even
    even = odd + odd + even
    three even is not here so he will choose two odd and one even
    so 34 + 19 + 21 or 34 + 19 + 23 or 34 + 19 + 25 or 34 + 21 + 23 or 34 + 23 + 25
    so 74 or 76 or 78 or 78 or 82
    now B has 37 or 38 or 39 or 41
    37 = 18 + 19 (19 is already used so not possible)
    38 is not pssible
    39 = 18 + 21 (this is possible combination)
    41= 23 + 18
    so 23 almond cookies are there


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


    Q10) Ann and Anne are in bumper cars starting 50 m apart. Each one approaches the other at a constant ground speed of 10 km/hr. A fly starts at Ann, flies to Anne, then back to Ann, and so on, back and forth until it gets crushed when the two bumper cars collide. When going from Ann to Anne, the fly flies at 20 km/hr; when going in the opposite direction the fly flies at 30 km/hr (thanks to a breeze). What is the total distance covered by the fly?


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