Quant Capsules by Shashank Prabhu - Set 20

  • CAT 100%iler, 5 times AIR 1, Director - Learningroots, Ex ITC, Pagalguy, TAS

    In how many ways can 6 letters A, B, C, D, E and F be arranged in a row such that D is always somewhere between A and B?

    Total ways is 6!
    For every arrangement, we will have an internal order of abd, adb, bad, bda, dab, and dba.
    Out of these, adb and bda are what we need.
    So, 1/3rd of the total arrangements.

    Fifty inabitants of the shire was surveyed to note down their posessions of Arkenstones, Elfstones and rings of power. Of then, 22 own an Arkenstone, 15 own an Elfstone and 14 own a ring of power. Nine of these inhabitants own exactly two items out of an Arkenstone, an Elfstone and a ring of power; and, one inhabitant owns all three. How many of the fifty inhabitants own none of three: Arkenstone, Elfstone or a ring of power?

    Total number of objects will be given by a+2b+3c where a is the number of people who possess only one object, b is the number of people who possess exactly two objects and c is the number of people who possess exactly three objects. So, in this case, total number of objects is 22+15+14=51. Number of people who have exactly 2 objects is 9 and so, 2b=18. Similarly, number of people who have exactly 3 objects is 1 and so, 3c=3. So, a+18+3=51 and so, a=30. So, a+b+c=30+9+1=40 and so, 10 do not have any objects

    A, B, C, D, E are five students who took CAT. Following are the sums of their overall scores, taken three at a time: 119, 121, 124, 125, 123, 126, 127, 128, 129 and 132. What is the highest and lowest score among the scores of A, B, C, D, E?

    Let a > b > c > d > e
    a + b + c = 132......(1)
    c + d + e = 119......(2)
    a + b + d = 129......(3)
    b + d + e = 121......(4)
    We know the values of all possible triplets. So, 5c3=10 cases in total. So, each element occurs 6 times.
    6a + 6b + 6c + 6d + 6e = 1254
    a + b + c + d + e = 209.....(5)
    From (1), (2) and (5)
    c = 42
    From (2), (3) and (5)
    d = 39... putting these values in (2)
    e = 38
    b = 44
    a = 46

    An institute conducts 32 tests. The number of tests attempted by three students studying in the class is as follows:
    Neil - 16
    Nitin - 18
    Mukesh - 20
    The number of tests written by more than one student is at least:
    a) 8
    b) 11
    c) 13
    d) 16
    e) 15

    Let a be the number of students who have taken exactly 1 test, b be the number of students who have taken exactly 2 tests, c be the number of students who have taken exactly 3 tests. Total number tests will be equal to a + 2b + 3c = 16 + 18 + 20 = 54
    Also, 32 tests have been conducted in total. So a+b+c=32
    From the two equations
    b + 2c = 22
    To minimize overlap, we need to maximize c.. so c=11.

    There are 49 zeros, 51 ones and 53 twos written on the board randomly. A student is blindfolded and then asked by his teacher to touch any two numbers on the board arbitrarily. The teacher deleted those two numbers and replaced them by a single number in the following manner:
    If the pair is Replaced by
    (0, 0) → 0
    (1, 1) → 0
    (2, 2) → 2
    (1, 2) → 1
    (0, 1) → 1
    (0, 2) → 0
    If they continued this process what was the number left on the board in the end?
    (a) 0
    (b) 1
    (c) 2
    (d) Cannot be determined

    Essentially, when we are replacing 2 zeros, we do not make any changes to the existing sum of all the elements on the board, when we replace 2 ones, we bring down the total by 2, when we replace 2 twos, we bring down the total by 2, when we replace a one and a two, we bring down the total by 2, when we replace a zero and a one, we do not change the total and when we replace a zero and a two, we bring down the total by 2. So, we bring down the total by a 0 or a 2 with each successive replacement. As the total at the beginning is 157 and we lose zeros or twos, the remaining number will be 1.

    Alternatively, you can actually cancel out pairs of zeros, ones, and twos and get the final answer.

    On a race track a maximum of 5 horses can race together at a time. There are a total of 25 horses. There is no way of timing the races but you can see the horses as they cross the finish line. What is the minimum number of races we need to conduct to get the top 3 fastest horses?
    A. 6
    B. 7
    C. 8
    D. 10

    5 winners from the first five races. Another race to determine who is the fastest. Now, positions 2 and 3 need to be figured out. The important thing to not here is that the horses who came 2nd and 3rd need not be the 2nd and 3rd fastest as they have competed with only the winner from the particular group and not the other horses. So, to be sure that a 'deserving' horse has got through to the final, we select those horses who have a chance of being the 2nd and 3rd fastest.

    Let the horses be represented by

    p1, p2.... p5
    p6, p7.... p10
    p21, p22.. p25

    Let p1, p6, p11, p16 and p21 win the races. As p16 and p21 have come 4th and 5th in this race, they cannot be among the three fastest horses. So, the entire branch of p16 and p21 is out. Now, p1 is the fastest horse in the first race and so, there is a chance that p2 and p3 are the second and the third fastest overall. Similarly, p6 is the fastest horse in the second race and there is a chance that p7 could be the third fastest (understand that p1 is the fastest and so, the best case scenario for p6 would be if he comes in the 2nd place and the best case scenario for p7 is when he comes in the 3rd place. As the 3 positions would have been taken, p8 cannot feature in the contenders' list). Finally, p11 would be a part of the final race having won his round and being in the 3rd position in the previous race. So, the final race will have p2, p3, p6, p7, p11 in it the top two of whom would win.

    Total of 5+1+1=7 races.

    There are one thousand students at the George Washington High School. Each student is assigned a locker, numbered 1 through 1000. On the first day of school each year, the students participate in an unusual ritual: All the lockers are closed in the beginning. The students then enter the school through one door, parade past the all the lockers, and then exit through another door. While in the school, the first student reverses the door position of each locker - if the door is open, he closes it, and if it is closed, he opens it. The second student reverses the door position of every other locker, starting with locker number 2. The third student reverses every third locker, starting with locker number 3, etc. After all 1000 students have completed this ritual, how many lockers will be left open?

    All the lockers are opened and closed the same number of times as they have factors. As they are all closed at the start, an odd number of operations have to be performed on the ones that are open at the end. So, if N = a^x * b^y * c^z... the number of factors will be (x+1)(y+1)(z+1)... now if this has to be odd, all of x+1, y+1, z+1 will be odd. So, x, y, z will be even. So, N has to be a perfect square. As there are 31 perfect squares less than 1000 (1-961), we understand that 31 lockers will be open after this exercise.

    There is a staircase of 10 steps. In how many ways can Amit climb the staircase if he can take a maximum of 3 steps at a time?
    a. 274
    b. 275
    c. 276
    d. 277

    If there were just 1 step, there would have been just 1 way to go up
    For 2 steps, either (1,1) or (2)
    For 3 steps, either (1,1,1)(2,1)(1,2)(3)
    For 4 steps, (1,1,1,1)(2,1,1)(1,2,1)(1,1,2)(2,2)(3,1)(1,3)
    If you look at it, to reach step 5, we need to be at either step 2, step 3 or step 4 before we make the last move. We could be at step 2 in 2 ways, at step 3 in 4 ways and at step 4 in 7 ways, So, s(5) = 2+4+7=13
    So basically, it is a Fibonacci sequence with each term being equal to sum of the three preceding terms
    1, 2, 4, 7, 13, 24, 44, 81, 149, 274

    If a clock gains 5 minutes every hour, how many times do the hands meet during the day ?

    The hands meet when the minute hand gains 360 degrees over the hour hand. 360/(11/2)= every 720/11 minutes . So we have to figure out how many such packets we have in a day. The clock goes ahead by 65 minutes for every 60 minutes that a normal clock would take. So, if the normal clock moves by 1440 minutes, the faulty clock would have moved 1440 * 65/60 = 1560 minutes.
    So 1560/(720/11)

    Amul and Cadbury run a 12km cycling race on a circular track of circumference 750metre.Amul can beat Cadbury in a race of 2,000metre by 500metre.If Amul gives a headstart of 500 metre to cadbury,then how many times will amul overtake cadbury in the race?

    Speed of Amul : Speed of Cadbury = 4/3
    For them to meet, Amul has to travel 750 m more
    (x + 750)/x = 4/3
    x = 2250. So, every time Amul covers 3000 m, he overtakes Cadbury. So, for 12 km, he will meet 4 times (that 500 m bit is required as the fourth time they meet, Amul will overtake Cadbury)
    If you calculate and check with Sa = 40 m/s and Sc=30 m/s you will get the first meeting at 50 seconds and every subsequent meeting after 75 seconds.

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