Quant Boosters - Shashank Prabhu, CAT 100 Percentiler - Set 3



  • Let abba and pqqp be the two numbers.
    The resultant five digit number has to start with 1 and so, will be of the form 1xyx1.
    1001a+110b+1001p+110q=10001+1010x+100y
    1001(a+p)+110(b+q)
    But as the last digit is 1, a+p has to be 11
    11011+110(b+q) will be a palindrome. So, b+q should not change the central values disproportionately. This is possible when b+q=0 or b+q=11. So, 11011 or 12221.



  • Q5) 9 contestants for miss India are seated around a round circular table. Aditya wants to date 3 of them such that he does not select any 2 neighbouring contestants. No. Of ways in which Aditya can do this is?



  • Any three in 9c3 ways.
    All three together in 9 ways.
    Two together in 5 ways per couple (From abcdefghi if he chooses ab, he can choose one among only defgh, similarly for the 8 other pairs).
    So, two together in 9 * 5 = 45 ways.
    So, no two together in 9c3 - 9 - 45 = 30 ways.



  • Q6) 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.



  • Q7) How many arrangements of of the letters of the word CATASTROPHE are there in which both the A's appear before both the T's?



  • 1 letters with two letters repeating twice in 11!/2!2! ways
    2 As and 2 Ts can be arranged in 4!/2!2!=6 ways. Only one among them AATT satisfies the condition. So, (1/6)*(11!/2!2!)=11!/4!
    PS: When I say AATT... i mean all cases where 2 As will come before 2 Ts. So, the arrangement could be something like CSARAOHTPET as well. I am just talking about how As and Ts appear among themselves. As there is no binding condition, the distribution would be even.



  • Q8) A, B, C, D, E are five students who took CAT 2015. 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



  • Q9) Find the number of zeroes at the end of 2180!



  • [2180/5] + [2180/5^2] + [2180/5^3] + [2180/5^4]
    = 436 + 87 + 17 + 3
    = 543



  • Q10) How many odd integers from 1000 to 8000 (inclusive) have distinct digits ?



  • The number has to be between 1000 and 8000. So, the first digit can take any value from 1-7 only

    _ _ _ 1... as 1 has been fixed, first digit in 6 ways, 2nd in 8 ways and third in 7 ways -> 6 * 8 * 7
    _ _ _ 3.. similar to the first case -> 6 * 8 * 7
    _ _ _ 5.. similar to the first case -> 6 * 8 * 7
    _ _ _ 7.. similar to the first case -> 6 * 8 * 7
    _ _ _ 9.. first digit can be taken in 7 ways, 2nd in 8 ways and 3rd in 7 ways
    6 * 8 * 7 * 4 + 7 * 8 * 7
    56 * 31=1736



  • Q11) How many unordered triplets of positive integers p, q, r exist such that p/q + q/r + r/p=2?
    a. 0
    b. 1
    c. 2
    d. More than 2



  • (p/q+q/r+r/p)/3 >= (p/q * q/r * r/p)^1/3
    (p/q+q/r+r/p) >= 3
    So, it can never be equal to 2.

    Can be done using simple substitution as well.



  • Q12) If set S= {1,2,3,4,5,6,7,8,9,10} then how many sunsets of S are possible so that the sum of the elements is greater than 27?



  • The simple approach is to figure out the total number of subsets possible i.e. 2^10. We know that the sum of all the elements will be 55. So, whatever set we form, if the sum is x, the sum of the elements in the complement set will be (55-x). So, half the cases will be greater than 55/2=27.5 and half will be lesser than 27.5. So, (2^10)/2 i.e 2^9=512 is our answer.

    Take a smaller case... {1,2,3,4,5} and you want the sets that have sum greater than 7. You can form 32 sets or 16 pairs of complements.
    {}...{1,2,3,4,5}
    {1}...{2,3,4,5}
    {2}...{1,3,4,5|
    {3}...{1,2,4,5}
    {4}...{1,2,3,5}
    {5}...{1,2,3,4}
    {1,2}...{3,4,5}
    {1,3}...{2,4,5}
    {1,4}...{2,3,5}
    {1,5}...{2,3,4}
    {2,3}...{1,4,5}
    {2,4}...{1,3,5}
    {2,5}...{1,3,4}
    {3,4}...{1,2,5}
    {3.5}...{1,2,4}
    {4,5}...{1,2,3}
    See that the second set is the complement of the first set. The sum total of the elements of the two sets will always be 15. If the first set totals to less than or equal to 7, the second will definitely by greater than 7. The elements are such that you will not have equal values in both the sets and so, you are left with half the cases. If the sum is an odd number and the total is +/-0.5 of the average then you can go with this method. Else, normal case making and permutations i guess



  • Q13) A total of 8000 people were present for an exhibition. The ratio of men to women was 2:3 and ratio of boys to girls was 4:5. What can be the maximum ratio of males to females?
    (1) 12:31
    (2) 59:81
    (3) 345:532
    (4) 23:57
    (5) None of these



  • Q14) A watch showed 10 minutes past 6 o’ clock on Thursday morning when the correct time was 6 o’ clock. It loses uniformly and was observed to be 15 minutes slow at 8 o’ clock on Saturday morning. When did the watch show correct time ?



  • As we need to increase the number of males, we need to increase boys instead of men. So, we need to find a multiple of (4+5) 9 which is close to 8000 so that the remaining number is a multiple of (2+3) 5.
    As we know that 7920 is a multiple of 9, we have a point from where we can start. So
    7929, 7938, 7947, 7956, 7965. We can't go above 7965. So the remaining 35 can be split in 2:3.
    This gives us Boys = (7965)4/9 = 3540
    And men = (35)2/5 = 14
    3540 + 14 = 3554
    8000 - 3554 = 4446
    Answer: 3554 : 4446 = 1777 : 2223

    This is the one wherein your worst nightmare comes true. Most of the time, we maintain that the fractions and ratios are there for a reason and by the end of it, things would cancel out. Well, this is the one where they don't. Beware of such questions. They might be rare but you will get at least a couple of these speed breakers.


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