Quant Boosters - Nitin Gupta - Set 1


  • QA/DILR Mentor | Be Legend


    Number of Questions - 30
    Topic - Quant Mixed Bag
    Solved ? - Yes
    Source - Selected questions/solutions from Nitin Gupta Sir (AlphaNumeric)


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    Q1) If x is a positive integer and f(1) = 13 and f(x + 1) – 4f(x) + 15 = –9x, then what is the value of f(101)?
    a) 4101 – 309
    b) 4101 + 309
    c) 4100 + 309
    d) 4100 – 309
    e) 4101 + 303


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    f(x + 1) – 4f(x) + 15 = –9x
    ∴ f(x + 1) = 4f(x) – 9x – 15
    Putting x = 1, we get,
    f(1 + 1) = 4f(1) – 9(1) – 15
    ∵ f(1) = 13
    ∴ f(2) = 4(13) – 9 – 15
    ∴ f(2) = 52 – 24
    ∴ f(2) = 28
    ∵ All the options contain one term in some or the other power of 4 and the second term in multiples
    of 3, we have to modify the above results accordingly.
    ∴ f(2) = 16 + 12
    ∴ f(2) = 42 + 3(1 + 3) ... (i)
    For x = 2,
    f(3) = 4f(2) – 9(2) – 15
    ∴ f(3) = 4(28) – 18 – 15
    ∴ f(3) = 112 – 33
    ∴ f(3) = 79
    ∴ f(3) = 64 + 15
    ∴ f(3) = 43 + 3(2 + 3) ... (ii)
    ∴ From (i) and (ii) we can conclude that, the following formula fits for the two values of x.
    f(x + 1) = 4(x + 1) + 3(x + 3)
    ∴ f(101) = 4101 + 309
    Hence, option 2.


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    Q2) A survey was conducted among some women to find out their liking for different soap brands - NTX, VTX and RTX. Those women who liked NTX and VTX were 32 in number. 15% of the women like only NTX and 56 liked NTX but did not like RTX. The number of women who liked only RTX was equal to the number of those who liked neither of the three. Of the total women surveyed 37% liked NTX, 51% liked VTX and 16 women liked VTX and RTX. In the survey it was found that 14% of them did not like any of the three soaps. What is the number of women who liked only VTX?
    a) 63
    b) 62
    c) 59
    d) 72
    e) None of these


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    Refer the Venn diagram, [Here, N represents NTX, V represents VTX and R represents RTX]
    0_1513862977311_a6dbf5e8-b770-471b-a2a9-afa19900aac5-image.png
    d + g = 32
    a = 15%
    a + d = 56
    a + d + g + e = 37%
    c + d + g + f = 51%
    g + f = 16
    b = 14%
    ∵ 14% didn’t like any soap, N ∪ V ∪ R = 100 − 14 = 86%
    ∴ N ∪ V = 86% – b = 86% – 14% = 72% = N + V – (N ∩ V)
    ∴ 72% = 37% + 51% − (N ∩ V)
    ∴ N ∩ V = d + g = 16% = 32
    ∴ 1% = 2
    ∴ a = 15% = 15 × 2 = 30
    ∵ a + d = 56
    ∴ d = 26
    Now, V = 51% = 2 × 51 = 102
    Number of women who like only VTX = c = 102 – (d + g + f ) = 102 – (26 + 16) = 60
    Hence, option 5.


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    Q3) Jack is in a game show. The host shows him 100 doors, and tells him that there is a car behind one of them, and nothing behind the rest. Jack has to choose just 1 door, and he'll get whatever is behind it. However, after he makes his choice, the host opens 98 of the 100 doors, leaving closed the door that Jack chose and one other door. There is nothing behind any of the opened doors. Now, the host asks Jack whether he would like to stick to his original choice or switch to the other closed door.
    What is the probability that Jack will win the car if he switches to the other door?
    a) 0.5
    b) 0.98
    c) 0.99
    d) 1.00
    e) None of these


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    The probability that the car is behind the door that Jack initially chose is just 0.01.
    The opening of 98 of the other 99 doors has no effect on this probability, as the event (Jack's initial choice) has already occurred.
    ∴ The probability that Jack will win the car by switching to the other closed door is 1 − 0.01 = 0.99
    In other words, the probability that the car is behind one of the 99 other doors is 0.99.
    After 98 of the remaining 99 doors are opened, this entire probability of 0.99 converges to the closed door that Jack may switch to.
    ∴ The required probability is 0.99.
    Hence, option 3.


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    Q4) What is the units digit of 1^1 + 2^2 + 3^3 + … + 2008^2008?
    a) 8
    b) 6
    c) 4
    d) 2
    e) 0


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    We consider the numbers 10 at a time, i.e. 11 + 22 + ... + 1010, and so on.
    Now, if we know the cyclicity of all the numbers from 0 to 9, we can conclude the following:
    All numbers ending in 1, 5, 6 or 0 will end with the same digits irrespective of the power they are raised to.
    All numbers ending in 4 are raised to an even power here, so their units digit will be 6.
    All numbers ending in 9 are raised to odd powers, so their units digit will be 9.
    For numbers ending in 2 or 8, the power is alternately of the form 4n or (4n + 2) (for example: 2, 22,42, ... and 18, 38, 58, ... are of the form (4n + 2), while 12, 32, 52, ... and 8, 28, 48, ... are of the form 4n).
    So, for numbers ending in 2, the units digit is alternately 4 or 6, while for the numbers ending in 8,the units digit is alternately 6 or 4.
    Similarly, for numbers ending in 3 or 7, the power is alternately of the form (4n + 3) or (4n + 1), and so the units digit is 7 and 3 respectively for numbers ending in 3, or 3 and 7 respectively for numbers ending in 7.
    ∴ For every 10 numbers, the sum of the units digits is
    1 + 4 + 7 + 6 + 5 + 6 + 3 + 6 + 9 + 0, or 1 + 6 + 3 + 6 + 5 + 6 + 7 + 4 + 9 + 0.
    In both cases, the sum of these 10 digits is 47.
    ∴ The sum from 1 to 2008 will be 47 × 2000 + (1 + 4 + 7 + 6 + 5 + 6 + 3 + 6) = 94038
    Since we are only concerned with the units digit, the answer is 8.
    Hence, option 1.


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    Q5) Akash runs a business of exporting mangoes. He has 8649 Alphonso mangoes and 7688 Kesar mangoes in his warehouse. He wants to pack these into boxes of uniform size in a single-layer square arrangement such that the number of mangoes in each row in a box should be equal to the number of mangoes in each column, and the total number of mangoes in a box should be equal for all boxes. Also, he wants to ensure that each box contains only one type of mango, and that no mangoes are left out at the end of this packing. How many mangoes will have to be packed into each box?
    a) 529
    b) 729
    c) 841
    d) 961
    e) None of these


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    The given conditions tell us that the number of mangoes in each box will have to be a common factor of 7688 and 8649, and since the mangoes have to be arranged in a square in each box, this number should be a perfect square.
    Instead of trying to factorize 7688 and 8649 directly, we can simplify our task if we notice that a common factor of 7688 and 8649 will also have to be a factor of their difference,
    i.e. (8649 − 7688) = 961
    The only factors that 961 has are 1, 31 and 961, and we can verify that 7688 = 961 × 8 and 8649 =961 × 9
    ∴ The required number is 961.
    Hence, option 4.


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    Q6) A cube of dimension 4 cm × 4 cm × 4 cm is painted red on all six faces. Now this cube is cut to form 1 cm × 1 cm × 1 cm identical cubes. What is the ratio of total area of painted surfaces to the total area of unpainted surfaces?
    a) 2/9
    b) 1/3
    c) 3/16
    d) 1/2
    e) None of these


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    As 4 cm × 4 cm × 4 cm cube is cut into 1 cm × 1 cm × 1 cm identical cubes.
    Number of cubes = 4 × 4 × 4 = 64
    Now, there are cubes out of which some cubes have 3 surface painted, some have 2 surface painted,some have 1 surface painted and remaining have no surface painted. We will find out each of them.
    Cubes with 3 surface painted = 8 (at 8 vertices)
    Cubes with 2 surface painted = 24 (2 each at 12 edges)
    Cubes with 1 surface painted = 24 (2 × 2 = 4 at centre of all 6 faces)
    Cubes with no surface painted = 8 (2 × 2 × 2)
    Total area of painted and unpainted surface = 64 × 6 = 384
    Area of painted surface = 3 × 8 + 2 × 24 + 1 × 24 + 0 × 8 = 96
    Area of unpainted surface = 384 − 96 = 288
    ∴ Ratio of painted surface to unpainted surface = 96/288 = 1/3
    Hence option 2.
    Alternatively,
    To cut the given cube in to 64 (4 × 4 × 4) smaller cubes, we have to make 3 cuts in all the 3 dimensions, i.e. a total of 9 cuts.
    Every cut will expose unpainted area equal to 2 faces of the original cube.
    So in total area of unpainted surfaces = 18 faces of original cube
    Area of painted surface = 6 surfaces of original cube
    ∴ Ratio of painted to unpainted surfaces = 6/18 = 1/3
    Hence, option 2.


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    Q7) What is the possible number of ways to select 3 numbers from the first 15 natural numbers, such that the selected numbers are in arithmetic progression?
    a) 36
    b) 42
    c) 24
    d) 45
    e) 49


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    These are the possible combinations, where d is the common difference.
    For d = 1; N = 13 (1, 2, 3); (2, 3, 4); (3, 4, 5); (4, 5, 6); (5, 6, 7); (6, 7, 8); (7, 8, 9); (8, 9, 10); (9, 10, 11); (10, 11, 12); (11,
    12, 13); (12, 13, 14); (13, 14, 15)
    For d = 2; N = 11 (1, 3, 5); (2, 4, 6); (3, 5, 7); (4, 6, 8); (5, 7, 9); (6, 8, 10); (7, 9, 11); (8, 10, 12); (9, 11, 13); (10, 12, 14);
    (11, 13, 15)
    For d = 3; N = 9 (1, 4, 7); (2, 5, 8); (3, 6, 9); (4, 7, 10); (5, 8, 11); (6, 9, 12); (7, 10, 13); (8, 11, 14); (9, 12, 15)
    For d = 4; N = 7 (1, 5, 9); (2, 6, 10); (3, 7, 11); (4, 8, 12); (5, 9, 13); (6, 10, 14); (7, 11, 15)
    For d = 5; N = 5 (1, 6, 11); (2, 7, 12); (3, 8, 13); (4, 9, 14); (5, 10, 15)
    For d = 6; N = 3 (1, 7, 13); (2, 8, 14); (3, 9, 15)
    For d = 7; N = 1 (1, 8, 15)
    ∴ Total number of ways = 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
    Hence option 5.


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    Q8) X is the set of all numbers n such that 10 < n ≤ 50 and the remainder when (n − 1)! is divided by n is not zero. The number of numbers in the Set X is equal to:
    a) 10
    b) 16
    c) 11
    d) 17
    e) 5


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    The remainder when (n − 1)! is divided by n is not zero only if n is a prime number.
    The prime numbers greater than 10 and less than 50 are (11,13, 17, 19, 23, 29, 31, 37, 41, 43, 47).
    The total number of numbers in the set is 11.
    Hence option 3.


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    Q9) Abhay took a novel to read from a bookseller and found that a page was missing. He went back to the bookseller and informed him about it, and also told him that the sum of the remaining pages is 10,000. Which of the following can be one of the page numbers missing from the novel?
    a) 5
    b) 11
    c) 77
    d) 153
    e) Option 1 or 3


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    The novel must have had consecutive page numbers. The sum of consecutive numbers is given by n(n + 1)/2.
    By trial and error, we can find that the value of n comes to 141 which gives us a total of 10011, or 142 which gives us 10153.
    ∴ The pages that were missing could be page numbers 5 and 6, or numbers 76 and 77.
    Hence option 5.


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    Q10) Given that log 3 = 0.477, log 7 = 0.845, log 2 = 0.301. Find the number of digits in y if y = 252^10
    a) 12
    b) 25
    c) 24
    d) 30
    e) Cannot be determined


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