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Number System Test - 7

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Number System Test - 7
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Weekly Quiz Competition
  • Question 1
    1 / -0
    How many numbers are there between 500 and 600 in which 9 occurs only once?
    Solution
    Considering the unit digit only in number like 509, 519, 529,...,589, 9 occurs 9 times .
    Considering the ten digit from 590 - 598, (590,591,592,...,598) 9 occurs 9 times.
    Hence, there are 9 + 9 = 18 numbers between 500 to 600 that contain the digit nine exactly.
  • Question 2
    1 / -0
    2! + 4! + 6! + 8! + 10! +.......100! when divided by 3, would leave remainder
    Solution
    All the number in this series except 2! are divisible by 3, therefore, reminder will be 2.
  • Question 3
    1 / -0
    For any natural number n, n4 + n2 + 1 is always
    Solution
    If n is even, then even + even + 1 = odd
    If n is odd, then odd + odd + 1 = even
    Therfore, n to the power 4 + n to the power 2 + 1 is always odd.
  • Question 4
    1 / -0
    Solve for x: .
    Solution
    This equation is done in two steps.
    (a) make the base equal
    (b) equate the power and solve

    25(x - 2) = 26- 3x
    5x - 10 = 6 - 3x
    8x = 16
    x = 2

  • Question 5
    1 / -0
    Let D be a rational number of the form D = 0. abcd abcd abcd ......, where digits a, b, c and d are integers lying between 0 to 9. At most three of these digits are zero. By which number should D be multiplied so that the result will be a natural number?
    Solution
    D = abcd / 9999, 49995 = 9999 * 5. Hence, once we multiply D with 9999 we will get a natural number.
  • Question 6
    1 / -0
    What is the greatest positive power of 5 that exactly divides 30!?
    Solution
    The total number of multiples of 5 in 30! = 6 (5, 10, ….., 30)
    The total number of multiples of 5 to the power 2 = 25 in 30! = 1 (25 only)
    Further powers of 5 cannot be these as 5 to the power 3 = 125 > 30
    Therefore, the greatest power of 5 that divides 30! exactly = 6 + 1 = 7
  • Question 7
    1 / -0
    The largest number that always divides the product of 3 consecutive multiples of 2 is
    Solution
    Suppose number are 2n, 2n + 2 and 2n + 4, where n is a whole number 2n (2n + 2) (2n + 4) = 2*2*2 (n) (n + 1) (n + 2) .
    Now, n (n + 1) (n + 2) is the product of three consecutive natural number and at least one of them will be divisble by 2 and at least one of them will be divisible by three.
    Therefore, n (n + 1) (n + 2) is divisible by 6.
    Therefore, 8 (n) (n + 1) (n + 2) will be always be divisble by 48.
  • Question 8
    1 / -0
    The sum of two natural numbers is 85 and their LCM is 102. Find the numbers.
    Solution
    Choices (b), (c) and (d) will have a factor of 5 in the LCM (they have multiple of 5 in number). So, the only logical choice is (a).
  • Question 9
    1 / -0
    By which smallest number must 21,600 be multiplied or divided in order to make it a perfect square?
    Solution
    21600 =
    We need another pair of 2 * 3, so that it becomes a perfect square. Hence, it needs to be multiplied or divided by 2 * 3 = 6.
  • Question 10
    1 / -0
    If we write all the natural numbers from 259 to 492 side by side, we shall get a very large natural number 259260261262…490491492. How many 8s will be used to write this large natural number?
    Solution
    From 259 to 458, there are two hundred natural numbers, so there will be 2 × 20 = 40 8s.
    From 459 to 492, we have 13 more 8s; so the answer is 40 + 13 = 53.
  • Question 11
    1 / -0
    When the sum of two natural numbers is multiplied by each number separately, the products obtained are 2418 and 3666. What is the difference between the two numbers?
    Solution
    Suppose the numbers are x and y, then:
    x(x + y) = 3666 … (1)
    y(x + y) = 2418 … (2)
    Adding 1 and 2, we get
    (x + y)2 = 6084
    Therefore, x + y = 78 … (3)
    Subtracting (2) from (1), we get
    x2 - y2 = 1248
    Or, (x + y)(x - y) = 1248
    Or, 78(x - y) = 1248
    Or, x - y = 16
    Hence, option 1 is correct.
  • Question 12
    1 / -0
    If 'p' be a number between 0 and 1, which one of the following is true?
    Solution
    Try with some natural numbers and get the answer.
  • Question 13
    1 / -0
    Dividing by 3/8 and then multiplying by 5/6 is equivalent to dividing by which number?
    Solution
    x / ( 3/8 ) * 5/6 = x* 8/3 *5/6 = x*40/18
    equivalent to dividing by 18/40 = 9/20
  • Question 14
    1 / -0
    A number when divided by 238, leaves a remainder 79. What will be the remainder when that number is divided by 17?
    Solution
    N = 238a + 79 =1714a + 174 + 11.
    N = 17(14a + 4) +11
    Hence, on dividing N by 17, the remainder is 11.
  • Question 15
    1 / -0
    What is the remainder when 17 to the power 23 is divided by 16?
    Solution
    17 when divided by 16 leaves 1 as remainder and we have 23 such 17s which when divided by 16 leave 1 as remainder.
  • Question 16
    1 / -0
    The product of two numbers is 16,200. If their LCM is 216, find their HCF.
    Solution
    The data is inconsistent as the LCM is always a multiple of HCF. The product of two numbers is always equal to the product of their LCM and HCF. Here, HCF = 75 and LCM = 216 do not show consistency, i.e. no such number is possible.
  • Question 17
    1 / -0
    The LCM of two numbers is 72 and their HCF is 12. If one of the numbers is 24, what is the other number?
    Solution
    72 * 12 = 24 * n ==> n = 36
  • Question 18
    1 / -0
    4a56 is a four digit number divisible by 33. What is the value of a?
    Solution
    By divisibility rule of 3, 4 + 5 + a + 6 = multiple of 3 a =3 or 6 or 9 by divisibility rule of 11 (4 + 5) - (6 + a) = multiple of 11n or zeroA = 3.
  • Question 19
    1 / -0
    A number when divided by 5 leaves a remainder 3. What is the remainder when the square of this number is divided by 5?
    Solution
    A number leaving 3 as remainder when divide by 5 ends either 3 or 8 .The number when squared will always have the last digit 4 or 9, thus leaving a remainder of 4 when divisible by 5.
  • Question 20
    1 / -0
    A two digit number is such that cube of its 24th part is same as the number obtained by interchanging the digits of the number. What is the number?
    Solution
    The best way is to check the options. (1) is not possible
    (3) 1/24 x 48 = 2, cube of 2 = 8 which is not a 2 digit number
    (2) 1/24 x 72 = 3, cube of 3 = 27 which is reverse of 72.
  • Question 21
    1 / -0
    What are the unit digits of 369, 6864 and 4725, respectively?
    Solution
    31 = 3, 32 = 9, 33 = 27, 34 = 81, 35 = 243, 36 = 729, 37 = 2167, 38 = 6501.......
    From the above expressions, we can observe that the unit digit of the result gets repeated after every four consecutive results. So, the unit digit of 368 will be 1. Hence, the unit digit of 369 will be 3.
    Similarly, the unit digit of 6864 will be unit digit of 2864 x 3864, which is 6 x 1 = 6.
    And, unit digit of 4725 will be 4 (as 725 = 2n + 1).
  • Question 22
    1 / -0
    If the last two digits of a four-digit number are interchanged, the new number obtained is greater than the original number by 54. What is the difference between the last two digits of the number?
    Solution
    Suppose that the last two digits of the four-digit number are c and d, where c is at the tens place and d is at the units place.
    We can ignore the first two digits.
    This gives us,
    (10c + d) - (10d + c) = 54
    9(c - d) = 54
    c - d = 6
  • Question 23
    1 / -0
    A number formed by writing any digit 6 times (say as 444444 or 999999) is always divisible by
    Solution
    aaaaaa = aaa * 1000 + aaa
    aaa (1000 + 1) = 1001(aaa)
    =7 * 11 *13 (aaa)
    This is obviously divisible by 7,11 and 13.
  • Question 24
    1 / -0
    A positive interger which is nearest to 1000 and divisible by 2, 3, 4, 5 and 6 is
    Solution
    LCM of 2, 3, 4, 5 and 6 = 60


    Closet to 1000, which is also divisible by 60 = 1020

  • Question 25
    1 / -0
    There are four prime numbers written in ascending order of magnitude. The product of the first three is 385 and that of the last three is 1001. Find the first number.
    Solution
    The best way is to factorise the number i.e 385 = 5 x 7 x 11
    1001 = 7 x 11 x 13
    So, the four prime numbers are 5, 7, 11 and 13.
    Hence, first prime number = 5
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