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Number System Test - 6

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Number System Test - 6
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Weekly Quiz Competition
  • Question 1
    1 / -0
    If n is any integer, which of the following statements is true?
    Solution
    If n is any odd number, it adds to 1 and becomes even, so it is divisible by 2.
    And if n is even, (n + 1) is odd, but when multiplied by n (even), it becomes even and then it is divisible by 2.
    So, if n is either even or odd, it is divisible by 2.
  • Question 2
    1 / -0
    Find the number of pairs of natural numbers with LCM equal to 56.
    Solution
    There are ten pairs (1, 56) (2, 56) (4, 56) (7, 56) (8, 56) (14, 56) (28, 56) (7, 8) (8, 28) and (8, 14) with LCM equal to 56.
  • Question 3
    1 / -0
    There are 5 consecutive natural numbers with L.C.M. 60. The product of the first two numbers is equal to the 5th number. What is the sum of the numbers?
    Solution
    Let n be 1st natural number.
    So, consecutive natural numbers are
    n, n + 1, n + 2, n + 3, n + 4
    According to question,
    n(n + 1) = n + 4
    n2 + n = n + 4
    n2 = 4
    n = 2
    – 2 cannot be taken as n is natural number. So, n = 2
    Numbers be 2, 3, 4, 5, 6
    Hence, sum of numbers = 2 + 3 + 4 + 5 + 6 = 20
  • Question 4
    1 / -0
    The numbers a and b have prime factors as 2 and 3 and their ratio is 36. Which of the following statements is true?
    Solution
    The two numbers with prime factors 2 and 3, and ratio 36 are 6 and 216.The HCF of 6 and 216 is 6.
  • Question 5
    1 / -0
    Find the remainder when 936 is divided by 730.
    Solution
    93 = 729
    936 = 72912

    So, remainder = 1
  • Question 6
    1 / -0
    In a division, the remainder was 71. With the same divisor, but twice the dividend, the remainder is 43. Which of the following numbers is the divisor?
    Solution
    Let the dividend be x and the remainder be 71, so that x + 71 is divisible by the divisor.
    When it is done, it comes twice, i.e. 2x + 142, which is also divisible.
    According to the question, it leaves the remainder 43.
    So, 2x + 142 – 43 = 2x + 99
    So, the divisor is 99.
  • Question 7
    1 / -0
    If n is a natural number, n4 + n2 + 1 is
    Solution
    n4 + n2 is always even.
    We know, even + odd = odd
    As 1 is odd
    So, n4 + n2 + 1 = odd
  • Question 8
    1 / -0
    The sum of all odd numbers of four digits which are divisible by 9 is _______.
    Solution
    Least 4 digit number divisible by 9 is 1008 and highest is 9999.
    Now, 9999 = 1008 + (n - 1)9
    9999 = 1008 + (n - 1)9
    = n - 1
    So, n - 1 = 999 or n = 1000
    So, number of odd numbers between 1008 and 9999 = = 500
    Sum of n numbers is Sn = (a + I).
    Here, n = 500, a = 1017, l = 9999 (Because least 4-digit odd number divisible by 9 is 1017.)
    = (1017 + 9999)
    = 250 (11,016) = 27,54,000
  • Question 9
    1 / -0
    How many zeroes are left in the end when we solve 75!?
    Solution
    There's a method to find the number of zeroes in a 'factorial of a number'.
    Formula : round + round + round +.......+ round
    Here, round function indicates that only the integer part of should be taken.
    Here in the above case, n = 75
    Number of zeroes = + 0 = 15 + 3 = 18
  • Question 10
    1 / -0
    Two out of six papers for an examination are of physics. What are the number of ways in which the papers can be arranged so that two physics papers are not together?
    Solution
    Arrangement of total number of papers = 6! or 720
    Number of ways when two physics papers are set together = 5! × 2 = 240
    So, required number of ways = 720 - 240 = 480
  • Question 11
    1 / -0
    If tn + 1 = 2tn + 1(n = 0, 1, 2, 3…) and t0 = 0, which of the following numbers is nearest to t10?
    Solution
    Given that tn + 1 = 2tn + 1 and t0 = 0
    Put n = 0.
    So, t0 + 1 = 2t0 + 1 or t1 = 0 + 1
    So, t1 = 1
    Now, put n = 1, so t1 + 1 = 2t1 + 1, t2 = 2(1) + 1 = 2 + 1 = 3
    Similarly, when n = 2, then t2 + 1 = 2t2 + 1 = t3 = 2(3) + 1 = 7
    Continuing these steps, we get t10 = 1023
  • Question 12
    1 / -0
    Given odd positive integers x, y and z, which of the following options is not necessarily true?
    Solution
    Since x and z are odd, therefore x4 and y2 will also be odd.
    So, when (odd × odd) = odd, then odd ÷ 2 is a decimal, and not an even number.
  • Question 13
    1 / -0
    If n is a natural number, which of the following options must always be odd?
    Solution
    Since n is an even number, so n2 + n will also be even. But when 1 is added to n2 + n, it becomes odd because even + odd = odd.
  • Question 14
    1 / -0
    The number 2.525252…. can be written as a fraction when reduced to its lowest form. The sum of the denominator and numerator is ______.
    Solution
    Let p = 2.525252…. i.e. (1)
    So, 100p = 252. (2)
    Now, subtract (1) from (2).
    ⇒ 99p = 250
    So,
  • Question 15
    1 / -0
    The sum of a 2 digit number and a number obtained by reversing the digits is a perfect square. How many such numbers exist?
    Solution
    Let two digit number = 10x + y and number on reversing the digits = 10y + x
    Sum of number and its reverse is 10x + y + 10y + x = 11 (x + y)
    Now, the number will be perfect square when x + y = 11
    The combinations with sum 11 are (2, 9) (3, 8) (4, 7) (5, 6) (6, 5) (7, 4) (8, 3) and (9, 2).
  • Question 16
    1 / -0
    Let x + y = 1. Find the minimum value of (x + )2 + (y + )2.
    Solution
    The minimum value for x and y will be 0.5.
    So putting x = y = 0.5, we get
    (0.5 + 2)2 + (0.5 + 2)2 = 12.5
    So, the correct option is 4.
  • Question 17
    1 / -0
    If a and b are real numbers such that |a| + |b| > |a + b|, which of the following options is true?
    Solution
    |a| + |b| > |a + b| is greater when a and b are of opposite signs. When a and b are of opposite signs, |a| and |b| both will become positive and add. But in |a + b|, they would subtract and we know sum of the given two numbers is always greater than their difference. Hence, option (3) is correct.
  • Question 18
    1 / -0
    Find the maximum power of 3 in 20!.
    Solution
    Highest power of prime number p in n! =
    where
    ● denotes the greatest integer less than or equal to x.
    In above case, n = 20 and p = 3. Highest power = = 6 + 2 = 8
  • Question 19
    1 / -0
    The remainder obtained when 784 is divided by 342 is
    Solution
    Divide the power of 7, i.e. 84 in parts of 3 because 73 when divided by 342 leaves 1 as a remainder.
    Now, 84 is divisible by 3.
    Therefore, 784, when divided by 342, leaves 1 as a remainder.
  • Question 20
    1 / -0
    If x is negative, y is greater than 0 and less than 1, and z is greater than 1, then which of the following statements is false?
    Solution
    Although x is a negative integer, x2 is a positive integer. Since z is greater than 1, z2 is always positive, but if z2 is greater than x2, x2 - z2 will be negative. Therefore, it cannot be positive every time.
  • Question 21
    1 / -0
    Find the number of positive integers which are not greater than 100 and are not divisible by 2, 3 or 5.
    Solution
    n(A) = number of numbers divisible by 2 = 50
    n(B) = number of numbers divisible by 3 = 33
    n(C) = numbers of numbers divisible by 5 = 20
    Using n(A ∪ B ∪ C) - (n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + a(A ∩ B ∩ C)), we get
    100 - (50 + 33 + 20 - 16 - 10 - 6 + 3)
    = 100 - 74
    = 26
  • Question 22
    1 / -0
    If x, y > 0, where x is a prime number and y is a composite number, which of the following statements is true?
    Solution
    y – x can be even if x = 2 and y is any composite even number.
    xy can be even if x = 2 and y is any composite even number.
    can also be even when x = 2 and y is any composite even number.
  • Question 23
    1 / -0
    Find the sum of two digit numbers which gives remainder 3 when divided by 7.
    Solution
    The smallest number is 10 which when divided by 7 gives a remainder 3.
    And the largest such number is 94.
    So, 94 = 10 + (n - 1)7
    ∴ n = 13
    S13 = (10 + 94) = 676
  • Question 24
    1 / -0
    What is the remainder when 496 is divided by 6?
    Solution
    has 4 as the remainder.

    has 4 as the remainder.

    has 4 as the reminder.

    has 4 as the remainder.

    So, 4 is the correct answer.
  • Question 25
    1 / -0
    The sum of the quotient and the remainder when a number is divided by 4 is 8 and the sum of their squares is 34. Find the number.
    Solution
    Now, Q + R = 8 or Q = 8 – R
    Q2 + R2 = 34
    (8 – R)2 + R2 = 34
    64 + R2 – 16R + R2 = 34
    2R2 – 16R + 30 = 0
    Or R2 – 8R + 15 = 0
    Or R2 – 5R – 3R + 15 = 0
    Or (R – 5) (R – 3) = 0
    ∴ R = 5, 3
    Case I:
    When R = 5, Q = 3
    3 × 4 + 5 = 17
    Number = 17
    Case II:
    When R = 3
    So, Q = 5 and 5 × 4 + 3 = 23
    Number = 23
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