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Theory of Numbers Test - 9

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Theory of Numbers Test - 9
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  • Question 1
    1 / -0
    is equal to
    Solution
    Let A =
    A = =
  • Question 2
    1 / -0
    If x = - 0.5, then which one of the following has the greatest value?
    Solution
    By using options, we can solve the question easily.
    Putting x = -, we get
    (1) = -2 (2) = 4
    (3) 2-1/2 = (4) 2-2 =
  • Question 3
    1 / -0
    The real number + is equal to
    Solution
    size=3> +
    = + -
    = 2 + -= 2
  • Question 4
    1 / -0
    Which of the following will be the smallest number to be multiplied with 134456 to make it a perfect cube?
    Solution
    134456 = 2 × 2 × 2 × 7 × 7 × 7 × 7 × 7
    So, we will multiply it by 7 to make a perfect cube.
  • Question 5
    1 / -0
    If 3x + 1 . 2n = 1728, then n/x is equal to
    Solution
    3x + 1 . 2n = 1728
    3x + 1 2n = 33 (26)
    So, n = 6 and x = 2, hence n/x = 3
  • Question 6
    1 / -0
    If a = - 30, b = - 20 and c = 50, then find a3 + b3 +c3.
    Solution
    Now, a + b + c = 3abc, when a + b + c = 0
    Thus, the required answer is 3(- 30)(- 20)(50) = 90,000
  • Question 7
    1 / -0
    If 4n - 4n -1 = 48, then nn is equal to
    Solution
    4n - 4n -1 = 48 or 22n - 22n-2 = 48 or y - y/4 = 48 (where 22n = y)
    Or y = 64
    Now, 22n = 64 or 2n = 6 or n = 3
    Thus, nn = 33 = 27
  • Question 8
    1 / -0
    The square root of 1 + y2 + is
    Solution
    Let x = (1 + y2) + = (1 + y2) +
    = (1 + y2) + = (1 + y2) +
    = [2(1 + y2) + 2
    = [( y2 + y + 1) + (y2 - y + 1)+ 2 ]
    = [ + + 2 ]
    = [ + ]2 or
    = [ + ]
  • Question 9
    1 / -0
    If x < y < z but x2 > y2 > z2 > 0, which of the following must be positive?
    Solution
    If x < y < z, then two cases arise.
    Case I: x = - 3, y = - 2, z = - 1
    Case II: x = - 3, y = - 2, z = 1
    From the above two cases, we conclude that for the expression to be positive, z must be raised
    to some even power.
    On checking the options, we find that option (2) has x and y raised to odd power, resulting in a
    positive expression.
  • Question 10
    1 / -0
    If 3a = 4b = 6c and a + b + c = 27, then is equal to
    Solution
    3a = 4b = 6c or a/4 = b/3 = c/2 = k (say)
    Thus, a = 4k, b = 3k and c = 2k
    Also, a + b + c = 27 or 9k = 27 or k = 3
    Hence, = = k = 3 . = 87
  • Question 11
    1 / -0
    How many positive numbers less than 100 have all even digits?
    Solution
    Single digit numbers are: 2, 4, 6, 8
    Two digit numbers are: 4 × 5 = 20
    (Because the first numeral can be any one from 2, 4, 6, 8 and the second numeral can be any
    one from 0, 2, 4, 6, 8.
    So, total 4 × 5 = 20 numbers). Required answer = 4 + 20 = 24.
  • Question 12
    1 / -0
    For how many two-digit numbers is the sum of the digits greater than the product of the digits?
    Solution
    If the two-digit number is ab, then a + b > ab
    This is possible, only if
    case 1: b is 0 and a ∈ [1, 9], such numbers are 10, 20……….., 90.
    case 2 : a is 1 and b ∈ [1, 9], such numbers are 11, 12………., 19.
    case 3 : b is 1 and a ∈ [2, 9], such numbers are 21, 31, 41…….., 91.
    So, such numbers are 10, 20……….., 90 and 11, 12………., 19 and 21, 31, 41…….., 91.
    So, total numbers are 9 + 9 + 8 = 26.
  • Question 13
    1 / -0
    Consider 4-digit numbers for which the first two digits and the last two digits are equal. How many such numbers are perfect squares?
    Solution
    Let the 4-digit number n2 be of the form aabb.
    Now, we find that the sum of digits at even places = sum of digits at odd places = a + b each
    Thus, the number is divisible by 11.
    This implies that 11 is a factor of n2
    Thus, 11 is a lso a factor of 'n'.
    Now, 1002 = 10000 which is the smallest 5-digit number
    Also, 312 = 961, and 322 = 1024
    Thus, n > 32 and n = 11b
    Trying out diferent values of 'n' by putting b = 3, 4, 5, 6, 7, 8 and 9, we find that n2 is of the form aabb only when b = 8 which gives us 'n' as 88 and n2 as 7744
    Thus, there is only one such 4-digit number.
  • Question 14
    1 / -0
    A yearly payment to a servant is Rs. 90 plus a turban. The servant leaves the job after 9 months and receives Rs. 65 and a turban. Find the price of the turban.
    Solution
    Let the cost of the turban be Rs. x.
    So, payment to the servant = 90 + x for 12 months
    For 9 months, servant receives Rs. 65.
    Thus, × (90 + x) = 65 + x
    x = Rs. 10
    Hence, price of turban = Rs. 10
  • Question 15
    1 / -0
    72 hens cost Rs. a96.7b, where two digits in place of 'a and b' are not visible as they are written in illegible handwriting. How much does each hen cost?
    Solution
    From the options, 5.51 × 72 = 396.72
  • Question 16
    1 / -0
    What is the smallest number which when multiplied by 1944 will make it a perfect square?
    Solution
    1944 = 23 × 35
    So, to make it a perfect square, we have to multiply it with 2 × 3 = 6.
  • Question 17
    1 / -0
    If nm leaves a remainder of 1 after being divided by 7 for all positive integers n that are not multiples of 7, then m could be equal to
    Solution
    Since n is not a multiple of 7, it can take the following six forms:
    (7k + 1), (7k + 2), (7k + 3), (7k + 4), (7k + 5) and (7k + 6)
    But (7k + 6) will give a remainder 1, when divided by 7 and it is raised to even powers. Thus, m is even.
    Thus, m can be either 2, 4 or 6.
    Now, (7k + 2) raised to 2 when divided by 7 becomes 22/7 and gives a remainder 4, not 1.
    Also, (7k + 2) raised to 4 when divided by 7 becomes 24/7 and gives a remainder 2, not 1.
    But, (7k + 2) raised to 6 when divided by 7 becomes 26/7 and gives a remainder 1.
    Hence, m = 6
  • Question 18
    1 / -0
    There are two lights. The first light blinks 3 times per minute, while the other blinks 5 times in two minutes. Find the number of times they will blink together in an hour.
    Solution
    The first light blinks after 20 seconds or 3 times in a minute.
    The second light blinks 5 times in two minutes or after a time period of 24 seconds.
    LCM of (20, 24) is 120.
    So, each of these lights will blink together after 2 minutes.
    So, in an hour, they will blink or 30 times.
  • Question 19
    1 / -0
    An electric wire is sold only in multiples of m and a customer requires several lengths of wire, each 85 cm long. To avoid any wastage and to minimise labour, he should purchase minimum lengths of
    Solution
    Take the LCM of 100 cm and 85 cm = 1700 cm = 17 m
  • Question 20
    1 / -0
    Three bells chime at intervals of 18 min, 24 min and 32 min, respectively. At a certain time, they begin to chime together. What length of time will elapse before they chime together again?
    Solution
    LCM of 18, 24 and 32 is 25 × 32 = 32 × 9 = 288
    Hence, bells will chime together at an interval of 288 minutes or 4 hours and 48 minutes.
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