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Real Numbers Test - 5

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Real Numbers Test - 5
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  • Question 1
    1 / -0
    Which of the following fractions will have a non-terminating decimal expansion?
    Solution
    A fraction has a terminating decimal expansion if its denominator is of the form 2n × 5m.

    Only option (1), i.e. = has a denominator which is not of the form 2n × 5m.

    Hence, it has a non-terminating decimal expansion.

    In option 4, square of 7 in denominator will cancel out with 49 in numerator and the resulting denominator will become in the form of 2n × 5m.
  • Question 2
    1 / -0
    A number when divided by 296 leaves the remainder 75. If the same number is divided by 37, then what will be the remainder?
    Solution
    Let x be the number.
    Given: x = 296n + 75 [where n is the quotient]
    x = 37(8n + 2) + 1
    Hence, when x is divided by 37, it gives 1 as the remainder.
  • Question 3
    1 / -0
    On rationalising the denominator of the expression , we get . What are the respective values of a and b?
    Solution


    =

    =

    ∴
  • Question 4
    1 / -0
    Which of the following is the correct fractional representation of ?
    Solution
    Let y = .... (1)
    ......(2)
    On subtracting equation (1) from (2) we have,
    99y = 35
    y =
  • Question 5
    1 / -0
    The increasing order of the numbers is

    Solution


    To compare, all the exponents should be the same.





    As, 35 < 56 < 330

    Arranging the numbers in increasing order, we get
  • Question 6
    1 / -0
    If x = (2 +), what is the value of x2 + ?
    Solution
    x = 2 +
    = = = 2 -
    Now, x2 + = - 2
    = (2 + + 2 - )2 - 2
    = (4)2 - 2
    = 16 - 2
    = 14
    Hence, option 4 is the correct option.
  • Question 7
    1 / -0
    What is the cube root of ?
    Solution
    The cube root of = =
  • Question 8
    1 / -0
    Which of the following statements is FALSE?
    Solution
    Let's take a non-zero rational number = 5 and an irrational number =
    Then, 5 = irrational
    And = irrational
    So, the product/division of a non-zero rational number and irrational number is always irrational.
  • Question 9
    1 / -0
    Find a and b, if .
    Solution
    =

    (Comparing this with a - b : a = 11, b = 6)
  • Question 10
    1 / -0
    Traffic lights at three consecutive road crossings change after every 48 seconds, 72 seconds and 108 seconds. If the lights change simultaneously at 7:00:00 a.m, at what time will they again change simultaneously?
    Solution
    Traffic lights change after every 48 seconds, 72 seconds and 108 seconds.
    ∴ The time when they will again change simultaneously
    = L.C.M. (48, 72, 108)
    48 = 2 × 2 × 2 × 2 × 3 = 24 × 3
    72 = 2 × 3 × 2 × 2 × 3 = 23 × 32
    108 = 3 × 3 × 2 × 2 × 3 = 33 × 22
    ∴ L.C.M. (48, 72, 108) = 24 × 33
    = 16 × 27
    = 432
    As 432 = 60 × 7 + 12
    432 seconds = 7 minutes and 12 seconds
    So, if the lights change simultaneously at 7:00:00 a.m, then they will again change simultaneously after 7 minutes 12 seconds, i.e. 7:07:12 a.m.
  • Question 11
    1 / -0
    Which of the following is a rational number?
    Solution


    = 5 - 7 = - 2

    (- 2) is a rational number.
  • Question 12
    1 / -0
    Which of the following fractions will have a terminating decimal expansion?
    Solution


    As the denominator in the given fraction has prime factors of the form 2n or 5m, so the given fraction is terminating.
  • Question 13
    1 / -0
    Euclid's division algorithm is a technique to compute the highest common factor of two given positive integers c and d , c > d following steps are to be followed : Apply Euclid's division lemma to c and d to get whole numbers q and r such that c = dq + r where
    Solution
    Euclid's division algorithm: Suppose the two positive numbers are 'a' and 'b' and are such that a > b. Then the H.C.F. of 'a' and 'b' can found by the following given below steps.
    (a) Apply the division lemma to find 'q' and 'r' , where a = bq + r, b > 0. 0 ≤ r < b.
    (b) If r = 0, then then H.C.F. is b. If r ≠ 0, then apply Euclid's lemma to find 'b' and 'r'.
    (c) Continue steps (a) and (b) till r = 0. The divisor at this state will be H.C.F. (a, b). Also , H.C.F. ( a , b) = H.C.F (b , r).
  • Question 14
    1 / -0
    Which of the following are non-terminating decimal expansions?

    (i)
    (ii)
    (iii)
    (iv)
    Solution
    A fraction has a terminating decimal expansion if its denominator is of the form 2n × 5m, where m and n are non negative integers.

    (i) [Terminating]
    (ii) = [Terminating]
    (iii) = [Non-terminating]
    (iv) = [Non-terminating]
  • Question 15
    1 / -0
    What is the greatest number that divides the numbers 38, 45 and 52 and leaves remainders 2, 3 and 4, respectively?
    Solution
    Let the required number be x.
    According to the question, when 38 is divided by x, we get remainder 2. This means (38 - 2) is completely divisible by x.
    This means x is a factor of (38 - 2).
    Similarly, x is a factor of (45 - 3) and (52 - 4).
    Since we need to find the largest common factor of 36, 42 and 48, we will find the HCF of (36, 42, 48).
    Now,
    Step 1: Using Euclid`s division algorithm on 36 and 42, we get
    42 = 36 × 1 + 6
    36 = 6 × 6 + 0
    HCF (36, 42) = 6
    Step 2: Using Euclid`s division algorithm on 6 and 48, we get
    48 = 6 × 8 + 0
    HCF (48, 6) = 6
    Hence, HCF (36, 42, 48) = 6
    ∴The greatest number is 6.
    Hence, (1) is the correct answer.

    Note : This can also be done by finding the HCF using division method.
  • Question 16
    1 / -0
    Bhumi is a teacher. She had to distribute books to her students. At first, she tried to distribute equally between two students, then among three, four, five and six students. But, every time, she was left with one book. When she finally tried to distribute equally among seven students, she succeeded. What would be the minimum number of books that she could have?
    Solution
    The LCM of 2, 3, 4, 5 and 6 is 60. The number of books will be of the form 60K + 1. We put various values to K so as to make it divisible by 7.
    Starting from K = 1,
    60(1) + 1 = 61, which is not divisible by 7.
    K = 2 ,
    60(2) + 1 = 121, which is not divisible by 7.
    K = 3
    60(3) + 1 = 181, which is not divisible by 7.
    K = 4
    60(4) + 1 =241, which is not divisible by 7.
    K = 5
    60(5) + 1 = 301, which is divisible by 7.
  • Question 17
    1 / -0
    Some soldiers were practising for the republic day parade. Three soldiers started together to march along the Rajpath at the same rate. The lengths of their march were 68 cm, 51 cm and 85 cm, respectively. How far did they go before their steps were in the same line together (in meters)?
    Solution
    The lengths of their march were 68 cm, 51 cm and 85 cm, respectively.
    Therefore, the required distance travelled by them so as their steps were in the same line together = LCM of 68 cm, 51 cm, and 85 cm.
    Now, the LCM of 68, 51, and 85 = 17 × 4 × 3 × 5 = 1020 cm = 10.2 m
  • Question 18
    1 / -0
    Rahul owns a garment shop. He sells t-shirts in his shop. He noticed that total number of t-shirts sold on Monday is the highest number, which divides 3838 and 5123 to leave remainders 3 and 13 respectively. Then find out total number of t-shirts sold on Monday.
    Solution
    According to question,
    When we divide 3838 by the required number, it leaves a remainder 3.
    Therefore, the number will completely divide 3835 (3838 - 3).
    Also, when we divide 5123 by the required number, it leaves a remainder 13.
    Therefore, the number will completely divide 5110 (5123 - 13 ).
    The total number of t-shirts sold on Monday = HCF of 3835 and 5110
    The HCF of 3835 and 5110 is 5.
  • Question 19
    1 / -0
    Aryan and Vishal are friends. Aryan notices that his age is divisible by 14 and his friend notices that it is divisible by 3. What could be the lowest possible age of Aryan?
    Solution
    As we know that the Aryan's age is divisible by 14.
    Also, Aryan`s friend notices that Aryan's age is divisible by 3.
    Therefore, the lowest possible age of Aryan = LCM of 14 and 3 = 42 years.
  • Question 20
    1 / -0
    Three tube lights are connected in such a manner that they glow for every 60 seconds, 48 seconds and 72 seconds, respectively. All of them glow at once at 10 a.m. When will they again glow simultaneously?
    Solution
    All the three tube lights glow at once at 10 a.m.
    The time when they glow simultaneously again = LCM (60,48,72) seconds = 720 seconds = 12 mins.
    Therefore, the time when three tube lights glow together again is 10:12:00 a.m.
  • Question 21
    1 / -0
    Which of the following statement is/are correct?

    (A) All natural numbers are whole numbers but all whole numbers are not natural numbers.
    (B) Every integer, natural and whole number is a irrational number.
    (C) All rational and all irrational numbers make the collection of real numbers.
    (D) The product of two consecutive positive integers is divisible by 3.
    Solution
    (A) Natural numbers:
    A set of counting numbers is called natural numbers.
    N = {1,2,3,4,5,...}
    Whole number :
    A set of natural numbers along with zero is called whole numbers.
    W= {0,1,2,3,4,5,....}
    So, All natural numbers are whole numbers but all whole numbers are not natural numbers.
    So, (A) is correct.
    (B) Every integer, natural and whole number is a rational number (not irrational numbers) as they can be expressed in terms of .
    So, (B) is incorrect

    (C) All rational and all irrational numbers make a collection of real numbers. It is denoted by the letter R .
    So, (C) is correct.

    (D) Let n – 1 and n be two consecutive positive integers.
    Then their product is n (n – 1) = n2 – n.
    We know that every positive integer is of the form 2q or 2q + 1 for some integer q.
    So, let n = 2q
    So,
    n2 – n
    = (2q)2– (2q)
    = 4q2 – 2q
    = 2q (2q – 1)
    = 2r [where r = q(2q – 1)]
    n2 - n is even and divisible by 2.
    Let n = 2q + 1
    So,
    n2 – n = (2q + 1) 2– (2q + 1)
    = (2q + 1) ((2q + 1) – 1)
    = (2q + 1) (2q)
    = 2r[r = q (2q + 1)]
    n2 – n is even and divisible by 2 and not by 3.
    So, (D) is incorrect.
  • Question 22
    1 / -0
    Fill in the blanks.

    (1) Every W represented by a numerical string of digits, possibly continuing forever. This is called the X representation of the number.
    (2) The set of all Y and fractional numbers between them comprise the set of Z numbers.

    W X Y Z
    (A) rational number natural real numbers natural
    (B) integer irrational natural numbers real
    (C) real number decimal integers rational
    (D) irrational number real rational numbers irrational
    Solution
    (1) Every real number (W)can be represented by a numerical string of digits, possibly continuing forever. This is called the decimal representation (X) of the number. Integers, like 3, -2, 0, 42, -5,658 and 142,235 are already in their decimal representation.The decimal representation of many numbers, such as π, have no pattern to the sequence of digits. Rational numbers are those that can be written as the quotient of two whole numbers, either have a terminating decimal representation or have a repeating representation.

    (2) The set of all integers (Y)and fractional numbers between them comprise the set of rational numbers (Z). The real numbers include all the rational numbers, such as the integer −5 and the fraction 4/3, and all the irrational numbers, such as √2 (1.41421356..., the square root of 2, an irrational algebraic number). Included within the irrationals are the transcendental numbers, such as π (3.14159265...).
  • Question 23
    1 / -0
    Match Column 1 with Column 2:

    Column 1 Column 2
    (A) Rational form of 0. 1.
    (B) Rational form of 2.
    (C) Rational form of 0.1 3.
    (D) Rational form of 1. 4.
    Solution
    (A) Let x = 0.
    x = 0.474747..... ----(1)
    Multiplying equation (1) with 100, we get
    100x = 100 × 0.474747.....
    100x = 47.474747......------(2)
    Subtracting (1) from (2) , we get 99x = 47
    x =
    Therefore ,
    x = 0. 474747.... =
    So, (A) = 2.

    (B) x =
    x = 0.233333......
    10x = 2.33333.....
    100x = 23.3333.....
    90x = 100x − 10x
    90x = (23.333.......) − (2.333333.....)
    90x = 21
    x = =
    So, (B) = 1

    (C) Let x = 0.1
    x = 0.1232323......
    10x = 1.232323.....
    1000x = 123.232323......
    1000x - 10x = (123.232323......) - (1.232323.....)
    990x = 122

    x =

    So, (C) = 4

    (D) x = 1.
    x = 1.232323......
    100x = 123.232323.....
    100x - x = (123.232323.....) - (1.232323......)
    99x = 122

    x =

    So, (D) = 3
  • Question 24
    1 / -0
    A cricket association distributes bats of different companies as gifts to players for their good performance in matches. The cricket association distributes 18 different bats of Reebok, 36 different bats of Adidas and 30 different bats of Puma to players. Each player is given maximum no. of bats of only one company of their interest and each player got equal number of bats.

    1. What is the maximum number of bats each player can get if all the players get equal number of bats and each player can have bats of only one brand?
    2. What will be the total number of players?
    Solution
    (1) We need to find G.C.F (H.C.F) for solving this question,
    H.C.F of 18,36 and 30
    18 = 3 × 6, 9 × 2, 18 × 1
    36 = 1 x 36, 2 x 18, 3 x 12, 4 x 9, 6 x 6.
    30 = 1 x 30, 2 x 15, 3 x 10, 5 x 6.
    So, H.C.F = 6
    Therefore, the number of bats each player received as gifts = 6.

    (2) No. of players who received Reebok bats = = 3

    No. of players who received Adidas bats = = 6

    No. of players who received Puma bats = = 5

    The total number of players who received bats = 3 + 6 + 5 = 14.
  • Question 25
    1 / -0
    Read the statements carefully and state 'T' for True and 'F' for False.

    (1) is a terminating decimal.

    (2) is a non-terminating decimal.

    (3) is a non-terminating decimal.

    (4) is a terminating decimal.

    (1) (2) (3) (4)
    A F F T T
    B F F F T
    C F F T F
    D T T F F
    Solution
    A fraction has a terminating decimal expansion if its denominator is of the form 2n × 5m or 2m or 5m, where m and n are non negative integers.

    (1) =

    The denominator of the above expression is not in the form of 2n × 5m. Therefore, is a non-terminating decimal.

    So, (1) is false.

    (2) =

    The denominator of above expression is in the form of 5m. Therefore, is a terminating decimal.

    So, (2) is false.

    (3)

    The denominator of the above expression is not in the form of 2n × 5m or 2n or 5m. Therefore, is a non-terminating decimal.

    So, (3) is true.


    (4)

    The denominator of above expression is in the form of 2n. Therefore, is a terminating decimal.

    So, (4) is true.
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