10th Maths - Book Back Answers - Chapter 2 Excercise 2.10 - English Medium Guides

 


    SSLC / 10th - Maths - Book Back Answers - Chapter 2 Excercise 2.10 - English Medium

    Tamil Nadu Board 10th Standard Maths - Chapter 2 Excercise 2.10: Book Back Answers and Solutions

        This post covers the book back answers and solutions for Chapter 2 Excercise 2.10 – Maths from the Tamil Nadu State Board 10th Standard Maths textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

        We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

        By going through this material, you’ll gain a strong understanding of Chapter 2 Excercise 2.10 along with the corresponding book back questions and answers (PDF format).

    Question Types Covered:

    • 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
    • 2 Mark Questions: Answer briefly 
    • 3, 4, and 5 Mark Questions: Answer in detail

    All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

    All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

    Chapter 2 Numbers and Sequences Ex 2.10

    1. Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….
    (1) 1 < r < b
    (2) 0 < r < b
    (3) 0 < r < 6
    (4) 0 < r < b
    Answer Key:

    (3) 0 < r < b
     
    2. Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….
    (1) 0, 1, 8
    (2) 1, 4, 8
    (3) 0, 1, 3
    (4) 1, 3, 5
    Answer Key:

    (1) 0, 1, 8
    Hint: Let the +ve integer be 1, 2, 3, 4 …………
    13 = 1 when it is divided by 9 the remainder is 1.
    23 = 8 when it is divided by 9 the remainder is 8.
    33 = 27 when it is divided by 9 the remainder is 0.
    43 = 64 when it is divided by 9 the remainder is 1.
    53 = 125 when it is divided by 9 the remainder is 8.
    The remainder 0, 1, 8 is repeated.
     
    3. If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is
    (1) 4
    (2) 2
    (3) 1
    (4) 3
    Answer Key:

    (2) 2
    Hint:
    H.C.F. of 65 and 117
    117 = 65 × 1 + 52
    65 = 52 × 1 + 13
    52 = 13 × 4 + 0
    13 is the H.C.F. of 65 and 117.
    65m – 117 = 65 × 2 – 117
    130 – 117 = 13
    m = 2
     
    4. The sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….
    (1) 1
    (2) 2
    (3) 3
    (4) 4
    Answer Key:

    (3) 3
    Hint: 1729 = 7 × 13 × 19
    Sum of the exponents = 1 + 1 + 1
    = 3
     
    5. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
    (1) 2025
    (2) 5220
    (3) 5025
    (4) 2520
    Answer Key:

    (4) 2520
    Hint:
    L.C.M. = 23 × 32 × 5 × 7
    = 8 × 9 × 5 × 7
    = 2520
     
    6. 74k ≡ ______ (mod 100)
    (1) 1
    (2) 2
    (3) 3
    (4) 4
    Answer Key:

    (1) 1
    Hint:
    74k ≡______ (mod 100)
    y4k ≡ y4 × 1 = 1 (mod 100)
     
    7. Given F1 = 1 , F2 = 3 and Fn = Fn-1 + Fn-2 then F5 is ………….
    (1) 3
    (2) 5
    (3) 8
    (4) 11
    Answer Key:

    (4) 11
    Hint:
    Fn = Fn-1 + Fn-2
    F3 = F2 + F1 = 3 + 1 = 4
    F4 = F3 + F2 = 4 + 3 = 7
    F5 = F4 + F3 = 7 + 4 = 11
     
    8. The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P
    (1) 4551
    (2) 10091
    (3) 7881
    (4) 13531
    Answer Key:

    (3) 7881
    Hint:
    t1 = 1
    d = 4
    tn = a + (n – 1)d
    = 1 + 4n – 4
    4n – 3 = 4551
    4n = 4554
    n = will be a fraction
    It is not possible.
    4n – 3 = 10091
    4n = 10091 + 3 = 10094
    n = a fraction
    4n – 3 = 7881
    4n = 7881 + 3 = 7884
    n =78844
    , n is a whole number.
    4n – 3 = 13531
    4n = 13531 – 3 = 13534
    n is a fraction.
    7881 will be 1971st term of A.P.
     
    9. If 6 times of 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P. is ………..
    (1) 0
    (2) 6
    (3) 7
    (4) 13
    Answer Key:

    (1) 0
    Hint:
    6 t6 = 7 t7
    6(a + 5d) = 7 (a + 6d)
    6a + 30d = 7a + 42d
    30 d – 42 d = 7a – 6a
    -12d = a
    t13 = a + 12d (12d = -a)
    = a – a = 0
     
    10. An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is
    (1) 16 m
    (2) 62 m
    (3) 31 m
    (4) 312
    m
    Answer Key:

    (3) 31 m
    Hint:
    t16 = m
    S31 =
    312(2a + 30d)
    =
    312(2(a + 15d))
    (
    t16 = a + 15d)
    = 31(t16) = 31m
     
    11. In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
    (1) 6
    (2) 7
    (3) 8
    (4) 9
    Answer Key:

    (3) 8
    Here a = 1, d = 4, Sn = 120
    Sn =
    n2[2a + (n – 1)d]
    120 =
    n2[2 + (n – 1)4] = n2[2 + 4n – 4)]
    =
    n2[4n – 2)] = n2× 2 (2n – 1)
    120 = 2n2 – n
    2n2 – n – 120 = 0 2n2 – 16n + 15n – 120 = 0
    2n(n – 8) + 15 (n – 8) = 0
    (n – 8) (2n + 15) = 0
    n = 8 or n =
    -152(omitted)
    n = 8
     
    12. A = 265 and B = 264 + 263 + 262 …. + 20 which of the following is true?
    (1) B is 264 more than A
    (2) A and B are equal
    (3) B is larger than A by 1
    (4) A is larger than B by 1
    Answer Key:

    (4) A is larger than B by
    A = 265
    B = 264+63 + 262 + …….. + 20
    = 2
    = 1 + 22 + 22 + ……. + 264
    a = 1, r = 2, n = 65 it is in G.P.
    S65 = 1 (265 – 1) = 265 – 1
    A = 265 is larger than B
     
    13. The next term of the sequence 316,18,112,118 is ………..
    (1)124

    (2)127

    (3)23

    (4)181

    Answer Key:

    (2)
    127
    Hint:

    316,18,112,118
    a=316 ,r=18÷316=18×163=23
    The next term is =118×23=127
     
    14. If the sequence t1,t2,t3 … are in A.P. then the sequence t6,t12,t18 … is
    (1) a Geometric Progression
    (2) an Arithmetic Progression
    (3) neither an Arithmetic Progression nor a Geometric Progression
    (4) a constant sequence
    Answer Key:

    (2) an Arithmetic Progression
    Hint:
    If t1, t2, t3, … is 1, 2, 3, …
    If t6 = 6, t12 = 12, t18 = 18 then 6, 12, 18 … is an arithmetic progression
     
    15. The value of (13 + 23 + 33 + ……. + 153) – (1 + 2 + 3 + …….. + 15) is …………….
    (1) 14400
    (2) 14200
    (3) 14280
    (4) 14520
    Answer Key:

    (3) 14280
    Hint:
    1202 – 120 = 120(120 – 1)
    120 × 119 = 14280

     

     


     


     

     

     

     






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