Every Book Back multiple-choice question from Numbers and Sequences (10th Standard Mathematics, Samacheer Kalvi) — each with the correct option highlighted and a clear, worked explanation. Free to read in English and Tamil.
Q1
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:
- A. \(1 < r < b\)
- B. \(0 < r < b\)
- C. \(0 \le r < b\)Correct
- D. \(0 < r \le b\)
Explanation. Euclid's division lemma states that the remainder \(r\) is always non-negative and strictly less than the divisor \(b\). This is expressed by the inequality \(0 \le r < b\).
Q2
Using Euclid's division lemma, if the cube of any positive integer is divided by \(9\) then the possible remainders are:
- A. \(0, 1, 8\)Correct
- B. \(1, 4, 8\)
- C. \(0, 1, 3\)
- D. \(1, 3, 5\)
Explanation. Any positive integer can be written as \(3k, 3k+1,\) or \(3k+2\). Cubing these forms gives expressions of the form \(9m, 9m+1,\) or \(9m+8\), leaving remainders of \(0, 1,\) or \(8\) when divided by \(9\).
Q3
If the HCF of \(65\) and \(117\) is expressible in the form of \(65m - 117\), then the value of \(m\) is:
- A. 4
- B. 2Correct
- C. 1
- D. 3
Explanation. The prime factorizations are \(65 = 5 \times 13\) and \(117 = 9 \times 13\), giving HCF \(= 13\). Setting \(65m - 117 = 13\) yields \(65m = 130\), which solves to \(m = 2\).
Q4
The sum of the exponents of the prime factors in the prime factorization of \(1729\) is:
- A. 1
- B. 2
- C. 3Correct
- D. 4
Explanation. The prime factorization of \(1729\) is \(7 \times 13 \times 19\). Each prime factor has an exponent of \(1\), so the sum of the exponents is \(1 + 1 + 1 = 3\).
Q5
The least number that is divisible by all the numbers from \(1\) to \(10\) (both inclusive) is:
- A. 2025
- B. 5220
- C. 5025
- D. 2520Correct
Explanation. The least common multiple (LCM) of the numbers \(1, 2, 3, 4, 5, 6, 7, 8, 9,\) and \(10\) is calculated by taking the highest powers of all prime factors, yielding \(2520\).
Q6
\(7^{4k} \equiv\) _____ \(\pmod{100}\):
- A. 1Correct
- B. 2
- C. 3
- D. 4
Explanation. Since \(7^4 = 2401 \equiv 1 \pmod{100}\), raising both sides to the power of \(k\) gives \((7^4)^k \equiv 1^k \equiv 1 \pmod{100}\). Thus, \(7^{4k} \equiv 1 \pmod{100}\).
Q7
Given \(F_1 = 1\), \(F_2 = 3\) and \(F_n = F_{n-1} + F_{n-2}\), then \(F_5\) is:
- A. 3
- B. 5
- C. 8
- D. 11Correct
Explanation. Calculating terms sequentially: \(F_3 = F_2 + F_1 = 3 + 1 = 4\); \(F_4 = F_3 + F_2 = 4 + 3 = 7\); \(F_5 = F_4 + F_3 = 7 + 4 = 11\).
Q8
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.?
- A. 4551
- B. 10091
- C. 7881Correct
- D. 13531
Explanation. The terms of this A.P. are of the form \(t_n = 1 + (n-1)4\), which means \(t_n \equiv 1 \pmod 4\). Testing the options, \(7881 \equiv 1 \pmod 4\), while others are congruent to \(3\).
Q9
If \(6\) times of \(6^{\text{th}}\) term of an A.P. is equal to \(7\) times the \(7^{\text{th}}\) term, then the \(13^{\text{th}}\) term of the A.P. is:
- A. 0Correct
- B. 6
- C. 7
- D. 13
Explanation. The relation \(6(a+5d) = 7(a+6d)\) simplifies to \(6a+30d = 7a+42d\), which gives \(a+12d = 0\). Since \(t_{13} = a+12d\), the \(13^{\text{th}}\) term is \(0\).
Q10
An A.P. consists of \(31\) terms. If its \(16^{\text{th}}\) term is \(m\), then the sum of all the terms of this A.P. is:
- A. 16 m
- B. 62 m
- C. 31 mCorrect
- D. \(\frac{31}{2}\) m
Explanation. The sum of \(31\) terms is \(S_{31} = \frac{31}{2}[2a + 30d] = 31(a + 15d)\). Since the \(16^{\text{th}}\) term \(t_{16} = a+15d = m\), the sum is \(31m\).
Q11
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 \(120\)?
- A. 6
- B. 7
- C. 8Correct
- D. 9
Explanation. Using \(S_n = \frac{n}{2}[2a + (n-1)d] = 120\) with \(a=1, d=4\) yields \(n(2n-1) = 120\). Solving \(2n^2 - n - 120 = 0\) gives \(n=8\) as the positive integer solution.
Q12
If \(A = 2^{65}\) and \(B = 2^{64} + 2^{63} + 2^{62} + \dots + 2^0\), which of the following is true?
- A. B is \(2^{64}\) more than A
- B. A and B are equal
- C. B is larger than A by 1
- D. A is larger than B by 1Correct
Explanation. \(B\) is a geometric series with sum \(\frac{2^{65}-1}{2-1} = 2^{65}-1\). Comparing \(A = 2^{65}\) and \(B = 2^{65}-1\) shows that \(A\) is larger than \(B\) by \(1\).
Q13
The next term of the sequence \(\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, \dots\) is:
- A. \(\frac{1}{24}\)
- B. \(\frac{1}{27}\)Correct
- C. \(\frac{2}{3}\)
- D. \(\frac{1}{81}\)
Explanation. This is a G.P. with common ratio \(r = \frac{1/8}{3/16} = \frac{2}{3}\). The next term is found by multiplying the fourth term by \(r\), giving \(\frac{1}{18} \times \frac{2}{3} = \frac{1}{27}\).
Q14
If the sequence \(t_1, t_2, t_3, \dots\) are in A.P. then the sequence \(t_6, t_{12}, t_{18}, \dots\) is:
- A. a Geometric Progression
- B. an Arithmetic ProgressionCorrect
- C. neither an Arithmetic Progression nor a Geometric Progression
- D. a constant sequence
Explanation. Since the terms are sampled at equal index intervals of \(6\), the common difference between consecutive terms is \(t_{12} - t_6 = 6d\) and \(t_{18} - t_{12} = 6d\), which forms an A.P.
Q15
The value of \((1^3 + 2^3 + 3^3 + \dots + 15^3) - (1 + 2 + 3 + \dots + 15)\) is:
- A. 14400
- B. 14200
- C. 14280Correct
- D. 14520
Explanation. We have \(\sum_{i=1}^{15} i = \frac{15 \times 16}{2} = 120\) and \(\sum_{i=1}^{15} i^3 = 120^2 = 14400\). The difference is \(14400 - 120 = 14280\).
