Chapter-1
Real Numbers
Q1. Two positive numbers have their HCF as 12 and their product as 6336. The number of pairs possible for the numbers is:
Solution Let numbers be \(12x\) and \(12y\) where x, y are co-primes. Product \(= 144xy = 6336 \Rightarrow xy = 44\). Possible pairs for 44: (1, 44), (4, 11). (2, 22) is rejected. So, 2 pairs.
Q2. The value of \((12)^{3x} + (18)^{3x}\), \(x \in N\), ends with the digit:
Solution For any \(x\), \((12)^{3x}\) ends in 8 or 2, and \((18)^{3x}\) ends in 2 or 8 respectively. Their sum ends in 0.
Q3. If \(n\) is an even natural number, then the largest natural number by which \(n(n+1)(n+2)\) is divisible is:
Solution Since \(n\) is even, \(n\) and \(n+2\) are consecutive even integers (product divisible by 8). \(n, n+1, n+2\) are 3 consecutive integers (divisible by 3). Total divisibility \(8 \times 3 = 24\).
Q4. If \(p_1\) and \(p_2\) are two odd prime numbers such that \(p_1 > p_2\), then \(p_1^2 - p_2^2\) is:
Solution Difference of squares of two odd numbers is always even. Example: \(5^2 - 3^2 = 25 - 9 = 16\) (Even).
Q5. The rational form of \(0.2\overline{54}\) is in the form of \(\frac{p}{q}\), then \((p+q)\) is:
Solution \(x = 0.25454... \Rightarrow x = \frac{14}{55}\). \(p+q = 14+55 = 69\).
Q6. If \(a = 2^3 \times 3\), \(b = 2 \times 3 \times 5\), \(c = 3^n \times 5\) and LCM(a, b, c) = \(2^3 \times 3^2 \times 5\), then \(n\) = ?
Solution LCM takes the highest power of each prime. The LCM has \(3^2\), so \(n\) must be 2.
Q7. Three sets of books (Maths: 240, Science: 960, Biology: 1024) are stacked subject-wise with equal height. The number of stacks for each are:
Solution HCF of 240, 960, 1024 is 16. Stacks: Maths = 240/16 = 15, Science = 960/16 = 60, Biology = 1024/16 = 64.
Q8. The rational number representing \(0.1\overline{34}\) is:
Solution \(0.1\overline{34} = \frac{134 - 1}{990} = \frac{133}{990}\).
Q9. If \(x\) and \(y\) are odd positive integers, then \(x^2 + y^2\) is:
Solution \(x^2+y^2 = 4(m^2+n^2+m+n)+2\). This is of form \(4k+2\), which is even but not divisible by 4.
Q10. The least number which is a perfect square and is divisible by each of 16, 20 and 24 is:
Solution LCM of 16, 20, 24 is 240. Multiples of 240 which are perfect squares: 3600 is the least one.
Q11. Which of the following rational numbers has a non-terminating repeating decimal expansion?
Solution All denominators are of form \(2^m \times 5^n\), so all are terminating.
Q12. When \(2^{256}\) is divided by 17, the remainder would be:
Solution \(2^{256} = (16)^{64}\). Since \(16 \equiv -1 \pmod{17}\), \((-1)^{64} = 1\).
Q13. The least number which when divided by 15, 25, 35 leaves remainders 5, 15, 25 respectively is:
Solution LCM(15, 25, 35) = 525. Difference is constant (10). Required number = \(525 - 10 = 515\).
Q14. In a factor tree, \(1001 \to 7, x\) and \(x \to 11, y\). The values of \(x\) and \(y\) are:
Solution \(1001 = 7 \times x \Rightarrow x = 143\). \(143 = 11 \times y \Rightarrow y = 13\).
Q15. The sum of three non-zero prime numbers is 100. One of them exceeds the other by 36. Then the largest number is:
Solution Sum is even, so one prime is 2. \(2+x+y=100 \Rightarrow x+y=98\). \(x-y=36\). Solving gives \(x=67\).
Q16. If \(P = 2 \times 4 \dots \times 20\) and \(Q = 1 \times 3 \dots \times 19\), then HCF of P and Q is:
Solution Min power of 3 in both is \(3^4\). Min power of 5 is \(5^2\). Min power of 7 is \(7^1\).
Q17. The number \(3^{13} - 3^{10}\) is divisible by:
Solution \(3^{13} - 3^{10} = 3^{10}(26) = 3^{10} \times 2 \times 13\). Divisible by 2, 3, 13.
Q18. Which of the following will have a terminating decimal expansion?
Solution \(\frac{23}{8} = \frac{23}{2^3}\). Denominator has only factor 2, so it terminates.
Q19. A number lying between 300 and 400. If added to its reverse, sum is 888. If unit and tens digit are swapped, new number is 9 more. The number is:
Solution For 345: \(3+5=8\). Check swap: 354. \(354-345=9\). Correct.
Q20. Two cyclists start together around a 360km circular field at 12 km/h and 15 km/h. They meet again at the start after:
Solution \(360/12 = 30h\), \(360/15 = 24h\). LCM of 30 and 24 is 120.