One question at a time. Build your Odisha BSE HSC Mathematics understanding.
First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
The lemma requires the remainder to be a whole number that is at least zero and strictly smaller than the divisor b. If r were equal to b or larger, the quotient could be increased by 1. Thus the defining condition is 0 ≤ r < b.
Dividing 360 repeatedly by primes gives 360 = 8 × 45 = 2³ × 9 × 5 = 2³ × 3² × 5. The Fundamental Theorem of Arithmetic guarantees that this product of primes is unique apart from the order of the factors. Hence the exponents must be 3, 2 and 1.
A rational number in lowest terms has a terminating decimal exactly when its denominator has no prime factors other than 2 and 5. Since 5/8 = 5/2³, its denominator contains only the prime 2, so the decimal terminates as 0.625. The other denominators contain the prime 3, so their decimals repeat.
For two positive integers, the product of the numbers equals HCF × LCM. Thus the unknown number is (12 × 2520) ÷ 180 = 30240 ÷ 180 = 168. Checking, HCF(180, 168) = 12 and LCM(180, 168) = 2520, so the value is consistent.
The sum of a rational number and an irrational number is irrational, and √5 is irrational because 5 is not a perfect square. Hence 3 + √5 cannot be written as a ratio of two integers. The other expressions simplify to 7, 0.5 and 22/7 respectively, all of which are rational.
Using a^(m/n) = (a^(1/n))^m, we get (27)^(2/3) = (∛27)² = 3² = 9. Also (16)^(1/2) = √16 = 4. Therefore the sum is 9 + 4 = 13.
Apply Euclid's algorithm repeatedly: 404 = 96 × 4 + 20, then 96 = 20 × 4 + 16, then 20 = 16 × 1 + 4, and 16 = 4 × 4 + 0. The last nonzero remainder is 4, so HCF(96, 404) = 4.
If p² = 2q², then p² is divisible by the prime 2. A prime dividing a square must divide the number itself, so 2 divides p, meaning p is even. Writing p = 2k then forces q also to be even, which contradicts the assumption that p and q are coprime.
The required number must divide both 245 − 5 = 240 and 1029 − 5 = 1024 exactly, so it is their HCF. Now 240 = 2⁴ × 3 × 5 and 1024 = 2¹⁰, so the only common prime factor is 2 and the HCF is 2⁴ = 16. Indeed, 245 and 1029 each leave remainder 5 on division by 16.
Since 3125 = 5⁵, the fraction has a terminating decimal. Multiply numerator and denominator by 2⁵ = 32: 13/3125 = (13 × 32)/(3125 × 32) = 416/100000 = 0.00416. The denominator becomes a power of 10, which fixes the decimal point.
Since 40 = 2^3 × 5, the denominator has only prime factors 2 and 5, so the decimal terminates. Multiplying numerator and denominator by 25 gives 7/40 = 175/1000 = 0.175, which has exactly three decimal places.
Any non-zero number raised to the power 0 equals 1. Therefore 5^0 = 1 and 3^0 = 1, so their sum is 2, and multiplying by 2^0 = 1 gives 2.
Write 36 = 2^2 × 3^2 and 84 = 2^2 × 3 × 7. The common prime factors appear as 2^2 × 3, so the HCF is 4 × 3 = 12.
Apply the algorithm: 867 = 255 × 3 + 102, then 255 = 102 × 2 + 51, and 102 = 51 × 2 + 0. The last non-zero remainder is 51, so the HCF is 51.
For two positive integers, HCF × LCM = product of the numbers. Here 9 × 108 = 972, so the other number is 972 ÷ 27 = 36. Indeed, HCF(27, 36) = 9 and LCM(27, 36) = 108.
Multiply numerator and denominator by the conjugate √7 + √6. The denominator becomes (√7)^2 − (√6)^2 = 7 − 6 = 1, so the expression simplifies to √7 + √6.
Suppose a + b were rational. Then b = (a + b) − a would be the difference of two rational numbers, hence rational, contradicting that b is irrational. Therefore a + b must be irrational.
If n = p^a q^b …, the number of positive divisors is (a + 1)(b + 1)…. To obtain 6 divisors, the smallest possibility is 2^2 × 3 = 12, whose divisors are 1, 2, 3, 4, 6, 12.
Since p is odd, p − 1 and p + 1 are consecutive even numbers, so one is divisible by 4 and their product is divisible by 8. Also, among p − 1, p, p + 1 one number is divisible by 3; p is not, so p^2 − 1 is divisible by 3. Hence it is divisible by 8 × 3 = 24.
Use fractional indices: 64^(2/3) = (∛64)^2 = 4^2 = 16, and (1/8)^(−2/3) = 8^(2/3) = (∛8)^2 = 4. Their product is 64. Also 16^(3/4) = (⁴√16)^3 = 2^3 = 8, so 64 ÷ 8 = 8.
The HCF is found by taking each prime common to both numbers and raising it to the lowest power that occurs. The common primes are 2 and 3, so HCF = 2² × 3² = 4 × 9 = 36.
This product has the difference-of-squares form (a + b)(a − b) = a² − b². Taking a = √7 and b = √3 gives 7 − 3 = 4.
Write 64 as 2⁶ and 81 as 3⁴. Then x = 6 and y = 4, so x + y = 10.
The next simultaneous toll occurs after the LCM of the three intervals. LCM(9, 12, 15) = 2² × 3² × 5 = 180 minutes = 3 hours, so the time is 7:00 a.m. + 3 hours = 10:00 a.m.
Simplify each surd first: √50 = 5√2, √18 = 3√2 and √8 = 2√2. Therefore 5√2 − 3√2 + 2√2 = 4√2.
Factorise 252 = 2² × 3² × 7. In a perfect square every prime exponent must be even; here only the exponent of 7 is odd, so one more factor of 7 is needed.
Rationalise 1/x as 1/(2 + √3) = (2 − √3)/((2 + √3)(2 − √3)) = (2 − √3)/(4 − 3) = 2 − √3. Hence x + 1/x = (2 + √3) + (2 − √3) = 4.
The digits repeat in a block of length 6: 142857. Since 100 = 6 × 16 + 4, the 100th digit is the 4th digit of the repeating block, which is 8.
Euclid's lemma states that if a prime p divides a product xy, then p divides x or p divides y. Taking x = y = a gives 5 | a² ⇒ 5 | a. Coprimality of a and b is used later in the proof, not at this step.
Let the integers be 6m and 6n with HCF(m, n) = 1. Then HCF × LCM = 6 × 210 = 1260, so 36mn = 1260 and mn = 35. The coprime pair giving both integers between 20 and 60 is m = 5, n = 7, giving 30 and 42.
A rational number in lowest terms has a terminating decimal exactly when its denominator has no prime factor other than 2 and 5. Here 12 = 2² × 3 includes the prime 3, so 5/12 produces a non-terminating recurring decimal. The denominators 8, 20 and 25 contain only 2s and 5s, so those fractions terminate.
Rewrite 2^(x−1) as 2^x/2 and 2^(x+1) as 2·2^x. Then (1/2 + 2)2^x = 40, so (5/2)2^x = 40, giving 2^x = 16 and x = 4.
For a × c to be a multiple of b, c must supply the missing prime powers from a compared with b. Since gcd(a, b) = 2² × 3 × 5 = 60, the least such c is b/gcd(a, b) = 540/60 = 9. Equivalently, a already has enough 2s and 5s but needs two more factors of 3, so c = 3² = 9.
Let x = 0.2777… Then 10x = 2.777… and 100x = 27.777…, so subtracting gives 90x = 25, that is x = 25/90 = 5/18. The shortcut rule is: (whole number up to the end of the first repeating block − non-repeating part) divided by as many 9s as there are repeating digits followed by as many 0s as there are non-repeating digits, here (27 − 2)/90. Reducing 25/90 by the common factor 5 gives 5/18.
Substitute x = 2 into p(x) = 2x² − 5x + 3, which gives 2(4) − 5(2) + 3 = 8 − 10 + 3 = 1. Only the variable x is squared; the coefficient 2 multiplies the whole term x². Hence p(2) = 1.
For an AP the nth term is aₙ = a + (n − 1)d, where a = 3 and d = 7 − 3 = 4. Thus a₁₀ = 3 + 9(4) = 3 + 36 = 39.
Write each surd in its simplest form: √50 = 5√2 and √18 = 3√2. Like surds are added by adding their coefficients, so 5√2 + 3√2 = 8√2. Radicands are never added directly.
For roots 3 and −4, the sum is 3 + (−4) = −1 and the product is 3 × (−4) = −12. A quadratic with these roots is x² − (sum)x + (product) = x² + x − 12 = 0.
From x − y = 1 we get x = y + 1. Substituting into 2x + 3y = 12 gives 2(y + 1) + 3y = 12, so 5y = 10, y = 2 and then x = 3. The pair (3, 2) satisfies both equations.
Multiply the numerator and denominator by the conjugate √7 + √5. The denominator becomes (√7)² − (√5)² = 7 − 5 = 2, so the expression equals (√7 + √5)/2.
Use Sₙ = (n/2)[2a + (n − 1)d] with n = 20, a = 5 and d = 3. Then S₂₀ = 10[10 + 19(3)] = 10[10 + 57] = 670.
By the remainder theorem, the remainder on division by x − 2 is p(2). Here p(2) = 8 + 8 − 10 + 6 = 12.
Any term of an AP equals the difference of two consecutive sums: aₙ = Sₙ − Sₙ₋₁. Here S₁₀ = 3(100) + 50 = 350 and S₉ = 3(81) + 45 = 288, so a₁₀ = 350 − 288 = 62.
From the coefficients, α + β = 5/2 and αβ = 3/2. Since α² + β² = (α + β)² − 2αβ, we get 25/4 − 2(3/2) = 25/4 − 3 = 13/4.
For a quadratic polynomial ax² + bx + c, the product of the zeroes is c/a. Here a = 4 and c = 2, so the product is 2/4 = 1/2. The sum of the zeroes would be −b/a = 3/4, which is a different quantity.
The discriminant is D = b² − 4ac. Substituting a = 1, b = −6 and c = 9 gives D = 36 − 36 = 0. A zero discriminant means the equation has two equal real roots.
Use the identity (a + b)(a − b) = a² − b². Here (√5)² − (√2)² = 5 − 2 = 3. The cross terms cancel because the signs are opposite.
Factorising, x² − 5x + 6 = (x − 2)(x − 3). Setting each factor equal to zero gives x = 2 or x = 3. These roots have sum 5 and product 6, matching the coefficients.
Infinitely many solutions occur when a₁/a₂ = b₁/b₂ = c₁/c₂. Here k/2 = 3/6 = 7/14, so k/2 = 1/2, giving k = 1. Both equations then represent the same line.
Here a = 9 and d = 8. Using Sₙ = n/2[2a + (n − 1)d] gives Sₙ = n/2(8n + 10) = n(4n + 5). Setting n(4n + 5) = 636 leads to 4n² + 5n − 636 = 0, whose positive root is n = 12.