One question at a time. Build your AP SSC 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.
Euclid’s division lemma requires the remainder to satisfy 0 ≤ r < b; the quotient q may be any integer, but r cannot equal b and cannot be negative. If r were allowed to equal b, uniqueness of q and r would fail because bq + b = b(q + 1) + 0.
Factor 360 step by step: 360 = 36 × 10 = (2² × 3²)(2 × 5) = 2³ × 3² × 5. The exponents must match the number of each prime in the product, and the factorisation is unique up to order.
For any two positive integers a and b, a × b = HCF(a,b) × LCM(a,b). Substituting the given values gives a × b = 6 × 72 = 432.
A rational number in lowest terms has a terminating decimal exactly when its denominator has no prime factors other than 2 and 5. For 5/12 the denominator is 12 = 2² × 3, and the factor 3 forces a non-terminating recurring expansion.
From 2q² = p², the prime 2 divides p², hence p is even; write p = 2k. Then 2q² = 4k², so q² = 2k², which makes q even too, contradicting that p and q are coprime.
Apply the algorithm: 867 = 255 × 3 + 102; 255 = 102 × 2 + 51; 102 = 51 × 2 + 0. The last nonzero remainder is 51, so HCF(867, 255) = 51.
The required number must be a multiple of 12, 15, and 18, so the least such number is their LCM. Prime factorisations give 12 = 2² × 3, 15 = 3 × 5, 18 = 2 × 3², so LCM = 2² × 3² × 5 = 180.
Use the difference of squares identity: (√5 + √3)(√5 − √3) = (√5)² − (√3)² = 5 − 3 = 2. This is a rational value despite the two irrational factors.
The repeating block 142857 has length 6. Since 100 = 6 × 16 + 4, the 100th digit is the fourth digit of the block, which is 8.
LCM(a, b) = 2³ × 3³ × 5 × 7, because each prime takes the larger exponent. A square divisor can use even exponents only: for 2, 0 or 2; for 3, 0 or 2; for 5, only 0; for 7, only 0. This gives 2 × 2 × 1 × 1 = 4 such divisors.
Use the product law a^m × a^n = a^(m+n) to combine the numerator, then the quotient law a^m ÷ a^n = a^(m−n). Here 2^3 × 2^5 = 2^8, and 2^8 ÷ 2^4 = 2^4 = 16.
For the HCF, take the smaller power of each common prime factor. Both numbers contain the primes 3 and 5, so HCF = 3^1 × 5^1 = 15.
The denominator 8 = 2^3 has only the prime factor 2, so the decimal terminates. Dividing 3 by 8 gives 0.375 exactly.
Euclid’s algorithm gives 252 = 105 × 2 + 42, then 105 = 42 × 2 + 21, then 42 = 21 × 2 + 0. The last non-zero remainder is 21, so HCF(252, 105) = 21.
The repeating block has two digits, so multiply by 100 and subtract the original number. This gives 99x = 36, hence x = 36/99 = 4/11.
If x + y were rational, then y = (x + y) − x would be the difference of two rational numbers and hence rational. This contradicts y being irrational, so x + y must be irrational.
The required number must divide both 245 − 5 = 240 and 1029 − 5 = 1024, so it is their HCF. Using Euclid’s algorithm, HCF(240, 1024) = 16.
A perfect square must have even exponents in its prime factorisation. In N, the exponents 4 and 2 are even, but 5 is odd, so multiplying by 3 changes 3^5 to 3^6.
Write a = 12m and b = 12n, where m and n are coprime. Then ab = 144mn = 4320, so mn = 30; each prime factor 2, 3, and 5 may be assigned to m or to n, giving 2^3 = 8 ordered pairs.
A number with exactly six divisors has form p^5 or p^2q for distinct primes. Below 100, p^5 gives only 32; counting p^2q for p = 2, 3, 5, 7 gives 8, 4, 2, and 1 numbers respectively, for a total of 16.
Natural numbers begin at 1, so 0 is never an element of a set built from N. The even natural numbers below 10 are 2, 4, 6 and 8, and the roster form lists each of them once inside braces, separated by commas.
Every listed element is a perfect square: 1 = 1², 4 = 2², 9 = 3², 16 = 4² and 25 = 5². The defining property is therefore x = n² where n runs over the natural numbers 1 to 5.
A number that leaves remainder 1 on division by 2 can never be exactly divisible by 2, so no number satisfies the condition and the set contains no elements. An empty set is written ∅ and has cardinal number 0.
The primes below 10 are 2, 3, 5 and 7, so n(A) = 4 and A has 2⁴ = 16 subsets in all. A proper subset is any subset other than A itself, so the count is 16 − 1 = 15.
The students who like at least one drink form the union, and n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting gives 25 + 20 − 10 = 35, the 10 being subtracted once because those students were counted in both groups.
De Morgan's law for sets states that the complement of a union is the intersection of the complements. An element lies outside A ∪ B exactly when it lies outside A and also outside B.
Two sets are equal when they contain exactly the same elements; the order in which the elements are written has no effect. Only {1, 2, 3} and {3, 2, 1} have identical elements, so only that pair is equal.
The difference A − B consists of the elements that belong to A but not to B. Removing the shared elements 3 and 4 from A leaves {1, 2}.
By inclusion–exclusion, n(A ∪ B ∪ C) = 25 + 30 + 20 − 10 − 8 − 5 + 3 = 55. Since 55 of the 60 people like at least one drink, 60 − 55 = 5 like none.
The condition restricts x to two-digit numbers whose digits add to 6, namely 15, 24, 33, 42, 51 and 60. Counting them gives n(A) = 6.
A set is infinite when its elements cannot be listed exhaustively by counting up to a last element. The natural-number multiples of 7 continue without end: 7, 14, 21, and so on. The other three sets have finitely many members.
In this syllabus N = {1, 2, 3, ...}, so 0 is not included. The positive multiples of 4 below 18 are 4, 8, 12 and 16, and the next multiple 20 is outside the condition.
For any two finite sets, n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Substituting gives 7 + 5 − 3 = 9, so the union contains 9 elements.
A − B contains elements of A not in B, so A − B = {1, 2, 3}; similarly B − A = {6, 7}. Their union is {1, 2, 3, 6, 7}, which lists each element once.
First form A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}. The complement in U consists of the elements of U not in this union, namely 9 and 10.
Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B). Thus 25 = 18 + 14 − n(A ∩ B), so n(A ∩ B) = 32 − 25 = 7.
If a set has n elements, its power set has 2^n elements because each element is either included or not included in a subset. Here n = 5, so 2^5 = 32.
The number who like at least one game is 28 + 22 − 10 = 40. Since the class has 50 students, those who like neither are 50 − 40 = 10.
Count multiples of 2 up to 30: there are 15, and multiples of 3: there are 10. Multiples of both 2 and 3 are multiples of 6, and there are 5; by inclusion-exclusion the count is 15 + 10 − 5 = 20.
By De Morgan's law, A′ ∩ B′ = (A ∪ B)′. Now n(A ∪ B) = 22 + 18 − 8 = 32, so the number outside A ∪ B is 40 − 32 = 8.
The degree of a polynomial in one variable is the highest power of the variable whose coefficient is non-zero. In 4x^3 - 2x^2 + 7x - 5 the powers present are 3, 2, 1 and 0, and the largest of these is 3, so the polynomial is cubic.
The value p(0) is obtained by substituting x = 0 in the polynomial. Every term that contains x vanishes, so only the constant term remains: p(0) = 0 - 0 + 6 = 6.
A point where the graph meets the x-axis has y = 0, so its x-coordinate is a zero of the polynomial. Two distinct points of intersection therefore give two distinct real zeros, and a quadratic cannot have more than two zeros in any case.
For a quadratic polynomial ax^2 + bx + c, the sum of the zeros is -b/a. Here a = 2 and b = -8, so the sum is -(-8)/2 = 4, which agrees with the zeros 1 and 3 obtained by factorising.
By the remainder theorem, dividing p(x) by (x - a) leaves the remainder p(a). Taking a = 2 gives p(2) = 8 - 12 + 8 - 5 = -1, so the remainder is -1.
The factor theorem states that (x - 3) is a factor exactly when p(3) = 0. Substituting gives 9 - 3k + 6 = 0, that is 15 - 3k = 0, so k = 5.
A quadratic with zeros alpha and beta is x^2 - (alpha + beta)x + alpha*beta. Here alpha + beta = 2 + (-5) = -3 and alpha*beta = -10, so the polynomial is x^2 - (-3)x + (-10), that is x^2 + 3x - 10.
From the coefficients, alpha + beta = 5 and alpha*beta = 6. Using the identity alpha^2 + beta^2 = (alpha + beta)^2 - 2(alpha*beta), we get 25 - 12 = 13, which matches the direct check with zeros 2 and 3.
Let the zeros be alpha and 3*alpha. Their sum is -(-8)/2 = 4, so 4*alpha = 4 and alpha = 1, making the zeros 1 and 3. Their product is k/2, so k/2 = 3 and therefore k = 6.
For a cubic ax^3 + bx^2 + cx + d, the sum of the products of the zeros taken two at a time is c/a. Here a = 1 and c = 11, so the value is 11, which is confirmed by the zeros 1, 2 and 3 giving 2 + 6 + 3 = 11.