answer:Let's denote the shares of Faruk, Vasim, and Ranjith as F, V, and R respectively. According to the problem, the ratio of their shares is 3:5:7. This means: F : V : R = 3 : 5 : 7 Let's assume the common multiple for the ratio is x. Then we can write their shares as: F = 3x V = 5x R = 7x The difference between Faruk's and Ranjith's shares is Rs. 1200, so: R - F = 1200 7x - 3x = 1200 4x = 1200 Now, we can solve for x: x = 1200 / 4 x = 300 Now that we have the value of x, we can find Vasim's share: V = 5x V = 5 * 300 V = 1500 So, Vasim's share is Rs. boxed{1500} .

answer:The original statement can be formulated as "If a quadrilateral is a square, then it is a parallelogram." The negation of the statement is "If a quadrilateral is not a square, then it is not a parallelogram." This statement is false because a quadrilateral can still be a parallelogram without being a square (e.g., a rectangle). The contrapositive of the statement is "If a quadrilateral is not a parallelogram, then it is not a square." This statement is true since any shape that is not a parallelogram cannot be a square, given that being a parallelogram is a necessary condition for being a square. Therefore, the correct answer is: [ boxed{A: One is true and one is false.} ]

answer:1. **Initial Clauses and Requirements:** - Given any rational number ( x ), we need to prove the existence of a sequence ( {x_n} ) of rational numbers with the properties: - (a) ( x_0 = x ); - (b) For every ( n geq 1 ), ( x_n ) must be either ( 2x_{n-1} ) or ( 2x_{n-1} + frac{1}{n} ); - (c) There exists some ( n ) such that ( x_n ) is an integer. 2. **Strategy and Approach:** - Think of the sequence as a process over time. - Show that at any given step ( t ), if the denominator of ( x_t ) has an odd prime power ( q = p^e ), you can eliminate one factor of ( p ) from the denominator while only adding powers of 2 to the denominator. - Essentially, we can remove every factor of any odd prime one by one and then multiply by 2 as needed. 3. **Detailed Step-by-Step Proof:** - Let us represent ( x_t ) in its simplest form ( frac{a}{b} ), where ( a ) and ( b ) are coprime integers, and ( b )’s prime factorization includes an odd prime ( p ). - To remove ( p ) from the denominator, we consider adding fractions of the form ( (2^kq)^{-1} ). 4. **Transforming the Denominators:** - Suppose ( t < 2^{r+1}q < 2^{r+2}q < cdots < 2^{r+m}q < n ). - For binary variables ( varepsilon_i in {0,1} ), reframe ( x_n ) as: [ x_n = 2^{n-t} x_t + c_1 cdot frac{varepsilon_1}{q} + c_2 cdot frac{varepsilon_2}{q} + cdots + c_m cdot frac{varepsilon_m}{q} ] where ( c_i ) is some power of 2. - The precise formulation for ( c_i = frac{2^{n-2^{r+i}q}}{2^{r+1}} ). 5. **Combining Fractions To Adjust the Denominator:** - If ( m ) is sufficiently large, the set ({0,c_1} + {0,c_2} + cdots + {0,c_m}) covers every possible congruence modulo ( p ). - Utilizing the Cauchy-Davenport theorem or the pigeonhole principle, when ( m = p ), you can always find such linear combinations that span every residue class modulo ( p ). 6. **Conclusion of Denominator Reduction:** - Accordingly, one can always find a combination of the ( varepsilon_i )s to ensure one factor of ( p ) is eliminated. - Repeat this process until there are no more odd prime factors in the denominator of ( x_k ). 7. **Ensuring ( x_n ) Eventually Becomes an Integer:** - After removing all odd primes, the denominator ( b ) will only contain powers of 2. - Continue doubling the value ( a ) until ( x_k ) becomes an integer. This doubling is permissible since it aligns with the operations allowed by ( x_n = 2x_{n-1} ). # Conclusion: Therefore, by systematically applying the steps above, the sequence ( {x_n} ) conforms to the requirements: [ (a) quad x_0 = x, ] [ (b) quad text{For every } n geq 1, text{x_n is either } 2x_{n-1} text{ or } 2x_{n-1} + frac{1}{n}, ] [ (c) quad text{There exists some } n text{ such that } x_n text{ is an integer.} ] [ boxed{} ]

answer:1. **Calculate ( f(1) )**: Start by setting ( x = 1 ) and ( y = 1 ) in the given functional equation: [ f(x f(y)) = frac{f(x)}{y} ] Substituting ( x = 1 ) and ( y = 1 ), we get: [ f(1 f(1)) = frac{f(1)}{1} = f(1). ] Thus, ( f(f(1)) = f(1) ). This implies that ( f(1) = 1 ) since ( f ) is a function from ( mathbb{Q}_{+}^{*} ) to ( mathbb{Q}_{+}^{*} ). 2. **Show that ( f(xy) = f(x) f(y) )**: Consider the equation again and set ( x = frac{1}{f(y)} ): [ fleft(frac{1}{f(y)} f(y)right) = frac{fleft(frac{1}{f(y)}right)}{y}. ] Since ( f(f(1)) = f(1) ) and ( f(1) = 1 ), this simplifies to [ f(1) = frac{fleft(frac{1}{f(y)}right)}{y}. ] Thus, [ 1 = frac{fleft(frac{1}{f(y)}right)}{y}, ] which implies [ fleft(frac{1}{f(y)}right) = y. ] Now set ( z = frac{1}{f(y)} ). Therefore, ( f(z) = y ). Hence, using the original functional equation and substituting back, we have: [ f(x y) = f(x f(z)) = frac{f(x)}{z} = f(x) f(y). ] 3. **Deduction for composite numbers**: As ( f ) is multiplicative, once we know ( f(p) ) for any prime ( p ), we can deduce ( f ) for any rational ( x ). Hence, it suffices to determine the values for ( f ) on prime arguments. 4. **Define ( f ) on primes split into two sets**: Partition the set of prime numbers ( mathcal{P} ) into two infinite subsets ( {p_1, p_2, ldots } ) and ( {q_1, q_2, ldots } ). Define ( f ) by: begin{align*} f(p_i) &= q_i, f(q_i) &= frac{1}{p_i}. end{align*} It is straightforward to check that this satisfies the functional equation: [ f(x f(y)) = frac{f(x)}{y}. ] For example, if ( x = p_i ) and ( y = q_j ), we compute [ f(p_i f(q_j)) = fleft(p_i cdot frac{1}{p_j}right) = fleft(frac{p_i}{p_j}right), ] and since ( frac{p_i}{p_j} ) is a rational number, by the multiplicative nature established in step 2, we get: [ fleft(frac{p_i}{p_j}right) = frac{f(p_i)}{f(p_j)} = frac{q_i}{f(p_j)}. ] Since ( f(p_j) = q_j ), we have: [ frac{q_i}{q_j} = frac{f(p_i)}{y}, ] which is consistent with the original equation. Hence, the function ( f ) as defined satisfies the equation. Thus, a function ( f ) satisfying the given functional equation is constructed: [ boxed{f(p_i) = q_i text{ and } f(q_i) = frac{1}{p_i}}. ]