💡First, I need to find the principal amount (P) using the given compound interest (CI) and simple interest (SI) for two years.I know that the formula for simple interest is:[ SI = frac{P times R times T}{100} ]And the formula for compound interest is:[ CI = P times left(1 + frac{R}{100}right)^T - P ]Given:[ CI = 11730 ][ SI = 10200 ][ T = 2 text{ years} ]The difference between compound interest and simple interest for two years is:[ CI - SI = P times left(frac{R}{100}right)^2 ][ 11730 - 10200 = P times left(frac{R}{100}right)^2 ][ 1530 = P times left(frac{R}{100}right)^2 ]From the simple interest formula:[ 10200 = frac{P times R times 2}{100} ][ 10200 = frac{P times R}{50} ][ R = frac{10200 times 50}{P} ]Substituting the value of ( R ) into the difference equation:[ 1530 = P times left(frac{10200 times 50}{P times 100}right)^2 ][ 1530 = P times left(frac{10200}{P}right)^2 times left(frac{50}{100}right)^2 ][ 1530 = P times frac{10200^2}{P^2} times frac{1}{4} ][ 1530 = frac{10200^2}{4P} ]Solving for ( P ):[ P = frac{10200^2}{4 times 1530} ][ P = frac{10200^2}{6120} ][ P = frac{10200}{6} ][ P = 1700 ]Therefore, the sum is Rs. 1700.

💡Okay, so I have this problem about a line passing through the point (1,2) and having equal intercepts on the x and y axes. There are two parts: first, finding the general equation of the line, and second, finding the minimum value of 3^a + 3^b where the point (a,b) lies on the line and the intercepts are not zero.Starting with part (1). The line has equal intercepts on both axes. That means the x-intercept and y-intercept are the same. Let me recall that the intercept form of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. Since they are equal, a = b. So, the equation becomes x/a + y/a = 1, which simplifies to x + y = a. So, the equation of the line is x + y = a.But wait, the line also passes through the point (1,2). So, substituting x = 1 and y = 2 into the equation, we get 1 + 2 = a, so a = 3. Therefore, the equation of the line is x + y = 3, or x + y - 3 = 0.But hold on, the problem says "general equation of line l." I think that might mean considering all possibilities. So, is there another case where the line could pass through the origin? Because if the line passes through the origin, then both intercepts are zero, which technically are equal. So, in that case, the equation would be y = mx, where m is the slope. Since it passes through (1,2), plugging in, we get 2 = m*1, so m = 2. Therefore, the equation is y = 2x.So, in summary, there are two possible equations for line l: one is y = 2x, which passes through the origin, and the other is x + y - 3 = 0, which has intercepts at (3,0) and (0,3). So, for part (1), the general equation is either y = 2x or x + y - 3 = 0.Moving on to part (2). It says that the intercepts are not zero, so we can disregard the case where the line passes through the origin. Therefore, the equation of the line is x + y - 3 = 0, which means a + b = 3, where a is the x-intercept and b is the y-intercept.Wait, no, actually, the point p(a,b) is on the line, so a and b are coordinates of a point on the line, not the intercepts. The intercepts are given to be non-zero, so the line is x + y = 3.So, since p(a,b) is on the line, we have a + b = 3. We need to find the minimum value of 3^a + 3^b.Hmm, okay. So, we have a constraint a + b = 3, and we need to minimize 3^a + 3^b.I remember that for expressions like this, the AM-GM inequality might be useful. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. But here, we have exponents. Maybe we can use the inequality for exponents.Alternatively, since 3^a and 3^b are both positive, we can apply the AM-GM inequality on them. Let me try that.By AM-GM, (3^a + 3^b)/2 ≥ sqrt(3^a * 3^b). Simplifying the right side, sqrt(3^{a+b}) = sqrt(3^3) = sqrt(27) = 3*sqrt(3). So, (3^a + 3^b)/2 ≥ 3*sqrt(3), which implies 3^a + 3^b ≥ 6*sqrt(3).So, the minimum value is 6*sqrt(3). But when does equality hold? In AM-GM, equality holds when all the terms are equal. So, 3^a = 3^b, which implies a = b. Since a + b = 3, we have a = b = 3/2.Therefore, the minimum value of 3^a + 3^b is 6*sqrt(3), achieved when a = b = 3/2.Wait, let me double-check. If a = b = 3/2, then 3^{3/2} + 3^{3/2} = 2*3^{3/2} = 2*(sqrt(3^3)) = 2*(3*sqrt(3)) = 6*sqrt(3). That seems correct.Alternatively, I could use calculus to verify. Let me express b in terms of a: b = 3 - a. Then, the expression becomes 3^a + 3^{3 - a}. Let me denote f(a) = 3^a + 3^{3 - a}.To find the minimum, take the derivative of f(a) with respect to a:f'(a) = ln(3)*3^a - ln(3)*3^{3 - a}.Set f'(a) = 0:ln(3)*3^a - ln(3)*3^{3 - a} = 0Divide both sides by ln(3):3^a - 3^{3 - a} = 0So, 3^a = 3^{3 - a}Since the bases are equal, the exponents must be equal:a = 3 - aSo, 2a = 3 => a = 3/2Therefore, b = 3 - 3/2 = 3/2So, the critical point is at a = 3/2, b = 3/2, which gives f(a) = 6*sqrt(3). To confirm it's a minimum, check the second derivative or test intervals.Alternatively, since the function is convex, the critical point is indeed a minimum.So, both methods give the same result. Therefore, the minimum value is 6*sqrt(3).**Final Answer**(1) The general equation of line ( l ) is (boxed{y = 2x}) or (boxed{x + y - 3 = 0}).(2) The minimum value of ( 3^{a} + 3^{b} ) is (boxed{6sqrt{3}}).

💡Okay, so I have this problem here where I need to figure out the relationship between log(1+x) and x/(1-x) when x is between 0 and 1. The options are whether they're equal, or which one is bigger. Hmm, let me think about this step by step.First, I remember that the logarithm function, log(1+x), is defined for x > -1, but in this case, x is between 0 and 1, so that's fine. I also recall that log(1+x) is an increasing function because its derivative, which is 1/(1+x), is always positive for x > -1. So, as x increases, log(1+x) increases too, but how does it compare to x/(1-x)?Let me think about the function x/(1-x). When x is between 0 and 1, the denominator (1-x) is positive and less than 1, so x/(1-x) is greater than x. For example, if x is 0.5, then x/(1-x) is 1, which is bigger than x=0.5. So, x/(1-x) seems to grow faster than x itself.But how does log(1+x) compare to x/(1-x)? Maybe I can use some approximations or series expansions to get a better idea. I remember the Taylor series expansion for log(1+x) around x=0 is x - x²/2 + x³/3 - x⁴/4 + ... and so on. This is an alternating series, so the terms get smaller in absolute value as x gets smaller.On the other hand, the expansion for x/(1-x) is a geometric series: x + x² + x³ + x⁴ + ... which is straightforward. Comparing these two series, for small x, log(1+x) is approximately x - x²/2, while x/(1-x) is approximately x + x². So, for small x, x/(1-x) is larger than log(1+x) because the x² term in x/(1-x) is positive, whereas in log(1+x), the x² term is negative.But does this hold for all x between 0 and 1? Maybe I should test some specific values to see.Let's try x=0.1:- log(1+0.1) = log(1.1) ≈ 0.09531- x/(1-x) = 0.1/(1-0.1) = 0.1/0.9 ≈ 0.1111So, 0.09531 < 0.1111, which means log(1+x) < x/(1-x) here.What about x=0.5:- log(1+0.5) = log(1.5) ≈ 0.4055- x/(1-x) = 0.5/(1-0.5) = 1Again, 0.4055 < 1, so log(1+x) < x/(1-x).How about x approaching 1? Let's say x=0.9:- log(1+0.9) = log(1.9) ≈ 0.6419- x/(1-x) = 0.9/(1-0.9) = 9Wow, 0.6419 is definitely less than 9. So, even as x approaches 1, log(1+x) is still much smaller than x/(1-x).Wait, but what if x is very close to 0? Let's say x=0.01:- log(1+0.01) ≈ 0.00995- x/(1-x) ≈ 0.01/0.99 ≈ 0.0101So, 0.00995 < 0.0101, which is still true. It seems consistent that log(1+x) is less than x/(1-x) for all x between 0 and 1.Maybe I can also think about the derivatives. The derivative of log(1+x) is 1/(1+x), and the derivative of x/(1-x) is (1)/(1-x)^2. Let's compare these derivatives.For x between 0 and 1, 1/(1+x) is always less than 1/(1-x)^2 because (1+x) is greater than (1-x)^2 when x is positive and less than 1. For example, at x=0.5:- 1/(1+0.5) = 2/3 ≈ 0.6667- 1/(1-0.5)^2 = 1/(0.5)^2 = 4So, 0.6667 < 4, which means the derivative of log(1+x) is less than the derivative of x/(1-x). This suggests that log(1+x) increases slower than x/(1-x) as x increases. Since both functions start at 0 when x=0, and x/(1-x) grows faster, log(1+x) remains below x/(1-x) for all x in (0,1).Therefore, I think the correct relationship is that log(1+x) is less than x/(1-x) when x is between 0 and 1. So, the answer should be option B.

💡First, I note that there are a total of 80 cupcakes ordered.Half of these cupcakes are gluten-free, which means there are 40 gluten-free cupcakes and 40 cupcakes that contain gluten.Out of the 40 cupcakes that contain gluten, 28 are non-vegan. This implies that there are 12 vegan cupcakes that contain gluten.Since half of the vegan cupcakes are gluten-free, the number of vegan cupcakes that are gluten-free is also 12.Therefore, the total number of vegan cupcakes is 12 (gluten-containing) + 12 (gluten-free) = 24.

💡First, I need to determine the total distance the train covers when it crosses the platform. This is the sum of the train's length and the platform's length.Next, I'll calculate the speed of the train in meters per second by dividing the total distance by the time taken to cross the platform.Finally, I'll convert the speed from meters per second to kilometers per hour using the appropriate conversion factor.

💡Okay, so I'm trying to figure out this problem about the two brothers selling sheep and dividing the money. Let me read it again to make sure I understand.They sold a flock of sheep, and for each sheep, they received as many rubles as there were sheep in the flock. So if there were, say, 10 sheep, they would get 10 rubles per sheep, totaling 100 rubles. Got it. So the total amount of money they received is the square of the number of sheep. Let's call the number of sheep "n," so the total money is n squared, or n² rubles.Now, they divided the money by taking turns: the elder brother took 10 rubles first, then the younger brother took 10 rubles, and so on. In the end, the younger brother lacked 10 rubles, so he took all the remaining small change. To make the division fair, the elder brother gave the younger brother his pocket knife. We need to find the value of the pocket knife.Alright, let's break this down step by step.First, the total amount of money is n². They are dividing this money by taking turns, each taking 10 rubles at a time. So, the elder brother takes 10, then the younger, then the elder, and so on. This process continues until there's less than 10 rubles left.At the end, the younger brother lacks 10 rubles, meaning that after all the 10-ruble turns, there wasn't enough for the younger brother to take his full 10 rubles. So, he took whatever was left, which was less than 10 rubles, and then the elder brother gave him a pocket knife to make it fair.So, the key here is to figure out how much money was left after all the 10-ruble turns, and then determine what the pocket knife was worth to balance it out.Let me think about how the division works. If they take turns taking 10 rubles, the total number of times they can do this is equal to the number of times 10 can be subtracted from n² before it goes below 10. So, the number of 10-ruble turns is floor(n² / 10). But since they alternate, the elder brother takes the first turn, then the younger, and so on.Wait, actually, it's more precise to say that the number of times each brother can take 10 rubles depends on whether the total number of 10-ruble turns is even or odd. If it's even, both brothers take the same number of turns. If it's odd, the elder brother takes one more turn than the younger brother.But in this case, the younger brother was left lacking 10 rubles, meaning that after all the full 10-ruble turns, there wasn't enough for the younger brother to take his next 10 rubles. So, the total number of 10-ruble turns must be odd, right? Because if it were even, both would have taken the same number of turns, and the remaining money would have been less than 10, but the younger brother wouldn't necessarily lack 10 rubles.Wait, maybe not. Let me think again. If the total number of 10-ruble turns is odd, the elder brother would have taken one more turn than the younger brother. So, the younger brother would have taken (total turns - 1)/2 turns, and the elder brother would have taken (total turns + 1)/2 turns.But the problem says that the younger brother lacked 10 rubles in the end, so he took all the remaining small change. That suggests that after the elder brother took his last 10 rubles, there wasn't enough for the younger brother to take another 10, so he took whatever was left.So, let's denote the total number of 10-ruble turns as k. Then, the total money taken in these turns is 10k rubles. The remaining money is n² - 10k rubles.Since the younger brother was the one who lacked 10 rubles, that means that after the elder brother took his k-th turn, the remaining money was less than 10 rubles. So, the younger brother took whatever was left, which was less than 10 rubles.Therefore, the total money can be expressed as:n² = 10k + rwhere r is the remaining money, which is less than 10.Now, the division process is such that the elder brother took the first 10 rubles, then the younger, and so on. So, if k is the total number of 10-ruble turns, then:- If k is even, both brothers took k/2 turns each.- If k is odd, the elder brother took (k + 1)/2 turns, and the younger brother took (k - 1)/2 turns.But in this case, since the younger brother lacked 10 rubles, it must be that k is odd, because if k were even, both would have taken the same number of turns, and the remaining money would have been less than 10, but the younger brother wouldn't necessarily lack 10 rubles.Wait, maybe not necessarily. Let me think again. If k is even, the elder brother took k/2 turns, the younger brother took k/2 turns, and then the remaining money is r. So, the younger brother would have taken k/2 turns of 10 rubles and then r rubles. But the problem says that the younger brother lacked 10 rubles, meaning that he couldn't take another 10 rubles after his last turn. So, if k is even, the last turn was taken by the younger brother, and then there was r rubles left, which he took. But he lacked 10 rubles, meaning that r was less than 10. So, in that case, the total money would be 10k + r, with r < 10.But if k is odd, the last turn was taken by the elder brother, leaving r rubles, which the younger brother took. So, in that case, the younger brother took (k - 1)/2 turns of 10 rubles and then r rubles. Since he lacked 10 rubles, r must be less than 10.Wait, so whether k is even or odd, the younger brother ends up taking r rubles, which is less than 10. But the problem says that the younger brother lacked 10 rubles in the end, so he took all the remaining small change. Hmm.I think the key is that the total number of 10-ruble turns is odd, so the elder brother took one more turn than the younger brother. Therefore, the elder brother took (k + 1)/2 turns, and the younger brother took (k - 1)/2 turns. Then, the remaining money r was taken by the younger brother, making his total (k - 1)/2 * 10 + r.But since the division was supposed to be fair, the elder brother gave the younger brother a pocket knife to balance it out. So, the value of the pocket knife must be the difference between what the younger brother should have received and what he actually received.Let me formalize this.Let’s denote:- Total money: N = n²- Number of 10-ruble turns: k- Remaining money: r = N - 10k, where 0 ≤ r < 10If k is odd:- Elder brother took (k + 1)/2 turns: 10*(k + 1)/2- Younger brother took (k - 1)/2 turns: 10*(k - 1)/2- Then, younger brother took r rubles: total for younger brother = 10*(k - 1)/2 + rTotal money: 10*(k + 1)/2 + 10*(k - 1)/2 + r = 10k + r = NNow, for the division to be fair, each brother should receive N/2 rubles.So, the elder brother received 10*(k + 1)/2, and the younger brother received 10*(k - 1)/2 + r.The difference between what the elder brother received and what he should have received is:Elder brother's share: 10*(k + 1)/2Fair share: N/2 = (10k + r)/2So, the difference is:10*(k + 1)/2 - (10k + r)/2 = [10(k + 1) - 10k - r]/2 = (10 - r)/2Similarly, the younger brother's share is 10*(k - 1)/2 + rDifference from fair share:(10*(k - 1)/2 + r) - (10k + r)/2 = [10(k - 1) + 2r - 10k - r]/2 = (-10 + r)/2So, the elder brother has an excess of (10 - r)/2, and the younger brother is short by (10 - r)/2.Therefore, to make the division fair, the elder brother should give the younger brother (10 - r)/2 rubles.But the problem says that instead of giving money, the elder brother gave his pocket knife. So, the value of the pocket knife must be (10 - r)/2 rubles.Now, we need to find r. Since r is the remaining money after k turns of 10 rubles, and r < 10.But we also know that N = n², so N must be a perfect square.Let’s think about the possible values of r. Since r = N - 10k, and N is a perfect square, r must be such that N ≡ r mod 10.So, the last digit of N is r. Therefore, r is the units digit of N, which is n².So, the units digit of n² must be r, which is less than 10.Now, the possible units digits of perfect squares are 0, 1, 4, 5, 6, 9.So, r can be 0, 1, 4, 5, 6, or 9.But since the younger brother took r rubles and was lacking 10 rubles, r must be less than 10, which it already is.But we also know that the elder brother gave the younger brother a pocket knife worth (10 - r)/2 rubles.So, (10 - r)/2 must be an integer because the value of the pocket knife should be a whole number of rubles.Therefore, (10 - r) must be even, meaning that r must be even.Looking back at the possible units digits of perfect squares: 0, 1, 4, 5, 6, 9.The even ones are 0, 4, 6.So, r can be 0, 4, or 6.But if r = 0, then the younger brother didn't lack any money, which contradicts the problem statement. So, r cannot be 0.Therefore, r is either 4 or 6.Now, let's consider both cases.Case 1: r = 4Then, the value of the pocket knife is (10 - 4)/2 = 6/2 = 3 rubles.Case 2: r = 6Then, the value of the pocket knife is (10 - 6)/2 = 4/2 = 2 rubles.So, the pocket knife could be worth either 2 or 3 rubles.But we need to determine which one it is.Let’s think about the total money N = n².If r = 4, then N ≡ 4 mod 10, so n² ≡ 4 mod 10. Therefore, n must end with 2 or 8 because 2² = 4 and 8² = 64.Similarly, if r = 6, then N ≡ 6 mod 10, but wait, perfect squares cannot end with 6. Wait, actually, 4² = 16, which ends with 6, and 6² = 36, which also ends with 6. So, n could end with 4 or 6.Wait, no, 4² is 16, which ends with 6, and 6² is 36, which ends with 6. So, if r = 6, n must end with 4 or 6.But let's check:If n ends with 2 or 8, n² ends with 4.If n ends with 4 or 6, n² ends with 6.So, both cases are possible.But we need more information to determine which one it is.Wait, let's think about the total number of 10-ruble turns, k.Since N = n² = 10k + r, and r is 4 or 6.So, k = (n² - r)/10.Since k must be an integer, (n² - r) must be divisible by 10.So, n² - r ≡ 0 mod 10.If r = 4, then n² ≡ 4 mod 10, which is possible if n ends with 2 or 8.If r = 6, then n² ≡ 6 mod 10, which is possible if n ends with 4 or 6.But we also know that the number of 10-ruble turns, k, must be such that the younger brother was the one who lacked 10 rubles, meaning that the total number of turns was odd.Because if k is odd, the elder brother took one more turn than the younger brother, leaving the younger brother to take the remaining r rubles.So, k must be odd.Therefore, k = (n² - r)/10 must be odd.So, let's see:If r = 4:k = (n² - 4)/10Since n² ends with 4, n² = ...4So, n² - 4 = ...0Therefore, k = (n² - 4)/10 is an integer.But is k odd?Let’s take an example.Suppose n = 12n² = 144r = 4k = (144 - 4)/10 = 140/10 = 14, which is even.But we need k to be odd.Another example: n = 22n² = 484r = 4k = (484 - 4)/10 = 480/10 = 48, which is even.Hmm, seems like when r = 4, k is even.Wait, let's check n = 32n² = 1024r = 4k = (1024 - 4)/10 = 1020/10 = 102, which is even.So, it seems that when r = 4, k is even.But we need k to be odd because the younger brother lacked 10 rubles, meaning that the last turn was taken by the elder brother, leaving the younger brother to take the remaining r.Therefore, k must be odd.So, let's check r = 6.If r = 6:k = (n² - 6)/10Since n² ends with 6, n² = ...6So, n² - 6 = ...0Therefore, k = (n² - 6)/10 is an integer.Now, let's see if k can be odd.Take n = 4n² = 16r = 6k = (16 - 6)/10 = 10/10 = 1, which is odd.Another example: n = 6n² = 36r = 6k = (36 - 6)/10 = 30/10 = 3, which is odd.Another example: n = 14n² = 196r = 6k = (196 - 6)/10 = 190/10 = 19, which is odd.Another example: n = 16n² = 256r = 6k = (256 - 6)/10 = 250/10 = 25, which is odd.So, when r = 6, k is odd, which fits the condition that the younger brother lacked 10 rubles because the elder brother took the last 10-ruble turn, leaving r = 6 for the younger brother.Therefore, r = 6.Thus, the value of the pocket knife is (10 - r)/2 = (10 - 6)/2 = 4/2 = 2 rubles.So, the pocket knife was worth 2 rubles.