answer：First, I need to identify the unknown fraction, which I'll call ( f ), and the given number, which is 90.The problem states that when 3 is added to half of this fraction of one-fifth of the number, the result is one-fifteenth of the number. I can translate this into an equation:[ 3 + frac{1}{2} cdot f cdot frac{1}{5} cdot 90 = frac{1}{15} cdot 90 ]Next, I'll simplify the equation step by step. Calculating one-fifth of 90 gives me 18, and one-fifteenth of 90 is 6. So the equation becomes:[ 3 + frac{1}{2} cdot f cdot 18 = 6 ]Simplifying further, half of 18 is 9, so the equation is now:[ 3 + 9f = 6 ]To isolate ( f ), I'll subtract 3 from both sides:[ 9f = 3 ]Finally, dividing both sides by 9 gives me the value of ( f ):[ f = frac{1}{3} ]Therefore, the fraction is ( frac{1}{3} ).

answer：To determine the number of terms in the arithmetic sequence (20), (25), (30), (dots), (145), (150), we start by identifying the first term and the common difference. The first term (a) is (20), and the common difference (d) is (5) (since each term increases by (5)).The general formula for the (n)-th term of an arithmetic sequence is:[a_n = a + (n-1)d]We need to find (n) such that (a_n = 150). Substituting the known values into the formula:[150 = 20 + (n-1) times 5]Solving for (n):[150 - 20 = (n-1) times 5 130 = (n-1) times 5 n - 1 = frac{130}{5} n - 1 = 26 n = 27]Thus, there are (27) terms in the sequence.

answer：Okay, so I have this problem about two infinite rows of evenly spaced dots. Arrows are pointing from every dot in the top row to some dot in the lower row, with two conditions: no two arrows point to the same dot, and no arrow can extend more than 1006 positions to the left or right. I need to show that at most 2012 dots in the lower row could have no arrow pointing to them.Hmm, let me try to understand this. So, we have two rows, top and bottom, each with infinitely many dots. Each dot in the top row has an arrow pointing to some dot in the bottom row. No two arrows point to the same dot, which means each dot in the bottom row can have at most one arrow pointing to it. Also, the arrows can't go more than 1006 positions left or right. So, if I'm at a particular dot in the top row, the arrow can only point to a dot in the bottom row that's within 1006 positions to the left or right.The question is about how many dots in the lower row could have no arrows pointing to them. So, we need to find the maximum number of such dots, which is 2012.Let me think about how to approach this. Maybe I can model this as a mapping from the top row to the bottom row, where each dot in the top maps to a dot in the bottom within a range of 1006 positions. Since the rows are infinite, but the arrows can't extend beyond 1006 positions, this creates some kind of constraint on how the dots can be mapped.If I consider the top row and the bottom row as two sequences, each with infinitely many elements, and each element in the top row maps to an element in the bottom row within a window of 2013 positions (1006 left, 1006 right, and the same position). So, each arrow can only point to a dot that's at most 1006 positions away in either direction.Now, if I think about the number of dots that can be missed, meaning no arrow points to them, I need to find the maximum number of such dots. Let's say there are M missed dots. Then, the number of target dots (dots that have arrows pointing to them) would be the total number of dots in the bottom row minus M. But since the rows are infinite, this might not be straightforward.Wait, maybe I should think about it in terms of intervals or blocks of dots. Suppose I take a large enough segment of the bottom row, say, 2013 dots. If I can show that within any such segment, there must be at least one dot that is a target, then the number of missed dots can't exceed 2012.Alternatively, maybe I can use the pigeonhole principle. If I have a certain number of arrows, each can only cover a limited number of positions, so the number of missed dots is limited by the number of positions each arrow can cover.Let me try to formalize this. Let's denote the top row as T and the bottom row as B. Each dot in T maps to a dot in B within 1006 positions. So, for each dot in T, the corresponding dot in B is within [T_i - 1006, T_i + 1006].Now, if I consider a window of 2013 consecutive dots in B, say from position 1 to 2013, how many dots in T can map to this window? Each dot in T can map to at most one dot in B, and each dot in B can be mapped to by at most one dot in T.But since the arrows can't extend more than 1006 positions, the dots in T that can map to the window in B must be within 1006 positions to the left or right of this window. So, the dots in T that can map to the window in B are from position (1 - 1006) to (2013 + 1006). But since the rows are infinite, we don't have to worry about the edges.Wait, maybe I should think about it differently. If I have a window of 2013 dots in B, how many dots in T can potentially map to this window? Each dot in T can map to at most one dot in B, and each dot in B can be mapped to by at most one dot in T.But the key is that each dot in T can only map to a dot in B within 1006 positions. So, for the window of 2013 dots in B, the number of dots in T that can map to this window is limited. Specifically, each dot in T can map to at most 2013 dots in B, but since the arrows can't extend beyond 1006 positions, the number of dots in T that can map to this window is also limited.Wait, maybe I'm complicating it. Let's think about it as a matching problem. Each dot in T is matched to a dot in B within a range of 1006 positions. So, the matching is constrained such that no two dots in T map to the same dot in B, and each mapping is within 1006 positions.If I consider the maximum number of missed dots, that would correspond to the maximum number of dots in B that are not matched by any dot in T. To maximize this, we need to arrange the mappings such that as many dots in B as possible are not covered.But due to the constraint on the arrow lengths, we can't have too many missed dots in a row. Specifically, if we have a sequence of missed dots, the arrows from the corresponding dots in T must point outside this sequence, but within 1006 positions.So, if we have a block of missed dots in B, the arrows from the corresponding dots in T must point outside this block, but within 1006 positions. Therefore, the size of such a block is limited by the range of the arrows.If we have a block of 2013 missed dots, then the arrows from the corresponding dots in T must point outside this block. However, since the arrows can only extend 1006 positions, the arrows can only point to dots within 1006 positions to the left or right of the block. But the block itself is 2013 dots long, so the arrows would have to point beyond the block, which is not possible because the arrows can't extend more than 1006 positions.Wait, that might be the key. If we have a block of 2013 missed dots, then the arrows from the corresponding dots in T would have to point outside this block, but the block is longer than the maximum arrow length, so it's impossible. Therefore, such a block cannot exist, meaning the maximum number of missed dots is 2012.Let me try to formalize this argument. Suppose, for contradiction, that there are 2013 or more missed dots in B. Then, there exists a contiguous block of 2013 missed dots. Let's denote this block as S in B. The dots in T that are above S must map to dots in B, but they can only map within 1006 positions. However, since S is 2013 dots long, the arrows from T can't reach beyond S because 1006 < 2013. Therefore, the arrows from T must map to dots within S, but S consists of missed dots, meaning no arrows point to them. This is a contradiction because the dots in T must map to some dots in B, but S cannot receive any arrows, and the arrows can't reach beyond S. Therefore, our assumption is wrong, and there can't be 2013 or more missed dots. Hence, the maximum number of missed dots is 2012.I think that makes sense. By assuming there are 2013 missed dots, we create a situation where the arrows from the corresponding top dots can't reach any dots in the lower row, which violates the problem's conditions. Therefore, the maximum number of missed dots must be 2012.Another way to think about it is using the concept of intervals and coverage. Each arrow can cover a range of 2013 dots (from -1006 to +1006). If we have a block of 2013 missed dots, then the arrows from the top row can't cover this block because they can't extend far enough. Hence, such a block can't exist, limiting the number of missed dots to 2012.I also recall something about the pigeonhole principle in similar problems. If you have more pigeons than holes, at least one hole must contain more than one pigeon. But in this case, it's more about the range constraints limiting the number of missed dots.To summarize, by considering a block of 2013 missed dots and analyzing the implications on the arrows from the top row, we reach a contradiction. Therefore, the maximum number of missed dots must be 2012.

answer：First, I need to find 25% of Rs. 600. To do this, I'll convert the percentage to its decimal form by dividing 25 by 100, which gives me 0.25.Next, I'll multiply the decimal form of the percentage by the total amount. So, I'll multiply 0.25 by 600.Finally, performing the multiplication will give me the result, which is Rs. 150.

answer：First, let's define the heights of the two trees based on the given ratio. Let the height of the shorter tree be (5x) feet and the height of the taller tree be (7x) feet.According to the problem, the taller tree is 20 feet higher than the shorter tree. This gives us the equation:[7x - 5x = 20]Simplifying the equation:[2x = 20][x = 10]Now, substitute (x = 10) back into the expression for the taller tree:[7x = 7 times 10 = 70]Therefore, the height of the taller tree is 70 feet.

answer：Okay, so Tim is trying to decide whether to go to a football game tomorrow. The game is definitely happening no matter what the weather is like. But Tim's decision to attend depends on the weather. If it rains, he has a 25% chance of going, and if it's sunny, his chance jumps up to 70%. The weather forecast says there's a 60% chance of rain. I need to figure out the overall probability that Tim will go to the game.Alright, let's break this down. First, there are two possible weather conditions: rain and sun. The forecast gives the probability of rain as 60%, which means the probability of it being sunny is 40% because those are the only two options and they have to add up to 100%.Now, for each weather condition, Tim has a different probability of attending. If it rains, he'll go with a 25% chance, and if it's sunny, he'll go with a 70% chance. So, I think I need to calculate the probability of Tim attending in both scenarios and then combine them somehow.I remember from probability that when you have two independent events, you can multiply their probabilities to find the combined probability. So, for the rain scenario, the probability of rain is 60%, and the probability Tim attends given it's raining is 25%. So, multiplying those together should give me the probability that it rains and Tim goes.Let me write that out: 0.60 (probability of rain) multiplied by 0.25 (probability Tim attends if it rains). That equals 0.15, or 15%. So, there's a 15% chance that it rains and Tim decides to go.Next, for the sunny scenario. The probability of sun is 40%, and the probability Tim attends if it's sunny is 70%. So, multiplying those: 0.40 times 0.70 equals 0.28, or 28%. So, there's a 28% chance that it's sunny and Tim goes.Now, to find the total probability that Tim attends the game, I think I just add these two probabilities together because they're mutually exclusive events—either it rains or it's sunny, not both. So, adding 15% and 28% gives me 43%. Therefore, there's a 43% chance that Tim will go to the football game.Wait, let me double-check that. So, 60% chance of rain times 25% attendance is 15%, and 40% chance of sun times 70% attendance is 28%. Adding them gives 43%. Yeah, that seems right. I don't think I missed anything here. It's just combining the probabilities for each scenario and then summing them up because those are the only two possibilities.I guess another way to think about it is using the law of total probability. The total probability of Tim attending is the sum of the probabilities of each weather condition multiplied by the probability of attending given that condition. So, that's exactly what I did. I think I'm confident with the 43% answer.