 on this thing, a pair of linear equations in two variables and anything. Yes, anyone wants to discuss anything he can, she can talk about this. Okay. So let's go back and start. So what is this? So this first of all, we'll discuss as we have been discussing portion which have been deleted. So cross multiplication method has been deleted. Okay. So there is no cross multiplication method later on anyways, in your 12th grade, you will be studying something called Kramer's rule, which is nothing but a generalized version of cross multiplication method. So there you will be anyways, coming back to solving or the determinant method of solving linear equations in either two variables or three variables, four variables, five variables, whatever, right? So cross multiplication method was this rule with only two variables. So unfortunately, that has been taken away. So good for you, bad for you, only time will tell, never mind. So, and this was one of the areas where a lot of people, no, determinant method is different. Matrix method is different. Matrix method is matrix transformation method, right? So you transfer, keep transforming matrix. So if you remember, we did this. So A matrix multiplied by X matrix. So this is X. So A matrix is equal to some constant matrix, if you remember. Yeah. So this is the matrix method where you transform the matrix and X is nothing but A inverse C. So this is what using matrix method. That was different than Kramer's rule was if you remember that determinant method where you wanted to find out D, DX, DY and X was DX upon D and Y was DY upon D. Do you remember that? We discussed it's there in our video modules as well on the Learnist, okay? So this has been removed. Anyways, this was not there in your syllabus. This was start in 11th and 12th, which in your grade, it is called cross multiplication method just to avoid some higher end terminologies. But you were doing this only in two variables. But anyways, that have been removed totally now. So you can just relax. Now, again, you must be, you know, thorough with this. So hence every, every time I come up with this slides, I show you this, right? So you must be very, very clear in your head. There are two sections, A and B. There are 32 marks in A, 48 marks in B, 32 marks, 16 plus 16, 16, one marker. And four case studies with five options out of which you have to do four, right? So this is part A and part B are typical two marker, three marker, five marker question. There are six, two marker, seven, three marker and three, five marker questions. So, and in, and you must also know that there are two questions of two marks where there will be options available to you, two questions of three marks. There will be option available to you and one question of five marks. There will be options available to you. So this is our structure. Now, let's quickly revise this thing. And I just want to take a survey here. I don't know whether I can do a poll. So what we are trying to do is we are trying to come up with how many of you are aware of shots in YouTube? I don't know if you're aware of. So what we are trying to do is we are coming up with all that we have done. Would you be, will this be helpful if we create a couple of minutes revision shots for you so that it becomes a very handy tool to revise quickly? So whatever we are discussing, we can come up with, let's say, in a short two minute video format, the entire slides which we have created so that you can do a quick revision because these things will be definitely helpful to you. Once you are also managing something else, something else, meaning, as I told you, we have to think about your futuristic goals. Okay. So but only three or four people are responding. So I'll give it to all these people. Surya, Harshini, Akshita, Aniket and, okay. Oh, now that I said, I will give only to those who started. Okay, never mind. So we will be, we are planning to do that and sooner it will come out. Anyways, so a quick revision equation. So what is an equation? A statement of equality of two algebraic expressions which involve one or more unknown quantities is an equation. So we have already seen quadratic equation. So there is an expression. One here and this is a mathematical expression and this is expression to, this is also a mathematical expression. Now it could be trigonometric quadratic exponential logarithmic, whatever, but we are restricting ourselves to polynomials and polynomial also in one variable and sorry, uh, in degree one, that is why it is called linear equation. So linear polynomial on the left side, linear polynomial on the right side, you quit them, you get a linear equation. So any question in which the maximum power of variable is one is called a linear equation. So there is nothing. Why is, why is a linear equation called a linear equation? The very first topic of our, you know, session was this. So why do we call a linear equation a linear equation? Yes. Anyone? Why do we call, you can just unmute forms or line when parted on the graph correct line, the graph of the equation is, oh, anyways, we'll see that good. So it comes out to be aligned. So that's the reason why, okay, very good. I question in the form of X plus B, this you learned in grade nine. Oh, sorry, AX plus B is equal to zero where X is a variable and we are real numbers and is not equal to zero is called a linear equation in one variable. Okay. X is the variable linear equation in two variables and equation of the form of AX plus B, Y plus C. So there is an introduction of a new variable Y where A, B and C are real numbers, A cannot be zero, B cannot be zero together in case of two, two variables and XY are variables is called linear equation in two variables, right? So in two variables, both of them, A and B cannot be zero. Okay. If anyone of A and B becomes zero, then the equation is reduced to a single variable equation. And if both of A and B are zero, then there is no equation. Okay. So this is preliminary concept. Now going to the next one, graph of a linear equation. Actually, you should not be talking about graph of a linear equation, but in your grade, it is, it is mentioned, discussed like that. We generally talk of graph of a function or a graph of a function. And especially in this case, we are talking about linear polynomials. So basically we plot linear polynomials, right? Y is a function of X or V is a function of T or A is a function of V, whatever. So let us draw a graph of equation this. So let's take an example to, you know, all, all of this. So what do you need to do? Express any Y in terms of X, let's say you can also express X in terms of Y, but usually the convention is Y is equal to express Y as in terms of the other variable. Then you randomly put values of X, you know, different, different values of X and get different values of Y corresponding values of Y. So for example, I have taken two zero one and two and corresponding values have of Y or Y one minus one and three. So this forms one pair, this forms one pair, this forms one pair. These are the pairs which you need to plot X and Y, right? On the graph. So you can see plotted here, here, here. And if you join them, in fact, the observation is all such points which are satisfying these two or this particular equation are lying on this particular line, right? And if you join all of them, you get a line. Hence the name linear equation, right? Linear equation. So plotting the points here on the graph paper, drawing the line, drawing them, you obtain the graph offline represented by a given equation as shown in the figure. So this is, right? Easy for you. Okay, pair of now, from one line, we are going to two pair of now. We are taking two pair of linear equation in two variables. So you can ask the question, why not three, why not four? You can take two and three and four, whatever number you wish, but in case the number of variables is lesser than, sorry, more than number of equations, you will not be able to solve the equation. So we'll see what does that mean. Okay, so a pair of linear equation in two variables is to form a system of simultaneous linear equation. This is, I hope this is clear, right? This is what we are, yeah, sorry, someone is saying something. Yep. Anything? No. So what is, what is the typical general form of the pair of linear equation? a1x plus b1y plus c1 is equal to zero and a2x plus b2y plus c2 is equal to zero. These are the two equations, right? So example, 2x minus 3y plus 7 is equal to zero and 7x plus 2y minus 4p is equal to zero. Now there could be multiple manifestations of such, you know, so don't think that okay, in this format only it has to be written. They can give you this as well, 3y minus 7. So this is also a linear equation in this, a different form. This could be written as 2y is equal to minus 7x plus 14 also, right? Or many times they can write 7 is equal to 3y minus 2x. So don't get confused. So same form can be written in multiple ways, multiple varieties. Why is this condition important? a1 square plus b1 square is equal to zero, not equal to zero. And a2 square plus b2 square should also not be equal to zero. It simply means that both of a1 and b1 and a2 and b2 can't be simultaneously zero, simultaneously zero. So one of them can be, so if a1 is zero, b1 cannot be. And if a2 is zero, b2 cannot be and vice versa like that. So both of them cannot be zero together. So this condition is fulfilled when, so this is zero, sum of two squares is zero, only when both of them are independently zero. So hence it is a standard way of putting this constraint. Okay, now, so examples are given now. Let's go to the, so, you know, first of all, yes. Okay, now pair of linear equation solving methods. So how do we solve? So what is the solution basically? Where did it go? The line is, anyways, so how many, how many solutions are possible for a linear equation in two variables? Only one. Why can't both be zero? Okay, so the question is, let me clarify this. Why can't both be zero, both a1 and b1 are zero, then you're saying a constant is equal to zero. So that, that means either you are asserting a wrong statement. So let's say if I have zero times x plus zero times y plus seven, can it be zero ever, cannot be zero ever, right? So this is not possible. Or if you write, if you say, no, the constant is also zero, then there is, there exists no equation, right? Isn't it understood, right? So hence both of a and b cannot be zero together. Point out so far. Okay. So, next, so I was, I was saying how many solutions are there for this equation? So this equation is y is equal to minus two x plus one. How many solutions are there? How many solutions are there for this equation? What is the solution, by the way? So solution is nothing but a pair of x and y. Again, for a equation in two variables, the solution will not be only one variable's value. No, it will have, so how many pairs of x comma y? This, this is called ordered pair. So how many pairs of x and y can be found out infinitely many, right? So one equation can have infinitely many pair of x and y, which satisfies that particular equation. Very good. Let's go to now solving. We are interested in solving. So, you know, you solved one linear equation in grade nine. Now we're solving pair of, right? So pair of meaning wherever the two lines are meeting, right? So there's a common point which is lying on both the lines. We are intending to find that particular point. When I say what is the point, what is the meaning of point point is nothing but p x and y coordinate of the point on the graph table. Okay. So how to do, first of all, graphical method, right? So graphical method, draw graphs of both equation. You know how to plot a graph, right? So, you know, express one in terms of the other, like that. And then generate two tables. If you notice, I have taken a third row, third column also. Sorry. While for creating a line, only two points are sufficient. Then why do I need to go for the third one? If only two points are good enough for, let's say, you know, getting a line, then there's absolutely no necessity of going for the third one. Don't you agree? So why do we need to do third one? Yeah. So basically, yes. So we just want to see that is if it is indeed a line. So if, you know, if there is a line, then all the three points must be collinear. So it's kind of acts as a check and doing one more point or finding one more point, make sure that you have not done any calculation error, right? So always go for more than two points, right? So I have drawn, I have taken the table. I have made the table and the graph and you can see it is intersecting at this point. And that is what is called the solution, right? So there was a question in 2019-20, no, 18-19, where two equations were given and you had to use the graph method to solve the linear equation. So it could be asked in your exam. Okay. Now, so you know how to do plot the first one for the second one, wherever is the point of intersection, try to find out the x and y coordinate of that, that is the solution of the equation. Next. Okay. So coming to algebraic method, now that's a, you know, the cross multiplication has been eliminated. So there are only two now. One is substitution and another one is elimination. So you know all of this already. Okay. So wait, people are joining in still. Okay. Never mind. So what I was saying is, so what do you do? I don't need to reiterate this. This you have done so many times, so many times, is it? So express one variable in terms of the other. Usually y is expressed in terms of x and then you plug in that value into the second equation. So second equation with meaning the equation, which we had not taken in the first place, first, first, you know, step, and then you substitute it and get, solve this. So solve the equation. So what is it? So let's say there are two equations to x minus y plus one equals zero and x plus y plus one is equal to zero. So what you need to do is here I have expressed y as two x plus one and then plugged it in the second equation. You can take it here. So this y was replaced by two x plus one. So what happens here? You convert one variable equation. You convert the equation into one variable equation. So that is easy to solve. You already know it. And then I got a value of x equals two minus two by three and then plug in this value back into any of the other, any of the equations. Usually we plug in back into the same. Well, you know, this, this, this equation where you had expressed one variable into the other and in form, in terms of the other, and then you find out the value. So this is easy for everyone. I believe we'll see problems on this and this is elimination. So in elimination, what do you need to do? So you have to equate the coefficient of any one of the variables. So you pick one, one of the variables, inspect first. So first of all, see the, the coefficients. Here it is two, here it is minus one. Here it is one, here it is one. So you pick a variable where you can quickly eliminate or equate the coefficient. So you can see already you have minus one and one. So the magnitude of the coefficient of y is same. So when that is same, you just simply add it and y will get eliminated. So, you know, so why are we equating the coefficients? So that I can either add or subtract the equation to eliminate the eliminate one variable so that the system of equation is reduced to equation one to only one equation in one variable. For example, let us say if you have 7x minus 2y plus 4 is equal to zero. This is equation number one. And let's say 3x minus, let's say 4y plus 2 is equal to zero. So what do you think, what should you do first? In this case, what should I, what should I, which, which variable should I eliminate first? What do you think, what will be your choice of variable? Yep, so which variable do you think you will be? Yeah, why is that? Ranjini Ghosh, yes, you're saying something Ranjini. You have raised your hand, Ranjini, can you hear me? Okay, so everyone is saying y has to be eliminated, why? Because you can see it is easier to equate the coefficient of y. What will you do to, what will you, what will be your next step? So let's say y has to be eliminated, what will you do? So what, what is the next step? So yeah, so yes, what is the next step in this case? So first equation by 2, very good. Why can I not divide the second equation by 2? Can I not divide the second equation by 2? It will be messy. Okay, yes, so multiplication is preferred. So hence multiplied by 2. Yep, very good. It will be into fractions. Yes, so avoid that route. So it will be clearly 14x. Minus 4y plus 8 is equal to 0. Now be careful while multiplying many eight times because of temptation of solving the equations quickly. People multiply the part which equates the coefficient of one variable that is there. They will write minus 4y, but they will make errors while writing 14 or 8. Okay, so don't do such kind of mistakes. Now, shall I add or subtract? What do I do? What is the next step for, this is equation number three. So what is the next step? What should I do? I should add. So Aditya says add. Okay, I should add or subtract. What do I do? Subtract. Yes, subtract here in this case. Why subtract? Because if the equations are of same sign, then subtract. Sorry, if the coefficients are of the same sign, then subtract. If the coefficients are of opposite sign, then add. For example, here, here opposite sign, right? Minus 1 plus 1, so hence added. But if the coefficients are of the same sign, then subtraction will help. So what will you do? Great. Okay, so what will you subtract it? So which one will you subtract? Second from third or third from second? Now these are all minor, minor details. Still you are careful. You can do whatever you want. So I will obviously, it's a natural choice or it's a very, what do you say? You have been habituated of subtracting a smaller one from the larger one. So 14x minus 3x, 11x and 8 plus 8 minus 2 will be 6. So be careful while, while you are doing it. And x is equal to minus 6 upon 11. Now the best part about linear equation is once you get the values, you can substitute it back to check whether you have done it correctly or not. So please make it a practice of checking it back while you are, after you are done with the problem solving. Now what do you do? Whether I should eliminate x separately or I have a choice of plugging this value into any of these equations. So plug it in here and find the value of y, is it? Everyone is clear with the elimination process or you want some more detail into it, right? Anyone? Any difficulty? Guys, elimination is easy. Now cross multiplication has been removed. So be happy. Now this one slide should be very clear to you. Again, typically the one marker will be of this, this will be from this particular slide. So what is that? So system of equations we have and there are three possibilities we have already learned and we are discussing here consistency of linear equation. Consistency meaning, so we can, there are solutions which are real. So when there are solutions which are real, we say the system of equation is consistent. If not, then we say it is inconsistent. So you can see consistency in this case. In the first column, consistent, consistent, inconsistent. Okay, what does it mean? So for consistency or for only one, if this criteria is fulfilled, that is the ratio of the coefficient of one variable is not equal to the ratio of the coefficient of other variable, then you will get one solution come what may and the physical significance is the lines are going to be intersecting. You can see. Isn't it? So lines are going to be intersecting and this is the condition. So a1 by a2 not equal to b1 by b2 is the condition. If all the three ratios, all the three ratios are equal, then we have coincident lines or infinity. Many solutions are there. So one line is on top of the other. Okay. And if this criteria is there, where the coefficient of the variables are in same ratio, but they are not equal to the constant ratio of the constant, then we say they are non-intersecting or inconsistent system or they are parallel lines. Right? I think this is pure to everyone. Now, the mistakes which people do here is this, that in again, in the, you know, in the rush of doing the sum, for example, the problem would be given like this. 2x minus 3y is equal to 7. And the other equation is let's say 4x minus 6y. Plus 14 is equal to 0. Is this system consistent? Not consistent. Is this consistent consistent or not? Which is there on the screen? Can you see that? 2x minus 3y is equal to 7. And 4x minus 6y plus 14 is equal to 0. Consistent or not consistent? Inconsistent. Very good. So here, usually this is, you know, this, this kind of a question actually creates a lot of trouble. Why? Because here 7 is on the right-hand side. So you have to make sure that these are in order. Okay. And then 4x minus 6y plus 14 is equal to 0. And you can clearly see 2 by 4 is equal to minus 3 by minus 6, but it is not equal to minus 7 by 14. So inconsistent, right? Many a times they will reverse the order of x and y. And here again, you realize after you come out of the examination hall that, hey, I read the question in, you know, incorrectly. So for example, they will say 2y minus 3x minus 7 is equal to 0. This is 1. And then they will write 6x minus 4y. Okay. And mine is equal to 14. Bolo. Consistent, inconsistent. Consistent or inconsistent? Consistent, inconsistent. I or C? Still which equation I'm talking about? We are talking about these two. Consistent or inconsistent? Consistent. Everyone, sure. No problems at all. Consistent. Consistent or inconsistent? I'm still asking. Inconsistent. See, people have changed their answer. Lots of people said consistent. Now, all of us are then inconsistent. Why? I didn't change anything. Now, you wanted where you will make a mistake? Everybody understood in just one example, it was made clearer to you that you will make mistakes in such kind of things, right? Despite the fact that all of you know what is the criteria. But everyone in the rush of, so it is highly inconsistent. Why? Because minus 7 by minus 14. It is 14 on the right-hand side. Okay. So minus 7 upon minus 14 will be half, which is not equal to minus 3 upon 6 or 2 upon minus 4. Okay. So I made my point. So please be careful while solving such equations. Okay. Majority of you said consistent. Okay. Guys. Sound Maya? Yes. No. Please be very, very, very, very thorough before you attempt such questions. Okay. Chalier. Next. Now let's go to our sample papers and the board papers. This is the first question. One marker you can see. There will be 1 plus 1 plus 5 in your board exam, mostly 1 plus 1 plus 5. So 7 marks from linear equation. Okay. So for what value of K, the pair of linear equations doesn't have a solution. It doesn't have a solution. So doesn't have a solution mean? Meaning what should be the criteria? So how should I, it is a one marker. So for no solution for no solution, you should write for no solution for no solution. Could you go back once just a minute? Let me finish this for no solution. It is A1 by A2 is equal to B1 by B2 must not be equal to C1 by C2. Right. This is a criteria. Is it? So what are A1 by A2 and all that here? So again, please be careful. So this is 3x plus y here. Both of the constants are on the right hand side. So you don't need to be worried about it. But it's always a good practice to standardize. So I am writing in general form like that. These are the two equations. So let's write because this C1 and C2 while we are writing, it means when all the three terms are on one side of the equality sign. So A1 by A2 is 3 upon 6 must be equal to y by k. That is 1 upon k and should not be equal to minus 3 upon minus 8. So k must be equal to 2. So this equality will give you k is equal to 2. Correct. So that's good enough. And k is not equal to, yes. So k is equal to no. So you know for one marker, Akshita, you don't need to write. You can just say when you're saying k is equal to 2, that's good enough. That even anyways means it is not equal to 8 by 3, isn't it? When you're saying, when you're answering the question, saying that k must be 2. Then obviously you are meaning that k is not equal to 8 by 3, isn't it? So hence only this is good enough. You don't need to write not equal to 8 by 3. Okay. Because I lost marks in the free words. No, but you must have lost it wrongly. Why? Because if k is, there has to be only one value of k for it to be of this nature. There is absolutely no other value of k which will help. k is equal to 4 will not help. k equal to 5 will not help. So k has only one value. So what is that question? What value of k will make it inconsistent? So only one answer is there and that's 2. And specifying that it is not equal to 8 by 3. Anyways is not the answer to this question. So they are not asking what it should not be. They're asking what it should be. So hence you could have challenged that. So please, this is good enough. So in your marking scheme for one marker also, you don't need to. Anyways, so let's go for this one. So if three chairs and one table cost rupees, this is like just write the equation. Don't solve it. Okay. So here, oh yes, could you go for the table? Which table? Yes, table. This is the table. Yes, Aditya, tell me. Yes. Any doubt? Can we proceed? Hello. Yeah. Okay. So chalo. So here in the temptation of solving, do not go and solve this. Just what is that form linear equations to represent the situation? Okay. So anyways, there is only one equation, so you'll not be able to solve, but so three chairs and one table cost to be 1560 and one table cost. Oh, there are two actually. So you can solve also. Yep. So done. Easy. So what will you say? You will say let the cost of one chair, one chair be X and let the cost of one table be Y rupees. Rupees. Rupees. Okay. Therefore, three X plus Y is equal to 1500 and six X plus Y 2400. Okay. Done. Next one marker, five marker. This is from the sample. So then you can see the combination is one plus one plus five to one marker. Obviously that will be around the value of or around inconsistency, consistency. And obviously the five marker will be a heavy duty one word problem. So a motor boat covers a distance of 16 kilometers upstream and 24 kilometers downstream in six hours. In the same time, it covers a distance of 12 kilometers upstream and 36 kilometers downstream. Find the speed of the boat in still water and that of the stream. Solve. Take for five marker, you can, you can take five minutes also. No problem. Take your time, but make sure that your answers are correct. How do you make sure that your answers are correct? Please check. Verify. Please verify your answer. I'm giving you four minutes of time. And those who have, who are, who will complete just say done. They don't, don't, you know, write your answers in the chat. Done. Our attendance is done. Very good. Okay. Three people have done it. Very good. Okay. Yep. So those who have, who have done can send me your answers individually. There's an option of sending privately. It's a five marker. So please make sure that your answers are correct and verified. Eight and four. Okay. Okay. Okay. Okay. Okay. Okay. Eight and four is the answer people are saying. No, no problem. My turn to solve it. So how will you solve it? Five marker, heavy duty, right? And do not miss the units. So you'll start with let the, so I mixed both of them now. Let the speed of boat in still water be x kmph, correct? This is what is, right? And let the speed of stream is equal to y kilometers per hour or be whichever. Right? So upstream speed, upstream speed of boat is equal to x minus y. Assuming the stream is the assumption hidden here. The stream is moving slowly with respect to the boat. Otherwise it will be different. Y minus, right? So kilometers, kilometers per hour. And downstream, downstream speed I'm writing in shorthand. Okay. You don't write like that. Kilometers per hour. Okay. Now the first question is a first equation, 16 kilometers upstream. So 16 upon x minus y plus 24 kilometers downstream, 24 upon x plus y, right? Time is distance by speed. So I did that. And this is equal to six hours. How many of you got this equation? So you will get half a mark if you get this equation. How many of you got this one? Am I writing correctly? Yes, 16 upstream plus 24, right? And then second one is 12 upstream. So 12x minus y plus 36 downstream, 36x plus, divide by x plus y. And the time is same. Right? In the same time, six hours. Equation number two. So one mark here. Okay. Now what do you need to do? Simplify. So you can see either you can take variables or let's say you can substitute as, okay, u as 1 by x minus y and all that. But I generally would first do this. But you have to be careful. So I will not substitute. I will show you how to do it with substitution as well. What will you substitute? One way is to take u is equal to 1 upon x minus y and v is equal to 1 upon x plus y. Like that. You can either do this or you could do directly also, but never mind. Let's do this. So you'll get 16u plus 24v is equal to 6 or you simplify this. You will get 8u plus 12v is equal to 3, right? Then second one 12u plus 36v is equal to 6. This implies 6u plus 18v is equal to 3. You call it 3 and 4. And then what do you need to do? Just eliminate. So how do you eliminate? You multiply first one by 3 and second one by 2 and eliminate v. Okay. So what am I saying? Multiply this by 2 and sorry 3 and this by 2 and you will get v eliminated. So what will happen? 24u plus 36v is equal to 9. This one 12u plus 36v is equal to 6. Then subtract 36v will get cancelled. So 12u is equal to 3. So u is equal to 1 upon 4. Okay. Yes. Correct so far. Any doubt? So u is equal to x plus 4. This implies x minus y is equal to 4. Okay. Now what will you do? You will find what happened? Wait. Now what do you do? The other part. So either you can put u is equal to 1 by 4 and solve for v. That is also easy. So let's do. So 8 into 1 upon 4 plus 12v is equal to 3. So this is 2. So 12v is equal to 1. Right. So v is equal to 1 upon 12. So x plus y is equal to 12. Right. So this is equation number 6. Oh I was using Roman numerals so 6. And this is equation number 5. Now add 5 and 6. So you'll write adding 5 and 6. What will you get? You'll get 2x is equal to 16. So x is 8 and hence y will be 4. Both kilometers per hour. So please write x is equal to 8 kmph and y is equal to 4 kmph. Correct. Any doubt? This will give you 5 marks. If you write like this, step by step. Okay. Or you can also write final line that the velocity of the boat in still water is 8 or and velocity of still water is 4 kmph. Don't miss the unit. Do not miss the unit. Okay. Next. This is actual boat paper. This year's boat paper. There were 20 MCQs this year solved. This should be quick and fast. Be alert. And if you notice, this is in the same format which we are talking about. See one is here on the left hand side. One constant is here. And here the constant is on the right hand side. I hope you have taken care of that. Easy. And this condition is find the value of k for no solution. Right. So you can have only one. Right. So k is equal to 2. How to do it? So that this thing is a1 by a2 is equal to b1 by b2 must not be equal to c1 by c2. Right. The ratios for no solution a1 by a2 is equal to b1 or b2 not equal to c1 by c2. So a1 by a2 is 1 upon 2 must be equal to 1 upon k. Right. Must not be equal to minus 4 upon minus 3. So there is no way this can be equal ever. No way. Right. So hence the only way k is equal to 2. Good. Next. Board paper. Second set. 1920. Now they have used the word inconsistent. And again, as I told you again see they are acting smartly here. Again they have segregated the constants. Okay. So first one x plus 2y minus 3 is equal to 0. And 5x plus ky minus sorry plus 7 is equal to 0. So 1 upon 5 must be equal to 2 upon k must not be equal to minus 3 upon 7 which will never be. So k has to be 10. Okay. So all of you are able to do it. Good. Next. Now this is a three marks question. So it could be that they can split. So they will reduce one two marker from one and then give you three plus two like that because totally is one plus one plus five seven marks linear equation questions. So you have to solve this. Solve this and not only solve this find the value of so here is another catch. So many people will just leave it at that. x is equal to this y is equal to this and they will lose marks. Solve. Which method will you adopt for this? Short shot. Very straightforward. Which method will you adopt substitution? Good. Eliminate quick. You can do it in your mind also but don't do that in there. x equals to what's x? x is 7. Cool. x is 7 and y is 9 is it? I have a doubt. Different values. 31 and minus 5 by 7. Oh, answer you are saying. Okay. x is so but the answer is 5 y minus 2 x. You have to find out 5 y minus 2 x and y y x minus 2. Okay. Cool. No problem. So it was again easy 2 x plus y equals 23. Write it equation number one. Then write 4 x minus y is equal to 19. Write it equation number two. Then simply add them. You write adding 1 and 2. Adding 1 and 2. You will get 2 x plus 4 x plus y minus y is equal to 23 plus 19. That's 6 x is equal to 42. So x is equal to 7. Okay. So hence y will be from 2. From 2. You can find out y is 4 x minus 19. So 4 times 7 minus 19 is 9. So y is 9 x is 7. So 5 y minus 2 x. 5 y minus 2 x is 5 times 9. Be careful while you are substituting the values. So 45 minus 14, which is 31. Very good. If you got it. And the second one is y upon x. So y is 9 upon x was 7 minus 2, which is 9 minus 14 upon 7, which is minus 5 upon 7. Correct. 3 marks to you folks. Next. So there was no 5 marker last year. So highest was 4. This could be a 5 marker for you this year. Swimming pool question. So you can take in the exam also at max 5 minutes for 5 marker. But usually you will be able to solve it within lesser time. Those who are done can send me your answers privately. Arya has solved it. Anyone else? Ranjini also. Okay. Akshita and Ranjini and Arya. Three people. Come on guys. Do it, do it. Oh, people are taking time in this question. Why? What happened? Hello. How many not able to? Three minutes. Oh, okay. Okay. Take, take time. Take time. Anyone else could solve? Aryan ten. Okay. Looks okay. Yes. Alisha, Aryan, Akshita, Arya. All your answers are correct. Shreyas, correct. I'm still waiting for the others to participate. Come on guys. What happened? Oh, you can also send me X privately. No problem. You can say not happening. Difficult explanation required. Number of error taken by larger pipe is this and smaller pipe is this. Yes, Satyam, correct. This is actual board paper. 1920 board paper. Prisham, calculation error beta. Check. First, correct. Second, wrong. You messed with some variables here and there. Some values. Do you? At least you tell me whether explanation required and I'll tell you how to approach such problems also. Those two. Come on. Only five to seven people, seven, eight people have responded so far. Okay. Shusti. Good. Good. Keep going. Anyone else? By your panel of India. Come on. Respond. What happened here? Where is the energy? What is the target? Target, target is what? What's target? Yes. Target, target, target, target should be there in the mind. So for that target, you need to do now the more time you have lots of time also. Correct. What happened to Surya Ranjini? Ranjini answered. Harsita, what happened to you? Come on. Answers please. Struggling. Chalo dekho. My turn. Now, how to go about it? It can take 12 hours to fill a swimming pool using two pipes. Yes Ananya, correct. If the pipe of larger diameter is used for four hours and the pipe of smaller dia for nine hours, only half of the pool can be filled. How long? Would it take for each pipe to fill the pool separately? So I'll start from that only. So I will say let the time taken by larger dia pipe. So I'm just writing in shorthand, right? Let the time taken by larger diameter pipe to fill the pool be XRs, okay? And let the time taken by smaller diameter pipe to fill the pool be YRs, right? So X hours are taken to fill complete pool. So in one hour, so part or fraction of pool or fraction of pool filled by as a nomenclature you can add A and B, right? So that you don't need to write again and again. So back it way, A and B you write. Fraction of pool filled by A in one hour is equal to one upon X. And fraction of pool or part of pool filled by B in one hour is equal to one upon Y, correct? Now let us say in, if both are acting together, then in one hour one upon X plus one upon Y, this much part of pool will be filled and together it is given how much 12 hours, right? So together they are taking 12 hours. So in one hour together they will be able to fill this much part of pool. Am I right? First equation correct or not? How many of you got this equation? This is the first equation is this one upon X plus one upon Y is 12, isn't it? So both pipes are open in one hour. One by X fraction of pool was filled in one hour. One by Y fraction of pool was filled. So put together totally you will get one by 12th of them. Why? Because the information is given that in 12 hours together they can fill one full, right? Now come to the second one. So larger diameter is used for four hours. That means what fraction will it fill? Four upon X. This is a fraction of the pool filled by the larger dia pipe individually, okay? Then now simultaneously after this is filled nine more hours of smaller dia pipe. So nine by Y this is what are we discussing about? This is a fraction of the pool which is being filled by four hours of larger dia, nine hours of smaller dia. And this is what? What are they saying? Half of the pool. So already they are saying half fraction one by two, right? So in these if this type of mechanism is done only this fraction of the pool will be filled. So you got the two equations. How many of you got the two equations? One mark for this. How many of you got the two equations? Yup, these are the two equations you have to solve. Now what to do? Multiply this one by four and simply subtract. You'll get the answer directly. So four into this. What will be the equations now? Four upon X plus four upon Y equals one upon three, correct? So this is one equation. Then second equation is four upon X plus nine upon Y equals one upon two, subtract these two. So what will you get? You'll get nine upon Y minus four upon Y is half minus one by three. So why one by two? Because this is a fraction, no half pool. What is LHS? Fraction of pool filled by both the pipes. One for four hours, another for nine hours. We are discussing in the equality is of what? We are talking about the fraction of the total pool being filled. Understood? Are you clear? So what are we saying? We are saying one upon X is the fraction of the pool filled in one hour. So four upon X will be the fraction of the pool filled by the larger diapide in four hours. So four upon X. So this LHS is fraction. So RHS also should be fraction, right? The total part of the pool filled by both the pipes. And what is that total? Half. They are only saying half, only half of the pool. Understood? So once you solve it, you will get five upon Y equals one upon six, right? So Y is equal to 30, correct? So if Y is equal to 30, X, whatever. So solve. So X, Y is equal to 30. So one upon one upon X plus one upon 30 equals one upon 12. So one upon X is one by 12 minus one by 30. 60 is the LCM. So five minus two. So X is equal to 20. R's. R's. Any doubt? Anyone? This kind of question, please practice. Work while our question. So if you want more questions on such topics, please let me know. Work and time. So these many workers working together, complete a task, these many R's, things like that. Understood? So this is where I have seen lots of people getting stuck. You know, getting stuck. Yes, time and okay. So let's complete this. Anyways, we will see if there is some more on this, there will be definitely. Shall we do it next paper? Same here. Other paper. Other set. So you know, CVSE conducts in multiple sets, multiple papers. Do this one. So the question is the value of K for which the pair of linear equations, this and this have infinitely many solutions, infinitely many solutions. One, two, three, four, five, six people. The value of K for which the pair of linear equations KX plus Y equals K squared. Please be careful. K squared is here. NX plus KY is equal to one have infinitely many solutions. Oh, now people have changed their options. Right? What? No, it's B. So someone is saying A, someone is saying B. What's happening? Okay, infinitely many solutions. How? So for infinitely many solutions, you must have A1 by A2 is equal to B1 by B2 is equal to C1 by C2. Am I right? First of all, make sure that this is a condition. Okay, so if you see the, what are the equations given KX plus Y minus K squared is equal to zero. And X plus KY minus one is equal to zero. Yep. Okay, now what? So you have to write K upon one must be equal to one upon K. And this must be equal to K squared upon one. Okay. Yes or no? So you'll get K squared is equal to one. Yep. Isn't it guys? And so this is from these two. And if you equate these two, but you don't require it actually K is equal to and what else K cube is one also. Yeah, so, yes, so K has to be, if you solve this, you'll get, you're getting plus and minus one. But minus one will not work here, isn't it? Minus one will not work here. Why? Because then this will become inconsistent. Why? Because then this will become inconsistent. Why? Because if you put K is equal to minus one, then what will happen? The first is minus one upon one. The second will be one upon minus one. Third will become minus one whole square Y one, which will not be equal. Right? So hence answer should be B. Clear? Yes or no? So if K is equal to minus one, K is equal to minus one is not satisfying this. Okay. Any doubt? Any doubt? Anyone? Below. So hence it's always advisable. So you can check here. Yeah, if K is equal to minus one, then equations are minus X plus Y minus one is equal to zero. And the second equation is X minus Y minus one is equal to zero. So these are, if K is equal to minus one, then these pair will come. This pair will come, which is inconsistent. Okay. So K has to be only one. Be careful. Okay. Now this one, this is again previous here. One mark. Pair of linear equation Y equals to zero, Y equals to minus six has no solution, no solution Y. Indeed Y is equal to zero is X axis. This is X, Y. So this one is, this itself is Y is equal to zero. And Y is equal to minus six will be somewhere here. Y is equal to minus six. Right. So they're parallel line, no solution, no solution. Okay. Good. Solve. Three marks, two minutes max. Done. Only people are still doing RA. You have to find out the value of 5X minus 3Y. Careful. Yeah. Most of you are getting it anyways. So what to do? I would solve directly no substitution of the 11 and 5 upon X minus 4 upon Y is equal to minus 7. So multiply this by 4 and multiply this by 3. Right. So you'll get 8 by X plus 12 upon Y is equal to 44. And this one 15 upon X minus 12 upon Y minus 21. Yep. Simply add. So what will happen 12 by Y minus 12 by Y will get cancelled. And say, hence what will you get? You'll get 15 plus 8 23 by X is equal to 44 minus 21 is 23. Oh, yep. So X is one. Okay. So one X is one. The moment X is one. What can you do? Put in any of these equations. So let's take the first equation. Let's take the first equation. So hence 2 upon 1 plus 3 upon Y is equal to 11. So hence 3 upon Y is equal to 9. So Y is equal to 3 upon 9, which is 1 upon 3. Okay, folks. So when is that done? What is to be found out? 5 X minus 5 X minus 3 Y. What is 5 X? 5 into X. X is how much? 1 minus 3 into 1 by 3. So 5 minus 1, 4. Cool. Any trouble? All okay. Those two, no doubt. So far so good. So far so good. SFSG. Guys, come on. Where is the energy? Josh, right? Energy. 2nd January, yeah. Abhi to 365 minus 2. 363 days. Still to go. So next year you will be in grade 11. Enjoying your life. Just imagine 10th is over. New set of new set of people around you, new friends, new task, new assignment at hand. 11th is nightmare. Don't enter any project or any, this thing with that mindset. Nothing is nightmare. People have done in the past. So why you guys cannot do it? So no such thing. You have to take it head on. You will do good. Yeah. Okay. Academy voluminous. So let it be voluminous. Life is voluminous. So many days to still survive. So don't worry about 11th and grade 12th. So do you want to, which video games is more, which video game is popular these days? PUBG is the one. So what is that? What else is there? Fortnite. Still there or some other video game is there? Yes or no? Minecraft. So hence will you play Minecraft only level one? Again and again. Or you will go to the next level. Now, I used to play Super Mario. I don't know if you've heard of it. Mario. Have you heard of Mario? Yeah. Very, very famous game Mario. We used to spend hours together. I don't know. Have you heard of that music? That's so, what do you say? Yeah. So I used to play Mario for days together. So in our IITs we used to have inter-hostile Mario contest like that. But later on more sophisticated games, age of empire and what is that? Counter strike. That was there. I don't know whether these exist these days. So hence they used to play with US universities. So US universities versus IITs and all that. Lot of fun. And they would put me as a target to practice their shots. So they would invite me just because they knew that I am pathetic in any type of video game. So they will call me and they will encourage me and saying that, okay, you will ace it, you have the talent and all that. And they will invite me in their game only to kill me multiple number of times. So anyways, so hence you need to go to the next level guys. Okay. So don't worry. Everything is fine. You will be able to manage it. Chalier, let's, right now let's manage linear equations. Okay. Taxi charges in a city consist of fixed charges and the remaining charges depend upon the distance traveled. This is the practical case for a journey of 10 kilometers. The charge paid is I don't know when I came to Bangalore. For the first time, the taxi charges, the auto charges were like that. So the first two kilometers, you'll have to pay rupees 15. And after that, for every 100 meter or so, the meter will go up. You see those old meters. It's like that even now. Do they use meters anymore? Oh, is it? Okay. So now everyone just books a roll on over and go. Okay. Anyways, so now do it. 20 rupees for first 13 kilometers. No worry. What are you saying? 1.3. Okay. So done. Are in, are in and has already done it. Chalier, taxi charges in a city consist of fixed charges and the remaining charges depend on the distance traveled. So fixed cost and variable cost. For a journey of 10 kilometers, the charge paid will be 75. And for a journey of 15, the charge paid is 110. Find the fixed charge and charge per kilometer. Hence find the charge of covering a distance of 35 kilometers. So you have to find many things here. Find the fixed charges and charges per kilometer. And then find the charge of covering a distance of 45 kilometers as well. Okay. Many people have come up with the right answer already. So let the, why am I writing here? Oh, let me take this here. So I'll start once again. Let the fixed charges be x rupees. And then say let the per kilometer charge be y rupees. Okay. So for a journey of 10 kilometer charge paid is 75. Right. So hence what will be the first equation? So x plus 10 y is equal to 75. This is equation number one. Then second one is x plus 15 y is equal to 110. So you can clearly eliminate x very easily. So do 2 minus 1 will give you 5 y. Barabar 35. So y is 7 rupees per kilometer. Right. So y is 7. So you can find out x also. So x will be 75 minus 10 y from 1. You can write. So this is 75 minus 10 into 7 which is 5. If it comes out to be negative, beware that you have done some mistake in understanding the question. So that is a check. And what is distance 35 kilometer for or distance of 35 kilometer? Charges to be paid is equal to 5 plus 35 into 7. Correct. Which happens to be 735. 735 is 5 and 31. Sorry, 24. 24 plus 5 rupees. Okay. So everywhere, wherever you are, if you're not writing 7 rupees per kilometer. Right. So RS. This is only rupees. Then next we are solving easily. Same here. Another set. These questions are easy where you don't need to find anything. Simply do 1 and done. Correct B. 1 upon 6 is not equal to 1 upon 2. So clearly A1 by A2 is equal to B1 by B2 is equal to C1 by C2. What is A1 by A2? 3 upon 2. Divide by 9 is equal to 5 upon 3. Divide by 10. Is it equal? Yes. This is equal. No. Why? Because this is 1 upon 6. And this is also 1 upon 6, which is not equal to minus 7 upon minus 14. So inconsistent. Very good. Fraction type. What question is if it was consistent, then two answers. They will not give you that. Ambiguous questions they will not give you. Consistent. So this could be possible. Consistent and consistent with one solution. Right. So this any ways this can be. So either of these will be. Don't try to eliminate here. Kitna time that. We will solve it. Be on the safer side. Don't need to do such things. Chalo. Do it. Aryan, Tandon, Akshita, Goyal and Ranjini Ghosh. Correct. Suresh, Sankaran, correct. Alisha, Meghna, Shobhita, Anushka. All correct. Arya, good. I don't know whether correct or not. At least same. Let's solve. Okay Prisham, correct. Looks like correct. You would have done it correctly only. Fraction becomes 1 by 3. So we will start with. Let the numerator be X. I usually prefer N, but anyways for you. Let the numerator be X. Let the denominator be Y. Okay. So what is it saying? A fraction becomes 1 by 3 when one is subtracted from the numerator. And it become very simple. So X minus Y upon Y is 1 upon 3. Equation number 1. And second one. So either you can simplify here itself. So it becomes 3X minus 3 is equal to Y. Good. Then second one. And it becomes 1 by 4 when 8 is added to its denominator. Right. It becomes 1 by 4. Isn't it? This is equation number 2. 8 is a check. Once again, 1 is subtracted from the numerator. So X minus Y divided by X, Y. So it will be 1 by 3. So 3X minus 3 is Y. Perfect. And then 1 by 4 when 8 is added to its denominator. So X, Y plus 8. Very good. So this is 4X is equal to Y plus 8. And Y to you already know, substitute it. So you will get 4X is equal to Y is 3X minus 3 plus 8. So this is X is equal to 5. So Y will be equal to 3X minus 3. So 15 minus 3, 12. So the fraction is required fraction is 5 upon, okay. Good. Age-related. Presentation. Here also you have to be little cost, be little causes. Here. Why? Because many a times you forget to add the same amount of years in both father and son and whatever relatives are there. So please be careful. Present age of a father is 3 years more than 3 times the age of his son. And 3 years hence, so here is the catch. 3 years hence. So when you're doing 3 years hence, the child's age will also go, you know, up by 3. So many a times that will be crucial. So decide then and there. So I'll do the rituals meanwhile. So I'll write let the age or rather present age, present, present age of father be X years. Let the present age of son be Y years, okay. Now the present age of a father. So X is equal to 3 years more than 3 times the age of his son, right. 3 years more than 3 times age of the son, okay. 3 years hence that means X will become X plus 3. The father's age will be 10 years more than, so 10 years more than twice the age of the son. So this is the second equation. So here this is what I was saying many a times people miss adding 3 to this part also. So after 3 years the age son of age of the son will be Y plus 3, right. Now solve X plus 3 is equal to 10 plus 2 Y plus 6. So that means X is equal to 2 Y and 10 plus 16 minus 3 plus 30. Now you can equate 1 and 2. So from 1 and 2, what do you get? 3 Y plus 3 is equal to 2 Y plus 13. So Y is equal to 10. So if Y is 10, what's X? X is equal to 2 Y plus 13. What's X? X is equal to 2 Y plus 13 which is 33 and you are right from 2. Done. Cool. Most of you have done it. Next one. 2 marks. Now this is 18, 19 guys. One year prior to this year. Previous year batch. One marker. Find C the system of equation has infinitely many solutions. Yeah, so this is the batch who graduated in the year 19, 2019. So those who are writing 12th J means now. They face this discussion. So one marker are typically easy, right. You can see all are predictable now. One marker has to be related to finding out these criteria for consistency. So find C if the system of equation has infinitely many solutions. So you have to write for infinitely many solutions. A1 by A2 is equal to B1 by B2. All equal. All equal infinite. All equal infinite. Okay, so C by 12 is equal to 3 by C. So C by 12 is equal to 3 by C. Yes. Is equal to 3 minus C upon minus C. Am I right? C upon 12. So C upon 12 is 3 upon C is 3 minus C by minus C. This is the equation. Okay, so you will get from this to equate you'll get C square is equal to 36. So C is coming out to be plus minus 6 from here. Okay, let's check. Now if let's check these two. So what will happen from here? You'll get minus 3C is equal to 3C minus C square. Am I right? Minus 3C is equal to 3C minus C square. Yes. So C square is equal to 6C. C square is equal to 6C. So this is going to give you C common C minus 6 is equal to 0. So from here either C is equal to 0, but C cannot be 0 and C is equal to 6. Right. So minus 6 is not helping. Okay. So C is equal to 6. C is equal to either 6. So this is a common solution. Right. So C is equal to 0 eliminated. Can't be why? Because if C is 0, then it is not a linear equation as in two variables. Yes or no? Below clear. So C cannot be 0. C can be 6, but C cannot can C be minus 6? No. Why? Because if C is minus 6, what happened? Wait a minute. Where is the pointer? Wait a minute here. Yeah. So if check, if C is minus 6, then what will happen? 3 by minus 6 is equal to 3 plus 6 by plus 6. Wrong. Not possible. So only possibility is C is equal to 6. So I try equating C by 12 to 3. What? C by 12. You get, what will you get? Equate that C by 12 is equal to 3 minus C. By minus C. So this is minus C square is equal to 36 minus 12 C. Correct. So this will be a quadratic equation. C square minus 12 C minus plus 36 equals to 0. So there is nothing 6 minus C minus 6 whole square is equal to 0. So C is equal to 6. Right. Anyway you will get. So you will just need to solve two equations. The third one will be redundant. You know, you just need to take care of any two. Third will automatically come. Okay. Done. Let's go. Next. Go back for a second. Here it is. Done. Can we proceed? Yeah. Here. For others age three times the sum of the age of his two children. After five years, this is, this we did actually, if you remember. In quadratic by mistake we had taken this question. Arian done, Harshita done, Akshita done. Very good. Pranav Vijay. Where are you? Where is the Jyots? Aditya Vishwanath done. Ranjini Ghosh. Arya Rao. Good. Come on fellas. Come on, come on, come on. Arun M. Gouda. Done. Satyam. Done. Ananya. Done. This question is there in every exam. Okay. Aditya. Done. Charon done. Okay. Come on guys. Looks like everyone has solved it. Fathers age is three times. So let the age of father X years. So present age. I missed the word present. So present age. Okay. Let the age of no sons B, Y and Z years. Okay. So what is given? Fathers age is three times the sum of the ages of his two children. So either you can take some of the ages as a variable. But the hardly matters. So present age X is equal to three times. Y plus Z. Right. After five years, why did I take Y and Z? Because this will eliminate some of the chances of making mistakes here. Why? So after five years, so X will become, immediately you add five to all. Whatever is the treatment, we'll see later. Sorry. Y plus five. Five. So all will go up by five. Okay. His age will be two times. So this is two times sum of their age. This is what they're saying. Correct. Find the, this is the equation. These are the two equations. I hope everybody got the same thing. So clearly you can say that Y plus Z is equal to X upon three. And now substitute here in the second equation. You'll get X plus five is equal to two times X plus Z plus 10. Right. So this means X plus five is equal to two times X by three plus 10. Done. And then solve for X. So what will I do? I will say, either you do that or yeah. So let us say X minus two X by three is equal to 20 minus five. Be careful while doing this. So this is X by three is equal to 15. Okay. This is how you'll have to do. Perfect. I hope this is clear. So I'm just, you know, moving forward. 18, 19, three marks. So this question and this question was, were in OR. So either you do this or you do this. But in the five marker type, you will get an OR mostly in heights and distance. I think so. So you'll have to anyway solve one linear equation question. Your answer appears to be a little different. Good. Very good. So you are solving three markers also within a minute time. I hope you're writing the steps there. That is crucial because as I told you, as I have been telling you, you will be knowing all the questions. There will not be a single question. You would not know how to proceed. Except maybe in cases like triangles, once in a while that is also, you know, but otherwise the questions will be very familiar to you. So you'll be using marks or you'll be away from center only because of your steps or errors. Okay. So a fraction becomes one upon three. So again, the customary part, let the numerator be X. Let the denominator. So you in the exam, please do not adopt any shortcuts rightful. So I am writing because just because to save time. So a fraction becomes one by three when two is subtracted from the numerator. So X minus two upon Y is one upon three. So you might like to simplify three X minus six is equal to Y. So this is one and it becomes one half then one subtracted from denominator. So X upon Y minus one half right. So this becomes two X is equal to Y minus one correct. So Y from here is two X plus one. And now we can equate these two right. So hence you'll get three X minus six is equal to two X plus one. And this is X is equal to seven right. And then Y is equal to clearly two into seven plus one 15. Okay. Check. So the best part, as I told you, we can check. So fraction is seven upon 15. So this is check. So on this on the, you know, right margins, you can always check like that. Check. Check. So fraction becomes one by three when two is subtracted from seven minus two by 15 is indeed one upon three. And when half and one is subtracted from the denominator. So seven by 15 minus one is indeed half. So true. True. So correct. Now, if you wanted these kind of questions, solve this. Two water taps together can fill a tank in one seven by eight hours. The tap with longer diameter takes two hours less than the tap with smaller one to fill the tank separately. Find the time in which in each tap can fill the tank separately. So again, you can see work time questions are there in boards every year, at least in the last couple of years, we have seen that. Yep. Done. Okay. Pranav is done. Others. Yep. Only two. Only two. Akshita. What happened? Answer. Pranav, your answer please. This was, this is simpler than the previous one because there's a direct link between the two variables. Very easy relation. Okay. So I will solve it now. Two water taps together can fill a tank in one seven upon eight hours. The top tap with longer diameter takes two hours less than the tap with smaller one. Okay. To fill the tank separately, find the time in which each tap can fill the tank separately. Yes, Satyam. Correct. So let these, let the time taken by the longer and smaller dia pipe time taken by the to fill the tank independently or the tank separately be X and Y hours respectively. Okay. So clearly it is, so, you know, X. So X and X plus Y. So, so I can write Y is equal to X plus two smaller dia will take more time. So Y is equal to X plus two. Okay. Now the second thing is together they fill a tank in one by one seven by eight hours. So that means together in one hour, the tank with larger dia will fill this in the, in one hour, the smaller dia tank will fill this much fraction. And together they are doing in eight 15 by eight hours. So it will be eight by 15. This is the equation and you have done right. So this implies one upon X plus one upon X plus two. Why did I do that? Again, reiterating this is a fraction of the tank filled by the larger dia pipe in one hour. This is the fraction of the tank filled by the smaller dia pipe in one hour. So together they will fill up this fraction of the tank and which is nothing but one upon whatever the total time taken by both. Right. So hence this is the equation. I hope you got the essence why it is like this and this is eight upon 15. And you have to now solve for X. It is, it is becoming a quadratic equation, is it? So there is no way now you can, you'll have to solve like this. Or, or, or, or, or, or, no, you can't do anything actually. Can you, can you avoid quadratic? So instead of breaking our head into this, we'll solve. So X plus two into X, sorry, plus X and the denominator X square plus two X is equal to eight upon 15. So this means this is two X plus two. Right. So I can simplify this. Two X plus one by, oh, no, this is coming out to be X into X plus two. Okay. X square plus two X is equal to eight by 15. So clearly this two, four. So 15 X plus 15 is equal to four X square plus eight X. So my equation is four X square minus seven X minus 15, zero, right, and 12 and five. So hence you'll, now let me solve it here. Sorry, lack of space. So you can write this as four X square minus 12 X plus five X minus 15 is equal to zero. Is equal to zero. So this means four X common X minus three plus five common X minus three. Correct. Is equal to zero. So this means X is equal to three. The other, other solution I'm discarding X cannot be equal to minus five by four. Time it is. So it cannot be this. Time cannot be negative. Time, time can't be less than zero. So X equals to three. If X equals to three, Y is equal to X plus two five hours. I hope you understood. Okay. So now the rest of the questions are again, I am giving you as an assignment. There are three, four more left. You can solve them. This is the typical, you can see the way they have written. Let the required two numbers be five X and six X given if seven is subtracted from both, you don't need to write this much actually, but they have, I don't know whether they have overdone it, but you know, try in the five marker, definitely write as much as you can. But for one marker and two markers, you can avoid writing this much thing, these many things. So, right? So you can start with this according to the question, this, right? Then solve, no, no need to write solving one and all. You can directly from here, right? This implies and go like this, this, this, these are all okay. So X equals to seven. Hence you can write the required number is five X, 35 and 40, like that. Good. This is another one. So the two equations they are solving to find X and Y. So, so hence multiplying this, this is a good, this is to be written. Then you eliminate and then you are finding Y is equal to one. So highlight that answer, not during the process while you are attempting all the questions. Once you are done, as I told you, do all this. Yeah, I will go to just a minute, Aditya. So while you're doing this, while you're revising, you can do all these highlighting and everything else. Usually when I write, when I finish, I just put a underline so that it draws attention of the examiner that here is the solution or answer, final answer. Like that. So nothing much to talk about here. So take care of. I told you what mistakes you will be doing. Yep. So this is for the tank question. Yeah. So the mistakes again will be in terms of transposing, right? From right hand side to left hand side, left hand side to right hand side. Please be extra process while you are transposing. You can do it by increasing the number of steps while writing it. And then, yes, since it is a linear equation, do always check, plug the values back into the equations, see if they are correct or not. That's the best way you can check. So at least in the exam, all those equations can be re-evaluated by plugging in the values. Okay. I hope you learned something today. Session was useful to you. This will be uploaded again in the Learnist portal. Please download these slides, if you wish, and solve the rest of the question, which are again from the year 1819, very recent. So you can practice them. Okay. Any other problem? Anyone? Just let me know. Right. So do not relax folks. And anyways, you have an upcoming PT, so you have to do that. So what is your suggestion? Do we continue with the same pace or we spread it out because now that you have time? So you can suggest. So earlier, we were finishing on 7th of Feb. But if you think it is too stressful, then we can spread it out so that we can take the entire month of Feb to cover the revision. Your choice, you can tell us. Okay. So we are contemplating on those things. So what is your suggestion on that? Please let me know. Okay. So finish it early. Okay. Same pace. Everyone revise again. My God. Surya, again, you are going in the wrong path. There is no need to revise 1,000 times. Yes, mocks will continue every week. You will have mocks. Right. So hence the problem is there is nothing to revise much with 30% reduced load. So it's like you will be wasting your time if you're right. So hence. Okay. Chalo never mind. Thanks for your time. See you again next week. Bye-bye. If at all there is a change in the schedule, we will let you know. Please keep checking the group. Bye-bye. Take care. Enjoy.