 Hello and welcome to the session. In this session we discussed the following question which says find the compounded ratio of the falling ratios. In the first part we are given three ratios which are 3 is to 4, 2 is to root 5 and root 5 is to 6. In the second part we are given two ratios which are x square minus 25 is to x square minus 7x plus 10 and x square minus 4 is to x square plus 5x plus 6. Before we move on to the solution let's discuss what is the compounded ratio. If we are given two ratios a is to b and c is to d so compounded ratio of a is to b and c is to d is given by the ratio of the product of the antecedents of the two ratios that is the product of the first terms of the two ratios which is a and c so ac is to the product of the consequence of the two ratios that is the second terms b and d which is bd and if we are given more than two ratios that is if we are given any number of ratios then also the compounded ratio would be the ratio of the product of the antecedents of the given ratios to the product of their consequence like compounded ratio of a is to b, c is to d and e is to f is given by ac e is to bdf. So in the same way we can find out the compounded ratio for any number of ratios. This is the key idea that we use for this question. Let's move on to the solution now. In the first part we have given the ratios 3 is to 4, 2 is to root 5 and root 5 is to 6. So we have to find the compounded ratios of the given three ratios thus the required compounded ratio is given by 3 multiplied by 2 multiplied by root 5 that is this is the product of the antecedents of the three ratios which are the first terms of the three ratios upon the product of their consequence which are 4 root 5 and 6. Now here root 5 cancels with root 5, 3 2 times is 6, this 2 cancels with 2 and in the denominator we are left with 4. So we get this is equal to 1 upon 4 or you can say that the compounded ratio of 3 is to 4, 2 is to root 5 and root 5 is to 6 is 1 is to 4. So this is the answer for the first part of the question. Let's move on to the second part of the question in which we have given the ratios x square minus 25 is to x square minus 7x plus 10 and x square minus 4 is to x square plus 5x plus 6. So here the required compounded ratio is equal to the product of the antecedents of the two ratios that is x square minus 25 the whole multiplied by x square minus 4 upon the product of their consequence that is x square minus 7x plus 10 the whole multiplied by x square plus 5x plus 6. Let us now factorize the polynomials in the numerator and the denominator. Now x square minus 25 is factorized as x plus 5 the whole multiplied by x minus 5. So this is equal to x plus 5 the whole multiplied by x minus 5 then x square minus 4 is factorized as x plus 2 the whole multiplied by x minus 2. So we write here x plus 2 whole multiplied by x minus 2 upon x square minus 7x plus 10 is factorized as x minus 5 the whole multiplied by x minus 2. So we write here x minus 5 whole multiplied by x minus 2 then the x polynomial that is to be factorized is x square plus 5x plus 6 which is factorized as x plus 3 the whole multiplied by x plus 2 minus 5 cancels with x minus 5 x minus 2 cancels with x minus 2 x plus 2 cancels with x plus 2 and we are left with x plus 5 upon x plus 3. Thus we can say that the required compounded ratio is equal to x plus 5 is to x plus 3. This is our final answer. This completes the session. Hope you have understood the solution for this question.