### Discussion :: Alligation or Mixture - General Questions (Q.No.14)

Sanjay said: (Aug 27, 2010) | |

Could anyone explain the method above? I couldn't make out why x is divided when it should have been divided by the quantity of water. |

Sourav said: (Aug 31, 2010) | |

How did quantity of water become x? |

Priya said: (Nov 29, 2010) | |

@sanjay Its a formula.... [a(1-X/a)n] a=initial quantity, X=quantity taken out, n=raised power on small bracket(no. of action done) |

Divya said: (Dec 28, 2010) | |

Thanks priya. |

Mamta said: (Dec 29, 2010) | |

Hi priya, thanks 4 giving formula can you explain why it is divded by x, why not by quantity of water. |

Mayank said: (Jan 19, 2011) | |

Please tell what is the mean of performs operations more then three times. And why multiply x. |

Sudheer said: (Jan 21, 2011) | |

Its 3 more times not "more then three times" so we used mul. |

Chitwanjit said: (Jan 29, 2011) | |

How come it get divided by x, While it should be quantity of water ? Alternate way can be dividing by x and putting in 16/97 instead of 16/81. |

Bappa said: (Jul 8, 2011) | |

Lets wine x after 1st time wine remain x-8 now ratio of wine:water x-8:8 so 2nd time wine removed=(x-8)*(8/x) after 2nd time wine remain=(x-8)-(x-8)*(8/x) =(x-8){1-8/x} =x(1-8/x)(1-8/x) =x(1-8/x)^2 ......... for n time removaline remain=x(1-8/x)^n. |

Himanshu said: (Jul 28, 2011) | |

Well explained Bappa. |

Rachit Sharma said: (Sep 4, 2011) | |

Intial wine = x. After 8 units are pulled out, (x-8) units of pure wine remains. In fractions, ((x-8)/8) = (1-(8/x)) of the initial content x remains after the first pullout. After the second pullout, (1-(8/x)) of the previous amount is left, i.e.- (1-(8/x)) of (1-(8/x)) of x = (1-(8/x))^2 of x. In the same way, after n such pullouts, the fraction of wine left will be (1-(8/x))^n. And the amount of pure wine left will be x*(1-(8/x))^n. This part is well understood. But why was the above quantity divided by x, while the question reported the ratio as that of leftover pure wine to water? I am finding it hard to understand the logic behind this. Or is it wrong? |

Shree said: (Sep 12, 2011) | |

Awesome rachit. |

Prajakt said: (Dec 4, 2011) | |

Can any one please explain ....How the quantity of water become x? |

Rattle said: (Jan 18, 2012) | |

@Prajakt the soln. assumes in the 1st line that the original quantity of wine be x. the question says the wine removed is replaced with water(obviously, in equal amount). Since we assume x as wine quantity, this replaced water quantity is also x. |

Sahil said: (Feb 13, 2012) | |

Since each time we are adding 8 litres of water and talking away 8 litres. So total composition is same only the proportions of individual componnts will change. So talking water as x litre is wrong. He has calculated the ratio of wine to whole soln. ! |

Sahil said: (Feb 13, 2012) | |

Equation should be equated with 16/97. Sine wine to water ratio is 16:81. So wine to total ratio is 16:97. Thats what we finding out actually by dividing it by x. |

Prateek said: (Jun 6, 2012) | |

I agree with Sahil. Another method: Amt. of wine left is given by x(1-8/x)^4. Amt. of water left is given by x- x(1-8/x)^4. So x(1-8/x)^4/[x-x(1-8/x)^4] = 16:81 With this method, i got answers to a couple of questions which were same as this question. But unable to get the answer to this one. I think my method is perfectly alright. |

Shaswat said: (Jul 3, 2012) | |

@Prateek. Here it is given that '8 liters are drawn from a cask full of wine' not 'wine and water' so you cannot subtract i.e., Amt. of water left is given by x- x(1-8/x)^4. So how can you subtract ? |

Tez said: (Jul 19, 2012) | |

Let the quantity of the wine in the cask originally be x liters. Then, quantity of wine left in cask after 4 operations. =[x (1- (8/8) ) ^4]. Hence, [{x (1- (8/x) ^4) ) }/x]=16/97. =>[ (x-8) /x]^4=16/97. => (x-8) /x=2/3.1 (since 97^1/4=3.1). =>x=22.5 Liters. Answer should be 22.5 Liters. |

Erik said: (Aug 15, 2012) | |

There can be no doubt about the fact that they got the last ratio wrong in the question. @rattle, sorry but you are right if the replacement stopped after the first round. After that when on the second round 8 litres is taken from the cask, it includes some wine and some water and then 8 litres of water in put in. This continues twice more. Sorry but we cannot assume x to be the amount of water in the cask after mixing is complete. |

Atchiya said: (Aug 31, 2012) | |

I think performs operation more than 3 times means it includes the first time which is in process + 3 more times. So 4 times. Please can anyone explain hoe that X came else explain this in alternative easy method. |

Richa said: (Jan 18, 2013) | |

let the quantity of the wine in the cask originally be x litres. Then, quantity of wine left in cask after 4 operations = x(1 -8/x)^4 litres. x(1 - (8/x))^4 = 16x/81 (1 -8/x)= 2/3 (x - 8)/x =2/3 3x - 24 = 2x x = 24. |

Jana said: (Jul 17, 2013) | |

Can anyone please explain me how "x" came in the denominator in the 3rd step? |

Pr!!!!! said: (Aug 22, 2013) | |

The ratio of wine to that of water is given as 16:81. The initial quantity of wine is not given and it is assumed as 'x'. And so the ratio becomes 16x:81. And hence for simplification 'x' came in denominator. |

Reeya said: (Aug 22, 2013) | |

I think the question is wrong. I know it can never be said as an excuse, but if the the ratio of wine remaining to that of water is 16:65, then we can take the denominator, wine originally as x. i.e., (x (1-8/x) ^4) / (x) = 16/ (16+65) = 16/81. Correct me if I am wrong. |

Dileep said: (Aug 28, 2013) | |

@Reeya. I didn't get you. what you are saying. From the given information we can write it as, x(1-8/x)^4 = 16/81. i.e x = 24. |

Karthiga said: (Sep 15, 2013) | |

The numerator 16(assume 16 liters) represents the quantity of remaining wine. The denominator (16+65=81)represents {the quantity of remaining wine +quantity of water added=initial quantity of wine(x)}. |

Shovik said: (Sep 19, 2013) | |

The real problem is that some of the initial steps are missing out here. Amt. of wine left is given by x(1-8/x)^4. Amt. of water left is given by x- x(1-8/x)^4. So x(1-8/x)^4/[x-x(1-8/x)^4] = 16:65. => x(1-8/x)^4/x = 16:81. |

Shanjev said: (Oct 6, 2013) | |

Why the quantity of wine left in cask after 4 op is DIVIDED BY X, can anyone explain me? |

Amit said: (Nov 8, 2013) | |

After the first step of adding 8 liters of water, the cask contains a mix of wine and water so the next time we take 8 liters out, it could have 75 % wine or 35 % wine or any other amount. So, how can we calculate the final proportion of wine? |

Kola.Shiva said: (Nov 11, 2013) | |

Simple method. 8 liters of original wine drawn from the cask. So question they had given at last the ratio is 16:65. So we add recent 8 liters to present ratio of wine. 16+8 = 24. Simple. |

Yogi Damle said: (Dec 21, 2013) | |

If the answer is 24 liters then we cannot take out 8 liters of wine 4 times. |

Murali said: (Jan 15, 2014) | |

After n operations the wine left is {(x-8)^n}/x^n-1. However, at the end of every cycle the total quantity is x. Therefore quantity of water left after n operations is x-quantity of wine left= x-{(x-8)^n}/x^n-1. Now, dividing quantity of wine left by quantity of water left we get: {(x-8)^n}/{x^n-(x-8)^n} = 16/65. Substitute each one of the options n see which one matches rather than going to solve this complex expression. |

Pravu said: (Jan 15, 2014) | |

Note that in each drawing, we take out 8/x part of the remaining wine and every time the total (wine +water) is x litre. In 1st draw, pure wine is (x-8) i.e. x(1-8/x). In 2nd draw, pure wine is (x-8)-(8/x)(x-8) i.e x(1-8/x)^2. Similarly in 4th draw pure wine is x(1-8/x)^4. Wine:water=16:65 or wine:total = 16:81. (note: 16+65 = 81). So, x(1-x)^4/x = 16/81. |

Pravu said: (Jan 20, 2014) | |

@Sanjay. Pure wine is divided by total amount in the vessel, i.e. x. So, pure wine: water = 16:65 OR it can be said as pure wine:total = 16:(16+65) = 16:81 Note that in each operation the total is always x litre. |

Pandit said: (Feb 13, 2014) | |

Lets, The volume of the drum is x. In 1st operation 8 liter is drawn. Remaining pure wine is x-8. Now. If we draw whole volume = the volume of wine is drawn is x-8. """"""""""''' 1 unit volume="""""""""""""""""""""""""" (x-8) /x. """"""""""'''8 unit volume=""""""""""""""""""""""""""8*(x-8) /x. The volume of wine remaining after 1st operation is (x-8) -{8* (x-8) /x} = { (x-8) ^2}/x... Similarly in 4th draw pure wine is x (1-8/x) ^4. |

Punith Mishra P said: (Aug 22, 2014) | |

It is like this: x(1-(8/x)^4)-(1) gives quantity of wine, whereas final ratio is 16/81. So final wine quantity is (16/81)*x which is equal to (1). |

Nishad Raut said: (Sep 9, 2014) | |

65 is given and 16 is assumed..so ans is assumed.. join my classes to become aptitude expert.. |

Pallavi Jha said: (Dec 12, 2014) | |

Can anyone explain why it is divided by x? |

Rupesh said: (Dec 18, 2014) | |

Answer for x. As the ratio is 16/65 (Wine:Water). So the amount wine/total mixture = 16/65+16 = 16/81. Here x = 81. |

Sibi said: (May 25, 2015) | |

I don't understand this problem. Please explain me clearly. |

Manju said: (Jun 2, 2015) | |

Not understood. Can anyone help? Why divide by x? Also shouldn't ratio be directly used as a fraction? |

Maram said: (Jun 13, 2015) | |

Hello anybody explain of to me. 81 is how its coming? |

Rahul said: (Jul 11, 2015) | |

Is another easy method for this problem? |

Garima said: (Jul 22, 2015) | |

It is possible that drawn out 8l wine from the mixture the answer can be 24l, because whenever the wine is drawn from the liquid it is again filled with the water in the same amount. |

Chatrapathi said: (Jul 30, 2015) | |

Let "x" liters of wine was present. This wine was replaced by 8 liters of water for 4 times. So, wine at present is x[1-(8/x)]^4. Also given that wine to water ratio at present is 16:65. So wine at present = 16 x/16+65. So x[1-(8/x)]^4 = 16 x/16+65. x = 24. |

Ultimate said: (Aug 7, 2015) | |

If you guys really do this maths then see this technique other than wrong methods. The question is - "How much wine did the cask hold originally?". So the method is : (liquid A left after n'th operation) / (liquid B left after n'th operation) =[ (1- (b/a))^n]/[1- (1- (b/a) ^n]. Here, a= total amount of vessel(a.k.a wine); b = amount of other liquid poured(a.k.a. water); n = no of operations had done. In this question they also mentioned that 3 more times it had done after it had changed with 8 ltr of water. So the n will b 4. Mind the tricks of the questions. Do a very sharp calculation you will get it? It is much easier than its look. |

Rahul.. said: (Aug 11, 2015) | |

Given ratio i.e. 16/65 is wine/water. Hence wine/(wine+water) is 16/81. Total quantity of wine+water will remain fixed i.e. equal to initial quantity of wine i.e. x. Hence final ratio of wine/(wine+water) will be (wine left/initial wine). Hence it will be x (1-8/x)^n/x equal to 16/81. |

Nitin said: (Nov 4, 2015) | |

Let wine left be x, 4 times 8 liter wine removed = 4*8 = 32. Therefore wine left = (x-32). Water filled 4 times = (4*8) = 32. Therefore ratio of (Quantity of wine left/Quantity of water). = (16/65). Therefore ( (x-32)/32) = (16/65). On solving we get x = 39.87~40. |

Ananta Chatterjee said: (Jan 8, 2016) | |

See the formula is well understood a(1-b/a)^n. Now even after adding and removing water we have the total amount o be the same that is x. So we have 16/81 of x = This is the part of wine left that should be equal to the formula. |

Sundar said: (Feb 6, 2016) | |

How can did wine left ratio is 16:65 to change 16:81 please explain anybody? |

Siddu said: (Mar 11, 2016) | |

If 16x/81 is written based on the LHS it converted to 2/3 then what about x. |

Xxx said: (Apr 14, 2016) | |

Step 1: (x - 8)^4/x = 16/81. Step 2: (x - 8)^4/x = (2/3)^4. Therefore, x - 8/x = 2/3. So, x=24. |

Vinod Reddy said: (Jun 14, 2016) | |

Hi, anyone clears my doubt. Logically, How can you remove only wine again after mixing 8 liters of water to the wine already present in the cask? Does anybody know the formula for calculating no of liters of wine present if the solution removed from second time onwards contain both wine and the water? |

Ayush Nigam said: (Jun 19, 2016) | |

I think most of you understood the solution but for those who didn't let me explain: First of all, There is a formula which says: Suppose container contains x of liquid from which if y units are taken out (of the whole liquid) and then replaced with water, and this process is simply repeated n no of times, then the quantity of original liquid present in the mixture now is: x[(1- y/x )^n]. Ok , now as the question says here we need to find the original quantity of the wine, let it be x. Now the quantity of liquid (not wine) removed is = 8 which is 'y' for the formula. This thing is done total of 4 times. So now the quantity of the wine present in the final mixture will be: x[(1 - 8/x) ^4]. No problem till here. Now as the problem says, the ratio of wine to water in the final mixture, we have calculated wine but we can't calculate the amount of water in the final mixture. So, what to do next. We are given the ratio of wine to water in the mixture and remember the water was being replaced each time, so total quantity of the mixture is still 'x'. Thus, if 16/65 was the given ratio of wine to water then, 16/(16+65) will be the ratio of wine to total mixture ie: x[(1-8/x)^n] / x = 16/81 I hope everyone get it now. |

Gaurav Gupta said: (Aug 27, 2016) | |

Good explanation @Ayush. You cleared the doubt everyone had. |

Hardeep Singh said: (Oct 1, 2016) | |

Let initial quantity of wine = x. After, 4 operations, quantity of wine left = x(1 - 8/x)^4. Also the quantity of wine in the jar = (16/81)x. Therefore, x(1-8/x)^4 = (16/81)x. On solving, x = 24. |

Srihu said: (Oct 26, 2016) | |

Thanks, @Pravu. You explained the derivation of the formula is very well. |

Vijay said: (Dec 16, 2016) | |

Thanks @Ayush. |

Lekhraj said: (Jan 7, 2017) | |

Suppose, the initial amount of mixture is X. Remaining quantity of wine after four operations ( P suppose)= X*(1-8/X)^4 = P(Suppose) The total quantity of mixture will always remain X because how much mixture we are removing same we are adding as water and quantity which is not of wine that is water so after 4 operations quantity of water in the mixture = X-P. Given: remaining wine/water=16/65= P/(X-P)...................(1). By solving equation (1) we will get P/X=16/81, that is the ratio of remaining wine and present water in the mixture. So this is the explanation why he divided by X. Further is simple as we supposed P = X * (1-8/X) ^4. So, P/X = (1-8/X)^4 = 16/81.............(2). Solving equation (2) we get X = 24. |

Vijay said: (Feb 2, 2017) | |

If this process done with 5 litres only one time, and the given ratio is 361:39. What will be the answer? |

Amit said: (Feb 5, 2017) | |

As 65:15 is wine as to water ratio is given means total solution was (65+16)=81 which was totally quantity of wine before removal of 8 liter wine that's why it is x = total =81. Where x=initial quantity of wine. W:T = x(1-(8/x))^4:x = 16/81. |

Dee said: (Mar 17, 2017) | |

Lets wine x. After 1st time wine remain x-8 Now, ratio of wine:water x-8:8 So 2nd time wine removed=(x-8) * (8/x) After 2nd time wine remain=(x-8) - (x-8) * (8/x). = (x-8){1-8/x} = x(1-8/x)(1-8/x) = x(1-8/x)^2. After 3rd time wine remain=[x(1-8/x)^2]-[x(1-8/x)^2]x/8. = [x(1-8/x)^2]{1-8/x} = [x(1-8/x)^2](1-8/x) = x(1-8/x)^3 and so on; For n operation remain=x(1-8/x)^n. |

Sobat Singh said: (Mar 30, 2017) | |

Let cask is full of wine = x lt. After 4 operations wine left= x(1-8/x)^4. Ratio of water to wine after operation = 16/65. From this ratio part of wine present after op= 16/81. Total amount of mixture is = x(as cask is full of x quantity). Now, x(1-8/x)^4. ---------------- = 16/81 X From this x = 24. |

Sajan said: (Apr 5, 2017) | |

Explanation for (x(1 - 8/x)pow 4)/x =16/81. In question it's given water/wine = 16/65, We assumed x= total wine in cask, That means total capacity of cask is x, So looking at ratio 16/65 we can see that total Capacity is (16+65) = 81. (x(1 - 8/x)pow 4) is nothing but the wine replaced by water, In other word, it is the quantity of water in a final mixture of water and wine. So, (x(1 - 8/x)pow 4)/x must be 16/81. |

Venky said: (Jul 3, 2017) | |

It should be 16:81. |

Aniket said: (Aug 21, 2017) | |

What if instead of wine it would have been 6 litres of solution of wine and water for second time? |

Jayaraj said: (Nov 24, 2017) | |

16:65 is wine to water ratio. That means if there is at total of 81 parts of solution 16 part will be wine and 65 part will be water (81=16+65). So wine to total solution amount ration would be 16:81. total solution amount =x. |

Prakash said: (Feb 18, 2018) | |

Use the formula (x-8/x)n = 16 / 81. where n is 4, x is original liquid (x-8) is the original liquid remaining after the exchange that is 16, x the original quantity will be 65+16=81. |

Purushuttam said: (May 21, 2018) | |

Have there any explanation of these laws [x(1-y/x)*n] where y<x? |

Surya said: (Aug 25, 2018) | |

Let the quantity of the wine in the cask originally be x litres. 8 litres of wine is removed from it and replaced with water. This operation is done a total 4 times. After 4 operations, Wine : Water = 16 : 65 or Wine : Total = 16 : 81. [(x-8)/x]4 = 16/81 (x-8)/x = 2/3 3x - 24 = 2x x = 24 The correct option is D. |

Shamika said: (Sep 3, 2018) | |

How come 81? |

Naeem said: (Sep 23, 2018) | |

How come 81 in the denominator? |

Jyoti said: (Sep 25, 2018) | |

@Naeem and @Shamika, Here 81 is the amount of total mixture (16+65=81). It is the ratio of wine to total mixture. Here, the water was being replaced each time, so the total quantity of the mixture is still 'x' which is 81. |

Moideen said: (Dec 30, 2018) | |

Well explained, Thanks @ Jayaraj @Jyoti. |

Venkat said: (Feb 16, 2019) | |

@All. 16x/81 is the quantity of wine in the mixture is given so equate to x(1-8/x)^4. (x is fixed litres of mixture). |

Vishnu Vardhan said: (Mar 20, 2019) | |

Yes, Total wine left = (16/81) * (initial litres of wine). |

Vivek Kumar said: (May 23, 2019) | |

From where 16:81 came, there is 16:65? can anyone explain. |

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