Problem Solving

Two vessels contain mixtures of spirit and water. In the first vessel the ratio of spirit to water is 8
: 3 and in the second vessel the ratio is 5 : 1. A 35-litre cask is filled from these vessels so as
to contain a mixture of spirit and water in the ratio of 4 : 1. How many litres are taken from the
first vessel ?

Here's my approach and it is wrong . Where am I committing a mistake?

4x+1x = 35
x=7. So the spirit is 28 liters and water 7 liters.
Hence (8/13)*28+(3/4)*7 = 22 liters (approx)

But OA is 11 liters

What is this intended to calculate? I could not figure out where you are getting 8/13 ratio or 3/4 ratio in this equation: (8/13)*28+(3/4)*7 = 22 liters (approx)

I can present my solution using allegation:
Spirit in Vessel 1 = 8/11
Spirit in Vessel 2 = 5/6
Spirit in 35L Casket = 4/5

Vess1 ---------Casket ------- Vess2
8/11 ------------------------- 5/6
--------- ------4/5 -------------
(4/5 - 5/6) ---------------- (4/5 - 8/11)
--1/30 ------------------------ 4/55
Ratio of quantities from V1:V2 = 11/330 : 24:330

Ratio from vessel 1 = 11/35

The total spirit in the 35 litres is 28 litres and water is 7 litres.
8x of spirit from vessel 1 and 5x of spirit in vessel 2 makes up 28 litres. Hence from spirit from vessel 1 is (8x/13x)*28. Similarly (3x/4x)*7

srcc25anu wrote:
Vess1 ---------Casket ------- Vess2
8/11 ------------------------- 5/6
--------- ------4/5 -------------
(4/5 - 5/6) ---------------- (4/5 - 8/11)
--1/30 ------------------------ 4/55

Plus arent u calculating just the spirit's proportion and not for both spirit and water from vessel 1?

Even if you do the similar calulations using water proportions in the two given mixtures, you will arrive at the same ratio of quantities for the 2 mixtures. The reason why this works even if we consider just one element is because we are looking at the ratios (not absolute amounts). Lets say a mixture is 1 consisting of salt and water only. if salt = 1/4, then water must be 3/4.

let me demonstrate what results if we consider water ratios for 2 mixtures.

Vess1 ---------Casket ------- Vess2
3/11 ------------------------- 1/6
--------- ------1/5 -------------
(1/5 - 1/6) ------------ (1/5 - 3/11)
--1/30 -------------------- 4/55
Ratio of quantities from V1:V2 = 11/330 : 24:330

Again Ratio from vessel 1 = 11/35 (same as if we consider spirit proportions only)

Allegation will be useful in solving this problem.

Go thru the attachment for solution

Regards,

Abhijit

Here is the question on similar concept
https://www.beatthegmat.com/unable-to-un ... 02593.html

Deepthi Subbu wrote:Two vessels contain mixtures of spirit and water. In the first vessel the ratio of spirit to water is 8
: 3 and in the second vessel the ratio is 5 : 1. A 35-litre cask is filled from these vessels so as
to contain a mixture of spirit and water in the ratio of 4 : 1. How many litres are taken from the
first vessel ?

Deepthi Subbu wrote:Solution A contains 18 ounces of alcohol and 32 ounces of water. Solution B contains 72 ounces of water and 8 ounces of alcohol. What percent of Solution B must be added to
Solution A so that the resulting mixture contains 5/7 water?

Let F = the first vessel, S = the second vessel, and M = the mixture.

The following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with fractions or percentages.
Thus, the ratios here must be converted.

Step 1: Convert the ratios to FRACTIONS.
F:
Since spirt:water = 8:3, and 8+3=11, water/total = 3/11.
S:
Since spirt:water = 5:1, and 5+1=6, water/total = 1/6.
M:
Since spirit:water = 4:1, and 4+1=5, water/total = 1/5.

Step 2: Put the fractions over a COMMON DENOMINATOR.
F = 3/11 = 90/330.
S = 1/6 = 55/330.
M = 1/5 = 66/330.

Step 3: Plot the 3 numerators on a number line, with the numerators of F and S (90 and 55) on the ends and M's numerator (66) in the middle.
F 90-----------------66------------------55 S

Step 4: Calculate the distances between the numerators.
F 90-------24--------66--------11-------55 S

Step 5: Determine the ratio in the mixture.
The ratio of F to S in the mixture is the RECIPROCAL of the distances in red.
F:S = 11:24.

Since F:S = 11:24, and the actual total volume = 35 liters, the actual volume of F = 11 liters, and the actual volume of S = 24 liters, so that F+S = 11+24 = 35 liters.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583

@GMATGuruNY :
I was able to understand alligation, but my point was to see if my method can actually work.
And you have also added another mixture question to help me relate. Wow, thanks

Deepthi Subbu wrote:@GMATGuruNY :
I was able to understand alligation, but my point was to see if my method can actually work.
And you have also added another mixture question to help me relate. Wow, thanks

You correctly determined that the final mixture is composed of 28 liters of spirit and 7 liters of water.
Here's how we could use this information to solve algebraically:

Spirit:
First vessel:
Since spirit:water = 8:3, and 8+3 = 11, the amount of spirit in the first vessel = (8/11)F.
Second vessel:
Since spirit:water = 5:1, and 5+1 = 6, the amount of spirit in the second vessel = (5/6)S.
Since the mixture contains 28 liters of spirit, we get:
(8/11)F + (5/6)S = 28.

Water:
First vessel:
Since spirit:water = 8:3, and 8+3 = 11, the amount of water in the first vessel = (3/11)F.
Second vessel:
Since spirit:water = 5:1, and 5+1 = 6, the amount of water in the second vessel = (1/6)S.
Since the mixture contains 7 liters of water, we get:
(3/11)F + (1/6)S = 7.

Multiplying the second equation by 5, we get:
(15/11)F + (5/6)S = 35.

Subtracting the first equation from the second, we get:
[ (15/11)F + (5/6)S ] - [ (8/11)F + (5/6)S ] = 35-28
(7/11)F = 7
F = 11.

Thus, the mixture contains 11 liters from the first vessel.