Allegation on 3+ values?

Question

Hi,

I've just learned the allgegation method for weighted average/mixture questions and it is very fast when computing mixture questions with two different values!
Does anyone know if this method works for 3 or more values?
If so, how?

Thanks

hsb91 · Answer

In what ratio must a person mix three kinds of metal costing Rs.60/kg, Rs.75/kg and Rs.100 /kg so that the resultant alloy when sold at Rs.96/kg results in a gain of 20%?
 (1) 1 : 2 : 4 (2) 3 : 7 : 6 (3) 1 : 4 : 2 (4) 1 : 4 : 1

Correct Answer - (3)

How can we solve this as it involves 3 quantities ?

argha · Answer

Hi,

Happy to help

The fastest way would be to plug in the values in the options:

1. gives (60*1+75*2+100*4)/7=610/7

2. gives (60*3+75*7+100*6)=1205/7

3. gives (60*1+75*4+100*2)=560/7

4. gives ((60*1+75*4+100*1)=460/7

Now, desired rate= Rs. 80. So option 3.

In case you are interested in a detailed mathematical solutioning, substitute the basic ratio as "x"

Thus in option 1, the ratios would be x, 2x and 4x and so on in the other options

Regards

Argha

hsb91 · Answer

Yes, agreed that plugging in the choices is the best method but using allegation technique to find ratios of metals
 M1 : M3= 1 : 1   and  M2 : M3 = 4 : 1  so according to that  M1 : M2 : M3 = 1: 4 : 1  should be the ratio of metals in the alloy ... could you point out what i'm missing ??

argha · Answer

Hi,

Sorry but did not understand your question.

Why are you assuming this: M1 : M3= 1 : 1 and M2 : M3 = 4 : 1???

Regards

Argha

hsb91 · Answer

M1 = metal costing 60/kg , M2 = metal costing 75/ kg and M3= 100/ kg.
sp = 96 
profit = 20%
thus cp = 96/1.2 = 80

Using allegation, M1 : M3 = (100-80)/(80-60) = 20/20= 1:1
similarly, M2 : M3 = (100-80)/ ( 80-75) = 20/5 = 4: 1

Thus, M1 : M2 : M3 = 1: 4 : 1
where am i wrong ?

itsmeatsri · Answer

M1 : M3 = 1 : 1 and
M2 : M3 = 4 : 1 
===> M1:M2:M3 = 1:4:2(1+1)

You should add it in 3 value allegation

ragic85 · Answer

I have the same doubt. Can @bunnel or  karishma  help with this?

Using allegation, M1 : M3 = (100-80)/(80-60) = 20/20= 1:1
similarly, M2 : M3 = (100-80)/ ( 80-75) = 20/5 = 4: 1

Thus, M1 : M2 : M3 = 1: 4 : 1

But actually, the common value is added in alligation. M1:M2:M3 = 1:4:2(1+1)

I am perplexed with this concept. Please help.

Henry S. Hudson Jr. · Answer

60---------1-------0-------->1

75---------0-------4-------->4

100---------1-------1-------->2

You can use allegation. You must first compare something below 80 to something above 80. So first we compare 60 to 100. One of each will average to 80. Then we compare 75 to 100. We must have them in a 4 to 1 ratio in order to average to 80. The above "chart" breaks it down.

nitesh50 · Answer

Can some expert please shed some light on this concept of using allegations in 3 variables.
 Bunuel ????

Allegation on 3+ values?

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