Aptitude - Alligation or Mixture - Discussion

Discussion :: Alligation or Mixture - General Questions (Q.No.12)

12. 

In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

[A]. 3 : 2
[B]. 3 : 4
[C]. 3 : 5
[D]. 4 : 5

Answer: Option A

Explanation:

S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%.

C.P. of 1 kg of the mixture = Rs. 100 x 68.20 = Rs. 62.
110

By the rule of alligation, we have:

Cost of 1 kg tea of 1st kind. Cost of 1 kg tea of 2nd kind.
Rs. 60 Mean Price
Rs. 62
Rs. 65
3 2

Required ratio = 3 : 2.


Sumeet said: (Jul 27, 2011)  
How the required ratio is obtained?

Arpit Jain said: (Sep 1, 2012)  
Ratio is obtained by difference between them...
62-60 = 2
65-62 = 3

Simit said: (Aug 3, 2013)  
Let us assume x/y ratio.

Now x+y = 1.

x/y+1 = 1/y ---- (1).

Now price,

60x+65y = 62.

60(x/y)+65 = 62[(x/y)+1].

Now this shows that difference become ratio.

Pandia said: (Sep 8, 2013)  
60*3+65*2 = 180+130.

310 this total 5 kg rate 3 60 kg and 2 65kg.

So one kg 310/5 = 62 this is cost price so the ratio of mixture is 3:2.

Swejal said: (May 17, 2014)  
Can someone please explain the 2nd step i.e C.P of 1 kg.

Rishabh said: (May 23, 2014)  
As we know %gain means %gain over CP =SP - CP.

Hence,10% * CP =SP - CP.

Upon solving you will get CP=100/110 * SP.

Usha said: (Aug 20, 2015)  
Can any one explain how 110/100*68.20 = Rs. 62?

Muna said: (Aug 20, 2015)  
How to find out mean price?

Sumanth Geras said: (Aug 20, 2015)  
Easy method applicable for all these kinds of problems.

(quantity 1*price 1+quantity 2*price 2)/quantity 1+quantity 2 = mean quantity.

Since it is given that he got 10 % profit so this 10% is for additional money which is added to mean value obtained (i.e 10% of 68.20 is 62 approximately).

(60(x)+65(y))/x+y = 62. If you solve it you get 3:2.

Krishna said: (Sep 22, 2015)  
It was said that there was 10% profit for selling a K.G of mixture for 68.20 rs.

But we generally know for any good of C.P 100 rs, S.P of that good becomes 110 rs in case of 10% profit.

Now, S.P - C.P.

110 - 100 => (100*68.20)/110 = 62.

68. 20 - ?

Now, by using the equation.

(60*(x)+65(y))/x+y = 62.

On substituting values from options we get x=3 y=2, ratio becomes 3:2.

Nitin said: (Nov 4, 2016)  
I think the ratio should be 2:3 not 3:2.

Correct me if I am wrong.

Nitesh said: (Jan 21, 2017)  
There will be profit, hence less price object will be in more than other.
Hence, 3 : 2 only one option.

Sam said: (Jul 16, 2017)  
I can't get it. Please explain.

Shubham Kumar said: (Mar 6, 2018)  
Please explain the 2nd line ratio.

Ashutosh said: (May 3, 2018)  
Anyone help me how to come 110?

Cosul said: (May 29, 2018)  
Easy method applicable for all these kinds of problems.

(quantity 1*price 1+quantity 2*price 2)/quantity 1+quantity 2 = mean quantity.

%gain over CP =SP - CP.
Hence,10% * CP =SP - CP.
Upon solving you will get CP=100/110 * SP. = 62 rs
(60(x)+65(y))/x+y = 62. If you solve it you get x/y =3/2.

Jyotsana said: (Apr 11, 2019)  
I think the ratio should be 2:3.

Zahir said: (Sep 25, 2019)  
We know,

Gain%= (SP-CP)/CP.
Let, the amount of tea Rs.65 is x kg.
And the amount of tea Rs.60 is y kg.

Now, CP= 60x+65y.
SP= 68.20(x+y),
So according to the gain% formula.
[{68.20(x+y)-(60x+65y)}/(60x+65y)]=10/100,
=> x/y=3/2.

Tamanna said: (Mar 30, 2020)  
I am not getting this, please anyone explain in detail.

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