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

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.