Aptitude - Alligation or Mixture - Discussion
Discussion Forum : Alligation or Mixture - General Questions (Q.No. 2)
2.
Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:
Answer: Option
Explanation:
Since first and second varieties are mixed in equal proportions.
| So, their average price = Rs. |  |
126 + 135 |  |
= Rs. 130.50 |
| 2 |
So, the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say, Rs. x per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find x.
By the rule of alligation, we have:
| Cost of 1 kg of 1st kind Cost of 1 kg tea of 2nd kind | ||
| Rs. 130.50 | Mean Price Rs. 153 | Rs. x |
| (x - 153) | 22.50 | |
 |
x - 153 | = 1 |
| 22.50 |
x - 153 = 22.50
x = 175.50
Discussion:
62 comments Page 3 of 7.
Ankan said:
1 decade ago
@Uma Upadhyay.
You are right and you maked its so easy. Thanks and also.
@Suren because you both are following similar process and right one.
You are right and you maked its so easy. Thanks and also.
@Suren because you both are following similar process and right one.
Ayesha Sultana said:
8 years ago
Problem is simple
(126*1 + 135*1 + 2x)/(1+1+2) = 153,
126+135+2x / 4 = 153,
261+2x= 153*4,
261+2x=612,
2x=612-261,
2x=351,
X =351/2,
X=175.5.
(126*1 + 135*1 + 2x)/(1+1+2) = 153,
126+135+2x / 4 = 153,
261+2x= 153*4,
261+2x=612,
2x=612-261,
2x=351,
X =351/2,
X=175.5.
(1)
Maya said:
1 decade ago
Three mixtures are in the ratio 1:1:2 and prices given that 126, 135, P.
1*126+1*135+2*P = (1+1+2)153.
On solving, we get P = 175.5.
1*126+1*135+2*P = (1+1+2)153.
On solving, we get P = 175.5.
Lokesh reddy said:
1 decade ago
126kg-1st 135kg-2nd xkg-3rd
1 : 1 : 2
(126*1 + 135*1 + 2*x )/ 4 = (153) total mixture.
By solving,
x=170.50.
1 : 1 : 2
(126*1 + 135*1 + 2*x )/ 4 = (153) total mixture.
By solving,
x=170.50.
Ashwini Sriram said:
1 decade ago
I agree with Suren. The problem is simpler than it is explained here.
It is just (126+135+2x)/4= 153.
Where x is the price per kg.
It is just (126+135+2x)/4= 153.
Where x is the price per kg.
Sucharitha said:
9 years ago
Can you please explain, if suppose the costs of three varieties are given but ratios are not given.
Then what will be the solution?
Then what will be the solution?
Suren said:
2 decades ago
Totally there are 4kg mixture which costs 153.
therfore,(126+135+x)/4=153
x=351.
Here x is the price for 2Kg.
therefore x=175.50
therfore,(126+135+x)/4=153
x=351.
Here x is the price for 2Kg.
therefore x=175.50
Preethi said:
1 decade ago
@suren.
In the question 1kg of the mixture is mentioned as Rs 153 na? how can you say 4kg costs 153.
Please explain.
In the question 1kg of the mixture is mentioned as Rs 153 na? how can you say 4kg costs 153.
Please explain.
Naveen said:
7 years ago
Simple formule:
(1:1:2)/4=153
(126+135+2X)/4 = 153,
261+2X = 4 * 153
261+2X = 612,
2X = 612-261,
2X = 351,
x = 175.5.
(1:1:2)/4=153
(126+135+2X)/4 = 153,
261+2X = 4 * 153
261+2X = 612,
2X = 612-261,
2X = 351,
x = 175.5.
(42)
Mounika said:
8 years ago
1 : 1 : 2 ratio.
So, we can take it as 1/4kg 1/4kg and 1/2 kg.
So, the answer ( 1/4)*126+(1/4)*135+(1/2)*x = 153.
So, we can take it as 1/4kg 1/4kg and 1/2 kg.
So, the answer ( 1/4)*126+(1/4)*135+(1/2)*x = 153.
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