Discussion
Home ‣ Aptitude ‣ Alligation or Mixture Comments
- Question
- A mixture of 150 liters of Petrol and Diesel contains 20% Diesel. How much more Diesel should be added so that Diesel becomes 25% of the new mixture?
Options- A. 20 liter
- B. 5 liter
- C. 10 liter
- D. 35 liter
- Correct Answer
- 10 liter
ExplanationLet us assume P liters of Diesel added to the mixture so that Diesel will be 25% in the new mixture.
According to Question,
Quantity of Diesel in 150 liters of the mixture = 20% of 150 = 150 x 20/100 = 30 liters.
After adding P liters of Diesel, The total quantity of Diesel becomes (30 + P) and total volume of mixture will be (150 + P).
Again According to Question,
After adding the P liters of Diesel in the mixture, the Diesel quantity becomes 25% of the new mixture.
Quantity of Diesel in new mixture = 25% of Total mixture.
? (30 + P) = (150 + P) x 25 %
? 30 + P = (150 + P) x 25/100
? 30 + P = (150 + P) x 1/4
? 120 + 4P = 150 + P
? 4P - P = 150 - 120
? 3P = 30
? P = 10 liters. - 1. How much chicory at Rs. 24 a kg should be added to 15 kg of tea at Rs. 60 a kg as to make the mixture worth Rs. 39 a kg?
Options- A. 21 kg
- B. 20 kg
- C. 27 kg
- D. 18 kg Discuss
- 2. 4 vessels of equal sizes contains mixture of spirit and water. The concentration of spirit in 4 vessels are 60 %, 70 %, 75 % and 80 % respectively. If all the 4 mixtures are mixed, find in the resultant mixture the ratio of spirit to water ?
Options- A. 57 : 13
- B. 23 : 57
- C. 57 : 23
- D. None of these Discuss
- 3. In a mixture of 60 liters, the ratio of milk and water is 2 : 1. If the ratio of the milk and water is to be 1 : 2, then the amount of water to be further added is ?
Options- A. 20 litres
- B. 30 litres
- C. 40 litres
- D. 60 litres Discuss
- 4. A petrol pump owner mixed leaded and unleaded petrol in such a way that the mixture contains 10 % unleaded petrol. What quantity of leaded petrol should be added to 1 liter mixtures, so that the percentage of unleaded petrol becomes 5 %?
Options- A. 1000 ml
- B. 900 ml
- C. 1900 ml
- D. 1800 ml Discuss
- 5. A person bought 60 quintals of rice of two different sorts for Rs. 4642·50. The better sort costs Rs. 80 per quintal and the worse Rs. 75·50 per quintal. How many quintals were there of each sort?
Options- A. 25 quintals, 35 quintals
- B. 20 quintals, 40 quintals
- C. 32 quintals, 28 quintals
- D. None of these Discuss
- 6. Two vessels A and B contain milk and water in the ratio 7 : 5 and 17 : 7 respectively. In what ratio mixtures from two vessels should be mixed to get a new mixture containing milk and water in the ratio 5 : 3?
Options- A. 1 : 2
- B. 2 : 1
- C. 2 : 3
- D. 3 : 2 Discuss
- 7. A person covers a distance of 100 kms in 10 hours, partly by walking at 7 km/hr and rest by running at 12 km/hr. Find the distance covered in each part.
Options- A. 48 kms
- B. 72 kms
- C. 108 kms
- D. 124 kms Discuss
- 8. From a cask of wine containing 25 litres, 5 litres are withdrawn and the cask is refilled with water. The process is repeated a second and then a third time. Find the quantity of wine left in the cask and also the ratio of wine to water in the resulting mixture.
Options- A. 61 : 64
- B. 16 : 46
- C. 46 : 16
- D. 64 : 61 Discuss
- 9. A vessel contains mixture of liquids A and B in the ratio 3: 2. When 20 litres of the mixture is taken out and replaced by 20 litres of liquid B, the ratio changes to 1: 4. How many litres of liquid A was there initially present in the vessel?
Options- A. 12 litres
- B. 18 litres
- C. 24 litres
- D. 22 litres Discuss
- 10. A piece of an alloy of two metals (A and B) weighs 15 gms and costs Rs. 150. If the weights of the two metals be interchanged, the new alloy would be worth Rs. 120. If the price of metal A is Rs. 6 per gm, find the weight of the other metal in the original piece of alloy.
Options- A. 5 gms
- B. 8 gms
- C. 10 gms
- D. 12 gms Discuss
Alligation or Mixture problems
Search Results
Correct Answer: 21 kg
Explanation:
Correct Answer: 57 : 23
Explanation:
Correct Answer: 60 litres
Explanation:
Correct Answer: 1000 ml
Explanation:
Correct Answer: 25 quintals, 35 quintals
Explanation:
Correct Answer: 2 : 1
Explanation:
First of all we write the fraction of milk present in three mixtures. In A : 7 /12
In B : 17 /24
In combination of A and B : 5 /8
We now apply allegation rule on these fractions from figure.
So, Ratio of A: B = 2 : 1
2nd Method
Let us assume P mixture taken from first vessel and Q mixture taken from second vessel to form a new mixture.
Part of Milk in P mixture from first vessel = 7P/12
Part of Milk in Q mixture from Second vessel = 17Q/24
Part of Water in P mixture from first vessel = 5P/12
Part of Milk in Q mixture from Second vessel = 7Q/24
According to question,
After mixing the P and Q, we will get mixture.
Milk in New Mixture / Water in New Mixture = 5/3
{(7P/12) + (17Q/24 ) } /{ (5P/12) + (7Q/24) } = 5/3
{(14P + 17Q)/24 } /{ (10P + (7Q)/24 } = 5/3
{(14P + 17Q) } /{ (10P + (7Q)} = 5/3
(14P + 17Q) x 3 = 5 x (10P + (7Q)
42P + 51Q = 50P + 35Q
51Q - 35Q = 50P - 42P
8P = 16Q
P = 2Q
P/Q = 2
P : Q = 2 : 1
Correct Answer: 72 kms
Explanation:
Average speed = (100 / 10) = 10 km/hr.
Ratio of time taken at 7 km/hr to 12 km/hr = 2 : 3
Time taken at 7 km/hr = 2 (2 + 3) × 10 = 4 hrs.
Distance covered at 7 km/hr = 7 × 4 = 28 km.
Distance covered at 12 km/hr = 100 ? 28 = 72 km.
Correct Answer: 64 : 61
Explanation:
Here, quantity of wine left after third operation
= [1 - (5 / 25)]3 x 25 = (4 / 5)3 x 25 = (64 / 125) x 25 = (64 / 5) = 12 4/5 liters.
Final ratio of wine to water = (64 / 125) / (1- 64 /125)
= (64 / 125) /(61 / 125)
Wine : Water = (64 / 61)
Correct Answer: 18 litres
Explanation:
21.% of liquid B initially present in the vessel = 2 / (3 + 2) × 100 = 40%
% of liquid B finally present in the vessel = 4 / (1 + 4) × 100 = 80%
The second solution is liquid B which is being mixed and it has 100% liquid B.
80% of liquid B present in the resultant mixture may be taken as average percentage. So, using rule of alligation on liquid B per cent, we can write,
or 1 : 2
The ratio of liquid left in the vessel to liquid B being mixed = 1: 2
Since the quantity of liquid B being mixed is 20 liters, the quantity of liquid left in the vessel is 10 liters.
Therefore, the total quantity of liquid initially present in the vessel = 10 + 20 = 30 liters
Quantity of liquid A = 3 / (2 + 3) × 30 = 18 liters.
Correct Answer: 10 gms
Explanation:
If the two alloys are mixed, the mixture would contain 15 gms of each metal and it would cost Rs. (150 + 120) = Rs. 270.
Cost of (15 gms of metal A + 15 gms of metal B) = Rs. 270
Cost of (1 gm of metal A + 1 gm of metal B) = Rs. (270 / 15) = Rs. 18
Cost of 1 gm of metal B = Rs. (18 ? 6) = Rs. 12
Average cost of original piece of alloy = (150 / 15) = Rs. 10 per gm.
Quantity of metal / A Quantity of metal B = (2 / 4) = (1 / 2)
Quantity of metal B = 2 (1 + 2) × 15 = 10 gms.
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.