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Quantitative Aptitude

MIXTURES AND ALLEGATIONS MCQs

Alligations And Mixtures

Total Questions : 245 | Page 20 of 25 pages
Question 191. 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?
  1.    2 : 3
  2.    1 : 2
  3.    2 : 1
  4.    3 : 2
 Discuss Question
Answer: Option C. -> 2 : 1

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 : 12nd MethodLet 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/12Part of Milk in Q mixture from Second vessel = 17Q/24Part of Water in P mixture from first vessel = 5P/12Part of Milk in Q mixture from Second vessel = 7Q/24According 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 + 35Q51Q - 35Q = 50P - 42P8P = 16QP = 2QP/Q = 2 P : Q = 2 : 1


Question 192. 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 ?
  1.    None of these
  2.    57 : 13
  3.    23 : 57
  4.    57 : 23
 Discuss Question
Answer: Option D. -> 57 : 23
Question 193. 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?
  1.    5 liter
  2.    35 liter
  3.    20 liter
  4.    10 liter
 Discuss Question
Answer: Option D. -> 10 liter

Let 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.


Question 194. 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?
  1.    12 litres
  2.    22 litres
  3.    18 litres
  4.    24 litres
 Discuss Question
Answer: Option C. -> 18 litres

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.


Question 195. 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.
  1.    10 gms
  2.    8 gms
  3.    5 gms
  4.    12 gms
 Discuss Question
Answer: Option A. -> 10 gms

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.


Question 196. A container of 90 liters with two liquids A and B, 60% of liquid A and 30% of liquid B are taken out of the vessel. This leaves the container 40% empty. Find the initial quantity of both liquids.
  1.    60 litres
  2.    54 litres
  3.    57 litres
  4.    50 litres
 Discuss Question
Answer: Option A. -> 60 litres

Here withdrawal of liquid A and B result into making the container empty. Hence percentage of two liquids withdrawn are two components of the percentage by which the container becomes empty. Applying the rule of alligation, we getA : B = 10 : 20 or 1 : 2 Quantity of liquid = 1 (1 + 2) × 90 = 30 liters Quantity of liquid B = 90 ? 30 = 60 liters.


Question 197. 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.
  1.    72 kms
  2.    108 kms
  3.    124 kms
  4.    48 kms
 Discuss Question
Answer: Option A. -> 72 kms

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.


Question 198. 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.
  1.    64 : 61
  2.    61 : 64
  3.    16 : 46
  4.    46 : 16
 Discuss Question
Answer: Option A. -> 64 : 61

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)


Question 199. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If mixture is worth Rs. 153 per kg, the price of the third variety per kg will be:
  1.    170
  2.    169.50
  3.    175.50
  4.    180
 Discuss Question
Answer: Option C. -> 175.50

Here first two varieties of tea are mixed in equal ratio.So their average price = (126 + 135) /2 = Rs. 130.50 Let price of the third variety per kg be Rs. A; then now mixture is formed by two varieties one at Rs. 130.50 per kg and other at Rs. A per kg in the same ratio 2 : 2 i.e, 1 : 1 By the rule of alligation,(A - 153) / (22.50) = 1 ? A ? 153 = 22.50 ? A = Rs. 175.50


Question 200. A jar full of whisky contains 40 % of alcohol. A part of this whisky is replaced by another containing 19 % alcohol and now the percentage of alcohol was found to be 26. The quantity of whisky replaced is ?
  1.    2 5
  2.    1 3
  3.    2 3
  4.    3 5
 Discuss Question
Answer: Option C. -> 2 3

Here first two varieties of tea are mixed in equal ratio.So their average price = (126 + 135) /2 = Rs. 130.50 Let price of the third variety per kg be Rs. A; then now mixture is formed by two varieties one at Rs. 130.50 per kg and other at Rs. A per kg in the same ratio 2 : 2 i.e, 1 : 1 By the rule of alligation,(A - 153) / (22.50) = 1 ? A ? 153 = 22.50 ? A = Rs. 175.50


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