Prayas_AllegationsMixtures
Quiz Summary
0 of 40 Questions completed
Questions:
Information
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading…
You must sign in or sign up to start the quiz.
You must first complete the following:
Results
Results
0 of 40 Questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 point(s), (0)
Earned Point(s): 0 of 0, (0)
0 Essay(s) Pending (Possible Point(s): 0)
Categories
 quant 0%
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 16
 17
 18
 19
 20
 21
 22
 23
 24
 25
 26
 27
 28
 29
 30
 31
 32
 33
 34
 35
 36
 37
 38
 39
 40
 Current
 Review
 Answered
 Correct
 Incorrect

Question 1 of 40
1. Question
12 litres of water and 28 litres of milk are blended by a delivery man. He contributes water to recharge the amount he has sold after selling onefourth of the mixture. Calculate the current milk percentage in the mixture.
CorrectIncorrectHint
Initially total mixture is of 40ltrs.
Ratio of milk : water = 28:12 = 7:3
After one fourth is finished, left will be 30 with milk:water = 21:9
Now, 10ltrs water is added, hence new quantity of milk and water becomes = 21:19
Required percentage = (21/40)*100 = 52.5% 
Question 2 of 40
2. Question
A container holds 16 litres of juice. 4L of juice was removed from this container and replaced with water. 4L of this mixture was removed and replaced with water. How much juice does the container now hold?
CorrectIncorrect 
Question 3 of 40
3. Question
There are two wine and water solutions, with a wine concentration of 0.3 and a water concentration of 0.7. Find the concentration of wine in the resultant solution if 7L of the first solution is mixed with 17L of the second solution.
CorrectIncorrectHint
Resultant concentration of wine:
= (0.3)(7) + (0.7)(17)/7+17
=14/21
=2/3
=0.67 
Question 4 of 40
4. Question
A mixture of 25 kg of spirit and water contains 20% water. How much water must be added to this mixture to raise the percentage of water to 33.33%?
CorrectIncorrectHint

Question 5 of 40
5. Question
Andrew was attempting to make water balls by heating a water and sugar solution and then evaporating the water. The mass of the solution is 3 kg, and it contains 90% water and 10% sugar by mass. He discovers that the solution contains 85 percent water after some time. What will the final solution’s mass (in kilogrammes) be?
CorrectIncorrectHint
TOTAL MASS SUGAR WATER
3 kg 0.3 kg 2.7
Water Evaporates 0 X
2 kg 0.3 kg
100% 15% 85%The mass of the solution in kg is 2 kg.

Question 6 of 40
6. Question
In a tanker, milk and water are mixed in a 7:3 ratio. This combination is drained and substituted with 3 litres of water. 2 litres of the blend is substituted with water once more. What is the final milktowater ratio in the tanker if it originally contained 10 litres of milkwater solution?
CorrectIncorrectHint
We know that Final quantity = Initial Quantity (1 – {quantity removed/total quantity})
In the first case, 3 litres are removed and replaced with water and in the second case 2 litres are removed and replaced with water. Also, we know that the total volume of the solution is 10 litres.
For milk,
Final proportion = 7/10 * (13/10)*(12/10)
Final proportion = 49/125
Therefore, ratio of milk and water = 49 : 76 
Question 7 of 40
7. Question
Some ethanol from a container having 50 litres of ethanol is drawn out and replaced with an equal amount of water. This process is repeated thrice and after that, only 25.6 litres of ethanol is left in the container. What is the amount of ethanol removed in each iteration?
CorrectIncorrectHint
Quantity left = Quantity present initially *{1Quantity removed/Total quantity}^n
where n is the number of iteration.
Let the quantity removed in each iteration be x litres
We get 25.6 = 50 * (1{50})^3
Solving this equation we can get x=10 litres 
Question 8 of 40
8. Question
When a 34 percent dairy and water solution is combined with a 51 percent milk solution, the resultant mixture has an equal amount of water and milk. Calculate the proportions in which the two solutions are combined.
CorrectIncorrectHint
Resultant mixture having equal amount of milk and water is equivalent to it having 50% milk.
50 = (34w1 + 51w2)/(w1 + w2) ?
16w1 = w2 i.e.
w1:w2 = 1:16 
Question 9 of 40
9. Question
How much Tea of variety A, costing Rs. 5 a kg should be added to 20 kg of Type B coffee at Rs. 12 a kg so that the cost of the two Tea variety mixture be worth Rs. 7 a kg?
CorrectIncorrectHint
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (127) : (75) = 5:2
Quantity of Variety A Tea that needs to be mixed ? 5:2 = x:20
? x =50 kg 
Question 10 of 40
10. Question
Water makes up 20% of a champagne and water mixture of 150 litres. How much more water should be added until the new mixture contains 25% water?
CorrectIncorrectHint
Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters
Let us Assume that another ‘P’ liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P)
Thus, (30 + P) = 25% of (150 + P)
Solving, we get P = 10 liters 
Question 11 of 40
11. Question
20 litres of milk and water in a 3:2 ratio are contained in a vessel. 10 litres of the mixture is removed and replaced with pure milk in an equal amount. Find the ratio of milk and water in the final mixture obtained if the process is repeated.
CorrectIncorrectHint
Milk = 3/5 x 20 = 12 liters, water = 8 liters
If 10 liters of mixture are removed, amount of milk removed = 6 liters and amount of water removed = 4 liters.
Remaining milk = 12 – 6 = 6 liters
Remaining water = 8 – 4 = 4 liters
10 liters of pure milk are added, therefore total milk = (6 + 10) = 16 liters.
The ratio of milk and water in the new mixture = 16:4 = 4:1
If the process is repeated one more time and 10 liters of the mixture are removed,
then amount of milk removed = 4/5 x 10 = 8 liters.
Amount of water removed = 2 liters.
Remaining milk = (16 – 8) = 8 liters.
Remaining water = (4 2) = 2 liters.
Now 10 lts milk is added => total milk = 18 lts
The required ratio of milk and water in the final mixture obtained
= (8 + 10):2 = 18:2 = 9:1. 
Question 12 of 40
12. Question
Only 8 litres of wine are in the diluted wine, with the rest being water. By replacing wine, a new mixture with a 30% concentration will be created. If the mixture originally contained 32 litres of water, how many litres should be replaced with pure wine?
CorrectIncorrectHint
Wine Water
8L 32L
1 4
20% 80%(Original ratio)
30% 70% (Required ratio)
In this case, the percentage of water being reduced when the mixture is being replaced with wine
So the ratio of left quantity to the initial quantity is 7:8
Therefore, 7/8=[1(k/40)]
>K= 5 Lit 
Question 13 of 40
13. Question
A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is
CorrectIncorrectHint
Profit (%) = 9.09 % = 1/11
Since the ratio of water and milk is 1 : 11,
Therefore the ratio of water is to mixture = 1:12
Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml 
Question 14 of 40
14. Question
Only 26% of the water in a milk and water mixture is water. The percentage of milk in the mixture increased to 76 percent after 7 litres of pure milk were added. The following is the amount of mixture:
CorrectIncorrectHint
Milk Water
76% 26% Initially
76% 24% After replacementLeft amount = Initial amount(1 replaced amount/total amount)
24 =26(17/k)
>k= 91 
Question 15 of 40
15. Question
A mixture of milk and water in the ratio of 4:5 is in a jar. If you add another 8 litres of milk to the container, it will be full, and the milktowater ratio will be 6:5. What is the bottle’s capacity?
CorrectIncorrectHint
Let the capacity of the bottle be ‘P’ litres.
Quantity of milk in the mixture before adding milk = 4/9 (P – 8)
After adding milk, quantity of milk in the mixture = 6/11 P.
6P/11 – 8 = 4/9(P – 8)
10P = 792 – 352 => P = 44.
The capacity of the bottle is 44 liters. 
Question 16 of 40
16. Question
There are two types of oil, each with a different price per litre, and each containing x and 120 litres in different vessels A and B. Now that 48 litres have been removed from both vessels, the oil from vessel A has been added to vessel B, and vice versa. As a result, the oil in both vessels now has the same price. Calculate the value of x.
CorrectIncorrectHint
For such cases when there are two solutions A and B having different and same quantity Q removed from both
and Q from A is added to B and visaversa such that now solutions have same price then,
Q=A*B/(A+B)
Now,
Here, we have Q =48
48 = [x*120]/[120+x]
48(120+x) = 120x
5760 + 48x = 120x
72x = 5760
x = 80 
Question 17 of 40
17. Question
Vanadium and aluminium are mixed in 3:4 and 1:2 ratios in two alloys. After mixing 14 kg of the first alloy with x kg of the second alloy, y kg of pure Vanadium is extracted from the alloy, resulting in a new alloy with an aluminium to Vanadium ratio of 2:1. Calculate the value of y.
CorrectIncorrect 
Question 18 of 40
18. Question
There are three types of milk, with milktowater ratios of 3:2, 4:3, and 5:4. By combining all of these milks, a new type of milk is created; determine the milktowater ratio in the final milktowater solution.
CorrectIncorrectHint
c1st type 3 : 2 (3+2=5)
2nd type 4 : 3 (4+3=7)
3rd type 5 : 4 (5+4=9)
L.C.M of 5, 7, 9= 315
Now
1st type (3 : 2)x63
2nd type (4 : 3)x45
3rd type (5 : 4)x351st type 189 : 126 (189+126=315)
2nd type 180 : 135 (180+135=315)
3rd type 175 : 140 (175+140=315)
Now, adding these three types =(189+180+175):(126+135+140)
= 544:401 
Question 19 of 40
19. Question
Chemical A costs Rs.250 per 10 gm, while chemical B costs Rs.330 per 10 gm. Rangeeta makes a mixture by combining chemical A and chemical B in a 3:5 ratio. What is the selling price (in Rs.) of 10 gm of mixture if Rangeeta sold it at a profit of 25%?
CorrectIncorrectHint
Since chemicals are in ratio 3 : 5
A : B = 3 : 5
Hence , mixture contain 37.5% chemical A and 62.5% chemical B.
Hence, 10 kg mixture contain 3.75 kg chemical A and 6.25 kgs chemical.Hence , cost of mixture = ( 3750 x 250 + 6250 x 330 ) x 1.25
10 10= Rs 3,75,000 per 10 kg
Hence cost of 10 gm mixture = Rs 375

Question 20 of 40
20. Question
Mr. White accidentally spilled dairy products into a container of honey in Mrs. White’s restaurant. To make up for the leak, he chose to modify 100 ml of the combination with 100 ml of pure honey, resulting in a 5:19 ratio of milk to honey in the mixture. How much milk was spilled into the jar if it originally contained 450 ml of pure honey?
CorrectIncorrectHint
Let the amount of milk that spilled into the jar be x ml.
Then the total volume of the mixture = (450 + x) ml
Total volume of honey in the jar after 100ml of mixture was taken out from the jar.
= [450 – (450/450+x) x 100] + 100
Also, total volume of milk in the jar in the final mixture = x – [x/(450+x)] x 100Ratio = x – x x 100
(450+ x) = 5/19
450 – 450 x 100 +100
450+xOn solving, we get , x = 150

Question 21 of 40
21. Question
Sanjay wants to convert a mixture of acid and water in the ratio 16 : 19 into a mixture of acid and water in the ratio of 19 : 16. Find the amount of acid as a percentage of initial solution that should be added to achieve the required ratio.
CorrectIncorrectHint
Initial ratio of acid and water is 16:19
Required ratio of salt and water is 19:16
Here the amount of water is not changing, so we can make the value of water same in both the cases.
Multiplying the initial ratio by 16 and required one by 19, we will get the ratios as
Initail – 256:304
Final 361:304
So we need to add 361256 =105 units of acid
So our required answer is 105/560 x 100 = 18.75% 
Question 22 of 40
22. Question
Twenty litres of a 50 percent milk solution with water and honey are mixed with ten litres of a 20% milk and water solution. The resulting solution is combined with 30 L of a honey and water solution containing 30% honey. Find the percentage of water in the final resultant mixture if the concentration of honey in the final resultant mixture is 20%.
CorrectIncorrectHint
Milk Water Honey
1st solution 2 8 0
2nd solution
10 X 10 – x
3rd solution 0 21 9
Total 12 29 + x 19 – xHoney is 20%
So, x = 7
Also, milk is 12 out of 60 which is 20% of the solution.
Therefore, water is ( 100 – 20 – 20) % = 60% 
Question 23 of 40
23. Question
A specific proportion of milk and water was combined. The profit obtained when the mixture was sold at the price of milk was Rs. 70. The loss incurred when the mixture was sold for the price of water was Rs. 56. What was the proportion of milk to water when it was mixed?
CorrectIncorrect 
Question 24 of 40
24. Question
Vessel J is filled with a 3:7 uniform mixture of two liquids A and B. Vessel J’s entire contents are divided into two bottles K and L in a 5:8 ratio. Vessel K’s entire contents are divided into two jars M and N in a 3:7 ratio. Liquid A has a volume of 45 ml in vessel M. How much liquid B was there in vessel J?
CorrectIncorrectHint
Let the volume of the mixture in the container J be 130x ml.
Therefore, Volume of mixture in containers K and L will be 50x ml and80x ml respectively.
Therefore, Volume of mixture in containers M and N will be 15x and 35x ml respectively.So, Volume of liquid A in container M = 3/10 * 15x = 45
x = 10
Total volume of mixture in container J was 1300 ml.
T 
Question 25 of 40
25. Question
Pulkit stole milk from a bottle that contained 80% of milk and he replaced what he had stolen with another milk solution having 30% milk. The bottle then contained only 60% milk. What percentage of the original milk solution did Pulkit steal?(M)
CorrectIncorrect 
Question 26 of 40
26. Question
A milkman mixes 12 liters of water with 28 liters of milk. After selling onefourth of this mixture, he adds water to replenish the quantity that he has sold. Find the current percentage of milk in the mixture.(M)
CorrectIncorrectHint
Initially total mixture is of 40ltrs.
Ratio of milk : water = 28:12 = 7:3
After one fourth is finished, left will be 30 with milk:water = 21:9
Now, 10ltrs water is added, hence new quantity of milk and water becomes = 21:19
Required percentage = (21/40)*100 = 52.5% 
Question 27 of 40
27. Question
An alloy consists of 80% Bronze and rest Brass. Bronze and Brass, themselves are alloys with Bronze having 60% Copper and the rest Zinc whereas Brass has Copper and Zinc in equal proportions. What percentage of the alloy of Bronze and Brass is Zinc?
CorrectIncorrect 
Question 28 of 40
28. Question
A medical center a saltwater solution that costs Rs. 7.62 per litre. This mixture has a salt content of 5%. Another 75 percent water mixture costs Rs. 7.82 per litre. What is the cost of 5 litres of a mixture containing 18% salt to the patient?
CorrectIncorrect 
Question 29 of 40
29. Question
A worker, working at an agricultural farm sells 75ml of pure ghee mixed with 25ml of honey to a juice seller. The juice vendor sells tea with ghee that contains 50% honey and the rest pure ghee . How much of the juice seller’s mixture comes from the worker, and how much of it is replaced with honey?
CorrectIncorrectHint
The mixture the worker sells to the juice seller has 75% pure ghee and 25% honey i.e. every 100 ml of the mixture has 75 ml ghee .
The final mixture the juice seller sells has 50% pure ghee i.e. every 100ml of the mixture has 50 ml of ghee.
Therefore, 25 ml of pure ghee for every 100 ml of the mixture has been replaced by the juice seller. Thus, the ratio of the mixture replaced = volume of ghee replaced per 100ml/total volume of ghee per 100ml = 25/75 _ 100% = 33.33% 
Question 30 of 40
30. Question
The ratio of ghee to honey in container X is 2:3. Another container Y, has a 4:7 ratio of ghee and honey. Both A and B are completely full. Both vessels are emptied into container Z, the third vessel. What is the honeytoghee ratio in container Z?
CorrectIncorrectHint
We have not been given anything about the ratio of volumes of container X and Y.
Hence we cannot determined the ratio of ghee and honey in container Z.
Note: Allegation rule can only be applied when volume of both the quantities are initially same. 
Question 31 of 40
31. Question
Aman has 3 beakers A, B, and C with him. All the beakers have 2 types of liquids – p and q. The ratio of p:q in beaker A, B, and C is 2:3, 3:4 and 5:3 respectively. Sameer takes 280 ml of liquid from each of the 3 beakers and pours them in a big vessel. How much liquid (in ml), having p and q in ratio 7:3, should he mix to the vessel so that the ratio of p:q in the vessel becomes 1:1?
CorrectIncorrectHint
We know that 280 ml of liquid is taken from each of the 3 beakers.
thus, in liquid from beaker a, quantity of p = 280*2/5 = 112 ml and q = 280112 = 168 ml.
in liquid from beaker b, p = 280*3/7 = 120 ml and q = 280120 = 160 ml
in liquid from beaker c, p = 280*5/8 = 175 ml and q = 280*3/8 = 105 ml
thus, the amount of p in the vessel = 112+120+175 = 407 ml the amount of q in vessel = 168+160+105 = 433 ml.
let the amount of liquid he adds to the mixture be x. thus, amount of p that he adds to the mixture = 0.7x and the amount of q that he adds to the mixture = 0.3x
We are given that after adding x, p=q.
Thus, 407+0.7x = 433+0.3x
26 = 0.4x
Thus, x = 65 ml.
Thus, aman needs to add 65 ml of liquid to the solution in the vessel. 
Question 32 of 40
32. Question
Alcohol is poured into a bucket to the brim. We empty half of it and then fill it with a litre of booze. We have 4 litres of alcohol in the container after performing this operation five times in a row. How much alcohol was there in the bucket at the start?
CorrectIncorrect 
Question 33 of 40
33. Question
Container J contains a homogenous mixture of two liquids A and B in the ratio 3:7. The entire contents of container J is divided into two containers K and L in the ratio 5: 8. The entire contents of container K is divided into two containers M and N in the ratio 3:7. The volume of liquid A in container M is 45 ml. What was the volume of liquid B in container J?
CorrectIncorrectHint
Let the volume of the mixture in the container J be 130x ml.
Therefore, Volume of mixture in containers K and L will be 50x ml and80x ml respectively.
Therefore, Volume of mixture in containers M and N will be 15x and 35x ml respectively.So, Volume of liquid A in container M = 3/10 * 15x = 45
x = 10
Total volume of mixture in container J was 1300 ml.
Therefore, Volume of liquid B in container J = 7/10 * 300 = 910 ml 
Question 34 of 40
34. Question
In a public library on 31st July, every person who came, read exactly 5 different books. Every book was read by exactly 60 persons. The public library had copy each of 15 different books. If the number of persons who visited the public library that day was 50% of the average number of persons who visit the public library per day in the month, then find the average number of persons who visit the public library per day.
CorrectIncorrectHint
The total number of readings = 60 x 15 =900
If every person reads 5 books each , the total number of persons who read the books = 900/5 = 180This forms 50% of the total persons, hence the number of persons who visit the public library = 360

Question 35 of 40
35. Question
David was trying to make Batasha by heating a solution of water and sugar and evaporating the water. The mass of the solution is 3 kg which contains 90% water and 10% sugar by mass. After sometime, he finds that the solution contains 85% water. What will be the mass (in kg) of the final solution?
CorrectIncorrectHint
TOTAL MASS SUGAR WATER
3 kg 0.3 kg 2.7
Water Evaporates 0 X
2 kg 0.3 kg
100% 15% 85%The mass of the solution in kg is 2 kg.

Question 36 of 40
36. Question
There are four beakers B1, B2, B3 and B4 with 2%, 4%, 6% and 8% milk solutions respectively.
From each of the four beakers, some amount of mixture is taken out and poured into an empty beaker B5. B5 now has a solution of 5% milk.
Read the two statements given below and mark the correct answer.
Statement 1. Amounts of mixtures taken from B1 and B4 are not equal.
Statement 2. Amounts of mixtures taken from B2 and B3 are equal.CorrectIncorrectHint
Since 5 is the average of 2,4,6,8 along with 2,8 and 4,6 , it can be considered that if the amounts of the mixtures taken from the B1 and B4 are equal, the amounts of mixtures taken from B2 and B3 must also be equal and vice versa.
This implies that if Statement I is false, Statement II has to be true. 
Question 37 of 40
37. Question
Aman wants to convert a mixture of acid and water in the ratio 16 : 19 into a mixture of acid and water in the ratio of 19 : 16. Find the amount of acid as a percentage of initial solution that should be added to achieve the required ratio.
CorrectIncorrectHint
nitial ratio of acid and water is 16:19
Required ratio of salt and water is 19:16
Here the amount of water is not changing, so we can make the value of water same in both the cases.
Multiplying the initial ratio by 16 and required one by 19, we will get the ratios as
Initail – 256:304
Final 361:304
So we need to add 361256 =105 units of acid
So our required answer is 105/560 x 100 = 18.75% 
Question 38 of 40
38. Question
A ten litre solution of 20% Solution A and Solution B is added to twenty litres of a 50% Solution A solution which also contains Solution B and Solution C. The resultant solution is mixed with 30 L of 30% Solution C and Solution B solution. If the concentration of Solution C in the final resultant mixture is 20%, then find the percentage of Solution B in final mixture.
CorrectIncorrectHint
Solution A Solution B Solution C
1st solution 2 8 0
2nd solution
10 X 10 – x
3rd solution 0 21 9
Total 12 29 + x 19 – xSolution C is 20%
So, x = 7
Also, Solution A is 12 out of 60 which is 20% of the solution.
Therefore, Solution B is ( 100 – 20 – 20) % = 60% 
Question 39 of 40
39. Question
Mustard and Walnut shake were mixed in a certain ratio. When the mixture was sold at the price of Mustard, the profit obtained was Rs. 70. When the mixture was sold at the price of Walnut shake, the loss incurred was Rs. 56. What is the ratio in which Mustard and Walnut shake was mixed?
CorrectIncorrect 
Question 40 of 40
40. Question
Aman went to Binoy to buy a drink and Binoy had 3 varieties, Apple Juice, Banana Juice and Cucumber Juice. One unit of Apple Juice is made by mixing 4 units of Banana Juice and 5 units of Cucumber Juice. One unit of Banana Juice is made by mixing 1 unit of Chick pea (X) and 4 units of Sambhar Masala (Y). One unit of Cucumber Juice is made by mixing 2 units of Chick pea and 6 units of Sambhar Masala . The weight of 1 unit each of Chick pea and Y is 5 kg and 3 kg respectively. What is the total quantity of Sambhar Masala required to make 1248 kg of Apple Juice?
CorrectIncorrectHint
Weight of 1 unit of Banana Juice = 1 x 5 + 4 x 3 = 17 kg
Weight of 1 unit of C = 2 x 5 + 6 x 3 = 28 kgTherefore, weight of 1 unit of Apple Juice= weight of 4 units of Banana Juice + weight of 5 units of C = 4*17 + 5*28 = 208 kg.
Therefore, 1040 kg of Apple Juicewill have 1248/208 = 6 units of Apple Juice
Therefore, B and C units = 4*6 and 5*6 respectively.
So, number of units of Y required to make 24 units of B and 30 units of C is 24*4 and 30*6 respectively.
Hence the weight of Y required = (24*4 + 30*6)*3 = 828 kgq