Cart 0
[vc_row][/vc_row][vc_column][/vc_column][vc_toggle title=”Explanation / Important formulas:” el_id=”1481454282462-60885454-0285″]Alligation: It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture of desired price.
Mean Price: The cost of a unit quantity of the mixture is called the mean price.
Rule of Alligation: If two ingredients are mixed, then:
(Quantity of cheaper/Quantity of dearer) = (C.P of dearer – Mean price/Mean price – C.P of cheaper)
We present as under:
- C.P. of a unit quantity of cheaper (c)
- C.P of unit quantity (d)
- Mean price (m)
- (Cheaper quantity) : (Dearer quantity) = (d–m) : (m – c)
Suppose a container contains x of liquid from which y units are taken out and replaced by water. After n operations, the quantity of pure liquid = [x(1-y/x)n] units
More information: https://en.wikipedia.org/wiki/Alligation[/vc_toggle][vc_toggle title=”Instructions to take Aptitude Test” el_id=”1481454372988-cd8f0f96-30f0″]
- Click on start to start taking the test.
- Click on the option (A, B, C or D) to figure out the right answer.
- You can answer multiple times till you get the right answer.
- Once you get the right answer, explanations (if any) for the same will be showcased down.
- On click of list, you get to see total no of questions, no of questions you answered and no of questions pending to answer.
- On click of question number you will go to that particular question.
- On click of END your test will end.
- On click of ‘Get Results’ you will get to see correct answer for each questions.
[/vc_toggle][vc_toggle title=”Test – Mixture and alligation” open=”true” el_id=”1481454458410-9a4a39ac-c8b7″]
Mixture and Alligation - Question and Answers
Congratulations - you have completed Mixture and Alligation - Question and Answers.
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1 |
Two vessels P and Q contain 62.5% and 87.5% of alcohol respectively. If 2 litres from vessel P is mixed with 4 litres from vessel Q, the ratio of alcohol and water in the resulting mixture is?
16:5 | |
14:5 | |
16:7 | |
19:5 |
Question 1 Explanation:
Quantity of alcohol in vessel P = 62.5/100 * 2 = 5/4 litres
Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5
Quantity of alcohol in vessel Q = 87.5/100 * 4 = 7/2 litres
Quantity of alcohol in the mixture formed = 5/4 + 7/2 = 19/4 = 4.75 litres
As 6 litres of mixture is formed, ratio of alcohol and water in the mixture formed = 4.75 : 1.25 = 19 : 5
Question 2 |
A vessel of capacity 90 litres is fully filled with pure milk. Nine litres of milk is removed from the vessel and replaced with water. Nine litres of the solution thus formed is removed and replaced with water. Find the quantity of pure milk in the final milk solution?
72 | |
72.9 | |
73.8 | |
74.7 |
Question 2 Explanation:
Let the initial quantity of milk in vessel be T litres
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively
Quantity of milk finally in the vessel is then given by [(T- y) /T]n * T
For the given problem, T = 90, y = 9 and n = 2
Hence, quantity of milk finally in the vessel
= [(90 - 9) /90]2 (90) = 72.9 litres
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively
Quantity of milk finally in the vessel is then given by [(T- y) /T]n * T
For the given problem, T = 90, y = 9 and n = 2
Hence, quantity of milk finally in the vessel
= [(90 - 9) /90]2 (90) = 72.9 litres
Question 3 |
In a mixture of milk and water, the proportion of milk by weight was 80%. If, in a 180 gm mixture, 36 gms of pure milk is added, what would be the percentage of milk in the mixture formed?
83.33% | |
100% | |
84% | |
87.5% |
Question 3 Explanation:
Percentage of milk in the mixture formed = [80/100 (180) + 36] / (180 + 36) * 100% = (144 + 36) /216 * 100% = 5/6 * 100% = 83.33%
Question 4 |
In a can, there is a mixture of milk and water in the ratio 4:5. If it is filled with an additional 8 litres of milk the can would be full and ratio of milk and water would become 6:5. Find the capacity of the can?
40 | |
44 | |
48 | |
52 |
Question 4 Explanation:
Let the capacity of the can be T litres
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44
Quantity of milk in the mixture before adding milk = 4/9 (T - 8)
After adding milk, quantity of milk in the mixture = 6/11 T
6T/11 - 8 = 4/9(T - 8)
10T = 792 - 352 => T = 44
Question 5 |
In what ratio should a variety of rice costing Rs. 6 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture costing Rs. 7.50 per kg?
5:6 | |
3:4 | |
7:8 | |
8:9 |
Question 5 Explanation:
Let us say the ratio of the quantities of cheaper and dearer varieties = x : y
By the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6
By the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6
Question 6 |
A mixture of 70 litres of milk and water contains 10% water. How many litres of water should be added to the mixture so that the mixture contains 12 1/2% water?
2 | |
8 | |
4 | |
5 |
Question 6 Explanation:
Quantity of milk in the mixture = 90/100 (70) = 63 litres
After adding water, milk would form 87 1/2% of the mixture
Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72
Hence 72 - 70 = 2 litres of water must be added
After adding water, milk would form 87 1/2% of the mixture
Hence, if quantity of mixture after adding x liters of water, (87 1/2) / 100 x = 63 => x = 72
Hence 72 - 70 = 2 litres of water must be added
Question 7 |
All the water in container A which was filled to its brim was poured into two containers B and C. The quantity of water in container B was 62.5% less than the capacity of container A. If 148 liters was now transferred from C to B, then both the containers would have equal quantities of water. What was the initial quantity of water in container A?
648 | |
888 | |
928 | |
1184 |
Question 7 Explanation:
B has 62.5% or (5/8) of the water in A. Therefore, let the quantity of water in container A(initially) be 8k
Quantity of water in B = 8k - 5k = 3k
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 litres
Quantity of water in B = 8k - 5k = 3k
Quantity of water in container C = 8k - 3k = 5k
Container: A B C
Quantity of water: 8k 3k 5k
It is given that if 148 liters was transferred from container C to container B, then both the containers would have equal quantities of water
5k - 148 = 3k + 148 => 2k = 296 => k = 148
The initial quantity of water in A = 8k = 8 * 148 = 1184 litres
Question 8 |
A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?
7 liters | |
15 liters | |
10 liters | |
9 liters |
Question 8 Explanation:
Number of liters of water in150 liters of the mixture = 20% of 150 = 20/100 * 150 = 30 liters
P liters of water added to the mixture to make water 25% of the new mixture
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P)
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters
P liters of water added to the mixture to make water 25% of the new mixture
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P)
(30 + P) = 25/100 * (150 + P)
120 + 4P = 150 + P => P = 10 liters
Question 9 |
A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. 10 liters of the mixture are removed and replaced with an equal quantity of pure milk. If the process is repeated once more, find the ratio of milk and water in the final mixture obtained?
9:1 | |
4:7 | |
7:1 | |
2:5 |
Question 9 Explanation:
Milk = 3/5 * 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 * 10 = 8 liters
Amount of water removed = 2 liters
Remaining milk = (16 - 8) = 8 liters
Remaining water = (4 -2) = 2 liters
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1
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 * 10 = 8 liters
Amount of water removed = 2 liters
Remaining milk = (16 - 8) = 8 liters
Remaining water = (4 -2) = 2 liters
The required ratio of milk and water in the final mixture obtained = (8 + 10):2 = 18:2 = 9:1
Question 10 |
In what ratio should two varieties of sugar of Rs.18 per kg and Rs.24 kg be mixed together to get a mixture whose cost is Rs.20 per kg?
1:3 | |
3:1 | |
1:2 | |
2:1 |
Question 10 Explanation:
1st variety Rs. 18/kg
Mixture cost:Rs. 20/kg = (20 - 18)
2nd variety Rs. 24/kg
Mixture cost Rs. 20/kg = (24 - 20)
=> 4:2 = 2:1
Mixture cost:Rs. 20/kg = (20 - 18)
2nd variety Rs. 24/kg
Mixture cost Rs. 20/kg = (24 - 20)
=> 4:2 = 2:1
Question 11 |
How many liters of oil at Rs.40 per liter should be mixed with 240 liters of a second variety of oil at Rs.60 per liter so as to get a mixture whose cost is Rs.52 per liter?
120 liters | |
180 liters | |
110 liters | |
160 liters |
Question 11 Explanation:
1st Variety Rs.40/ltr
Mixture cost : Rs. 52/ltr = (52 - 40)
2nd Variety Rs.60/ltr
Mixture cost : Rs. 52/ltr = (60 - 52)
= 8:12 = 2:3
The two varities of oil should be mixted in the ratio 2:3. So, if 240 liters of the 2nd variety are taken, then the 1st variety should be taken as 160 liters
Mixture cost : Rs. 52/ltr = (52 - 40)
2nd Variety Rs.60/ltr
Mixture cost : Rs. 52/ltr = (60 - 52)
= 8:12 = 2:3
The two varities of oil should be mixted in the ratio 2:3. So, if 240 liters of the 2nd variety are taken, then the 1st variety should be taken as 160 liters
Question 12 |
Two varieties of wheat - A and B costing Rs. 9 per kg and Rs. 15 per kg were mixed in the ratio 3 : 7. If 5 kg of the mixture is sold at 25% profit, find the profit made?
Rs. 13.50 | |
Rs. 14.50 | |
Rs. 15.50 | |
Rs. 16.50 |
Question 12 Explanation:
Let the quantities of A and B mixed be 3x kg and 7x kg.
Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66
Profit made in selling 5 kg of the mixture = 25/100
(cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50
Cost of 3x kg of A = 9(3x) = Rs. 27x
Cost of 7x kg of B = 15(7x) = Rs. 105x
Cost of 10x kg of the mixture = 27x + 105x = Rs. 132x
Cost of 5 kg of the mixture = 132x/10x (5) = Rs. 66
Profit made in selling 5 kg of the mixture = 25/100
(cost of 5 kg of the mixture) = 25/100 * 66 = Rs. 16.50
Question 13 |
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
1/3
| |
1/4 | |
1/5 | |
1/7 |
Question 13 Explanation:
Suppose the vessel initially contains 8 litres of liquid.
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (3-3x/8 +x) litres
Quantity of syrup in new mixture = (5-5x/8) litres
(3 - 3x/8 + x) = (5-5x/8)x
5x + 24 = 40 - 5x 10x = 16
x = 8/5 So, part of the mixture replaced = (8/5 x 1/8) = 1/5
Let x litres of this liquid be replaced with water.
Quantity of water in new mixture = (3-3x/8 +x) litres
Quantity of syrup in new mixture = (5-5x/8) litres
(3 - 3x/8 + x) = (5-5x/8)x
5x + 24 = 40 - 5x 10x = 16
x = 8/5 So, part of the mixture replaced = (8/5 x 1/8) = 1/5
Question 14 |
In what ratio should a variety of rice costing Rs. 6 per kg be mixed with another variety of rice costing Rs. 8.75 per kg to obtain a mixture costing Rs. 7.50 per kg?
5:6 | |
3:4 | |
7:8 | |
8:9 |
Question 14 Explanation:
Let us say the ratio of the quantities of cheaper and dearer varieties = x : y
By the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6
By the rule of allegation, x/y = (87.5 - 7.50) / (7.50 - 6) = 5/6
Question 15 |
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?
Rs. 169.50 | |
Rs. 170 | |
Rs. 175.50 | |
Rs. 180 |
Question 15 Explanation:
Explanation: Since first and second varieties are mixed in equal proportions.
So, their average price = Rs. (126 + 135)/2 = Rs. 130.50
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 allegation, 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
So, their average price = Rs. (126 + 135)/2 = Rs. 130.50
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 allegation, 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
Question 16 |
A can contains a mixture of two liquids A and B is the ratio 7:5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7:9. How many litres of liquid A was contained by the can initially?
10 | |
20 | |
21 | |
25 |
Question 16 Explanation:
Explanation: Suppose the can initially contains 7x and 5x of mixtures A and B respectively.
Quantity of A in mixture left = (7x - 7/12x 9)litres = (7x - 21/4) litres
Quantity of B in mixture left = (5x - 5/12x 9)litres = (5x - 15/4) litres.
(7x - 21/4) / ((5x - 15/4) + 9) = 7/9
(28x - 21) / (20x + 21) = 7/9
252x - 189 = 140x + 147 112x = 336
x = 3. So, the can contained 21 litres of A.
Quantity of A in mixture left = (7x - 7/12x 9)litres = (7x - 21/4) litres
Quantity of B in mixture left = (5x - 5/12x 9)litres = (5x - 15/4) litres.
(7x - 21/4) / ((5x - 15/4) + 9) = 7/9
(28x - 21) / (20x + 21) = 7/9
252x - 189 = 140x + 147 112x = 336
x = 3. So, the can contained 21 litres of A.
Question 17 |
In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 kg?
3 : 7 | |
5 : 7 | |
7 : 3 | |
7 : 5 |
Question 17 Explanation:
Explanation: By the rule of allegation:
Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind
Rs. 15 Mean Price
Rs. 16.50 Rs. 20
3.50 1.50
Required rate = 3.50 : 1.50 = 7 : 3
Cost of 1 kg pulses of 1st kind Cost of 1 kg pulses of 2nd kind
Rs. 15 Mean Price
Rs. 16.50 Rs. 20
3.50 1.50
Required rate = 3.50 : 1.50 = 7 : 3
Question 18 |
A dishonest milkman professes to sell his milk at cost price but he mixes it with water and thereby gains 25%. The percentage of water in the mixture is?
4% | |
6 1/4% | |
20% | |
25% |
Question 18 Explanation:
Let C.P. of 1 litre milk be Re. 1Then, S.P. of 1 litre of mixture = Re. 1, Gain = 25%.
C.P. of 1 litre mixture = Re. ((100 x1) /125 = 4/5
By the rule of alligation, we have:
C.P. of 1 litre of milk C.P. of 1 litre of water
Re. 1 4/5 Mean Price Re. 4/5 0 1/5
Ratio of milk to water = 4/5 : 1/5 = 4 :1
5 5 Hence, percentage of water in the mixture = (1x 100) % = 20%
C.P. of 1 litre mixture = Re. ((100 x1) /125 = 4/5
By the rule of alligation, we have:
C.P. of 1 litre of milk C.P. of 1 litre of water
Re. 1 4/5 Mean Price Re. 4/5 0 1/5
Ratio of milk to water = 4/5 : 1/5 = 4 :1
5 5 Hence, percentage of water in the mixture = (1x 100) % = 20%
Question 19 |
A container contains 40 litres of milk. From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
26.34 litres | |
27.36 litres | |
28 litres | |
28 litres |
Question 19 Explanation:
Amount of milk left after 3 operations = [40(1 - 4/40) cube] litres
= (40 x 9/10 x 9/10 x 9/10)
= 29.16 litres.
= (40 x 9/10 x 9/10 x 9/10)
= 29.16 litres.
Question 20 |
A jar full of whisky contains 40% 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:
1/3 | |
2/3 | |
2/5 | |
3/5 |
Question 20 Explanation:
By the rule of allegation, we have:
Strength of first jar Strength of 2nd jar
40% - 7
Mean Strength26%
19% - 14
So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
Required quantity replaced = 2/3
Strength of first jar Strength of 2nd jar
40% - 7
Mean Strength26%
19% - 14
So, ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2
Required quantity replaced = 2/3
Question 21 |
In what ratio must water be mixed with milk to gain 16 % on selling the mixture at cost price?
1:6 | |
6:1 | |
2:3 | |
4:3 |
Question 21 Explanation:
Let C.P. of 1 litre milk be Re. 1
S.P. of 1 litre of mixture = Re.1, Gain = 50/3 %
C.P. of 1 litre of mixture = (100 x 3/350 x 1) = 6/7
By the rule of allegation, we have:
C.P. of 1 litre of water C.P. of 1 litre of milk
0
1
7
Mean Price
Re. 6/7
Re. 1
6/7
Ratio of water and milk = 1:6 = 1:6.
S.P. of 1 litre of mixture = Re.1, Gain = 50/3 %
C.P. of 1 litre of mixture = (100 x 3/350 x 1) = 6/7
By the rule of allegation, we have:
C.P. of 1 litre of water C.P. of 1 litre of milk
0
1
7
Mean Price
Re. 6/7
Re. 1
6/7
Ratio of water and milk = 1:6 = 1:6.
Question 22 |
8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16:65. How much wine did the cask hold originally?
18 litres | |
24 litres | |
32 litres | |
42 litres |
Question 22 Explanation:
Let the quantity of the wine in the cask originally be x litres.
Then, quantity of wine left in cask after 4 operations = [x(1 - 8/x) to the power 4] litres
(x(1- (8/x)to the power 4/x) = 16/81
(1-8/x)to the power 4 = (2/3) to the power 4
(x-8)/x = 2/3
3x-24 = 2x
x = 24
Then, quantity of wine left in cask after 4 operations = [x(1 - 8/x) to the power 4] litres
(x(1- (8/x)to the power 4/x) = 16/81
(1-8/x)to the power 4 = (2/3) to the power 4
(x-8)/x = 2/3
3x-24 = 2x
x = 24
Question 23 |
How many litres of a 12 litre mixture containing milk and water in the ratio of 2 : 3 be replaced with pure milk so that the resultant mixture contains milk and water in equal proportion?
1.0 litres | |
1.5 litres | |
2.0 litres | |
4.0 litres |
Question 23 Explanation:
The mixture contains 40% milk and 60% water in it
That is 4.8 litres of milk and 7.2 litres of water
Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%
That is we will end up with 6 litres of milk and 6 litres of water
Water gets reduced by 1.2 litres
To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove 1.20.61.20.6 litres of the mixture = 2 litres
That is 4.8 litres of milk and 7.2 litres of water
Now we are replacing the mixture with pure milk so that the amount of milk and water in the mixture is 50% and 50%
That is we will end up with 6 litres of milk and 6 litres of water
Water gets reduced by 1.2 litres
To remove 1.2 litres of water from the original mixture containing 60% water, we need to remove 1.20.61.20.6 litres of the mixture = 2 litres
Question 24 |
A zookeeper counted the heads of the animals in a zoo and found it to be 80. When he counted the legs of the animals he found it to be 260. If the zoo had either pigeons or horses, how many horses were there in the zoo?
30 | |
40 | |
50 | |
60 |
Question 24 Explanation:
Let the number of horses = x
Then the number of pigeons = 80 − x
Each pigeon has 2 legs and each horse has 4 legs
Therefore, total number of legs = 4x + 2(80−x) = 260
⇒ 4x + 160 − 2x = 260
⇒ 2x = 100
⇒ x = 50
Then the number of pigeons = 80 − x
Each pigeon has 2 legs and each horse has 4 legs
Therefore, total number of legs = 4x + 2(80−x) = 260
⇒ 4x + 160 − 2x = 260
⇒ 2x = 100
⇒ x = 50
Question 25 |
A container contains 40 litres of milk.From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container?
26.34 litres | |
27.36 litres
| |
28 litres | |
29.16 litres |
Question 25 Explanation:
Amount of milk left after 3 operations
[40(1-)]
[40(1-(4/40)3)] = (40 x 9/10 x 9/10 x 9/10) = 29.16 litres
[40(1-(4/40)3)] = (40 x 9/10 x 9/10 x 9/10) = 29.16 litres
Question 26 |
In a mixture of petrol and kerosene, petrol is only 99 litres. If this same quantity of petrol would be presented in another mixture of petrol and kerosene where total volume would be 198 litres less than the actual mixture then the concentration of petrol In the present mixture would have been 13.33% point less than that. What is the concentration of petrol in actual mixture?
20% | |
16.66% | |
26.66% | |
8.33% |
Question 26 Explanation:
Petrol kerosene Total Volume
99 x (x+99)
Petrol kerosene Total volume 99 (x-198) (x-99)
now; [99/(x-99)*100]-[99/(x = 99)*100] = 13.33
by simplifying [9900*198/(x^2-99^2)] = 40/3
x^2-99^2 = 147015 147015 can be written as (99^2*15)
Therefore x^2 = 99^2*16
x = 396
now; [ 99/(99+396)] *100 = 20%
Petrol kerosene Total volume 99 (x-198) (x-99)
now; [99/(x-99)*100]-[99/(x = 99)*100] = 13.33
by simplifying [9900*198/(x^2-99^2)] = 40/3
x^2-99^2 = 147015 147015 can be written as (99^2*15)
Therefore x^2 = 99^2*16
x = 396
now; [ 99/(99+396)] *100 = 20%
Question 27 |
The amount of water (in ml) that should be added to reduce 9 ml lotion, containing 50% alcohol, to a lotion containing 30% alcohol, is?
1 ml | |
6 ml | |
3 ml | |
9 ml |
Question 27 Explanation:
50% of 9ml = 30% of xml
50/100*9 = 30/100*x
x = 15
since it is reduced by 9ml we have
15ml - 9ml = 6ml
50/100*9 = 30/100*x
x = 15
since it is reduced by 9ml we have
15ml - 9ml = 6ml
Question 28 |
If 729 kg of mixture contains two types of wheat in the ratio 7:2. How much more type2 is to be added to get a new mixture containing type1 and type2 in the ratio 7:3?
81 kg | |
121 kg | |
98 kg | |
102 kg |
Question 28 Explanation:
Ratio of type1 and type2 in 729 kg = 7:2
Then type1 in 729 kg of mixture = (7/9 x 729 ) kg = 567 kg
And type2 in 729 kg of mixture = 729 – 567 = 162 kg
Let X be the quantity of type2 added to new mixture with the ratio 7:3
Quantity of type2 in the new mixture = (162 + X ) kg
Then 7 / 3 = 567 / (X + 162)
7(162 + X) = 3(567)
1134 + 7X = 1701
7X = 1701 - 1134
7X = 1134
X = 567/7 = 81 kg
Quantity of type2 added to new mixture = 81 kg
Then type1 in 729 kg of mixture = (7/9 x 729 ) kg = 567 kg
And type2 in 729 kg of mixture = 729 – 567 = 162 kg
Let X be the quantity of type2 added to new mixture with the ratio 7:3
Quantity of type2 in the new mixture = (162 + X ) kg
Then 7 / 3 = 567 / (X + 162)
7(162 + X) = 3(567)
1134 + 7X = 1701
7X = 1701 - 1134
7X = 1134
X = 567/7 = 81 kg
Quantity of type2 added to new mixture = 81 kg
Question 29 |
The proportion of milk and water in 3 samples is 2:1, 3:2 and 5:3. A mixture comprising of equal quantities of all 3 samples is made. The proportion of milk and water in the mixture is?
2:1 | |
5:1 | |
99:61
| |
227:133 |
Question 29 Explanation:
Proportion of milk in 3 samples is 2/3, 3/5, 5/8
Proportion of water in 3 samples is 1/3, 2/5, 3/8
Since equal quantities are taken,
Total proportion of milk is 2/3 + 3/5 + 5/8 = 227/120
Total proportion of water is 1/3 + 2/5 + 3/8 = 133/120
Proportion of milk and water in the solution is = 227:133
Proportion of water in 3 samples is 1/3, 2/5, 3/8
Since equal quantities are taken,
Total proportion of milk is 2/3 + 3/5 + 5/8 = 227/120
Total proportion of water is 1/3 + 2/5 + 3/8 = 133/120
Proportion of milk and water in the solution is = 227:133
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 29 questions to complete.
[/vc_toggle][vc_empty_space][vc_btn title=”More Aptitude Question” color=”warning” size=”lg” align=”center” button_block=”true” link=”url:https%3A%2F%2Fjumpwhere.com%2Faptitude%2F||”]
