Practice Alligation and mixtures - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) How much water must be added to a cask which contains 40 litres of milk at cost price Rs. 3.5/litres so that the cost of milk reduces to Rs. 2/litre?
(a)
(b)
(c)
(d)
(e)
This question can be solved in so many different ways. But the method of alligation method is the simplest of all the methods. We will apply the alligation on price of milk, water and mixture.
∴ ratio of milk and water should be 2 : 15 = 4 : 3
∴ added water = ${40}/4$ × 3 = 30 litres
Q-2) In a mixture of 60 litres, the ratio of milk to water is 2 : 1. If the ratio of milk to water is to be 1 : 2, then amount of water to be further added is ___________.
(a)
(b)
(c)
(d)
(e)
Apply the alligation on fraction of milk in each mixture.
Ratio of mixture to water = 1 : 1
Therefore, if there is 60 liture of solution, 60 litres of water should be added.
Q-3) A mixture of certain quantity of milk with 16 litres of water is worth 90 P per litre. If pure milk be worth Rs. 1.08 per litre, how much milk is there in the mixture?
(a)
(b)
(c)
(d)
(e)
The mean value is 90 P and the price of water is 0 P.
By the Alligation Rule, Milk and water are in the ratio of 5 : 1.
∴ quantity of milk in the mixture = 5 × 16 = 80 litres.
Q-4) A person has a chemical of Rs. 25 per litre. In what ratio should water be mixed in that chemical so that after selling the mixture at Rs. 20/litre he may get a profit of 25%?
(a)
(b)
(c)
(d)
(e)
In this question, the alligation method is applicable on prices,
so we should get the average price of mixture.
SP of mixture = Rs.20/litre; profit = 25%
∴ average price = 20 × ${100}/{125}$ = Rs. 16/litre
Q-5) A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can in filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
(a)
(b)
(c)
(d)
(e)
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively.
Quantity of A mixture left = $(7x - {7/{12}} × 9) litres = (7x - {{21}/4})$
Quantity of B in mixture left = $(5x - {5/{12}} × 9)$ litres
= $(5x - {{15}/4})$ litres.
∴ ${(7x-{{21}/4})}/{(5x-{15/4}) + 9} = 7/9⇒{28x - 21}/{20x + 21} = 7/9$
⇒252x – 189 = 140x + 147
⇒112 x = 336⇒x = 3.
So, the can contained 21 litre
Q-6) How many kg of salt at 42 P per kg must a man mix with 25 kg of salf at 24 P per kg so that he may, on selling the mixture at 40 P per kg gain 25% on the outlay?
(a)
(b)
(c)
(d)
(e)
Cost price of mixture = 40 × ${100}/{125}$P = 32P per kg
By the rule of fraction
Ratio = 4 : 5
Thus, for every 5 kg of salt at 24 P, 4 kg of salt at 42P is used.
∴ the required no. of kg = 25 × $4/5$ = 20.
Q-7) 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 ___________.
(a)
(b)
(c)
(d)
(e)
We will apply alligaton on % profit. If he sells the milk at CP,
he gains 0%. But if he sells water at CP, he gains 100%.
Ratio of milk to water in the mixture should be 3 : 1
∴ % of water in mixture = $1/{3 + 1}$ × 100 = 25%
Q-8) In a mixture of milk and water the proportion of water by weight was 75%. If in 60 gm of mixture 15 gm water was added, what would be the percentage of water? (Weight in gm)
(a)
(b)
(c)
(d)
(e)
Initially water (weight) = 45 gm & milk 15 gm. After added 15 gm
water the percentage of water = ${\text"weight of water"}/{\text"total weight of mixture"}$
= ${60}/{75}$ × 100 = 80%
Q-9) Jayashree purchased 150 kg of wheat of the rate of Rs. 7 per kg. She sold 50 kg at a profit of 10%. At what rate per kg should she sell the remaining to get a profit of 20% on the total deal?
(a)
(b)
(c)
(d)
(e)
Selling price of 150 kg wheat at 20% profit
= 150 × 7 $({120}/{100})$ = Rs. 1260
Selling price of 50 kg wheat at 10% profit
= 50 × 7 $({110}/{100})$ = Rs. 385
∴ Selling price per kg of remaining 100 kg wheat
= ${1260 - 385}/100$ = Rs.8.75
Q-10) A can contains a mixture of two liquids A and B in proportion 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the proportion of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?
(a)
(b)
(c)
(d)
(e)
Apply alligation on fraction of A in each mixture.
Ratio of original mixture to B = $7/{16} : 7/{48}$ = 3 : 1
When 9 litres of B is mixed, original mixture should be $9/1$ × 3
= 27 litres.
Therefore initial quantity in can = 27 + 9 = 36 litres.
Q-11) A jar contains a mixture of two liquids A and B in the ratio 4 : 1. When 10 litres of the mixture is taken out and 10 litres of liquid B is poured into the jar, the ratio becomes 2 : 3. How many litres of liquid A was contained in the jar?
(a)
(b)
(c)
(d)
(e)
In original mixture, % of liquid B = $1/{4 + 1}$ × 100 = 20%
In the resultant mixture, % of liquid B = $3/{2 + 3}$ × 100 = 60 %
Replacement is made by the liquid B, so the % of B in second mixture = 100%
Then, by the method of Alligation :
∴ Ratio in which first and second mixtures should be added is 1 : 1. What does it imply? It simply implies that the reduced quantity of the first mixture and the quantity of mixture B which is to be added are the same.
∴ Total mixture = 10 + 10 = 20 litres
and liquid A = ${20}/5$ × 4 = 16 litres
Q-12) Gold is 19 times as heavy as water and copper 9 times. In what ratio should these metals be mixed so that the mixture may be 15 times as heavy as water?
(a)
(b)
(c)
(d)
(e)
∴ Gold : Copper = 6 : 4 = 3 : 2
Q-13) In what ratio must water be mixed with milk costing Rs. 12 per litre to obtain a mixture worth of Rs. 8 per litre?
(a)
(b)
(c)
(d)
(e)
By the rule of alligation:
Ratio of water to milk = 4 : 8 = 1 : 2
Q-14) Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.
(a)
(b)
(c)
(d)
(e)
By the rule of alligation:
∴ Required rate = 60 : 90 = 2 : 3
Q-15) A container contained 80 kg of milk. From this container, 8 kg of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container?
(a)
(b)
(c)
(d)
(e)
Amount of liquid left after n operations, when the container
originally contains x units of liquid from which y units in
taken out each time is x $({x - y}/x)^n$ units.
Thus, in the above case, amount of milk left
= 80$[{80 - 8}/80]^3$ kg = 58.32 kg
Q-16) In what ratio must tea at Rs. 62 per kg be mixed with tea at Rs. 72 per kg so that the mixture must be worth Rs. 64.50 per kg?
(a)
(b)
(c)
(d)
(e)
By the rule of alligation:
∴ Required ratio = 750 : 250 = 3 : 1.
Q-17) In what ratio should milk and water be mixed so that after selling the mixture at the cost price a profit of 2 16 % 3 is made?
(a)
(b)
(c)
(d)
(e)
Short-Cut-Method : In such questions the ratio is
water : milk = 16$2/3$ : 100 = 1 : 6
Q-18) 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 per kg?
(a)
(b)
(c)
(d)
(e)
By the rule of alligation:
∴ Required rate = 3.50 : 1.50 = 35 : 15 = 7 : 3
Q-19) A butler stores wine from a butt of sherry which contained30% of spirit and he replaced what he had stolen by wine containing only 12% of spirit. The butt was then 18% strong only. How much of the butt did he steal?
(a)
(b)
(c)
(d)
(e)
By the alligation rule, we find that wine containing 30% of spirit and wine containing 12% of spirit should be mixed in a ratio 1 : 2 to produce a mixture containing 18% of spirit.
Ratio = 6 : 12 = 1 : 2
This means that $1/3$ rd of the butt of sherry was left, i.e. to
say, the butler drew out $2/3$ rd of the butt.
∴ $2/3$ rd of the butt was stolen.
Q-20) In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?
(a)
(b)
(c)
(d)
(e)
S.P. of 1 kg of the mixture = Rs. 68.20, Gain = 10%
C.P. of 1 kg of the mixture = Rs. $({100}/{110} × 68.20)$ = Rs. 62.
By the rule of alligation, we have
∴ Required ratio = 3 : 2