ALIGATIONS AND MIXTURES EXAMPLE QUESTIONS
1)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 the water is 16 : 65. How much wine the cask hold originally?
A) 18 litres B) 24 litres
C) 32 litres D) 42 litres
Answer: B) 24 litres
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)^4]litres
∴⎡⎣x(1−8x)^4/x⎤⎦ = 16/81
⇒[1−8/x]^4=(2/3)^4
⇒x=24
2)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 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?
A) 10 B) 20
C) 21 D) 25
Answer: C) 21
Explanation:
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left =(7x−7/12*9)=(7x−21/4) litres.
Quantity of B in mixture left = (5x−5/12*9)=(5x−15/4)litres.
(7x−21/4)/(5x−15/4)+9=7/9
⇒28x−21/20x+21=7/9
⇒x=3
So, the can contained 21 litres of A.
3)In what proportion water must be added to spirit to gain 20% by selling it at the cost price ?
Answer:
Let the C.P of spirit be = Rs.10 per litre
Then S.P l litre of the mixture = Rs. 11
Profit = 20%
Therefore, C.P of the mixture = Rs. [100120×10] = Rs. 25/3 per litre
Therefore, Ratio of water and spirit = 1 : 5
4)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?
A) 10 liters B) 20 liters
C) 30 liters D) 40 liters
Answer: A) 10 liters
Explanation:
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
5)From a container, 6 liters milk was drawn out and was replaced by water. again 6 liters of mixture was drawn out and was replaced by the water.Thus the quantity of milk and water in the container after these two operations is 9:16.
The quantity of mixture is:
A) 15 B) 16
C) 25 D) 31
Answer: A) 15
Explanation:
Let quantity of mixture be x liters.
Suppose a container contains x units of liquid from which y units are taken out and replaced by Water.
After operations , the quantity of pure liquid = x(1−y/x)^n units, Where n = no of operations .
So, Quantity of Milk = x(1−6/x)^2
Given that, Milk : Water = 9 : 16
=> Milk : (Milk + Water) = 9 : (9+16)
=> Milk : Mixture = 9 : 25
Therefore, x(1−6/x)^2/x=9/25
=> x = 15 liters
6)The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene,the ratio become 2:3. Then total quantity of the mixture in the container is:
A) 25 B) 30
C) 45 D) cannot be determined
Answer: B) 30
Explanation:
pool : kerosene
3 : 2(initially)
2 : 3(after replacement)
Remaining Quantity/Initial Quantity=(1−Replaced Quantity/Total Quantity)
(for petrol) 2/3=(1−10/k)
=> K = 30
Therefore the total quantity of the mixture in the container is 30liters.
7)The diluted wine contains only 8 liters of wine and the rest is water.A new mixture whose concentration is 30%, is to be formed by replacing wine.How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture ?
A) 4 B) 5
C) 8 D) None of these
Answer: B) 5
Explanation:
Wine Water
8L 32L
1 : 4
20 % 80% (original ratio)
30 % 70% (required ratio)
In ths 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
8)A container contains 50 litres of milk. From that 8 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 ?
A) 24.52 litres B) 29.63 litres
C) 28.21 litres D) 25.14 litres
Answer: B) 29.63 litres
Explanation:
Given that container has 50 litres of milk.
After replacing 8 litres of milk with water for three times, milk contained in the container is:
⇒[50(1−8/50)^3]
⇒(50×42/50×42/50×42/50) = 29.63 litres.
9)A vessel is filled with liquid, 3 parts of which are water and 5 parts of 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?
A) 1/3 B) 1/4
C) 1/5 D) 1/7
Answer: C) 1/5
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)
=> 5x + 24 = 40 - 5x
=> 10x = 16 => x = 8/5
So, part of the mixture replaced = (8/5×1/8) = 1/5.
10)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 :
A) 83.33 ml B) 90.90 ml
C) 99.09 ml D) can't be determined
Answer: A) 83.33 ml
Explanation:
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
11)From a container of wine, a thief has stolen 15 liters of wine and replaced it with same quantity of water.He again repeated the same process. Thus in three attempts the ratio of wine and water became 343:169. The initial amount of wine in the container was:
A) 75 liters B) 100 liters
C) 150 liters D) 120 liters
Answer: D) 120 liters
Explanation:
Wine (left)/Water (added)=343/169
It means Wine (left)/Wine(initial amount)=343/512 [∵ 343+169 = 512]
Thus ,
343x=512x(1−15/K)^3
⇒343/512=(7/8)^3=(1−15/k)^3
=> K = 120
Thus the initial amount of wine was 120 liters.
12)In a mixture of milk and water, there is only 26% water. After replacing the mixture with 7 liters of pure milk , the percentage of milk in the mixture become 76%. The quantity of mixture is:
A) 65 liters B) 91 liters
C) 38 liters D) None of these
Answer: B) 91 liters
Explanation:
Milk Water
74% 26% (initially)
76% 24% ( after replacement)
Left amount = Initial amount (1−replaced amount/total amount)
24 = 26(1−7/k)
=> k = 91
13)A sum of Rs.118 was divided among 50 boys and girls such that each boy received Rs.2.60 and each girl Rs.1.80.Find the number of boys and girls ?
Average money received by each = 118/50 = Rs. 2.36
Therefore, Ratio of No.of boys and girls = 56 : 24 = 7 : 3
Therefore, Number of boys = 50 x (7/10) = 35
Number of girls = 50 - 35 = 15
14)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) 80% B) 70%
C) 75% D) 62%
Answer: A) 80%
Explanation:
Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.
After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and weight of a mixture = 60 + 15 = 75 gm.
So % of water = 100 x 60/75 = 80%.
15)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.
A) 26.34 litres B) 27.36 litres
C) 28 litres D) 29.16 litres
Answer: D) 29.16 litres
Explanation:
Amount of milk left after 3 operations
(40(1−4/40)^3)=(40*9/10*9/10*9/10)=29.16 litres
16)A jar was full with honey. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar,the rest part of the jar was filled with the sugar solution. The initial amount of honey in the jar was filled with the sugar solution. The initial amount of honey in the jar was:
A) 1.25 kg B) 1 kg
C) 1.5 kg D) None of these
Answer: A) 1.25 kg
Explanation:
Let the initial amount of honey in the jar was K, then
512=K(1−1/5)^4 [∵ 20% =20/100=1/5]
or
512=K(4/5)^4
Therefore, K = 1250
Hence initially the honey in the jar= 1.25 kg
17)A man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km /hr.Find the distance travelled on foot ?
Let the Journey on foot be x km
Journey on cycle = (80 – x) km
∴x/8+80−x/16= 7
=> x = 32km
Therefore, Distance covered on foot in 4 hrs = ( 4 x 8) = 32 km
18)Equal quantities of three mixtures of milk and water are mixed in the ratio 1:2, 2:3 and 3:4. The ratio of water and milk in the mixture is ?
A) 193 : 122 B) 97 : 102
C) 115 : 201 D) 147 : 185
Answer: A) 193 : 122
Explanation:
Given the three mixtures ratio as (1:2),(2:3),(3:4)
(1+2),(2+3),(3+4)
Total content = 3,5,7
Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105
105/3 = 35 , 105/5 = 21 , 105/7 = 15
Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)
i.e (35,70), (42,63), (45,60)
So overall mixture ratio of milk and water is
35+42+45 : 70+63+60
122:193
But in the question asked the ratio of water to milk = 193 : 122
19)In a pot, 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 pot would be full and ratio of milk and water would become 6 : 5. Find the capacity of the pot ?
A) 11 lit B) 22 lit
C) 33 lit D) 44 lit
Answer: B) 22 lit
Explanation:
Let the capacity of the pot 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 pot is 44 liters.
20)From a tank of petrol , which contains 200 liters of petrol, the seller replaces each time with kerosene when he sells 40 liters of petrol(or mixture).Everytime he sells out only 40 liters of petrol (pure or impure). After replacing the petrol with kerosene 4th time,the total amount of kerosene in the mixture is
A) 81.92L B) 96L
C) 118.08L D) None of these
Answer: C) 118.08L
Explanation:
The amount of petrol left after 4 operations = 200×(1−40/200)^4=81.92
Hence the amount of kerosene = 200 - 81.92 = 118.08 liters
21)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 ?
A) 6 ml B) 11 ml
C) 15 ml D) 9 ml
Answer: A) 6 ml
Explanation:
Let us assume that the lotion has 50% alcohol and 50% water.
ratio = 1:1
As the total solution is 9ml
alcohol = water = 4.5ml
Now if we want the quantity of alcohol = 30%
The quantity of water = 70%
The new ratio = 3:7
Let x ml of water be added
We get,
4.5/4.5+x = 3/7
=> x=6
Hence 6ml of water is added.
22)4 kg of a metal contains 1/5 copper and rest in Zinc.Another 5 kg of metal contains 1/6 copper and rest in Zinc.The ratio of Copper and Zinc into the mixture of these two metals:
A) 49 : 221 B) 39:231
C) 94:181 D) None of these
Answer: A) 49 : 221
Explanation:
Copper in 4 kg = 4/5 kg and Zinc in 4 kg = 4 x (4/5) kg
Copper in 5 kg = 5/6 kg and Zinc in 5 kg = 5 x (5/6) kg
Therefore, Copper in mixture = 4/5+5/6=49/30kg
and Zinc in the mixture = 16/5+25/6=221/30kg
Therefore the required ratio = 49 : 221
23)In what ratio should two varieties of sugar at Rs.18 per kg and Rs.24 Kg be mixed together to get a mixture whose cost is Rs.20 per kg ?
Answer:
= 4 : 2 = 2 : 1
Thus, the two varieties should be mixed in the ratio 2 : 1
24)Milk and water are mixed in a vessel A as 4:1 and in vessel B as 3:2.
For vessel C, if one takes equal quantities from A and B, find the ratio of milk to water in C.
A) 1:9 B) 9:1
C) 3:7 D) 7:3
Answer: D) 7:3
Explanation:
Ratio of Milk and water in a vessel A is 4 : 1
Ratio of Milk and water in a vessel B is 3 : 2
Ratio of only milk in vessel A = 4 : 5
Ratio of only milk in vessel B = 3 : 5
Let 'x' be the quantity of milk in vessel C
Now as equal quantities are taken out from both vessels A & B
=> 4/5 : 3/5 x
3/5-x x - 4/5
=> 3/5−x/x−4/5 = 1/1 (equal quantities)
=> x = 7/10
Therefore, quantity of milk in vessel C = 7
=> Water quantity = 10 - 7 = 3
Hence the ratio of milk & water in vessel 3 is 7 : 3
25)From the 50 liters of milk, 5 liters of milk is taken out and after it 5 liters of water is added to the rest amount of milk.Again 5 liters of milk and water is drawn out and it was replaced by 5 liters of water. If this process is continued similarly for the third time, the amount of milk left after the third replacement:
A) 45L B) 36.45L
C) 40.5L D) 42.5L
Answer: B) 36.45L
Explanation:
General Formula:
Final or reduced concentration = initial concentration x (1−amount being replaced in each operation/total amount)^n
where n is the number of times the same operation is being repeated.
The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture.
Here amount being replaced denotes the quantity which is to be withdrawn in each time.
Therefore, 50×(1−5/50)^3
= 36.45 L
26)One type of liquid contains 25 % of benzene, the other contains 30% of benzene.
A can is filled with 6 parts of the first liquid and 4 parts of the second liquid.
Find the percentage of benzene in the new mixture.
A) 27 % B) 26 %
C) 29 % D) 21 %
Answer: A) 27 %
Explanation:
Let the percentage of benzene = X
(30 - X)/(X- 25) = 6/4 = 3/2
=> 5X = 135
X = 27
So, required percentage of benzene = 27 %
27)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 maximum whose cost is Rs.52 per liter ?
Answer:
Apply Allegation Method and first calculate the ratio in which they have to be mixed.
= 8 : 12 = 2 : 3
Thus, the two varieties of oil should be mixed in the ratio 2 : 3. So, if 240 liters of the second variety are taken,
then 160 liters of the first variety should be taken.
28)A milkman claims to sell milk at its cost price, still, he is making a profit of 30% since he has mixed some amount of water in the milk.What is the % of milk in the mixture?
A) 71.02% B) 76.92%
C) 63.22% D) 86.42%
Answer: B) 76.92%
Explanation:
Let the milk he bought is 1000 ml
Let C.P of 1000 ml is Rs. 100
Here let he is mixing K ml of water
He is getting 30% profit
=> Now, the selling price is also Rs. 100 for 1000 ml
=> 100 : K%
= 100 : 30
10 : 3 is ratio of milk to water
=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%
29)A mixture of 70 litres of Fruit Juice and water contains 10% water.How many litres of water should be added to the mixture so that the mixture contains 12.5% water ?
A) 2 lit B) 4 lit
C) 1 lit D) 3 lit
Answer: A) 2 lit
Explanation:
Quantity of fruit juice in the mixture = 70 - [70 x (10/100) ]= 63 litres.
After adding water, juice would form 87.5% of the mixture.
Hence, if quantity of mixture after adding x liters of water, [(87.5) /100 ]*x = 63 => x = 72
Hence 72 - 70 = 2 litres of water must be added.
30)In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3.
If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?
A) 1:1 B) 2:1
C) 3:1 D) 1:2
Answer: A) 1:1
Explanation:
Quantity of alohol in the mixture = 40 x 5/8 = 25 lit
Quantity of water = 40 - 25 = 15 lit
According to question,
Required ratio = 20 − (40 x 20/100 x 5/8)/15 − (40 x 20/100 x 3/8) + 40 x 20/100 = 20/15 − 3 + 8 = 1 : 1
31)In a 48 ltr mixture, the ratio of milk and water is 5:3.How much water should be added in the mixture so as the ratio will become 3:5 ?
A) 24 lit B) 16 lit
C) 32 lit D) 8 lit
Answer: C) 32 lit
Explanation:
Given mixture = 48 lit
Milk in it = 48 x 5/8 = 30 lit
=> Water in it = 48 - 30 = 18 lit
Let 'L' lit of water is added to make the ratio as 3:5
=> 30/(18+L) = 3/5
=> 150 = 54 + 3L
=> L = 32 lit.
32)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 ?
A) 5 lit B) 10 lit
C) 15 lit D) 20 lit
Answer: B) 10 lit
Explanation:
Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 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 x (150 + P)
120 + 4P = 150 + P => P = 10 liters.
33)From a tank of petrol, which contains 200 litres of petrol, the seller replaces each time with kerosene when he sells 40 litres of petrol(or its mixture). Everytime he sells out only 40 litres of petrol(pure or impure). After replacing the petrol with kerosen 4th
the total amount of kerosene in the mixture is
A) 81.92L B) 96L
C) 118.08L D) None of these
Answer: C) 118.08L
Explanation:
The amount of petrol left after 4 operations
= 200 × (1−40/200)^4
= 200 × (4/5)^4
= 200 × (256/625)
= 81.92 litres
Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres
34)If a man buys 1 lt of milk for Rs.12 and mixes it with 20% water and sells it for Rs.15, then what is the percentage of gain ?
A) 25% B) 30%
C) 17% D) 19%
Answer: A) 25%
Explanation:
He has gain = 15 - 12 = 3,
Gain% = (3/12) x 100 = (100/4) = 25.
He has 25% gain.
35)There are two mixtures of honey and water in which the ratio of honey and water are as 1:3 and 3:1 respectively. Two litres are drawn from first mixture and 3 litres from second mixture, are mixed to form another mixture.What is the ratio of honey and water in it ?
A) 111:108 B) 11:9
C) 103:72 D) None
Answer: B) 11:9
Explanation:
From the given data,
The part of honey in the first mixture = 1/4
The part of honey in the second mixture = 3/4
Let the part of honey in the third mixture = x
Then,
1/4 3/4
x
(3/4)-x x-(1/4)
Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3
=> 3/4−x/x−14=2/3
=> Solving we get the part of honey in the third mixture as 11/20
=> the remaining part of the mixture is water = 9/20
Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .
36)A man pays Rs. 6.40 per litre of milk. He adds water and sells the mixture at Rs. 8 per litre,thereby making 37.5% profit. The proportion of water to milk received by the customers is
A) 1 : 10 B) 10 : 1
C) 9 : 11 D) 11 : 9
Answer: A) 1 : 10
Explanation:
Customer ratio of Milk and Water is given by
Milk :: Water
6.4 0
8/1 + 3/8 =64/11
64/11 64/10 − 64/11
=> Milk : Water = 110 : 11 = 10 : 1
Therefore, the proportionate of Water to Milk for Customer is 1 : 10
37)The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
A) 8:3 B) 6:7
C) 7:5 D) 11:7
Answer: C) 7:5
Explanation:
Milk in 1-litre mixture of A = 4/7 litre.
Milk in 1-litre mixture of B = 2/5 litre.
Milk in 1-litre mixture of C = 1/2 litre.
By rule of alligation we have required ratio X:Y
X : Y
4/7 2/5
\ /
(Mean ratio)
(1/2)
/ \
(1/2 – 2/5) : (4/7 – 1/2)
1/10 1/1 4
So Required ratio = X : Y = 1/10 : 1/14 = 7:5
38)
A tin a mixture of two liquids A and B in the proportion 4 : 1.If 45 litres of the mixture is replaced by 45 litres of liquid B, then the ratio of the two liquids becomes 2 : 5.How much of the liquid B was there in the tin? What quantity does the tin hold?
A) 58 l B) 65 l
C) 50 l D) 62 l
Answer: C) 50 l
Explanation:
Let the tin contain 5x litres of liquids
(5x − 45) * 4/5/(5x − 45) * 1/5 + 45 = 2/5
=> 5(4x - 36) = 2(x + 36)
=> 20x - 180 = 2x + 72
=> x = 14 litres
Hence, the initial quantity of mixture = 70l
Quantity of liquid B
= (70 − 45) * 1/5 + 45
= 50 litres.
39)640 ml of a mixture contains milk and water in ratio 6:2. How much of the water is to be added to get a new mixture containing half milk and half water ?
A) 360 ml B) 320 ml
C) 310 ml D) 330 ml
Answer: B) 320 ml
Explanation:
Here total parts of milk and water in the solution is 6+2 = 8 parts.
1part = 640/8 = 80
old mixture contains 6parts of milk and 2 parts of water(6:2).
To get new mixture containing half milk and half water,
add 4parts of water to the old mixture then 6:(2+4) to make the ratio same.
i.e, 4 x 80 = 320 ml.
40)A jar was full with honey. A person used to draw out 20% of the honey from the jar and replaced it with sugar solution. He has repeated the same process 4 times and thus there was only 512 gm of honey left in the jar,the rest part of the jar was filled with the sugar solution. The initial amount of honey in the jar was:
A) 1.25 kg B) 1 kg
C) 1.5 kg D) None of these
Answer: A) 1.25 kg
Explanation:
Let the initial amount of honey in the jar was k, then
512 = k(1 − 1/5)^4 or 512 = k(4/5)^4
=> k = 512 × 625/256
=> k = 1250
Hence initially the honey in the jar = 1.25 kg
41)In a mixture, the ratio of juice and water is 4 : 3.By adding 6 litre water the ratio of juice and water will be 8 : 7. What is the amount of juice in the original mixture ?
A) 38 lit B) 96 lit
C) 48 lit D) 52 lit
Answer: C) 48 lit
Explanation:
Let the amount of juice and water in original mixture '4x' litre and '3x' litre respectively.
According to given data,
4x/3x+6 =8/7
28x=24x+48
28x–24x=48
4x = 48
x = 12
Amount of juice = 4x = 4×12 = 48 litre.
42)An alloy contains gold and silver in the ratio 5 : 8 and another alloy contains gold and silver in the ratio 5 : 3. If equal amount of both the alloys are melted together, then the ratio of gold and silver in the resulting alloy is ?
A) 113/108 B) 105/103
C) 108/115 D) 103/113
Answer: B) 105/103
Explanation:
As given equal amounts of alloys are melted, let it be 1 kg.
Required ratio of gold and silver = 5/13 + 5/8/8/13 + 3/8 = 105/103.
Hence, ratio of gold and silver in the resulting alloy = 105/103.
43)In what ratio must wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg so that the mixture be worth Rs. 3/kg?
A) 1:2 B) 3:2
C) 2:1 D) 2:3
Answer: C) 2:1
Explanation:
Given rate of wheat at cheap = Rs. 2.90/kg
Rate of wheat at cost = Rs. 3.20/kg
Mixture rate = Rs. 3/kg
Ratio of mixture =
2.90 3.20
3
(3.20 - 3 = 0.20) (3 - 2.90 = 0.10)
0.20 : 0.10 = 2:1
Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.
44)An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?
A) 45 gm B) 50 gm
C) 55 gm D) 60 gm
Answer: B) 50 gm
Explanation:
Initial quantity of copper =80/100 x 50 = 40 g
And that of Bronze = 50 - 40 = 10 g
Let 'p' gm of copper is added to the mixture
=> (50 + p) x 90/100 = 40 + p
=> 45 + 0.9p = 40 + p
=> p = 50 g
Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.
45)The concentration of glucose in three different mixtures (glucose and alcohol) is 1/2,3/5 and 4/5 respectively.If 2 litres, 3 litres and 1 litre are taken from these three different vessels and mixed. What is the ratio of glucose and alcohol in the new mixture?
A) 3:2 B) 4:3
C) 2:3 D) 3:4
Answer: A) 3:2
Explanation:
Concentration of glucose are in the ratio = 1/2:3/5:4/5
Quantity of glucose taken from A = 1 liter out of 2
Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit
Quantity of glucose taken from C = 0.8 lit
So, total quantity of glucose taken from A,B and C = 3.6 lit
So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit
Ratio of glucose to alcohol = 3.6/2.4 = 3:2
46)Two bottles contains mixture of milk and water.First bottle contains 64% milk and second bottle contains 26% water. In what ratio these two mixtures are mixed so that new mixture contains 68% milk?
A) 3 : 2 B) 2 : 1
C) 1 : 2 D) 2 : 3
Answer: A) 3 : 2
Explanation:
% of milk in first bottle = 64%
% of milk in second bottle = 100 - 26 = 74%
Now, ATQ
64% 74%
68%
6 4
Hence, by using allegation method,
Required ratio = 3 : 2
47)A mixture contains 25% milk and rest water.What percent of this mixture must taken out and replaced with milk so that in mixture milk and water may become equal.
A) 31.8% B) 31%
C) 33.33% D) 29.85%
Answer: C) 33.33%
Explanation:
Now, take percentage of milk and applying mixture rule
25 100
50
50 25 = 2 : 1
Hence required answer = 1/3 or 33.33%
48)In what ratio must a merchant mix two varieties of oils worth Rs. 60/kg and Rs. 65/kg,so that by selling the mixture at Rs. 68.20/kg, he may gain 10% ?
A) 2:3 B) 4:3
C) 3:4 D) 3:2
Answer: D) 3:2
Explanation:
Let he mixes the oils in the ratio = x : y
Then, the cost price of the oils = 60x + 65y
Given selling price = Rs. 68.20
=> Selling price = 68.20(x+y)
Given profit = 10% = SP - CP
=> 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y)
=> 6x + 6.5y = 8.20x + 3.20y
=>2.2x = 3.3y
=> x : y = 3 : 2
49)Ratio of water and milk in a container is 2 : 3.If 40 liter mixture removed from the container and same quantity of milk is added to it then the ratio of water to milk becomes 1 : 4.
Find the initial quantity of mixture ?
A) 75 lit B) 80 lit
C) 85 lit D) 90 lit
Answer: B) 80 lit
Explanation:
From the given data,
let the initial quantity of the mixture = 5x
Then,
2x − 16/3x − 24 + 40 = 1/4
8x − 64 = 3x + 16
5x = 80
x = 16 lit
Then the initial quantity of the mixture = 5x = 5 x 16 = 80 lit.
50)A container contains 120 lit of Diesel. From this container, 12 lit of Diesel was taken out and replaced by kerosene. This process was further repeated for two times. How much diesel is now there in the container ?
A) 88.01 lit B) 87.48 lit
C) 87.51 lit D) 87.62 lit
Answer: B) 87.48 lit
Explanation:
For these type of problems,
Quantity of Diesel remained = (q(1−p/q)^n)
Here p = 12 , q = 120
=> (120(1−12/120)^3) = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.
51)In a 100 litre of mixture the ratio of milk and water is 6:4. How much milk must be added to the mixture in order to make the ratio 3 : 1?
A) 85 B) 60
C) 55 D) 45
Answer: B) 60
Explanation:
Let M litres milk be added
=> 60 + M/40 = 3/1
=> 60 + M = 120
=> M = 60 lit.
52)Manideep purchases 30kg of barley at the rate of 11.50/kg and 20kg at the rate of 14.25/kg. He mixed the two and sold the mixture in the shop. At what price per kg should he sell the mixture to make 30% profit to him ?
A) 15.84 B) 14.92
C) 13.98 D) 16.38
Answer: D) 16.38
Explanation:
Given, Manideep purchases 30kg of barley at the rate of 11.50/kg nad 20kg at the rate of 14.25/kg.
Total cost of the mixture of barley = (30 x 11.50) + (20 x 14.25)
=> Total cost of the mixture = Rs. 630
Total kgs of the mixture = 30 + 20 = 50kg
Cost of mixture/kg = 630/50 = 12.6/kg
To make 30% of profit
=> Selling price for manideep = 12.6 + 30% x 12.6
=> Selling price for manideep = 12.6 + 3.78 = 16.38/kg.
53)Shiva purchased 280 kg of Rice at the rate of 15.60/kg and mixed it with 120 kg of rice purchased at the rate of 14.40/kg. He wants to earn a profit of Rs. 10.45 per kg by selling it. What should be the selling price of the mix per kg?
A) Rs. 22.18 B) Rs. 25.69
C) Rs. 26.94 D) Rs. 27.54
Answer: B) Rs. 25.69
Explanation:
Rate of rice of quantity 280 kg = Rs. 15.60/kg
Rate of rice of quantity 120 kg = Rs. 14.40/kg
He want to earn a profit of Rs. 10.45/kg
Rate of Mix to sell to get profit of 10.45 =
280 x 15.60 + 120 x 14.40/280 + 120 + 10.45
4368 + 1728/400 + 10.45
=> 15.24 + 10.45
= 25.69
54)A milkman has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk.If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, the profit % ?
A) 22.5% B) 25%
C) 20% D) 33.33%
Answer: B) 25%
Explanation:
As water costs free, water sold at cost price of milk gives the profit.
Required profit % = 5/20 x 100 = 5 x 5 = 25%.
55)From container A containing 54 liter of mixture of milk and water in ratio of 8 : 1 , 18 liter of the mixture is taken out and poured into container B in which ratio of milk to water is 3 : 1. If difference between total milk and total water in container B is 30 liter then find the quantity of initial mixture in container B.
A) 30 Liter B) 28 Liter
C) 32 Liter D) 36 Liter
Answer: C) 32 Liter
Explanation:
Let initially milk and water in container B is 3x liter and x liter respectively
Now, 3x + (8/9) × 18 – x – (1/9) × 18 = 30
3x + 16 – x – 2 = 30
x = 8
Initial quantity is container B = 8 (3 + 1) = 32 Liter.
56)8 litres are drawn from a flask containing milk and then filledwith water. The operation is performed 3 more times. The ratio of the quantity of milk left and total solution is 81/625.How much milk the flask initially holds?
A) 10 ltr B) 20 ltr
C) 30 ltr D) 40 ltr
Answer: B) 20 ltr
Explanation:
Let initial quantity be Q, and final quantity be F
F = Q(1 - 8/Q)^4
=> Q = 20
57)In a mixture of 35 litres, the ratio of milk and water is 4:1. If 7 litres of water is added to the mixture,the ratio of milk and water of the new mixture will be
A) 2:1 B) 1:2
C) 4:5 D) 5:4
Answer: A) 2:1
Explanation:
Ratio of milk andwater = 4:1
Quantity of water = 35/5 = 7 litres
Quantity of milk = 35 X 4/5 = 28 litres
If 7 litre of water is added, new quantity of water = 14 litre
New ratio of milk and water = 28:14 = 2:1
58)The ratio of milk to water is 5:4. If two litres of water is added, the ratio becomes 10:9, then find the new amount of water in the mixture.
A) 12 litres B) 14 litres
C) 16 litres D) 17 litres
Answer:
Explanation:
Let initial quantity of milk & water in the mixture are 5x & 4x
2 litres water is added.
Hence,
New quantity of milk & water will be 5x & (4x+2)
ATQ,
5x/(4x+2) = 10/9
x/(4x+2) = 2/9 9x = 8x + 4
x=4
Hence,
New amount of water in the mixture = 4x+2 = (4*4) + 2 = 18 litres
59)A vessel, full of water, weighs 24 kg. When the vessel is 1/4 full, it weighs 9 kg. Find the weight of empty vessel.
A) 4 kg B) 5 kg
C) 8 kg D) 3 kg
Answer: A) 4 kg
Explanation:
Let the weight of vessel be x
Weight of water = 24 – x
⇒ (24 – x)/4 + x = 9
⇒ 24 – x + 4x = 36
⇒ 3x = 12
⇒ x = 4 kg
60)Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4.
Jar B which had 20 litres of mixture of milk and water, was emptied into jar A, and as a result in jar A, the respective ratio of milk and water becomes 5: 3. What was the quantity of water in jar B?
A) 5 lit B) 3 lit
C) 8 lit D) 2 lit
Answer: A) 5 lit
Explanation:
Jar A has 36 litres of mixture of milk and water in the respective ratio
of 5 : 4 => Quantity of milk in Jar A = 5/9 x 36 = 20 litres Quantity of
water in Jar A = 36 - 20 = 16 litres Let quantity of water in Jar B = x
litres => Quantity of milk in Jar B = (20 - x) litres Acc. to ques,
=>[20 + (20-x)]/(16+x) = 5/3 => 120−3x = 80+5x => 5x +3x = 120−80 => 8x = 40
=> 5 litres.
61)In a mixture of 60 litres, the ratio of milk and water is 2:1.What amount of water must be added to make the ratio of milk and water as 1:2?
A) 42 Litres B) 56 Litres
C) 60 Litres D) 77 Litres
Answer: C) 60 Litres
Explanation:
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 – 20 = 60 litre water to be added.
62)In a container, there is 960 ltr of pure milk from which 48 ltr of milk is replaced with 48 ltr of water,again 48 ltr milk is replaced by same amount of water, as this process is done once more. Now, what is the amount of pure milk?
A) 901.54 ltr B) 821.54 ltr
C) 719.64 ltr D) 823.08 ltr
Answer: D) 823.08 ltr
Explanation:
Amount of pure milk= a(1 –b/a)n(n = 3, a = pure milk and b = amount
replaced)= 960 (1 –48/960)^3= 960(1-1/20)^3= 960 * 19/20 * 19/20*19/20
= 823.08 lt.
63)Two different kinds of rice costing Rs. 20 per kg and 35 per kg are mixed to get a mixture that costs Rs. 25 per kg. In what ratio the two kinds of rice have been mixed.
Applying the rule of mixture and Alligation:
(35-25) / (25-20) = 10/5 = 2/1
Hence, the required ratio is 2:1.
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