1.How much pure alcohol must be added to 400 ml of a solution containing 16% of alcohol to change the concentration of alcohol in the mixture to 40%
- 160 ml
- 68 ml
- 100 ml
- 128 ml
Explanation:
Solution quantity = 400ml
Let the quantity of pure alcohol to be added in 400ml be A ml.
alcohol in 400ml solution = 16 × 400/100 = 64ml.
Then,
⇒ 400 × 16/100 + A = (400 + A) × 40/100
⇒ 64 + A = 160 + 2A/5
⇒ 3A/5 = 96
⇒ A = 96 × 5/3
⇒ A = 160
Alternate Method
The ratio of solution to pure alcohol = 60 ∶ 24 = 5 ∶ 2
5 units → 400 ml
Then, 2 units → 400/5 × 2 = 160 ml
∴ 160ml pure alcohol to be added to make 40% alcohol in solution.
2. In a particular type of alloy the ratio of iron to carbon is 2 : 3. The amount of iron that should be added to 15 kg of this material to make the ratio of the contents1 : 1 is
- 2 Kg
- 3 Kg
- 6 Kg
- 9 Kg
Explanation:
Given, iron : carbon = 2 : 3
In 15 kg of this alloy, we have
amount of iron
amount of carbon
Hence, the amount of iron to be added to this mixture to make the ratio of the contents 1:1 is 3 kg of iron.- 1:2
- 3:2
- 2:5
- 1:6
Given:
Assam Tea costing Rs. 300 per kg be mixed with Darjeeling Tea costing Rs. 400 per kg.
Selling the mixture at Rs. 408 per kg.
There is gain of 20%.
Concept used:
Basic concept of ratio of distribution.
Calculation:
From the given data,
Selling price of mixture is Rs. 408 per kg
Profit% = (selling price/Cost price – 1) × 100
⇒ 20 = {(408/Cost price) – 1} × 100
⇒ 0.2 + 1 = 408/Cost price
⇒ Cost price of mixture = 408/1.2 = Rs.340 per kg
∴ Ratio of Assam Tea to the Darjeeling Tea = (400 - 340) : (340 - 300) = 60 : 40 = 3 : 2
∴ The ratio should be 3 : 2.
Alternate Method
Earned 20% on Rs. 408
Cost price of sold team is,
⇒ 100/120 × 408
⇒ Rs. 340
By allegation method,
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⇒ 60 : 40
⇒ 3 : 2
∴ The ratio is 3 : 2.
4.A mixture is made by mixing alcohol and water in the ratio 9 ∶ 7. If ‘x’ litres of alcohol and ‘3x’ litres of water is mixed in 80 litres of mixture, the new ratio becomes 13 ∶ 14. Find the quantity of new mixture.
- 88 L
- 96 L
- 100 L
- 108 L
Sum of first ratio = 9 + 7 = 16
Quantity of alcohol in 80 litres of mixture = 9/16 × 80 = 45 litres
Quantity of water in 80 litres of mixture = 7/16 × 80 = 35 litres
When ‘x’ litres of alcohol and ‘3x’ litres of water is added,
Quantity of alcohol in new mixture = 45 + x
Quantity of water in new mixture = 35 + 3x
But, ratio of new mixture = 13 ∶ 14
⇒ (45 + x)/(35 + 3x) = 13/14
⇒ 14(45 + x) = 13(35 + 3x)
⇒ 630 + 14x = 455 + 39x
⇒ 39x – 14x = 630 – 455
⇒ x = 175/25 = 7
∴ Total quantity of new mixture = 45 + x + 35 + 3x = 80 + 4(7) = 80 + 28 = 108 litres
5. 20 L of a mixture contains alcohol and water in the ratio 2 : 3. If 4 L of water is mixed in it, the percentage of alcohol in the new mixture will be
- 50
- 25
- 20
- 33.33%
Explanation:
Amount of alcohol in the mixture = 2/5 × 20 = 8 L
And amount of water in the mixture = 20 – 8 = 12 L
Given, 4 L of water is added. Hence,
Amount of water in the new mixture = 16 L
Hence, the percentage of alcohol in the new mixture =
- 7:9
- 5:7
- 15:13
- 9:11
Let ‘x’ quantity of each of three mixtures to be mixed
Quantity of milk in the new mixture = (1/4)x + (3/8)x + (11/16)x = 21x/16
Quantity of water in the new mixture = (3/4)x + (5/8)x + (5/16)x = 27x/16
∴ Required ratio = (21x/16) ∶ (27x/16) = 7 ∶ 9
Alternate Method
1 + 3 = 4, 3 + 5 = 8, 11 + 5 = 16
Let the quantity of each sample LCM(4, 8, 16) = 16
Milk in the first sample
⇒ 16 × 1/(1 + 3)
⇒ 4
Water in the first sample = 16 - 4 = 12
Milk in the second sample
⇒ 16 × 3/8
⇒ 6
Water in the second sample = 16 - 6 = 10
Milk in the third sample
⇒ 16 × 11/16
⇒ 11
Water in the second sample = 16 - 11 = 5
Ratio of milk and water in the new mixture
⇒ ( 4 + 6 + 11)/(12 + 10 + 5)
⇒ 21/27 = 7/9
∴ Required ratio is 7 : 9
7.Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively. Find the ratio in which these mixtures be mixed to obtain a new mixture in vessel C containing spirit and water in the ratio 8 : 5 ?
- 12:17
- 8:9
- 8:3
- 7:9
Given:
Two vessels A and B contain spirit and water in the ratio 5 : 2 and 7 : 6 respectively.
In a certain ratio these mixtures be mixed to obtain a new mixture in vessel C containing spirit and water in the ratio 8 : 5
Calculation:
Quantity of spirit in vessel A = 5/7
Quantity of spirit in vessel B = 7/13
Quantity of spirit in vessel C = 8/13
By rule of alligation,
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⇒ Mixture from Vessel A : Mixture from Vessel B
- 7:12
- 5:12
- 5:7
- 7:5
- 5:9
Let amount of the mixture of salt and sugar is 1 kg
Mixture contains 4 parts of salt and 5 parts of sugar
⇒ Amount of salt in the mixture = 4/9 kg
⇒ Amount of sugar in the mixture = 5/9 kg
After replacing 1/4th part with salt,
⇒ Amount of salt in the new mixture = (4/9 – 4/9 × ¼ + 1/4) kg = (16 – 4 + 9) /36 kg = 7/12 kg
⇒ Amount of sugar in the new mixture = (5/9 – 5/9 × 1/4) kg = 5/12 kg
∴ Required ratio = 7/12 ∶ 5/12 = 7 ∶ 5
Alternate Method
Let the total quantity of mixture be 36 kg [for ease of calculation - lcm of total ratio 9 and portion replated (1/4) so 4]
Amount of salt in the mixture = 36 × (4/9) = 16 kg
- Amount of sugar in the mixture = 36 - 16 = 20 kg
Amount of mixture replaced with salt = 36 × (1/4) = 9 kg
In the 9 kg, amount of salt = 9 × (4/9) = 4 kg
In the 9 kg, amount of sugar = 9 - 4 = 5 kg
Now, total amount of salt = 16 - 4 + 9 = 21 kg
Amount of sugar in the final mixture = 20 - 5 = 15 kg
Final ratio of salt to sugar = 21 ∶ 15 = 7 ∶ 5.
9.Two litres of a mixture of spirit and water contains 12% water. This is added to 3 litres of another mixture of spirit and water containing 5% of water. Find the percentage of water in the resulting mixture.
- %
- %
- %
- %
Given, two litres of a mixture of spirit and water contains 12% water.
Volume of water = 12% of 2 = 0.24 litres
Now, this is added to 3 litres of another mixture of spirit and water containing 5% of water.
Total volume of water = 0.24 + 5% of 3 = 0.39 litres
∴ % of water in the mixture
10.A mug contained 90 liter alcohol. From this mug 9 liter of alcohol was taken out and replaced by water. This process was repeated once more times. How much alcohol is now contained in the mug?
- 72.9 Liters
- 73.9 Liters
- 70 Liters
- 74.6 Liters
Given:
The total quantity of alcohol = 90 liter
Quantity of alcohol taken out = 9 liter
Number of times operation repeated = 2
Concept Used:
Amount of alcohol after nth operation = a × {1 – (b/a)}n
Where,
a = quantity of alcohol
b = quantity taken out in each operation
n = number of times operation repeated
Calculation:
Amount of alcohol after the 2nd operation
⇒ a × {1 – (b/a)}n
⇒ 90 × {1 - (9/90)}2
⇒ 90 × {(90 - 9)/90}2
⇒ 90 × (81/90) × (81/90)
⇒ 72.9 litres
∴ The alcohol contained in the mug is 72.9 litres.
Initial quantity of alcohol = 90
After first process remaining alcohol = 90 × (90 - 9)/90 = 81
After second process remaining alcohol = 81 × (90 - 9)/90 = 72.9
∴ The alcohol contained in the mug is 72.9 litres.
11.A vendor sells potatoes at a cost price, but he mixes some rotten potatoes and thereby gains 25%. The percentage of rotten potatoes in the mixture is?
- 25%
- 8%
- 22%
- 20%
Let cost price of 1 kg of pure potato in Rs. be ‘x’
Let quantity of pure potato in the mixture be ‘m’ kg
⇒ Cost price of 1 kg of the mixture in Rs. = mx
∵ selling price of the mixture = cost price of pure potatoes
⇒ (Selling price – Cost price) /Cost price × 100% = 25%
⇒ (x – mx) /mx = 0.25
⇒ 1 – m = 0.25m
⇒ m = 0.80
⇒ Amount of rotten potatoes in the mixture = (1 – 0.80) kg = 0.20 kg
∴ Required percentage = 0.20/1 × 100% = 20%
Alternate Method:
let the cost price of 100 kg potato is Rs. 100
according to question,
The selling price of 100 kg potato is Rs.100
But profit percent = 25%
So, cost price × 125% = 100
So, cost price = Rs. 80
And, we know that in Rs. 80 we can buy only 80 kg potato
So, The quantity of rotten potato is 100 - 80 = 20 kg
So, the percentage of rotten potato is 20/100 × 100%
therefore, the required percentage is 20%
12.Two equal glasses filled with mixture of milk and water in the proportions of 2 : 1 and 1 : 1 respectively are emptied into a third glass. What is the proportion of milk and water in the third glass?
- 3:7
- 5:7
- 7:5
- 4:7
Given:
Two equal glasses filled with mixture of milk and water in the proportions of 2 : 1 and 1 : 1 respectively are emptied into a third glass.
Calculation:
Let the volume of equal glasses be ‘a’.
Volume of milk in first glass = 2a/3
Volume of water in first glass = a/3
Volume of milk in second glass = a/2
Volume of water in second glass = a/2
Ratio of milk to water in the third glass
Therefore the correct answer is 7 : 5.
13.6 kg sugar costing Rs. 10/kg is mixed with 4 kg sugar costing Rs. 15/kg. What is the average cost of the mixture per kilogram?
- 12 Rs
- 15 Rs
- 10 Rs
- 13 Rs
Quantity of sugar = 6 kg
Price of sugar per kg = Rs. 10
∴ Total cost = 10 × 6 = Rs. 60
Another quantity of sugar = 4 kg
Price of sugar per kg = Rs. 15
∴ Total cost = 15 × 4 = Rs. 60
⇒ Total cost of 10 kg of sugar = 60 + 60 = Rs. 120
14.According to an instruction a mixture of colour and Turpentine containing half part of each is perfect for painting a wall. A painter is provided a mixture, 3 parts of which are colour and 5 parts are Turpentine. How much of the mixture drawn off and replaced with colour so that the mixture becomes perfect?
- 4/5th part
- 2/3rd part
- 1/5th part
- 5/8th part
Let total amount of the mixture be 1 litre
3 parts of the mixture are colour and 5 parts are Turpentine
⇒ Amount of colour in the mixture = 3/8 litre
⇒ Amount of turpentine in the mixture = 5/8 litre
Let amount of mixture removed by the painter be ‘x’ litre
⇒ Amount of colour in the new mixture = (3/8 – 3x/8 + x) litre
⇒ Amount of turpentine in the new mixture = (5/8 – 5x/8) litre
The mixture will be perfect if, Amount of colour in the new mixture = Amount of turpentine in the new mixture
⇒ 3/8 – 3x/8 + x = 5/8 – 5x/8
⇒ 3 – 3x + 8x = 5 – 5x
⇒ 10x = 2
⇒ x = 1/5
∴ 1/5th part of the mixture should be replaced- 2:1
- 19:9
- 20:11
- 21:10
Given:
The ratio of Iron, Silver and Bronze = 7 : 2 : 3
The ratio of other alloy Silver and Bronze = 1 : 2
Calculation:
Let the first alloy be M kg
And N kg of second alloy is taken.
Amount of Iron in first alloy = 7M/(7 + 2 + 3) kg = 7M/12 kg
Amount of Silver in first alloy = 2M/(7 + 2 + 3) kg = 2M/12 kg
Amount of Silver in second alloy = N/3 kg
According to the question:
Ratio of Iron and Silver in mixture is 5 : 3.
⇒ (7M/12)/(2M/12 + N/3) = 5/3
⇒ (7M/12)/[(2M + 4N)/12)]
⇒ [(7M/(2M + 4N)] = 5/3
⇒ 21M = 10M + 20N
⇒ 21M – 10M = 20N
⇒ 11M = 20N
⇒ M/N = 20/11
∴ The ratio will be 20 : 11.- 46
- 42
- 32
- 38
Given:
Wheat costing Rs. 30/kg, Rs. 35/kg and a third variety of wheat are mixed in the ratio of 3 ∶ 4 ∶ 2.
The mixture costs Rs. 34/kg.
Concept :
If N is divided into two part in the ratio a : b, then
First part = N × a/(a + b)
Second part = N × b/(a + b)
Calculation:
Let the quantity of three varieties of wheat be 3x, 4x and 2x and cost of third variety of wheat be Rs. a/kg
According to the question,
⇒ (3x × 30) + (4x × 35) + (2x × a) = 34 × (3x + 4x + 2x)
⇒ 90x + 140x + 2xa = 306x
⇒ 2xa = 76x
⇒ a = 38
∴ The cost of the third variety of wheat is Rs. 38/kg- 6 : 25 : 41
- 25 : 41 : 6
- 25 : 6 : 41
- 4 : 5 : 7
In first alloy, ratio of iron and zinc is 4 : 5
In second alloy, ratio of iron, copper and zinc is 3 : 2 : 7
Suppose, amount T of both alloys is taken.
⇒ Amount of iron in first alloy = (4/ (4 + 5)) × T = (4/9) T
Amount of zinc in first alloy = (5/ (4 + 5)) × T = (5/9) T
Amount of iron in Second alloy = (3/ (3 + 2 + 7)) × T = (1/4) T
Amount of Copper in second alloy = (2/ (3 + 2 + 7)) × T = (1/6) T
Amount of zinc in second alloy = (7/ (3 + 2 + 7)) × T = (7/12) T
After mixing
Total amount of Iron = (4/9) T + (1/4) T = (25/36) T
Total amount of copper = (1/6) T = (6/36) T
Total amount of zinc = (5/9) T + (7/12) T = (41/36) T
⇒ Ratio of Iron, Copper and Zinc = (25/36) T : (6/36) T : (41/36) T
= 25 : 6 : 41- 106 : 69
- 103 : 72
- 89 : 86
- 11:9
Given:
The ratio of honey and water in the two mixtures are 1 : 3 and 3 : 1
2 litres are drawn from the first mixture
3 litres are drawn from the second mixture
Calculation:
The percentage of honey in the first mixture = 1/4 × 100 = 25%
The percentage of honey in the second mixture = 3/4 × 100 = 75%
According to the question
Percentage of honey in the new mixture
Then, the ratio of honey and water in the new mixture would be 55 : (100 – 55)
⇒ 55 : 45 = 11 : 9
∴ The ratio of honey and water in it is 11 : 9.
19.Type 1 rice of Rs. 56 per kg and type 2 of Rs. 72 per kg are mixed and their average price is Rs. 68 per kg. If 60 kg of type 2 was present, what is the amount of type 1 rice?
- 16 Kg
- 20Kg
- 24 Kg
- 26 Kg
GIVEN :
Type 1 rice = Rs. 56 per kg
Type 2 rice = Rs. 72 per kg
Price of mixture = Rs. 68 per kg
CONCEPT :
Alligation Method
CALCULATION :
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= 4 ∶ 12 = 1 ∶ 3
Type 2 = 60 kg
Type 1 = x kg
∴ x ∶ 60 = 1 ∶ 3
∴ 3x = 60
∴ x = 20 kg
20.There are two vessels A and B. Vessel A is containing 60L of Pure milk and vessel B is containing 28L of pure water. From vessel A, 8L of milk is taken out and poured into vessel B. Then, 6L of mixture (milk and water) is taken out from vessel B and poured into vessel A. What is the ratio of the quantity milk in vessel A to the quantity of pure water in vessel B?
- 14:9
- 16:7
- 24:13
- 5:6
GIVEN :
There are two vessels A and B. Vessel A is containing 60L of Pure milk and vessel B is containing 28L of pure water.
From vessel A, 8L of milk is taken out and poured into vessel B. Then, 6L of mixture (milk and water) is taken out from vessel B and poured into vessel A.
CALCULATION :
Initial quantity of milk in vessel A = 60L
Initial quantity of water in vessel B = 28L
Now, 8L milk of is poured to vessel B
Quantity of milk left in vessel A = 60 - 8 = 52 L
Quantity of milk in vessel B = 8L
Quantity of water in vessel B = 28L
(Milk ∶ water) in vessel B = 8 ∶ 28 = 2 ∶ 7
Now, 6L of mixture (milk and water) is taken out from vessel B and poured into vessel A.
Quantity of milk in 6L = 6 × 2/9 = 4/3
Quantity of water in 6L = 6 × 7/9 = 14/3
Now, Quantity of milk in vessel A = 52 + 4/3 = 160/3
Quantity of water in vessel B = 28 - 14/3 = 70/3
∴ Required Ratio = 160/3 ∶ 70/3 = 16 ∶ 7
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