Formulae
1. 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.
2. Mean Price:
The cost of a unit quantity of the mixture is called the mean price.
3. Rule of Alligation:
If two ingredients are mixed, then
\( {{Quantity of cheaper} \over {Quantity of dearer}} \) = \( {{C.P. of dearer - Mean Price} \over {Mean price - C.P. of cheaper}} \)
We present as under:
∴(Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).
4. 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 = \( [x {({1 - {y \over x}})^n}] \) units.
Exercise : Alligation or Mixture MCQ Questions and Answers
1. How much coffee 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 coffee variety mixture be worth Rs. 7 a kg?
| A. |
| B. |
| C. |
| D. |
Correct Answer : D. 50 kg
Description :
As per the rule of alligation,
Quantity of Dearer: Quantity of Cheaper = (12-7) : (7-5) = 5:2
Quantity of Variety A coffee that needs to be mixed ⇒ 5:2 = x:20
⇒ x =50 kg
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2. The ratio of milk and water in three samples is 1 ∶ 3, 3 ∶ 5 and 11 ∶ 5. A mixture containing equal quantities of all three samples is made. What will be the ratio of milk and water in the new mixture?
| A. |
| B. |
| C. |
| D. |
Correct Answer : A. 7 : 9
Description :
Method 1 -
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
Method 2 -
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.
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3. A vessel full of orange juice contains 40% orange pulp. A part of juice is replaced by another juice containing 19% orange pulp and now the percentage of orange pulp is found to be 26%. What quantity of juice is replaced?
| A. |
| B. |
| C. |
| D. |
Correct Answer : A. 2:3
Description :
Concentration of orange pulp in 1st vessel = 40% Concentration of orange pulp in 2nd vessel = 19% After the mixing, Concentration of orange pulp in the mixture = 26% By rule of alligation,
Concentration of orange pulp in 1st Concentration of orange pulp in 2nd
Vessel Vessel
40% 19%
26%
26-19=7 40-26=14
Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2 i.e., 2/(1+2)=2/3 part of the juice is replaced.
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4. Cost of two types of pulses is Rs.15 and Rs, 20 per kg, respectively. If both the pulses are mixed together in the ratio 2:3, then what should be the price of mixed variety of pulses per kg?
| A. |
| B. |
| C. |
| D. |
Correct Answer : C. Rs. 18 per kg
Description :
Let the cost of mixed variety of pulse be Rs. x
As per the alligation rule,
2:3 = (20-x) : (x-15)
⇒ 2x+3x = 60+30
⇒ 5x = 90
⇒ x = 18
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5. 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. |
| B. |
| C. |
| D. |
Correct Answer : D. 10
Description :
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.
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6. Two solutions S1 and S2 contain whisky and soda in the ratio 2 : 5 and 6 : 7 respectively. In what ratio these solutions be mixed to get a new solution S3, containing whisky and soda in the ratio 5 : 8 ?
| A. |
| B. |
| C. |
| D. |
Correct Answer : A. 7:9
Description :
Let the amount taken from S1 be 7x
And amount taken from S2 be 13y
(2x + 6y)/(5x + 7y) = 5/8
16x + 48y = 25x + 35y
9x = 13y
x/y = 13/9
Actual ratios of amounts
= 7x/13y
= (7/13) * (13/9)
= 7/9
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7. 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.
| A. |
| B. |
| C. |
| D. |
Correct Answer : A. 108 litres
Description :
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
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8. 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. |
| B. |
| C. |
| D. |
Correct Answer : D. 76.92 %
Description :
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 %
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9. From a homogeneous mixture of salt and sugar containing 4 parts of salt and 5 parts of sugar, 1/4th part of the mixture is drawn off and replaced with salt. The ratio of salt and sugar in the new mixture is?
| A. |
| B. |
| C. |
| D. |
Correct Answer : B. 7 ∶ 5
Description :
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
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10. 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?
| A. |
| B. |
| C. |
| D. |
Correct Answer : D. 1/5th
Description :
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
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