Problems on Mixture and Alligation - Solved Examples
 1. 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 litres B. 29.16 litres C. 28 litres D. 28.2 litres

Explanation:

Suppose a container contains $x$ units of a liquid from which $y$ units are taken out and replaced by water. After $n$ operations, quantity of pure liquid
$=x\left(1-\dfrac{y}{x}\right)^n$ units.

milk contained by the container now
$=40\left(1-\dfrac{4}{40}\right)^3\\= 40\left(1-\dfrac{1}{10}\right)^3 \\ =40\times\dfrac{9}{10}\times\dfrac{9}{10}\times\dfrac{9}{10}\\= \dfrac{4\times9\times9\times9}{100}\\= 29.16$

 2. Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed with a third variety of tea in the ratio 1 : 1 : 2. If the mixture is worth Rs. 153 per kg, what is the price of the third variety per kg ? A. Rs.182.50 B. Rs.170.5 C. Rs.175.50 D. Rs.180

Explanation:

Tea worth Rs. 126 per kg and Rs. 135 per kg are mixed in the ratio 1 : 1
So their average price $=\dfrac{\left(126 + 135 \right)}{2} = 130.5$

Hence let's consider that the mixture is formed by mixing two varieties of tea, one at Rs. 130.50 per kg and the other at Rs. $x$ per kg in the ratio 2 : 2, i.e., 1 : 1. Now let's find out $x$.

By rule of alligation,

 Cost of 1 kg of 1st kind of tea Cost of 1 kg of 2nd kind of tea 130.50 $x$ Mean Price 153 $(x-153)$ 22.50

$(x-153):22.5=1:1\\ \Rightarrow x-153=22.50\\ \Rightarrow x=153+22.5=175.5$

 3. A milk vendor has 2 cans of milk. The first contains 25% water and the rest milk. The second contains 50% water. How much milk should he mix from each of the containers so as to get 12 litres of milk such that the ratio of water to milk is 3 : 5? A. 5litres, 7 litres B. 7litres, 4 litres C. 6litres, 6 litres D. 4litres, 8 litres

Explanation:

Solution 1

Let $x$ and $(12-x)$ litres of milk be mixed from the first and second container respectively.

Amount of milk in $x$ litres of the the first container $=.75x$
Amount of water in $x$ litres of the the first container $=.25x$

Amount of milk in $(12-x)$ litres of the the second container $=.5(12-x)$
Amount of water in $(12-x)$ litres of the the second container $=.5(12-x)$

Ratio of water to milk
$=[.25x+.5(12-x)]:[.75x+.5(12-x)]$ $=3:5$

$\Rightarrow\dfrac{\left(.25x+6-.5x\right)}{\left(.75x+6-.5x\right)}=\dfrac{3}{5}\\ \Rightarrow\dfrac{\left(6-.25x\right)}{\left(.25x+6\right)}=\dfrac{3}{5}\\ \Rightarrow 30-1.25x=.75x+18\\ \Rightarrow 2x=12\\ \Rightarrow x=6$

Since $x=6, 12-x=12-6=6$

Hence 6 and 6 litres of milk should mixed from the first and second container respectively.

Solution 2

Let cost of 1 litre milk be Rs.1
Milk in 1 litre mix in 1st can $=\dfrac{3}{4}$ litre
Cost Price(CP) of 1 litre mix in 1st can = Rs. $\dfrac{3}{4}$

Milk in 1 litre mix in 2nd can $=\dfrac{1}{2}$ litre.
Cost Price(CP) of 1 litre mix in 2nd can = Rs. $\dfrac{1}{2}$

Milk in 1 litre of the final mix $=\dfrac{5}{8}$
Cost Price(CP) of 1 litre final mix =Rs. $\dfrac{5}{8}$
=> Mean price $=\dfrac{5}{8}$

By rule of alligation,

 CP of 1 litre mix in 2nd can CP of 1 litre mix in 1st can $\dfrac{1}{2}$ $\dfrac{3}{4}$ Mean Price $\dfrac{5}{8}$ $\dfrac{3}{4}-\dfrac{5}{8}=\dfrac{1}{8}$ $\dfrac{5}{8}-\dfrac{1}{2}=\dfrac{1}{8}$

=> mix in 2nd can :mix in 1st can $=\dfrac{1}{8}:\dfrac{1}{8}=1:1$

ie, from each can, $\dfrac{1}{2} \times 12 = 6$ litre should be taken.

 4. 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 ? A. 3: 4 B. 4 : 3 C. 9 : 7 D. 7 : 9

Explanation:

Let Cost Price(CP) of 1 litre spirit be Rs.1

Quantity of spirit in 1 litre mixture from vessel A $=\dfrac{5}{7}$
Cost Price(CP) of 1 litre mixture from vessel A = Rs. $=\dfrac{5}{7}$

Quantity of spirit in 1 litre mixture from vessel B $=\dfrac{7}{13}$
Cost Price(CP) of 1 litre mixture from vessel B = Rs. $=\dfrac{7}{13}$

Quantity of spirit to be obtained in 1 litre mixture from vessel C $=\dfrac{8}{13}$
Cost Price(CP) of 1 litre mixture from vessel C(Mean Price) = Rs. $=\dfrac{8}{13}$

By rule of alligation,

 CP of 1 litre mixture from vessel A CP of 1 litre mixture from vessel B $\dfrac{5}{7}$ $\dfrac{7}{13}$ Mean Price $\dfrac{8}{13}$ $\dfrac{8}{13}-\dfrac{7}{13}=\dfrac{1}{13}$ $\dfrac{5}{7}-\dfrac{8}{13}=\dfrac{9}{91}$

=> Mixture from Vessel A : Mixture from Vessel B
$=\dfrac{1}{13}:\dfrac{9}{91}=7:9$

 5. The cost of Type 1 material is Rs. 15 per kg and Type 2 material is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then what is the price per kg of the mixed variety of material? A. Rs. 19 B. Rs. 16 C. Rs. 18 D. Rs. 17

Explanation:

Solution 1

Cost Price(CP) of Type 1 material is Rs. 15 per kg
Cost Price(CP) of Type 2 material is Rs. 20 per kg

Type 1 and Type 2 are mixed in the ratio of 2 : 3.

Hence Cost Price(CP) of the resultant mixture
$=\dfrac{(15 \times 2)+(20 \times 3)}{(2+3)}$
$=\dfrac{\left(30 +60\right)}{5} = \dfrac{90}{5} = 18$

=> Price per kg of the mixed variety of material = Rs.18

Solution 2

Cost Price(CP) of Type 1 material is Rs. 15 per kg
Cost Price(CP) of Type 2 material is Rs. 20 per kg
Let Cost Price(CP) of resultant mixture be Rs.$x$ per kg

By rule of alligation,

 CP of Type 1 material CP of Type 2 material 15 20 Mean Price $x$ $(20-x)$ $(x-15)$

=> Type 1 material : Type 2 material $=(20-x):(x-15)$

Given that Type 1 material : Type 2 material = 2 : 3
$\Rightarrow (20-x) : (x-15) = 2 : 3\\ \Rightarrow \dfrac{(20-x)}{(x-15)} = \dfrac{2}{3}\\ \Rightarrow 3(20-x) = 2(x-15)\\ \Rightarrow 60 - 3x = 2x - 30\\ \Rightarrow 90 = 5x\\ \Rightarrow x = \dfrac{90}{5} = 18$

=> price per kg of the mixed variety of material = Rs.18

 6. 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. 4 : 3 B. 3 : 4 C. 2 : 3 D. 3 : 2

Explanation:

CP of 1kg 1st kind rice = Rs.7.20
CP of 1kg 2nd kind rice = Rs.5.70
CP of 1kg mixed rice = Rs.6.30

By rule of alligation,

 CP of 1kg 1st kind rice CP of 1kg 2nd kind rice 7.2 5.7 Mean Price 6.3 6.3 - 5.7 = .6 7.2 - 6.3 = .9

Required Ratio = .6 : .9 = 6:9 = 2:3

 7. 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 did the cask originally hold? A. 30 litres B. 26 litres C. 24 litres D. 32 litres

Explanation:

Let initial quantity of wine $=x$ litre

After a total of 4 operations, quantity of wine
$=x\left(1-\dfrac{y}{x}\right)^n = x\left(1-\dfrac{8}{x}\right)^4$

Given that after a total of 4 operations, the ratio of the quantity of wine left in cask to that of water = 16 : 65

$\Rightarrow \dfrac{x\left(1-\dfrac{8}{x}\right)^4}{x} = \dfrac{16}{81}\\ \Rightarrow \left(1-\dfrac{8}{x}\right)^4 = \left(\dfrac{2}{3}\right)^4\\ \Rightarrow \left(1-\dfrac{8}{x}\right) = \dfrac{2}{3}\\ \Rightarrow \left(\dfrac{x-8}{x}\right) = \dfrac{2}{3}\\ \Rightarrow 3x-24=2x\\ \Rightarrow x=24$

 8. A jar full of whiskey 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 A. $\dfrac{4}{3}$ B. $\dfrac{3}{4}$ C. $\dfrac{3}{2}$ D. $\dfrac{2}{3}$

Explanation:

Concentration of alcohol in 1st Jar = 40%
Concentration of alcohol in 2nd Jar = 19%
After the mixing, Concentration of alcohol in the mixture = 26%

By rule of alligation,

 Concentration of alcohol in 1st Jar Concentration of alcohol in 2nd Jar 40% 19% Mean 26% 26-19=7 40-26=14

Hence ratio of 1st and 2nd quantities = 7 : 14 = 1 : 2

i.e., $\dfrac{2}{1+2}=\dfrac{2}{3}$ part of the whisky is replaced.

 9. How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per kg? A. 60 kg B. 63 kg C. 58 kg D. 56 kg

Explanation:

Selling Price(SP) of 1 kg mixture= Rs. 9.24

Profit = 10%

Cost Price(CP) of 1 kg mixture $=\dfrac{100}{(100+\text{Profit}\%)}\times \text{SP}$
$=\dfrac{100}{(100+10)}\times 9.24\\ = \dfrac{100}{110}\times 9.24=\dfrac{92.4}{11}=\text{ Rs.} 8.4$

By rule of alligation,

 CP of 1 kg sugar of 1st kind CP of 1 kg sugar of 2nd kind Rs. 9 Rs. 7 Mean Price Rs.8.4 8.4 - 7 = 1.4 9 - 8.4 = 0.6

i.e., to get a cost price of 8.4, the sugars of kind1 and kind2 should be mixed in the ratio 1.4 : 0.6 = 14 : 6 = 7 : 3

Suppose $x$ kg of kind1 sugar is mixed with 27 kg of kind2 sugar.
then $x$ : 27 = 7 : 3
$\Rightarrow 3x=27×7\\ \Rightarrow x=9×7=63$

 10. In what ratio should rice at Rs.9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the mixture be worth Rs.10 per kg ? A. 7 : 8 B. 8 : 7 C. 6 : 7 D. 7 ; 6

Explanation:

By rule of alligation,

 Cost of 1 kg rice of 1st kind Cost of 1 kg rice of 2nd kind 9.3 10.80 Mean Price 10 10.8-10 = .8 10 - 9.3 = .7

Required ratio = .8 : .7 = 8 : 7.

 11. In what ratio must tea worth Rs. 60 per kg be mixed with tea worth Rs. 65 a kg such that by selling the mixture at Rs. 68.20 a kg ,there can be a gain 10%? A. 3 : 2 B. 2 : 3 C. 4 : 3 D. 3 : 4

Explanation:

Cost Price(CP) of 1 kg mixture = Rs. 68.20

Profit = 10%

Cost Price(CP) of 1 kg mixture $=\dfrac{100}{(100+\text{Profit}\%)}\times \text{SP}$
$=\dfrac{100}{(100+10)}\times 68.20 \\= \dfrac{100}{110}\times 68.20=\dfrac{682}{11}= \text{Rs. }62$

By rule of alligation

 CP of 1 kg tea of 1st kind CP of 1 kg tea of 2nd kind 60 65 Mean Price 62 65 - 62 = 3 62 - 60 = 2

Hence required ratio = 3 : 2

 12. A container contains a mixture of two liquids P and Q in the ratio 7 : 5. When 9 litres of mixture are drawn off and the container is filled with Q, the ratio of P and Q becomes 7 : 9. How many litres of liquid P was contained in the container initially? A. 23 B. 21 C. 19 D. 17

Explanation:

Let initial quantity of P in the container be $7x$
and initial quantity of Q in the container be $5x$

Now 9 litres of mixture is drawn off from the container.
Quantity of P in 9 litres of the mixture drawn off
$=9\times \dfrac{7}{12} = \dfrac{63}{12}=\dfrac{21}{4}$
Quantity of Q in 9 litres of the mixture drawn off
$=9\times \dfrac{5}{12} = \dfrac{45}{12}=\dfrac{15}{4}$

Hence,
Quantity of P remaining in the mixture after 9 litres is drawn off
$=7x-\dfrac{21}{4}$
Quantity of Q remaining in the mixture after 9 litres is drawn off
$=5x-\dfrac{15}{4}$

Since the container is filled with Q after 9 litres of mixture is drawn off, quantity of Q in the mixture
$= 5x-\dfrac{15}{4}+9=5x+\dfrac{21}{4}$

Given that the ratio of P and Q becomes 7 : 9

$\Rightarrow \left(7x-\dfrac{21}{4}\right) : \left(5x+\dfrac{21}{4}\right) = 7 : 9\\ \Rightarrow 9\left(7x-\dfrac{21}{4}\right)=7\left(5x+\dfrac{21}{4}\right)\\ \Rightarrow 63x-\left(\dfrac{9\times 21}{4}\right)=35x+\left(\dfrac{7\times 21}{4}\right)\\ \Rightarrow 28x=\left(\dfrac{16\times 21}{4}\right)\\ \Rightarrow x=\left(\dfrac{16\times 21}{4\times 28}\right)$

Litres of P contained in the container initially
$=7x = \left(\dfrac{7 \times16\times 21}{4\times 28}\right)\\= \dfrac{16\times 21}{4\times 4}= 21$

 13. A vessel is filled with liquid, 3 parts of which are water and 5 parts are 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. $\dfrac{1}{3}$ B. $\dfrac{1}{4}$ C. $\dfrac{1}{5}$ D. $\dfrac{1}{6}$

Explanation:

Let the quantity of the liquid in the vessel = 8 litre. Then,
quantity of water in the liquid = 3 litre,
and quantity of syrup in the liquid = 5 litre.

Suppose $x$ litre of the mixture is drawn off and replaced with water. Then,
Quantity of water in the new mixture
$=3-\dfrac{3x}{8}+x$
Quantity of syrup in the new mixture
$=5-\dfrac{5x}{8}$

Given that in the new mixture, quantity of water = quantity of syrup
$\Rightarrow 3 - \dfrac{3x}{8} + x = 5 - \dfrac{5x}{8}\\ \Rightarrow \dfrac{10x}{8} = 2 \\ \Rightarrow \dfrac{5x}{4} = 2\\ \Rightarrow x = \dfrac{8}{5}$

i.e., if the quantity of the liquid is 8 litre, $\dfrac{8}{5}$ litre of the mixture needs to be drawn off and replaced with water so that the mixture may be half water and half syrup.

It means $\dfrac{1}{5}$ of the mixture needs to be drawn off and replaced with water so that the mixture may be half water and half syrup.

 14. In what ratio must water be mixed with milk costing Rs.12 per litre in order to get a mixture worth of Rs.8 per litre? A. 1 : 3 B. 2 : 2 C. 1 : 2 D. 3 : 1

Explanation:

By rule of alligation,

 Cost Price of 1 litre water Cost Price of 1 litre milk 0 12 Mean Price 8 12-8=4 8-0=8

Required Ratio = 4 : 8 = 1 : 2

 15. 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. 1 : 2 B. 2 : 1 C. 3 : 1 D. 1 : 3

Explanation:

By rule of alligation,

 Cost of 1 kg of 1st kind tea Cost of 1 kg of 2nd kind tea 62 72 Mean Price 64.5 72-64.5=7.5 64.5-62=2.5

Required Ratio = 7.5 : 2.5 = 3 : 1

 16. In what ratio must a grocer mix two varieties of pulses costing Rs.15 and Rs. 20 per kg respectively to obtain a mixture worth Rs.16.50 per kg? A. 1 : 2 B. 2 : 1 C. 3 : 7 D. 7 : 3

Explanation:

By rule of alligation,

 CP of 1 kg of 1st variety pulse CP of 1 kg of 2nd variety pulse 15 20 Mean Price 16.5 20-16.5 = 3.5 16.5-15=1.5

Required Ratio = 3.5 : 1.5 = 35 : 15 = 7 : 3

 17. A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The quantity sold at 18% profit is A. 300 B. 400 C. 600 D. 500

Explanation:

By rule of alligation,

 Profit% by selling 1st part Profit% by selling 2nd part 8 18 Net % profit 14 18-14=4 14-8=6

=> Quantity of part1 : Quantity of part2 = 4 : 6 = 2 : 3

Total quantity is given as 1000 kg. So quantity of part2 (quantity sold at 18% profit)
$=1000 \times \dfrac{3}{5}$ = 600 kg

 18. A dishonest milkman sells his milk at cost price but he mixes it with water and thereby gains 25%. What is the percentage of water in the mixture? A. 25% B. 20% C. 22% D. 24%

Explanation:

Solution 1

If a trader professes to sell his goods at cost price, but uses false weights, then
Gain% $=\left[\dfrac{\text{Error}}{(\text{True Value- Error})} \times 100 \right]\%$

Here Gain= 25%
error = quantity of water he mixes in the milk $=x$
true value = true quantity of milk = T

So the formula becomes, $25=\dfrac{x}{(T - x)} \times 100$
$\Rightarrow 1= \dfrac{x}{(T - x)}\times 4\\ \Rightarrow T - x = 4x\\ \Rightarrow T = 5x$

Percentage of water in the mixture
$=\dfrac{x}{T}\times 100 = \dfrac{x}{5x}\times 100 \\= \dfrac{1}{5}\times 100 = 20\%$

Solution 2

Let CP of 1 litre milk = Rs.1
SP of 1 litre mixture = CP of 1 litre milk = Rs.1
Gain = 25%

Hence CP of 1 litre mixture
$=\dfrac{100} {\left(100 + \text{Gain} \% \right)}\times \text{SP}\\ = \dfrac{100}{(100+25)}\times 1 = \dfrac{100}{125}= \dfrac{4}{5}$

 By rule of alligation, CP of 1 litre milk CP of 1 litre water 1 0 CP of 1 litre mixture $\dfrac{4}{5}$ $\dfrac{4}{5}-0=\dfrac{4}{5}$ $1-\dfrac{4}{5}=\dfrac{1}{5}$

=> Quantity of milk : Quantity of water $=\dfrac{4}{5}:\dfrac{1}{5}=4:1$

Hence percentage of water in the mixture $=\dfrac{1}{5}\times100=20\%$

 19. In what ratio must water be mixed with milk to gain $16\dfrac{2}{3}\%$ on selling the mixture at cost price? A. 6 : 1 B. 1 : 6 C. 1 : 4 D. 4 : 1

Explanation:

Let CP of 1 litre milk = Rs.1

SP of 1 litre mixture = CP of 1 litre milk = Rs.1

Gain $=16\dfrac{2}{3}\%=\dfrac{50}{3}\%$

CP of 1 litre mixture $=\dfrac{100}{\left(100 + \text{Gain}\% \right)}\times \text{SP}$
$=\dfrac{100}{\left(100+\dfrac{50}{3}\right)}\times 1=\dfrac{100}{\left(\dfrac{350}{3}\right)}\\=\dfrac{300}{350}=\dfrac{6}{7}$

 By rule of alligation, CP of 1 litre water CP of 1 litre milk 0 1 CP of 1 litre mixture $\dfrac{6}{7}$ $1-\dfrac{6}{7}=\dfrac{1}{7}$ $\dfrac{6}{7}-0=\dfrac{6}{7}$

Quantity of water : Quantity of milk $=\dfrac{1}{7}:\dfrac{6}{7}=1:6$

 20. In what ratio must rice at Rs.7.10 be mixed with rice at Rs.9.20 so that the mixture may be worth Rs.8 per kg? A. 5 : 4 B. 2 : 1 C. 3 : 2 D. 4 : 3

Explanation:

By rule of alligation,

 CP of 1 kg rice of 1st kind CP of 1 kg rice of 2nd kind 7.1 9.2 Mean Price 8 9.2 - 8 = 1.2 8 - 7.1 = .9

Required ratio = 1.2 : .9 = 12 : 9 = 4 : 3

 21. How many kg of rice at Rs.6.60 per kg be mixed with 56 kg of rice at Rs.9.60 per kg to get a mixture worth Rs.8.20 per kg? A. 56 kg B. 52 kg C. 44 kg D. 49 kg

Explanation:

By rule of alligation,

 Cost of 1 kg of 1st kind rice Cost of 1 kg of 2nd kind rice 6.6 9.6 Cost of 1 kg of the mixture 8.2 9.6 - 8.2 = 1.4 8.2 - 6.6 = 1.6

Quantity of 1st kind rice : Quantity of 2nd kind rice = 1.4 : 1.6 = 7 : 8
=> Quantity of 1st kind rice : 56 = 7 : 8
=> Quantity of 1st kind rice $=56 \times \dfrac{7}{8} = 49$

 22. How many litres of water must be added to 16 liters of milk and water containing 10% water to make it 20% water in it? A. 4 litre B. 2 litre C. 1 litre D. 3 litre

Explanation:

By rule of alligation,

 % Concentration of water in pure water (100) % Concentration of water in the given mixture (10) Mean % concentration (20) 20 - 10 = 10 100 - 20 = 80

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here quantity of the mixture = 16 litres
=> Quantity of water : 16 = 1 : 8
=> Quantity of water $=16 \times \dfrac{1}{8} = 2$ litre

 23. We have a 630 ml mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3? A. 70 ml B. 60 ml C. 80 ml D. 50 ml

Explanation:

concentration of water in mixture1 $=\dfrac{2}{9}$ (since the ratio of milk and water = 7:2)     ...(1)

concentration of water in pure water= 1      ...(2)

Now the above mentioned items are mixed to form mixture2 where milk and water ratio = 7 : 3
=> concentration of water in mixture2 $=\dfrac{3}{10}$

By rule of alligation,

 concentration of water in mixture1 $\left(\dfrac{2}{9}\right)$ concentration of water in pure water (1) Mean concentration $\left(\dfrac{3}{10}\right)$ $1-\dfrac{3}{10}=\dfrac{7}{10}$ $\dfrac{3}{10}-\dfrac{2}{9}=\dfrac{7}{90}$

=> Quantity of mixture1 : Quantity of water
$=\dfrac{7}{10}:\dfrac{7}{90}=\dfrac{1}{10}:\dfrac{1}{90}=1:\dfrac{1}{9}$

Given that Quantity of mixture1 = 630 ml
=> 630 : Quantity of water $=1:\dfrac{1}{9}$
=> Quantity of water $=630 \times\dfrac{1}{9}=70$ ml

 24. 3 litre of water is added to 11 litre of a solution containing 42% of alcohol in the water. The percentage of alcohol in the new mixture is A. 25% B. 20% C. 30% D. 33%

Explanation:

Solution 1

We have a 11 litre solution containing 42% of alcohol in the water.
=> Quantity of alcohol in the solution $=\dfrac{11 \times 42}{100}$

Now 3 litre of water is added to the solution.
=> Total quantity of the new solution = 11 + 3 = 14

Percentage of alcohol in the new solution $=\dfrac{\dfrac{11 \times 42}{100}}{14}\times 100$
$=\dfrac{11 \times 3}{100} = 33\%$

Solution 2

%Concentration of alcohol in pure water = 0
%Concentration of alcohol in mixture = 42
Quantity of water : Quantity of mixture = 3 : 11
Let the %concentration of alcohol in the new mixture $=x$

By rule of alligation,

 %Concentration of alcoholin pure water (0) %Concentration of alcohol in mixture(42) Mean %concentration $(x)$ $42-x$ $x-0=x$

But $(42 - x) : x = 3 : 11$
$\Rightarrow 11(42-x)=3x\\ \Rightarrow 42 \times 11 - 11x = 3x\\ \Rightarrow 14x = 42 \times 11 \\ \Rightarrow x = 3\times 11 = 33$

i.e., Percentage of alcohol in the new mixture is 33%

 25. Rs.460 was divided among 41 boys and girls such that each boy got Rs.12 and each girl got Rs.8. What is the number of boys? A. 33 B. 30 C. 36 D. 28

Explanation:

Solution 1

Assume that the number of boys = b and number of girls = g

number of boys + number of girls = 41
=> b + g = 41     ...(1)

Given that each boy got Rs.12 and each girl got Rs.8. Then the total amount is Rs.460
=> 12b + 8g = 460     ...(2)

Now we need to solve these equations to get b and g.

(1) × 8 => 8b + 8g = 8 × 41 = 328    ...(3)
(2) - (3) => 4b = 460 - 328 = 132
=> b $=\dfrac{132}{4}=33$

Solution 2

Given that amount received by a boy = Rs.12,
Amount received by a girl =Rs.8

Total amount = 460

Given that number of boys + number of girls = 41
Hence, mean amount $=\dfrac{460}{41}$

 By rule of alligation, Amount received by a boy (12) Amount received by a girl(8) Mean amount $\dfrac{460}{41}$ $\dfrac{460}{41}-8=\dfrac{132}{41}$ $12-\dfrac{460}{41}=\dfrac{32}{41}$

Number of boys : Number of girls
=$\dfrac{132}{41}:\dfrac{32}{41}=132:32\\=66:16=33:8$

Given that number of boys + number of girls = 41

Hence number of boys $=41 \times \dfrac{33}{41} = 33$

 26. A trader has 1600 kg of sugar. He sells a part at 8% profit and the rest at 12% profit. If he gains 11% on the whole , find the quantity sold at 12%. A. 1200 kg B. 1400 kg C. 1600 kg D. 800 kg

Explanation:

By rule of alligation,

 % Profit by selling part1 % Profit by selling part2 8 12 Net % Profit 11 12 - 11 = 1 11 - 8 = 3

=> Quantity of part1 : Quantity of part2 = 1 : 3

Given that total quantity = 1600 kg

Hence, quantity of part2 (quantity sold at 12% profit)
$=1600 \times \dfrac{3}{4} = 1200$

 27. In 40 litres of a mixture, the ratio of milk to water is 7:1. In order to make the ratio of milk to water as 3:1, the quantity of water that should be added to the mixture will be A. $5 \dfrac{2}{3}\text{ litre}$ B. $4 \dfrac{1}{3}\text{ litre}$ C. $6 \dfrac{2}{3}\text{ litre}$ D. $6\text{ litre}$

Explanation:

By rule of alligation,

 Concentration of waterin pure water : 1 Concentration of waterin mixture : $\dfrac{1}{8}$ Concentration of water in the final mixture : $\dfrac{1}{4}$ $\dfrac{1}{4}-\dfrac{1}{8}=\dfrac{1}{8}$ $1-\dfrac{1}{4}=\dfrac{3}{4}$

Quantity of water : Quantity of mixture $=\dfrac{1}{8}:\dfrac{3}{4}=1:6$

Given that quantity of mixture = 40 litre
=>Quantity of water : 40 = 1 : 6
=> Quantity of water $=40 \times \dfrac{1}{6} = 6\dfrac{2}{3}$ litre

 28. Some amount out of Rs.7000 was lent at 6% per annum and the remaining was lent at 4% per annum. If the total simple interest from both the fractions in 5 years was Rs.1600, the sum lent at 6% per annum was A. Rs. 2400 B. Rs. 2200 C. Rs. 2000 D. Rs. 1800

Explanation:

Total simple interest received , I = Rs.1600
Principal , p = 7000
period, n = 5 years
Rate of interest, r = ?

Simple Interest, $I=\dfrac{pnr}{100}$
$\Rightarrow 1600 = \dfrac{7000 \times 5 \times r}{100}\\ \Rightarrow r = \dfrac{1600 \times 100}{7000 \times 5} = \dfrac{160}{35} = \dfrac{32}{7}\%$

By rule of alligation,

 Rate of interest % from part1 Rate of interest % from part2 6 4 Net rate of interest % $\dfrac{32}{7}$ $\dfrac{32}{7}-4=\dfrac{4}{7}$ $6-\dfrac{32}{7}=\dfrac{10}{7}$

=> Part1 : part2 $=\dfrac{4}{7}:\dfrac{10}{7}=4:10=2:5$

Given that total amount is Rs.7000. Therefore, the amount lent at 6% per annum (part1 amount)
$=7000 \times \dfrac{2}{7} = \text{Rs. }2000$

 29. In 1 kg mixture of iron and manganese, 20% is manganese. How much iron should be added so that the proportion of manganese becomes 10% A. 1.5 kg B. 2 kg C. .5 kg D. 1 kg

Explanation:

By rule of alligation,

 Percentage concentration of manganese in the mixture : 20 Percentage concentration of manganese in pure iron : 0 Percentage concentration of manganese in the final mixture 10 10 - 0 = 10 20 - 10 = 10

=> Quantity of the mixture : Quantity of iron = 10 : 10 = 1 : 1

Given that quantity of the mixture = 1 kg

Hence quantity of iron to be added = 1 kg

 30. John bought 20 kg of wheat at the rate of Rs.8.50 per kg and 35 kg at the rate of Rs.8.75 per kg. He mixed the two. Approximately at what price per kg should he sell the mixture to make 40% profit at the cost price? A. Rs.12 B. Rs.8 C. Rs.16 D. Rs.20

Explanation:

CP $=20×8.5+35×8.75$
$=170+306.25=476.25$

Profit = 40%

$\text{SP }=\dfrac{(100+\text{Profit%})}{100}\times\text{ CP}\\= \dfrac{(100+40)}{100}\times 476.25 \\= \dfrac{140}{100}\times 476.25 \\ = \dfrac{140}{4}\times 19.05 = 35 \times 19.05$

Total quantity = 20 + 35 = 55 kg

SP per kg $=\dfrac{35 \times 19.05}{55} = \dfrac{7 \times 19.05}{11}$
$\approx \dfrac{7 \times 19}{11} \approx \dfrac{133}{11} \approx 12$

sreejothi
2015-03-26 11:45:20
in what ratio water is to added to 90% milk solution so that it becomes 60% milk solution?

Dev
2015-04-05 20:11:35
 Percentage concentration of milk in the given solution (90) Percentage concentration of milk in water (0) Percentage concentration of milk in the mixture (60) 60-0=60 90-60=30

Quantity of water : Quantity of milk solution
= 30 : 60 = 1 : 2
sonal
2015-02-17 07:36:27
a color dye is prepared by adding red and yellow colors in the ratio 2:5.If quantity of red color in the dye is increased by 25%,determine the percentage of yellow color in the new dye formed?
shubham pathak
2015-07-26 17:10:44
Suppose that red and yellow colour = 2x and 5x
red is increased by 25%so red=2x *125/100=5x/2
but steel yellow is 5x
so yellow % is=5x/(5x+5x/2)=2/3
so yellow % is=2/3*100
Subeer Goswami
2015-03-01 20:04:34
Take quantity of red and yellow components as 2 gm and 5 gm. Total is 7g

quantity of the red component becomes 2*125/100 = 2.5 gm
Total quantity becomes 7.5

Required percentage = 5*100/7.5 = 66.67%
2015-02-11 08:24:49
A milk man mixed 1 : 4 solution of milk and water with another 1 : 2 solution of milk and water in the volume of ratio 3 : 2. If the profit earned by selling the first solution was 20% and the mixture was sold at the same price, what is the profit or loss percentage? You have to assume that water comes free of cost.
a. 5.26% profit
b. 5.25% loss
c. 6.25% loss
d. None of these
Jay
2015-02-12 20:10:29
Consider 90 litre of solution1 and 60 litre of solution2

90 litre of solution1 contains 90*1/5=18 litre of milk and (90-18)=72 litre of water
60 litre of solution2 contains 60*1/3=20 litre of milk and (60-20)=40 litre of water

Mixture will have 18+20=38 litre of milk 72+40=112 litre of water
i.e., Quantity of milk : Quantity of water = 38 : 112 = 19 : 56

Take 75 litre of solution1.  Quantity of milk = 75*1/5=15 litre. Profit earned in selling solution is 20%
assume cost price of 1 litre of milk = Rs.1
Cost price of 75 litre of solution1 = Rs.15.
Selling price of 75 litre of solution1 = 15*120/100 = Rs.18

Take 75 litre of mixture.  Quantity of milk = 75*19/75=19 litre.
Cost price is Rs.19
Selling price will be same and is Rs.18
so there will be a loss
Loss = Rs.1 in selling 75 litre of the mixture
Percentage loss = 1*100/19 = 5.25% loss
navraj devkota
2015-02-06 09:05:05
A mixture of 30 litres of milk and water contains 30% of water. The new mixture is formed by adding 10 lit of water. What is the percentage of water in the new mixture?
Jay
2015-02-11 21:28:55
Quantity of water in the 30 litre mixture = 30*30/100 = 9 litre

After adding 10 litre of water, quantity of water becomes 19 litre and total quantity becomes 40 litre
percentage of water = 19*100/40 = 47.5%
Nitasha Mehmi
2015-01-30 11:35:32
One litre of water was mixed to 6 litres of sugar solution containing 4% of sugar.What is the percentage of sugar in the solution?
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