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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. 29.16 litres | B. 28 litres |

C. 28.2 litres | D. 26 litres |

Discuss |

answer with explanation

Answer: Option A

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.

$=x\left(1-\dfrac{y}{x}\right)^n$ units.

milk contained by the container now

$=40\left(1-\dfrac{4}{40}\right)^3\\= 40\left(1-\dfrac{1}{10}\right)^3 \\=40×\dfrac{9}{10}×\dfrac{9}{10}×\dfrac{9}{10}\\= \dfrac{4×9×9×9}{100}\\= 29.16$

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.175.50 | B. Rs.180 |

C. Rs.182.50 | D. Rs.170.5 |

Discuss |

answer with explanation

Answer: Option A

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. 4litres, 8 litres |

C. 6litres, 6 litres | D. 7litres, 4 litres |

Discuss |

answer with explanation

Answer: Option C

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, $\dfrac{1}{2} × 12 = 6$ 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. 7 : 9 | B. 3: 4 |

C. 9 : 7 | D. 4 : 3 |

Discuss |

answer with explanation

Answer: Option A

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. 17 | B. Rs. 18 |

C. Rs. 19 | D. Rs. 16 |

Discuss |

answer with explanation

Answer: Option B

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

$=\dfrac{(15 × 2)+(20 × 3)}{(2+3)}$

$=\dfrac{\left(30 +60\right)}{5} = \dfrac{90}{5} = 18$

=> 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

Shwetha Suresh

2020-01-18 07:03:58

From a jar of wine containing $32$ litres, $4$ litres is drawn out, and the jar is filled up with water. If the same proportion of wine is further drawn out two more times, what proportion of wine to water will be there in the resulting mixture?

dev

2020-01-18 18:54:49

This is similar to question no. $1.$ After third iteration, quantity of wine

$=32\left(1-\dfrac{4}{32}\right)^3\\=32×\left(\dfrac{7}{8}\right)^3=\dfrac{10976}{512}$

Quantity of water at this point $=32-\dfrac{10976}{512}=\dfrac{5408}{512}$

Required proportion $=10976:5408$

I guess, the question was intended to find proportion of wine in the resulting mixture, after third iteration.

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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

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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?

please can anyone solve this?

please can anyone solve this?

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

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

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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%

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%

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Madhu

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

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

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

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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?

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