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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

Ajay

2013-11-04 18:26:17

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

if x:y is the ratio,to get the quantity of x,the formula is x/(x+y),and to get the quantity of y,the formula is y/(x+y).

Here, second quantity replaces the first . Hence, the 2/3 is the answer

if x:y is the ratio,to get the quantity of x,the formula is x/(x+y),and to get the quantity of y,the formula is y/(x+y).

Here, second quantity replaces the first . Hence, the 2/3 is the answer

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shashank

2013-08-31 21:06:49

ashish i will explain what is the concept behind this.

suppose there is no such restriction of the word drawn off means it will become simple alligation question, like in a mixture of milk and water which in the ratio of 7:5, 9 litre of water is added and it becomes 7:9, find the initial volume of mixture, so this is straight forward question answer is 27 litre. now apply restriction of the word remember 9 litre were drawn off initially so it will be 27+9=36 litre of mixture. and you have to find amount of p which is (36*7)/(7+5)=21 litre.

suppose there is no such restriction of the word drawn off means it will become simple alligation question, like in a mixture of milk and water which in the ratio of 7:5, 9 litre of water is added and it becomes 7:9, find the initial volume of mixture, so this is straight forward question answer is 27 litre. now apply restriction of the word remember 9 litre were drawn off initially so it will be 27+9=36 litre of mixture. and you have to find amount of p which is (36*7)/(7+5)=21 litre.

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palak

2013-06-27 17:56:42

a milkman mixes 10 litres of water to 50 litres of milk of rs.16 per litre then the cost price of mixture per litre is what?

i want answer of above question

i want answer of above question

Nayana

2013-06-30 14:34:49

Cost price of water = 0

Cost price of 10 L water = 10 * 0 = 0

Cost price of milk = 16 per litre

Cost price of 50 L milk = 16 * 50

Cost price of the mixture = 0 + (16 * 50) = (16 * 50)

cost price of mixture per litre = (16 * 50)/60 = Rs. 13.33

Cost price of 10 L water = 10 * 0 = 0

Cost price of milk = 16 per litre

Cost price of 50 L milk = 16 * 50

Cost price of the mixture = 0 + (16 * 50) = (16 * 50)

cost price of mixture per litre = (16 * 50)/60 = Rs. 13.33

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Manju

2013-05-06 04:48:41

The ratio of milk and water in a mixture is 4:3. If 14 litres of water is added to the mixture, the ratio of milk and water is 3:4. The quantity of milk in the mixture is:?

shashank

2013-08-31 20:35:44

answer is 24 litre

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prasad

2013-04-14 19:34:58

in what proportion must water be mixed with spirit to gain 16 2/3% by selling at its cost price?

shashank

2013-08-31 20:40:05

6:1 ratio

suppose initial cost price is x rs. and we are mixing the water has cost price is equal to 0 rs.

because there is a gain of 16.67% means 1/6 overall becomes 7/6 for selling price, and this will become 6/7 foe cp and now then you have all value to apply method of alligation

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Support Team, careerbless.com

2013-04-15 21:49:02

Dear Prasad, Its almost the same as question 19. Please revert for clarification

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