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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

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%

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%

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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?

Sundeep

2015-10-13 14:50:30

Answer is 3.4%

first 4% of 6 is 6/25

After the mixture 6+1 = 7

Concentration of sugar is same

let x be the percent we have to find

so, x% of 7 is 6/25

After solving this equation you will get 3.4 %

first 4% of 6 is 6/25

After the mixture 6+1 = 7

Concentration of sugar is same

let x be the percent we have to find

so, x% of 7 is 6/25

After solving this equation you will get 3.4 %

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viksa bansal

2015-08-30 20:45:01

Suppose 6000ml and sugar is 4%.

So sugar is 240 out of new mixture 7000

so $\dfrac{240}{7000}×100=3.42\%$

So sugar is 240 out of new mixture 7000

so $\dfrac{240}{7000}×100=3.42\%$

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Raj

2015-02-09 21:36:16

6 litres of sugar solution contains $\dfrac{6×4}{100}=0.24$ litre of sugar

quantity of water = $\dfrac{6×96}{100}=5.76$

After adding one litre of water, its quantity becomes 6.76

Required percentage of sugar = $\dfrac{0.24×100}{0.24+ 6.76}=\dfrac{24}{7}=3.42\%$

quantity of water = $\dfrac{6×96}{100}=5.76$

After adding one litre of water, its quantity becomes 6.76

Required percentage of sugar = $\dfrac{0.24×100}{0.24+ 6.76}=\dfrac{24}{7}=3.42\%$

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Blanco

2015-01-26 17:41:17

Two vessels contain mixture of milk and water in the ratio 5: 1 and 9 :1. They are mixed together in the ratio 1 : 5. Find the ratio of milk and water in the final mixture.

Raj

2015-02-09 20:06:28

Concentration of milk in these vessels will be 5/6 and 9/10.

quantity of milk in 1 litre of first vessel = 1*5/6 = 5/6

quantity of milk in 5 litres of second vessel = 5*9/10 = 9/2

quantity of milk in 6 litres of the mixture = 5/6+9/2 = 32/6

quantity of water = 6-32/6 = 4/6

Required ratio = 32/6 : 4/6 = 8 : 1

quantity of milk in 1 litre of first vessel = 1*5/6 = 5/6

quantity of milk in 5 litres of second vessel = 5*9/10 = 9/2

quantity of milk in 6 litres of the mixture = 5/6+9/2 = 32/6

quantity of water = 6-32/6 = 4/6

Required ratio = 32/6 : 4/6 = 8 : 1

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sam

2016-02-04 06:23:19

@shefali, total quantiy is 6 litre in which quantity of milk is 32/6.

Hence 6-32/6 is the quantity of water

Hence 6-32/6 is the quantity of water

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shefali

2016-01-12 05:54:57

How 6-32/6 please explain

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bilu

2015-01-26 17:31:32

Amount used to purchase 1 litre of honey can be used to purchase 3 l of milk or 5l of water. Total amount spent is 1530 what will be the amt spent on honey if the total amt is spent to purchase equal volumes of honey and water only.

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