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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

Neeraj Jain

2014-07-19 21:51:12

cost of milk per litre Rs. 10.50 | cost of water per litre 0 | ||||||||

Cost of mixture per litre Rs. 9 | |||||||||

9-0=9 | 10.5-9=1.5 |

Quantity of milk : Quantity of water = 9 : 1.5 = 6 : 1

Total quantity = 56 litre

Quantity of water = 56 * 1/7 = 8 litre

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Atul

2014-05-04 23:56:14

Please Provide Solution For This :-

cost price of three types of cashew nuts are Rs. 482 , Rs. 578 , Rs. 698 Respect.In what proportion these should be mixed , so that the mixture costs Rs. 566 Per Kg ?

Ajay

2014-05-05 15:56:15

Since there are more than two quantities, multiple ratios are possible where the resultant mixture will cost Rs.566

For example, take the required ratio as x:y:x

then, (482x + 578y + 698z)/(x+y+z) = 566

on simplification, this becomes y + 11z = 7x

if we take y=z,

12y = 7x

y=7 and x=12 satisfies this. z=y=7

__Required ratio = 12 : 7 : 7__

See that (482x + 578y + 698z)/(x+y+z) = 566 for these values of x,y,z

similarly you can get different ratios which also satisfies this.

For example, take the required ratio as x:y:x

then, (482x + 578y + 698z)/(x+y+z) = 566

on simplification, this becomes y + 11z = 7x

if we take y=z,

12y = 7x

y=7 and x=12 satisfies this. z=y=7

See that (482x + 578y + 698z)/(x+y+z) = 566 for these values of x,y,z

similarly you can get different ratios which also satisfies this.

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andy

2014-02-08 10:04:15

400 gm spirit solution has 40% spirit in it. How may grams of spirit should be added to make it 60 % in the solution?

Jay

2014-02-11 19:01:28

% concentration of spirit in the solution 40% | % concentration of spirit in pure spirit 100% | ||||||||

Mean Concentration 60% | |||||||||

100-60=40 | 60-40=20 |

Since the quantity of the solution is 400 gm, amount of spirit needs to be added = 400/2 =

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Mohan

2013-11-16 11:27:43

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

Hence 20 * 8.50 + 35 * 8.75 is done

Hence 20 * 8.50 + 35 * 8.75 is done

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Mononit

2013-11-15 16:42:00

The proportion of 1st and 2nd liquids in a mixture is 2:3 while in another mixture their ratio is 5:4. In what ratio should the two mixture be mixed, so that the new mixture contains equal amount of the two liquids?

Ajay

2013-11-15 18:35:04

Let x liter of the first mixture needs to be mixed with y litre if the second mixture so that the resultant mixture will have both the liquids as 1 : 1

Quantity of 1st liquid in the resultant mixture = 2x/5 + 5y/9

Quantity of 2nd liquid in the resultant mixture = 3x/5 + 4y/9

2x/5 + 5y/9 = 3x/5 + 4y/9

=>18x + 25y = 27x + 20y

=> 9x = 5y

=> x/y = 5/9

=> x:y = 5:9

Required ratio is 5:9

Quantity of 1st liquid in the resultant mixture = 2x/5 + 5y/9

Quantity of 2nd liquid in the resultant mixture = 3x/5 + 4y/9

2x/5 + 5y/9 = 3x/5 + 4y/9

=>18x + 25y = 27x + 20y

=> 9x = 5y

=> x/y = 5/9

=> x:y = 5:9

Required ratio is 5:9

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