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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

Raj

2015-02-09 19:56:14

Let cost of 1 litre of honey = x

Cost of 5 litre of water = x. => Cost of 1 litre of water = x/5

Cost of 1 litre of honey : Cost of 1 litre of water = x : x/5 = 5 : 1

ie, amount should be in the ration 5 : 1 to purchase equal quantity of honey and water

Amount spent on honey = 1530 * 5/6 =Rs.1275

Cost of 5 litre of water = x. => Cost of 1 litre of water = x/5

Cost of 1 litre of honey : Cost of 1 litre of water = x : x/5 = 5 : 1

ie, amount should be in the ration 5 : 1 to purchase equal quantity of honey and water

Amount spent on honey = 1530 * 5/6 =Rs.1275

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Ratan

2015-01-24 02:30:43

How many kilograms of sugar must be
added to 20 kilograms of 20% solution of sugar and water to increase the
concentration of sugar to 50% of solution?

Arun Kumar

2016-07-21 10:50:55

The concentration of sugar in solution =20*20% = 4 kg

rest solution is 16 kg

Now solution is 16: 4 (kgs)

Need to increase concentration of sugar up to 50%, means equal quantity of sugar in solution.

4 kg already, +12 kg add

Answer: 12 kg

rest solution is 16 kg

Now solution is 16: 4 (kgs)

Need to increase concentration of sugar up to 50%, means equal quantity of sugar in solution.

4 kg already, +12 kg add

Answer: 12 kg

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Jay

2015-01-24 11:24:55

Percentage concentration of sugar in pure sugar(100) | Percentage concentration of sugar in the solution (20) | ||||||||

50 | |||||||||

50-20=30 | 100-50=50 |

Quantity of sugar : Quantity of solution should be 30 : 50 = 3:5

Quantity of sugar : 20 = 3:5

Quantity of sugar = 20*3/5=12 kg

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Chandra

2015-01-05 11:29:33

The weight of an empty bottle is one sixth of the bottle full of water. Certain percentage of water is removed from the bottle and weighed. The weight of the bottle is now one third of the bottle full of water. What is the percentage of water removed?

Jay

2015-01-11 19:45:25

Let weight of the full bottle is 60 Kg

Then, weight of the empty bottle is 1/6 of 60 which 10 Kg

Weight of water = 60-10= 50 Kg

After removing some amount of water, weight of the bottle = 1/3 of 60 kg which is 20 Kg

Weight of water present in the bottle = 20 -10 = 10 Kg

Weight of water removed = 50-10 = 40 Kg

Required percentage = 40*100/50 = 80%

Then, weight of the empty bottle is 1/6 of 60 which 10 Kg

Weight of water = 60-10= 50 Kg

After removing some amount of water, weight of the bottle = 1/3 of 60 kg which is 20 Kg

Weight of water present in the bottle = 20 -10 = 10 Kg

Weight of water removed = 50-10 = 40 Kg

Required percentage = 40*100/50 = 80%

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mahender

2014-12-24 20:40:31

A sells milk which contains 5% of water. What quantity of pure milk should be added to 2 litre of milk(contains 5% water) so that proportion of water becomes 4%?

Ankita Gupta

2014-12-28 14:37:55

Water component in 2lts of milk(contains 5% water) is 2*5/100 = 0.1 litre

and milk component is 2-0.1 = 1.9 litre

Let required quantity of pure milk added is x litre

Total quantity of pure milk will be (1.9+x) litre

Total quantity of the solution will be (2+x) litre

Total quantity of water will be same as 0.1 litre

(2+x) * 4/100 = 0.1

2+x = 2.5

x = 0.5 litre

and milk component is 2-0.1 = 1.9 litre

Let required quantity of pure milk added is x litre

Total quantity of pure milk will be (1.9+x) litre

Total quantity of the solution will be (2+x) litre

Total quantity of water will be same as 0.1 litre

(2+x) * 4/100 = 0.1

2+x = 2.5

x = 0.5 litre

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arun kumar

2016-07-21 11:50:54

Let's do it in alligation rule

Apply alligation rule on those components are added or subtracted.

Here is Milk

(old) (every thing itself 100%)

95% 100%

96% (New proportion)

4 1

Divided by 2 by making old quantity 2 litre.

2 .5

Added milk is 0.5 litre

Apply alligation rule on those components are added or subtracted.

Here is Milk

(old) (every thing itself 100%)

95% 100%

96% (New proportion)

4 1

Divided by 2 by making old quantity 2 litre.

2 .5

Added milk is 0.5 litre

0
0

Saurabh

2014-12-24 07:33:49

Pure milk costs 16 per litre. After adding water the milkman sells the mixture

15 per litre and thereby makes a profit of 25%. In what respective ratio does

he mix milk with water?

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