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21. How many kg of rice at Rs.6.60 per kg be mixed with 56 kg of rice at Rs.9.60 per kg to get a mixture worth Rs.8.20 per kg? | |

A. 56 kg | B. 52 kg |

C. 44 kg | D. 49 kg |

Discuss |

answer with explanation

Answer: Option D

Explanation:

By rule of alligation,

Cost of 1 kg of 1st kind rice | Cost of 1 kg of 2nd kind rice | |||||||||

6.6 | 9.6 | |||||||||

Cost of 1 kg of the mixture | ||||||||||

8.2 | ||||||||||

9.6 - 8.2 = 1.4 | 8.2 - 6.6 = 1.6 |

Quantity of 1st kind rice : Quantity of 2nd kind rice = 1.4 : 1.6 = 7 : 8

=> Quantity of 1st kind rice : 56 = 7 : 8

=> Quantity of 1st kind rice $=56 × \dfrac{7}{8} = 49$

22. How many litres of water must be added to 16 liters of milk and water containing 10% water to make it 20% water in it? | |

A. 3 litre | B. 2 litre |

C. 4 litre | D. 1 litre |

Discuss |

answer with explanation

Answer: Option B

Explanation:

By rule of alligation,

% Concentration of water in pure water (100) | % Concentration of water in the given mixture (10) | |||||||||

Mean % concentration (20) | ||||||||||

20 - 10 = 10 | 100 - 20 = 80 |

=> Quantity of water : Quantity of the mixture = 10 : 80 = 1 : 8

Here quantity of the mixture = 16 litres

=> Quantity of water : 16 = 1 : 8

=> Quantity of water $=16 × \dfrac{1}{8} = 2$ litre

23. We have a 630 ml mixture of milk and water in the ratio 7:2. How much water must be added to make the ratio 7:3? | |

A. 60 ml | B. 70 ml |

C. 50 ml | D. 80 ml |

Discuss |

answer with explanation

Answer: Option B

Explanation:

concentration of water in mixture1 $=\dfrac{2}{9}$ *(since the ratio of milk and water = 7:2)* ...(1)

concentration of water in pure water= 1 ...(2)

Now the above mentioned items are mixed to form mixture2 where milk and water ratio = 7 : 3

=> concentration of water in mixture2 $=\dfrac{3}{10}$

By rule of alligation,

concentration of water in mixture1 $\left(\dfrac{2}{9}\right)$ | concentration of water in pure water (1) | |||||||||

Mean concentration | ||||||||||

$\left(\dfrac{3}{10}\right)$ | ||||||||||

$1-\dfrac{3}{10}=\dfrac{7}{10}$ | $\dfrac{3}{10}-\dfrac{2}{9}=\dfrac{7}{90}$ |

=> Quantity of mixture1 : Quantity of water

$=\dfrac{7}{10}:\dfrac{7}{90}=\dfrac{1}{10}:\dfrac{1}{90}=1:\dfrac{1}{9}$

Given that Quantity of mixture1 = 630 ml

=> 630 : Quantity of water $=1:\dfrac{1}{9}$

=> Quantity of water $=630 ×\dfrac{1}{9}=70$ ml

24. 3 litre of water is added to 11 litre of a solution containing 42% of alcohol in the water. The percentage of alcohol in the new mixture is | |

A. 20% | B. 33% |

C. 30% | D. 25% |

Discuss |

answer with explanation

Answer: Option B

Explanation:

**Solution 1**

We have a 11 litre solution containing 42% of alcohol in the water.

=> Quantity of alcohol in the solution $=\dfrac{11 × 42}{100}$

Now 3 litre of water is added to the solution.

=> Total quantity of the new solution = 11 + 3 = 14

Percentage of alcohol in the new solution $=\dfrac{\dfrac{11 × 42}{100}}{14}× 100$

$=\dfrac{11 × 3}{100} = 33\%$

**Solution 2**

%Concentration of alcohol in pure water = 0

%Concentration of alcohol in mixture = 42

Quantity of water : Quantity of mixture = 3 : 11

Let the %concentration of alcohol in the new mixture $=x$

By rule of alligation,

%Concentration of alcohol in pure water (0) | %Concentration of alcohol in mixture(42) | |||||||||

Mean %concentration $(x)$ | ||||||||||

$42-x$ | $x-0=x$ |

But $(42 - x) : x = 3 : 11$

$\Rightarrow 11(42-x)=3x\\\Rightarrow 42 × 11 - 11x = 3x\\\Rightarrow 14x = 42 × 11 \\\Rightarrow x = 3× 11 = 33$

i.e., Percentage of alcohol in the new mixture is 33%

25. Rs.460 was divided among 41 boys and girls such that each boy got Rs.12 and each girl got Rs.8. What is the number of boys? | |

A. 30 | B. 28 |

C. 33 | D. 36 |

Discuss |

answer with explanation

Answer: Option C

Explanation:

**Solution 1**

Assume that the number of boys = b and number of girls = g

number of boys + number of girls = 41

=> b + g = 41 ...(1)

Given that each boy got Rs.12 and each girl got Rs.8. Then the total amount is Rs.460

=> 12b + 8g = 460 ...(2)

Now we need to solve these equations to get b and g.

(1) × 8 => 8b + 8g = 8 × 41 = 328 ...(3)

(2) - (3) => 4b = 460 - 328 = 132

=> b $=\dfrac{132}{4}=33$

Given that amount received by a boy = Rs.12,

Amount received by a girl =Rs.8

Total amount = 460

Given that number of boys + number of girls = 41

Hence, mean amount $=\dfrac{460}{41}$

By rule of alligation, | ||||||||||

Amount received by a boy (12) | Amount received by a girl(8) | |||||||||

Mean amount | ||||||||||

$\dfrac{460}{41}$ | ||||||||||

$\dfrac{460}{41}-8=\dfrac{132}{41}$ | $12-\dfrac{460}{41}=\dfrac{32}{41}$ |

Number of boys : Number of girls

=$\dfrac{132}{41}:\dfrac{32}{41}=132:32\\=66:16=33:8$

Given that number of boys + number of girls = 41

Hence number of boys $=41 × \dfrac{33}{41} = 33$

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