A) 1:3 | B) 2:3 |

C) 3:4 | D) 4:5 |

Explanation:

By the rule of alligation:

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

Required ratio = 60 : 90 = 2 : 3

A) 45 gm | B) 50 gm |

C) 55 gm | D) 60 gm |

Explanation:

Initial quantity of copper =$\frac{\mathbf{80}}{\mathbf{100}}\mathbf{}\mathbf{x}\mathbf{}\mathbf{50}$ = **40 g **

And that of Bronze = **50 - 40 = 10 g**

Let** 'p' gm** of copper is added to the mixture

=> $\left(\mathbf{50}\mathbf{}\mathbf{+}\mathbf{}\mathbf{p}\right)\mathbf{}\mathbf{x}\mathbf{}\frac{\mathbf{90}}{\mathbf{100}}$ **= 40 + p**

=> 45 + 0.9p = 40 + p

=> p = 50 g

Hence, **50 gms** of copper is added to the mixture, so that the copper is increased to 90%.

A) 1 : 10 | B) 10 : 1 |

C) 9 : 11 | D) 11 : 9 |

Explanation:

Customer ratio of Milk and Water is given by

**Milk :: Water**

6.4 0

$\frac{8}{1+{\displaystyle \frac{3}{8}}}=\frac{64}{11}$

**$\frac{64}{11}$ $\frac{64}{10}-\frac{64}{11}$**

=> **Milk : Water** = 110 : 11 = *10 : 1*

*Therefore, the proportionate of Water to Milk for Customer is** 1 : 10*

A) 1:1 | B) 2:1 |

C) 3:1 | D) 1:2 |

Explanation:

Quantity of alohol in the mixture = 40 x 5/8 = 25 lit

Quantity of water = 40 - 25 = 15 lit

According to question,

Required ratio = $\frac{\mathbf{20}\mathbf{}\mathbf{-}\mathbf{}\left(\mathbf{40}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{20}}{\mathbf{100}}}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{5}}{\mathbf{8}}}\right)}{\mathbf{15}\mathbf{}\mathbf{-}\mathbf{}\left(\mathbf{40}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{20}}{\mathbf{100}}}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{3}}{\mathbf{8}}}\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{40}\mathbf{}\mathbf{x}\mathbf{}{\displaystyle \frac{\mathbf{20}}{\mathbf{100}}}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{20}}{\mathbf{15}\mathbf{}\mathbf{-}\mathbf{}\mathbf{3}\mathbf{}\mathbf{+}\mathbf{}\mathbf{8}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{:}\mathbf{}\mathbf{1}$

A) 85 | B) 60 |

C) 55 | D) 45 |

Explanation:

Let M litres milk be added

=> $\frac{\mathbf{60}\mathbf{}\mathbf{+}\mathbf{}\mathbf{M}}{\mathbf{40}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{3}}{\mathbf{1}}$

=> 60 + M = 120

=> M = 60 lit.

A) 31.8% | B) 31% |

C) 33.33% | D) 29.85% |

Explanation:

Now, take percentage of milk and applying mixture rule

25 100

50

50 25 = 2 : 1

**Hence required answer = 1/3 or 33.33%**

A) 3:2 | B) 4:3 |

C) 2:3 | D) 3:4 |

Explanation:

Concentration of glucose are in the ratio = $\frac{1}{2}:\frac{3}{5}:\frac{4}{5}$

Quantity of glucose taken from A = 1 liter out of 2

Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit

Quantity of glucose taken from C = 0.8 lit

So, total quantity of glucose taken from A,B and C = 3.6 lit

So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit

Ratio of glucose to alcohol = 3.6/2.4 = 3:2

A) 71.02% | B) 76.92% |

C) 63.22% | D) 86.42% |

Explanation:

Let the milk he bought is 1000 ml

Let C.P of 1000 ml is Rs. 100

Here let he is mixing K ml of water

He is getting 30% profit

=> Now, the selling price is also Rs. 100 for 1000 ml

=> 100 : K%

= 100 : 30

10 : 3 is ratio of milk to water

=> Percentage of milk = 10 x 100/13 = 1000/13 = 76.92%

A) 8:3 | B) 6:7 |

C) 7:5 | D) 11:7 |

Explanation:

Milk in 1-litre mixture of A = 4/7 litre.

Milk in 1-litre mixture of B = 2/5 litre.

Milk in 1-litre mixture of C = 1/2 litre.

By rule of alligation we have required ratio X:Y

X : Y

4/7 2/5

\ /

(Mean ratio)

(1/2)

/ \

(1/2 – 2/5) : (4/7 – 1/2)

1/10 1/1 4

So Required ratio = X : Y = 1/10 : 1/14 = **7:5**