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

C) 5:6 | D) 7:9 |

Explanation:

Let the C.P. of spirit be Re. 1 litre.

Spirit in 1 litre mix. of A = 5/7 litre, C.P. of 1 litre mix.

in A = Re. 5/7

Spirit in 1 litre mix. of B = 7/13 litre, C.P. of 1 litre mix.

in B = Re. 7/13

Spirit in 1 litre mix. of C = 8/13 litre, Mean price = Re. 8/13.

By the rule of alligation, we have:

Cost of 1 litre mixture in A Cost of 1 litre mixture in B

Required ratio = 1/13 : 9/91 = 7:9.

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**

A) 60 lit | B) 55 lit |

C) 45 lit | D) 50 lit |

A) 24 lit | B) 16 lit |

C) 32 lit | D) 8 lit |

Explanation:

Given mixture = 48 lit

Milk in it = 48 x 5/8 = 30 lit

=> Water in it = 48 - 30 = 18 lit

Let 'L' lit of water is added to make the ratio as 3:5

=> 30/(18+L) = 3/5

=> 150 = 54 + 3L

=> L = 32 lit.

A) 88.01 lit | B) 87.48 lit |

C) 87.51 lit | D) 87.62 lit |

Explanation:

For these type of problems,

Quantity of Diesel remained = $\left(q{\left(1-\frac{p}{q}\right)}^{n}\right)$

Here p = 12 , q = 120

=> $\left(120{\left(1-\frac{12}{120}\right)}^{3}\right)$ = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.