A) 25 | B) 30 |

C) 45 | D) cannot be determined |

Explanation:

pool : kerosene

3 : 2(initially)

2 : 3(after replacement)

(for petrol)

Therefore the total quantity of the mixture in the container is 30 liters.

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 =

Here p = 12 , q = 120

=>

=> 120 x 0.9 x 0.9 x 0.9

=> 120 x 0.729

= 87.48 lit.

A) 15.84 | B) 14.92 |

C) 13.98 | D) 16.38 |

Explanation:

Given, Manideep purchases 30kg of barley at the rate of 11.50/kg nad 20kg at the rate of 14.25/kg.

Total cost of the mixture of barley = (30 x 11.50) + (20 x 14.25)

=> Total cost of the mixture = Rs. 630

Total kgs of the mixture = 30 + 20 = 50kg

Cost of mixture/kg = 630/50 = 12.6/kg

To make 30% of profit

=> Selling price for manideep = 12.6 + 30% x 12.6

=> Selling price for manideep = 12.6 + 3.78 = 16.38/kg.

A) 111:108 | B) 11:9 |

C) 103:72 | D) None |

Explanation:

From the given data,

The part of honey in the first mixture = 1/4

The part of honey in the second mixture = 3/4

Let the part of honey in the third mixture = x

Then,

1/4 3/4

x

(3/4)-x x-(1/4)

Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3

=>

=> Solving we get the part of honey in the third mixture as 11/20

=> the remaining part of the mixture is water = 9/20

Hence, the ratio of the mixture of honey and water in the third mixture is **11 : 9** .