A) 1/3 | B) 1/4 |

C) 1/5 | D) 1/7 |

Explanation:

Suppose the vessel initially contains 8 litres of liquid.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = litres.

Quantity of syrup in new mixture = litres.

=> 5x + 24 = 40 - 5x

=> 10x = 16 => x = 8/5

So, part of the mixture replaced = = 1/5.

A) 193 : 122 | B) 97 : 102 |

C) 115 : 201 | D) 147 : 185 |

Explanation:

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

(35,70), (42,63), (45,60)

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

But in the question asked the ratio of water to milk = 193 : 122

A) 80% | B) 70% |

C) 75% | D) 62% |

Explanation:

Water in 60 gm mixture=60 x 75/100 = 45 gm. and Milk = 15 gm.

After adding 15 gm. of water in mixture, total water = 45 + 15 = 60 gm and

weight of a mixture = 60 + 15 = 75 gm.

So % of water = 100 x 60/75 = 80%.

A) 6 ml | B) 11 ml |

C) 15 ml | D) 9 ml |

Explanation:

Let us assume that the lotion has 50% alcohol and 50% water.

ratio = 1:1

As the total solution is 9ml

alcohol = water = 4.5ml

Now if we want the quantity of alcohol = 30%

The quantity of water = 70%

The new ratio = 3:7

Let x ml of water be added

We get,

=> x=6

Hence 6ml of water is added.

A) 25% | B) 30% |

C) 17% | D) 19% |

Explanation:

He has gain = 15 - 12 = 3,

Gain% = (3/12) x 100 = (100/4) = 25.

He has 25% gain.

A) 27 % | B) 26 % |

C) 29 % | D) 21 % |

Explanation:

Let the percentage of benzene = X

(30 - X)/(X- 25) = 6/4 = 3/2

=> 5X = 135

X = 27

So, required percentage of benzene = 27 %