A) 15 kg | B) 25 kg |

C) 35 kg | D) 45 kg |

Explanation:

Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x.

Let k be the quantity of third variety of oil in the final mixture.

Let the ratio of initial quantity of oils be 'y'

From given details,

4x + 5 = 5y ..... (1)

5x + 10 = 7y .....(2)

8x + k = 9y ......(3)

By solving (1) & (2), we get

x = 5 & y = 5

From (3) => k = 5

Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.

A) 2 | B) 2.5 |

C) 3 | D) 3.5 |

Explanation:

Given number of boxes = 14

Number of workers = 4

Now, number of whole boxes per worker = 14/4 = 3.5

Hence, number of whole boxes per each coworker = **3**

A) 2 | B) 1.5 |

C) 1.25 | D) 2.5 |

Explanation:

Given Five boxes of bananas sell for Rs. 30.

=> 1 Box of Bananas for **= 30/5 = Rs. 6**

Then, for Rs. 9

**=> 9/6 = 3/2 = 1.5**

Hence, for Rs. 9, 1.5 box of bananas can buy.

A) 30 | B) 40 |

C) 25 | D) 20 |

Explanation:

The third proportional of two numbers p and q is defined to be that number r such that

**p : q = q : r.**

Here, required third proportional of 10 & 20, and let it be 'a'

=> 10 : 20 = 20 : a

10a = 20 x 20

=> **a = 40**

Hence, third proportional of 10 & 20 is** 40.**

A) 54.21 kgs | B) 51.07 kgs |

C) 52.66 kgs | D) 53.45 kgs |

Explanation:

Given total number of passengers in the bus =** 45**

First average weight of **45** passengers =** 52 kgs**

Average weight of **5** passengers who leave bus =** 48**

Average weight of **5 **passengers who joined the bus = **54**

Therefore, the net average weight of the bus is given by

$\mathbf{=}\mathbf{}\frac{\mathbf{45}\mathbf{}\mathbf{x}\mathbf{}\mathbf{52}\mathbf{}\mathbf{-}\mathbf{}\left(\mathbf{5}\mathbf{}\mathbf{x}\mathbf{}\mathbf{48}\right)\mathbf{}\mathbf{+}\mathbf{}\left(\mathbf{5}\mathbf{}\mathbf{x}\mathbf{}\mathbf{54}\right)}{\mathbf{45}}\mathbf{}\phantom{\rule{0ex}{0ex}}=\frac{2370}{45}\phantom{\rule{0ex}{0ex}}=\frac{158}{3}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{}\mathbf{52}\mathbf{.}\mathbf{66}\mathbf{}\mathbf{kgs}\mathbf{.}$

A) 4 | B) 6 |

C) 2 | D) 3 |

Explanation:

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> **(12 - b) x 900 = 9000**

=> **b = 2**

**Hence, **the number of boys who dropped out is **2.**

A) 2212 | B) 2154 |

C) 2349 | D) 2679 |

Explanation:

Let x, x+2, x+4, x+6, x+8 and x+10 are six consecutive odd numbers.

Given that their average is **52**

Then, x + x+2 + x+4 + x+6 + x+8 + x+10 = 52×6

6x + 30 = 312

x = 47

So Product = 47 × 57 = 2679

A) Rs. 1398.96 | B) Rs. 1457.09 |

C) Rs. 1662.35 | D) Rs. 1536.07 |

Explanation:

Let average expenditure was **Rs. R**

**13 x 79 + 6x(R + 4) = 19R**

=> R = Rs. 80.84

Total money = * 19 x 80.84 *= Rs. 1536.07.

A) 12 | B) 10 |

C) 8 | D) 6 |

Explanation:

Let the number of boys = x

From the given data,

=> [21x + 16(18)]/(x+16) = 19

=> 21x - 19x = 19(16) - 16(18)

=> 2x = 16

=> x = 8

**Therefore, the number of boys in the class = 8.**