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) 3:2 | B) 2:3 |

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

Explanation:

Let the number of boys = b

Let the number of girls = g

From the given data,

81b + 83g = 81.8(b + g)

81.8b - 81b = 83g - 81.8g

0.8b = 1.2g

b/g = 1.2/0.8 = 12/8 = 3/2

**=> g : b = 2 : 3**

Hence, ratio between the number of girls to the number of boys =** 2 : 3.**

A) 120 | B) 160 |

C) 80 | D) 60 |

Explanation:

Let the three numbers be x, y, z.

From the gien data,

**x = 2y ....(1)**

**x = z/2 => z = 2(2y) = 4y .....(From 1) ...........(2)**

Given average of three numbers = 56

Then,

$\frac{\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{}\mathbf{+}\mathbf{}\mathbf{z}}{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{56}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{2}\mathbf{y}\mathbf{}\mathbf{+}\mathbf{}\mathbf{y}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{y}}{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{56}(\mathrm{From}12)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}7\mathrm{y}=56\mathrm{x}3\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{24}$

Now,

**x = 2y => x = 2 x 24 = 48**

**z = 4y = 4 x 24 = 96**

Now, the highest number is z = 96 & smallest number is y = 24

Hence, required sum of highest number and smallest number

**= z + y **

**= 96 + 24 **

**= 120.**

A) 4 | B) 15 |

C) 3 | D) 50 |

Explanation:

Given that,

Number of windows = 50

Each window covering covers 15 windows

=> 50 windows requires 50/15 window coverings

= 50/15 = 3.333

Hence, more than 3 window coverings are required. In the options 4 is more than 3.

Hence, 4 window coverings are required to cover 50 windows of each covering covers 15 windows.

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