A) 2/5 | B) 4/7 |

C) 4/9 | D) 5/11 |

Explanation:

One way to deal with fractions is to convert them all to decimals.

In this case all you would need to do is to see which is greater than 0.5.

Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7.

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.**