A) 79 | B) 77 |

C) 76 | D) 73 |

Explanation:

Assuming Sripad has scored the least marks in subject other than science,

Then the marks he could secure in other two are 58 each.

Since the average mark of all the 3 subject is 65.

i.e (58+58+x)/3 = 65

116 + x = 195

x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

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.**