A) 1 | B) 2 |

C) 3 | D) 4 |

Explanation:

Total age of 4 members, 10 years ago = (24 x 4) years = 96 years.

Total age of 4 members now = [96 + (10 x 4)] years = 136 years.

Total age of 6 members now = (24 x 6) years = 144 years.

Sum of the ages of 2 children = (144 - 136) years = 8 years.

Let the age of the younger child be x years.

Then, age of the elder child = (x+2) years.

So, x+(x+2) =8 <=> x=3

Age of younger child = 3 years.

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.**

A) 40.36 | B) 41.24 |

C) 41.92 | D) 42.05 |

Explanation:

Given,

mean of 100 observations is 40

=> Total of 100 observations = 40 x 100 = 4000

84 is misread as 48

=> Difference = 84 - 48 = 36

=> Now, new total of 100 observations = 4000 + 36 = 4036

Correct Mean = 4036/100 = **40.36**

A) 97 | B) 98 |

C) 92 | D) 93 |

A) Rs. 175 | B) Rs. 220.75 |

C) Rs. 218.55 | D) Rs. 656.25 |

Explanation:

Assume Hebah has Rs. M

Since 25% more money at Poonam

=> Money at Poonam = M + (25% of M)

=> Money at Poonam = M + 0.25M = Rs. 1.25M

Money with Navaneet is thrice the money with Poonam,

=> Money at Navaneet = 3(1.25M) = Rs. 3.75M

Now, sum of all Money = (M + 1.25M + 3.75M) = Rs. 6M

But given the average of the money is Rs. 350

=> 6M/3 = 350

=> 2M = 350

=> M = 350/2 = Rs. 175

=> Amount of money Navaneet has = Rs. 3.75M = 3.75 x 175 = Rs. 656.25.