Questions and Answers Type: | MCQ (Multiple Choice Questions). |

Main Topic: | Quantitative Aptitude. |

Quantitative Aptitude Sub-topic: | Average Aptitude Questions and Answers. |

Number of Questions: | 10 Questions with Solutions. |

- \(24.2\)
- \(22.2\)
- \(22.5\)
- \(23.5\)

Answer: (c) \(22.5\)

Solution: Given values, \(15, 20, 25, 30\) then according to average formula- $$ Average = \frac{k_1 + k_2 + k_3 + k_4}{4} $$ $$ = \frac{15 + 20 + 25 + 30}{4} = \frac{90}{4} = 22.5 $$

Solution: Given values, \(15, 20, 25, 30\) then according to average formula- $$ Average = \frac{k_1 + k_2 + k_3 + k_4}{4} $$ $$ = \frac{15 + 20 + 25 + 30}{4} = \frac{90}{4} = 22.5 $$

- \(70.05 \ km/hr\)
- \(60 \ km/hr\)
- \(65.67 \ km/hr\)
- \(66.67 \ km/hr\)

Answer: (d) \(66.67 \ km/hr\)

Solution: Given values, \(x_1 = 50 \ km/hr\), and \(x_2 = 100 \ km/hr\) then $$ Average \ speed = \frac{2 \ x_1 \ x_2}{x_1 + x_2} $$ $$ = \frac{2 \times 50 \times 100}{50 + 100} = \frac{10000}{150} = 66.67 \ km/hr $$

Solution: Given values, \(x_1 = 50 \ km/hr\), and \(x_2 = 100 \ km/hr\) then $$ Average \ speed = \frac{2 \ x_1 \ x_2}{x_1 + x_2} $$ $$ = \frac{2 \times 50 \times 100}{50 + 100} = \frac{10000}{150} = 66.67 \ km/hr $$

- \(3.23 \ km/hr\)
- \(2.1429 \ km/hr\)
- \(2 \ km/hr\)
- \(3.1429 \ km/hr\)

Answer: (b) \(2.1429 \ km/hr\)

Solution: Given values, \(d_1 = d_2 = 150 \ km\), \(t_1 = 60 \ minutes\), and \(t_2 = 80 \ minutes\) then $$ Average \ speed = \frac{d_1 + d_2}{t_1 + t_2} $$ $$ = \frac{150 + 150}{80 + 60} = \frac{300}{140} = 2.1429 \ km/min $$

Solution: Given values, \(d_1 = d_2 = 150 \ km\), \(t_1 = 60 \ minutes\), and \(t_2 = 80 \ minutes\) then $$ Average \ speed = \frac{d_1 + d_2}{t_1 + t_2} $$ $$ = \frac{150 + 150}{80 + 60} = \frac{300}{140} = 2.1429 \ km/min $$

- \(170\)
- \(166.67\)
- \(165.67\)
- \(155.58\)

Answer: (b) \(166.67\)

Solution: $$ Average = \frac{k_1 + k_2 + k_3}{3}$$ $$ = \frac{100 + 150 + 250}{3} $$ $$ = \frac{500}{3} = 166.67 $$

Solution: $$ Average = \frac{k_1 + k_2 + k_3}{3}$$ $$ = \frac{100 + 150 + 250}{3} $$ $$ = \frac{500}{3} = 166.67 $$

- \(50\)
- \(42\)
- \(40\)
- \(35\)

Answer: (c) \(40\)

Solution: First number divisible by \(5\) after \(20\) is \(25\) and last number divisible by \(5\) before \(60\) is \(55\), then $$ Average = \frac{25 + 55}{2} = \frac{80}{2} = 40 $$

Solution: First number divisible by \(5\) after \(20\) is \(25\) and last number divisible by \(5\) before \(60\) is \(55\), then $$ Average = \frac{25 + 55}{2} = \frac{80}{2} = 40 $$

- \(65\)
- \(67\)
- \(70\)
- \(68\)

Answer: (a) \(65\)

Solution: Average marks obtained by the student $$ Average = \frac{k_1 + k_2 + k_3 + k_4}{4} $$ $$ = \frac{50 + 60 + 70 + 80}{4} = \frac{260}{4} = 65 \ marks $$

Solution: Average marks obtained by the student $$ Average = \frac{k_1 + k_2 + k_3 + k_4}{4} $$ $$ = \frac{50 + 60 + 70 + 80}{4} = \frac{260}{4} = 65 \ marks $$

- \(20\)
- \(26\)
- \(25\)
- \(28\)

Answer: (a) \(20\)

Solution: Seven numbers average is \(50\) and after one number excluded then six numbers average is \(55\) then $$Average \ increases = 55 - 50 = 5 $$ $$ Total \ increased \ number = 5 \times 6 = 30 $$ it means excluded number is \(30\) less than average of seven numbers \(50\) $$ then \ excluded \ number = 50 - 30 = 20 $$

Solution: Seven numbers average is \(50\) and after one number excluded then six numbers average is \(55\) then $$Average \ increases = 55 - 50 = 5 $$ $$ Total \ increased \ number = 5 \times 6 = 30 $$ it means excluded number is \(30\) less than average of seven numbers \(50\) $$ then \ excluded \ number = 50 - 30 = 20 $$

- \(44\)
- \(55\)
- \(60\)
- \(58\)

Answer: (c) \(60\)

Solution: First number divisible by \(3\) after \(40\) is \(42\) and last number divisible by \(3\) before \(80\) is \(78\), then $$ Average = \frac{42 + 78}{2} = \frac{120}{2} = 60 $$

Solution: First number divisible by \(3\) after \(40\) is \(42\) and last number divisible by \(3\) before \(80\) is \(78\), then $$ Average = \frac{42 + 78}{2} = \frac{120}{2} = 60 $$

- \(56 \ km/hr\)
- \(60 \ km/hr\)
- \(65 \ km/hr\)
- \(62 \ km/hr\)

Answer: (b) \(60 \ km/hr\)

Solution: Given values, \(d_1 = d_2 = 300 \ km\), \(t_1 = 4 \ hours\), and \(t_2 = 6 \ hours\) then $$ Average \ speed = \frac{d_1 + d_2}{t_1 + t_2} $$ $$ = \frac{300 + 300}{4 + 6} = \frac{600}{10} = 60 \ km/hr $$

Solution: Given values, \(d_1 = d_2 = 300 \ km\), \(t_1 = 4 \ hours\), and \(t_2 = 6 \ hours\) then $$ Average \ speed = \frac{d_1 + d_2}{t_1 + t_2} $$ $$ = \frac{300 + 300}{4 + 6} = \frac{600}{10} = 60 \ km/hr $$

- \(50 \ km/hr\)
- \(45.67 \ km/hr\)
- \(46.67 \ km/hr\)
- \(57.67 \ km/hr\)

Answer: (c) \(46.67 \ km/hr\)

Solution: $$Average \ speed = \frac{20 + 50 + 70}{3}$$ $$ = \frac{140}{3} = 46.67 \ km/hr $$

Solution: $$Average \ speed = \frac{20 + 50 + 70}{3}$$ $$ = \frac{140}{3} = 46.67 \ km/hr $$