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. |

- Find out the average of six consecutive odd numbers starting from seven?
- \(10\)
- \(14\)
- \(12\)
- \(15\)

Answer: (c) \(12\)

Solution: Six odd numbers starting from seven are \(= 7, 9, 11, 13, 15\). Then $$ Average = \frac{7 + 9 + 11 + 13 + 15}{5} $$ $$ = \frac{72}{6} = 12 $$

Solution: Six odd numbers starting from seven are \(= 7, 9, 11, 13, 15\). Then $$ Average = \frac{7 + 9 + 11 + 13 + 15}{5} $$ $$ = \frac{72}{6} = 12 $$

- Find out the average of five consecutive even numbers starting from \(10 \ ?\)
- \(12\)
- \(13\)
- \(14\)
- \(16\)

Answer: (c) \(14\)

Solution: Five even numbers starting from \(10\) are \(10, 12, 14, 16, 18\), then $$Average = \frac{10 + 12 + 14 + 16 + 18}{5}$$ $$ = \frac{70}{5} = 14 $$

Solution: Five even numbers starting from \(10\) are \(10, 12, 14, 16, 18\), then $$Average = \frac{10 + 12 + 14 + 16 + 18}{5}$$ $$ = \frac{70}{5} = 14 $$

- If the average of \(5\) numbers is \(3\) then find the average after multiplying the number by \(6 \ ?\)
- \(16\)
- \(20\)
- \(14\)
- \(18\)

Answer: (d) \(18\).

Solution: Given average of \(5\) numbers = \(3\)

Total value of \(5\) numbers = \(5 \times 3 = 15\)

Total value of five numbers after multiplying the numbers by \(6\) = \(15 \times 6 = 90\)

then the new average \(= \frac{90}{5} = 18\)

Solution: Given average of \(5\) numbers = \(3\)

Total value of \(5\) numbers = \(5 \times 3 = 15\)

Total value of five numbers after multiplying the numbers by \(6\) = \(15 \times 6 = 90\)

then the new average \(= \frac{90}{5} = 18\)

- If the average of \(7\) numbers is \(5\) then find the average after multiplying the numbers by \(9 \ ?\)
- \(35\)
- \(40\)
- \(45\)
- \(42\)

Answer: (c) \(45\)

Solution: Given average of \(7\) numbers = \(5\)

Total value of \(7\) numbers = \(7 \times 5 = 35\)

Total value of \(7\) numbers after multiplying the numbers by \(9\) = \(35 \times 9 = 315\)

then the new average \(= \frac{315}{7} = 45\)

Solution: Given average of \(7\) numbers = \(5\)

Total value of \(7\) numbers = \(7 \times 5 = 35\)

Total value of \(7\) numbers after multiplying the numbers by \(9\) = \(35 \times 9 = 315\)

then the new average \(= \frac{315}{7} = 45\)

- If a train running between two stations and train covers first \(50 \ km\) at the average speed of \(20 \ km/hr\), second \(50 \ km\) at the average speed of \(40 \ km/hr\), and last \(50 \ km\) at the average speed of \(50 \ km/hr\). Find the average speed of the train during whole journey?
- \(30.05 \ km/hr\)
- \(31.56 \ km/hr\)
- \(32.56 \ km/hr\)
- \(33.26 \ km/hr\)

Answer: (b) \(31.56 \ km/hr\)

Solution: $$Average \ Speed = \frac{Total \ distance \ covered}{Total \ time \ taken}$$ $$Average \ Speed = \frac{50 + 50 + 50}{\frac{50}{20} + \frac{50}{40} + \frac{50}{50}}$$ $$= \frac{150 \times 4}{19} = \frac{600}{19} = 31.56 \ km/hr$$

Solution: $$Average \ Speed = \frac{Total \ distance \ covered}{Total \ time \ taken}$$ $$Average \ Speed = \frac{50 + 50 + 50}{\frac{50}{20} + \frac{50}{40} + \frac{50}{50}}$$ $$= \frac{150 \times 4}{19} = \frac{600}{19} = 31.56 \ km/hr$$

- If a man travels first \(100 \ km\) at the average speed of \(60 \ km/hr\), second \(150 \ km\) at the average speed of \(80 \ km/hr\), and last \(200 \ km\) at the average speed of \(100 \ km/hr\). Find the average speed of the man during the whole journey?
- \(81.203\)
- \(80.305\)
- \(82.205\)
- \(83.203\)

Answer: (a) \(81.203\)

Solution: $$Average \ Speed = \frac{Total \ distance \ traveled}{Total \ time \ taken}$$ $$Average \ Speed = \frac{100 + 150 + 200}{\frac{100}{60} + \frac{150}{80} + \frac{200}{100}}$$ $$= \frac{450 \times 24}{133} = \frac{10800}{133} = 81.203 \ km/hr$$

Solution: $$Average \ Speed = \frac{Total \ distance \ traveled}{Total \ time \ taken}$$ $$Average \ Speed = \frac{100 + 150 + 200}{\frac{100}{60} + \frac{150}{80} + \frac{200}{100}}$$ $$= \frac{450 \times 24}{133} = \frac{10800}{133} = 81.203 \ km/hr$$

- If the average age of three men is \(50\) years and the ratio of their ages are \(5 : 7 : 8\) then find the age of oldest man?
- \(50 \ years\)
- \(65 \ years\)
- \(60 \ years\)
- \(65 \ years\)

Answer: (c) \(60 \ years\)

Solution: Let the ages of men are \(5x, 7x,\) and \(8x\) years. then $$ 50 = \frac{5x + 7x + 8x}{3} $$ $$ 20x = 150 $$ $$ x = 7.5 $$ Then the age of oldest man \(= 8 \times 7.5 = 60 \ years\)

Solution: Let the ages of men are \(5x, 7x,\) and \(8x\) years. then $$ 50 = \frac{5x + 7x + 8x}{3} $$ $$ 20x = 150 $$ $$ x = 7.5 $$ Then the age of oldest man \(= 8 \times 7.5 = 60 \ years\)

- If the average weight of five students is \(40 \ kg\) and their weight ratio are \(2 : 4 : 5 : 6 : 8\) then find the students who have the lowest and highest weight?
- \(16 \ and \ 62 \ kg\)
- \(16 \ and \ 64 \ kg\)
- \(18 \ and \ 66 \ kg\)
- \(15 \ and \ 68 \ kg\)

Answer: (b) \(16 \ and \ 64 \ kg\)

Solution: Let the weight of the students are \(2x, 4x, 5x, 6x, \ and \ 8x \ kg\), then $$ 40 = \frac{2x + 4x + 5x + 6x + 8x}{5} $$ $$ 40 = \frac{25x}{5} $$ $$ 25x = 200 $$ $$ x = 8 $$ then lowest weight student \(= 2x = 2 \times 8 = 16 \ kg\)

highest weight student \(= 8x = 8 \times 8 = 64 \ kg\)

Solution: Let the weight of the students are \(2x, 4x, 5x, 6x, \ and \ 8x \ kg\), then $$ 40 = \frac{2x + 4x + 5x + 6x + 8x}{5} $$ $$ 40 = \frac{25x}{5} $$ $$ 25x = 200 $$ $$ x = 8 $$ then lowest weight student \(= 2x = 2 \times 8 = 16 \ kg\)

highest weight student \(= 8x = 8 \times 8 = 64 \ kg\)

- Find the average of first six odd numbers after multiplying by \(2 \ ?\)
- \(12\)
- \(14\)
- \(16\)
- \(18\)

Answer: (a) \(12\)

Solution: $$ Average = \frac{2 \ (1 + 3 + 5 + 7 + 9 + 11)}{6} $$ $$ = \frac{72}{6} = 12 $$

Solution: $$ Average = \frac{2 \ (1 + 3 + 5 + 7 + 9 + 11)}{6} $$ $$ = \frac{72}{6} = 12 $$

- If there are four numbers and average of all four numbers is \(100\) but the average of first three numbers is \(70\), then find the fourth number?
- \(210\)
- \(190\)
- \(160\)
- \(170\)

Answer: (b) \(190\)

Solution: Sum of all four numbers = \(4 \times 100 = 400\)

sum of the first three numbers = \(3 \times 70 = 210\)

Then the fourth number = \(400 - 210 = 190\)

Solution: Sum of all four numbers = \(4 \times 100 = 400\)

sum of the first three numbers = \(3 \times 70 = 210\)

Then the fourth number = \(400 - 210 = 190\)

Lec 1: Introduction to Average
Exercise-1
Lec 2: Weighted Average
Exercise-2
Exercise-3
Exercise-4
Exercise-5