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. |

- If the distance between two stations \(A\) and \(B\) is \(200 \ km\). A train covers the distance from \(A\) to \(B\) At the average speed of \(100 \ km/hr\) and returns from \(B\) to \(A\) at the average speed of \(x \ km/hr\). The average speed of the train during the whole journey is \(70 \ km/hr\), find the value of \(x \ ?\)
- \(53 \ km/hr\)
- \(52.84 \ km/hr\)
- \(53.84 \ km/hr\)
- \(55.23 \ km/hr\)

Answer: (c) \(53.84 \ km/hr\)

Solution: Given values, \(x_1 = 100 \ km/hr\), \(x_2 = x \ km/hr\), and the average speed \(= 70 \ km/hr\), then $$ Average \ speed = \frac{2 \ x_1 \ x_2}{x_1 + x_2} $$ $$ 70 = \frac{2 \times 100 \times x}{100 + x} $$ $$ 7000 + 70 \ x = 200 \ x $$ $$ 130 \ x = 7000 $$ $$ x = \frac{7000}{130} = 53.84 \ km/hr $$

Solution: Given values, \(x_1 = 100 \ km/hr\), \(x_2 = x \ km/hr\), and the average speed \(= 70 \ km/hr\), then $$ Average \ speed = \frac{2 \ x_1 \ x_2}{x_1 + x_2} $$ $$ 70 = \frac{2 \times 100 \times x}{100 + x} $$ $$ 7000 + 70 \ x = 200 \ x $$ $$ 130 \ x = 7000 $$ $$ x = \frac{7000}{130} = 53.84 \ km/hr $$

- A man travels first \(60 \ km\) in \(1 / hr\) and second \(50 \ km\) in \(x \ hr\). If the average speed of the man is \(40 \ km/hr\) then find the value of \(x \ ?\)
- \(1.05 \ hr\)
- \(1.25 \ hr\)
- \(1.50 \ hr\)
- \(1.75 \ hr\)

Answer: (d) \(1.75 \ hr\)

Solution: Given values, \(d_1 = 60 \ km\), \(d_2 = 50 \ km\), \(t_1 = 1 \ hr\), \(t_2 = x \ hr\) and average speed \(= 40 \ km/hr\) then, $$ Average \ speed = \frac{d_1 + d_2}{t_1 + t_2} $$ $$ 40 = \frac{60 + 50}{1 + x} $$ $$ 40 + 40 \ x = 110 $$ $$ x = \frac{70}{40} = 1.75 \ hr $$

Solution: Given values, \(d_1 = 60 \ km\), \(d_2 = 50 \ km\), \(t_1 = 1 \ hr\), \(t_2 = x \ hr\) and average speed \(= 40 \ km/hr\) then, $$ Average \ speed = \frac{d_1 + d_2}{t_1 + t_2} $$ $$ 40 = \frac{60 + 50}{1 + x} $$ $$ 40 + 40 \ x = 110 $$ $$ x = \frac{70}{40} = 1.75 \ hr $$

- Find the average of numbers between \(10\) to \(30\) divisible by \(3 \ ?\)
- \(18.5\)
- \(19.5\)
- \(20.5\)
- \(16.5\)

Answer: (b) \(19.5\).

Solution: First number divisible by \(3\) after \(10\) is \(12\).

last number divisible by \(3\) before \(30\) is \(27\). Then $$ Average = \frac{12 + 27}{2} $$ $$ = \frac{39}{2} = 19.5 $$

Solution: First number divisible by \(3\) after \(10\) is \(12\).

last number divisible by \(3\) before \(30\) is \(27\). Then $$ Average = \frac{12 + 27}{2} $$ $$ = \frac{39}{2} = 19.5 $$

- The average of five numbers is \(70\), if one number is excluded then Average becomes \(80\). Find the excluded number?
- \(30\)
- \(32\)
- \(35\)
- \(36\)

Answer: (a) \(30\)

Solution: Average increases = \(80 - 70 = 10\)

Total increased number = \(10 \times 4 = 40\)

It means is \(40\) less the average, then excluded number = \(70 - 40 = 30\)

Solution: Average increases = \(80 - 70 = 10\)

Total increased number = \(10 \times 4 = 40\)

It means is \(40\) less the average, then excluded number = \(70 - 40 = 30\)

- A bus covers \(100 \ km\) distance, first \(30 \ km\) at the speed of \(50 \ km/hr\), second \(30 \ km\) at the speed of \(60 \ km/hr\) and last \(40 \ km\) at the speed of \(x \ km/hr\). If the average speed of the bus during the whole journey is \(65 \ km/hr\), then find the value of \(x \ ?\)
- \(86\)
- \(85\)
- \(87\)
- \(88\)

Answer: (b) \(85\)

Solution: $$ Average \ speed = \frac{k_1 + k_2 + k_3}{3} $$ $$ 65 = \frac{50 + 60 + x}{3} $$ $$ 195 = 110 + x $$ $$ x = 85 \ km/hr $$

Solution: $$ Average \ speed = \frac{k_1 + k_2 + k_3}{3} $$ $$ 65 = \frac{50 + 60 + x}{3} $$ $$ 195 = 110 + x $$ $$ x = 85 \ km/hr $$

- Find the average of first four even natural numbers after multiplying by \(5\) and divide by by \(2 \ ?\)
- \(13.5\)
- \(14.5\)
- \(12.5\)
- \(10.5\)

Answer: (c) \(12.5\)

Solution: First four even natural numbers = \(2, 4, 6, 8\)

First four even natural numbers after multiplying by \(5\) and divide by \(2\) = \(5, 10, 15, 20\) then, $$ Average = \frac{5 + 10 + 15 + 20}{4} $$ $$ = \frac{50}{4} = 12.5 $$

Solution: First four even natural numbers = \(2, 4, 6, 8\)

First four even natural numbers after multiplying by \(5\) and divide by \(2\) = \(5, 10, 15, 20\) then, $$ Average = \frac{5 + 10 + 15 + 20}{4} $$ $$ = \frac{50}{4} = 12.5 $$

- If the average of \(8\) numbers is \(30\) then find the average after multiplying the numbers by \(5 \ ?\)
- \(150\)
- \(160\)
- \(158\)
- \(156\)

Answer: (a) \(150\)

Solution: Average of \(8\) numbers is \(30\).

total value of \(8\) numbers = \(8 \times 30 = 240\)

total value of \(8\) numbers after multiplying the numbers by \(5\) = \(240 \times 5 = 1200\) then, $$ Average = \frac{1200}{8} = 150 $$

Solution: Average of \(8\) numbers is \(30\).

total value of \(8\) numbers = \(8 \times 30 = 240\)

total value of \(8\) numbers after multiplying the numbers by \(5\) = \(240 \times 5 = 1200\) then, $$ Average = \frac{1200}{8} = 150 $$

- If the average weight of four men is \(65 \ kg\) and the ratio of their weight is \(2 : 4 : 3 : 5\), then find the weight of lowest weight man?
- \(35.50 \ kg\)
- \(37.25 \ kg\)
- \(37.14 \ kg\)
- \(41.26 \ kg\)

Answer: (c) \(37.14 \ kg\)

Solution: Let the weight of men = \(2x, 4x, 3x, 5x\) then, $$ 65 = \frac{2x + 4x + 3x + 5x}{4} $$ $$ 14x = 260 $$ $$ x = 18.57 $$ the weight of lowest weight man = \(2 \times 18.57 = 37.14 \ kg\)

Solution: Let the weight of men = \(2x, 4x, 3x, 5x\) then, $$ 65 = \frac{2x + 4x + 3x + 5x}{4} $$ $$ 14x = 260 $$ $$ x = 18.57 $$ the weight of lowest weight man = \(2 \times 18.57 = 37.14 \ kg\)

- Find the average of first three natural numbers of multiplying by \(10\) and divide by \(5 \ ?\)
- \(2\)
- \(4\)
- \(6\)
- \(8\)

Answer: (b) \(4\)

Solution: First three natural numbers = \(1, 2, 3\)

First three natural numbers after multiplying by \(10\) and divide by \(5\) = \(2 , 4, 6\) then, $$ Average = \frac{2 + 4 + 6}{3} = \frac{12}{3} = 4 $$

Solution: First three natural numbers = \(1, 2, 3\)

First three natural numbers after multiplying by \(10\) and divide by \(5\) = \(2 , 4, 6\) then, $$ Average = \frac{2 + 4 + 6}{3} = \frac{12}{3} = 4 $$

- If a man travels three consecutive distances \(100 \ km\), \(150 \ km\), and \(200 \ km\), at the speed of \(45 \ km/hr\), \(55 \ km/hr\), and \(x \ km/hr\). If the average speed of the man during the whole journey is \(60 \ km/hr\), then find the value of \(x \ ?\)
- \(75 \ km/hr\)
- \(88 \ km/hr\)
- \(82 \ km/hr\)
- \(80 \ km/hr\)

Answer: (d) \(80 \ km/hr\)

Solution: $$ Average \ speed = \frac{k_1 + k_2 + k_3}{3} $$ $$ 60 = \frac{45 + 55 + x}{3} $$ $$ x = 80 \ km/hr $$

Solution: $$ Average \ speed = \frac{k_1 + k_2 + k_3}{3} $$ $$ 60 = \frac{45 + 55 + x}{3} $$ $$ x = 80 \ km/hr $$

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