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 there are two classes in a school \(A\) and \(B\), consisting of \(50\) and \(60\) students, respectively. If the average weight of the students of class \(A\) and \(B\) are \(55 \ kg\) and \(65 \ kg\), respectively. Find the average weight of the both the classes combined?
- \(45.124 \ kg\)
- \(55.417 \ kg\)
- \(50.417 \ kg\)
- \(60.124 \ kg\)

Answer: (b) \(55.417 \ kg\)

Solution: Given values, \(x_1 = 50\), \(x_2 = 60\), \(y_1 = 55\), \(y_2 = 65\), then average weight of the students of both classes combined, $$Average \ weight = \frac{x_1 \ y_1 + x_2 \ y_2}{y_1 + y_2}$$ $$ = \frac{50 \times 55 + 60 \times 65}{55 + 65} = \frac{2750 + 3900}{120} $$ $$ = \frac{6650}{120} = 55.417 \ kg $$

Solution: Given values, \(x_1 = 50\), \(x_2 = 60\), \(y_1 = 55\), \(y_2 = 65\), then average weight of the students of both classes combined, $$Average \ weight = \frac{x_1 \ y_1 + x_2 \ y_2}{y_1 + y_2}$$ $$ = \frac{50 \times 55 + 60 \times 65}{55 + 65} = \frac{2750 + 3900}{120} $$ $$ = \frac{6650}{120} = 55.417 \ kg $$

- If there are three different sections in the Army \(A\), \(B\), and \(C\), consisting \(50\), \(60\), and \(70\) members, respectively. If the average female members in section \(A\), \(B\), and \(C\) are \(20\), \(25\), and \(30\), respectively. Find the average female members in the army?
- \(61.34\)
- \(63.34\)
- \(53.63\)
- \(60.35\)

Answer: (a) \(61.34\)

Solution: Given values, \(x_1 = 50\), \(x_2 = 60\), \(x_3 = 70\), \(y_1 = 20\), \(y_2 = 25\), \(y_3 = 30\) then average female members in the Army, $$Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3}$$ $$ = \frac{50 \times 20 + 60 \times 25 + 70 \times 30}{20 + 25 + 30}$$ $$ = \frac{1000 + 1500 + 2100}{75} = \frac{4600}{75} = 61.34 $$

Solution: Given values, \(x_1 = 50\), \(x_2 = 60\), \(x_3 = 70\), \(y_1 = 20\), \(y_2 = 25\), \(y_3 = 30\) then average female members in the Army, $$Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3}$$ $$ = \frac{50 \times 20 + 60 \times 25 + 70 \times 30}{20 + 25 + 30}$$ $$ = \frac{1000 + 1500 + 2100}{75} = \frac{4600}{75} = 61.34 $$

- There are two groups of studentsin a class, If average age of first group students is \(15\) and that of second group students is \(18\). If group first consists of \(75\) students and group second consists of \(80\) students then find the average age of the students of the class?
- \(76.25\)
- \(77.73\)
- \(78.73\)
- \(75.05\)

Answer: (b) \(77.73\)

Solution: Given values, \(x_1 = 75\), \(x_2 = 80\), \(y_1 = 15\), \(y_2 = 18\), then average age of the students of the class, $$Average \ age = \frac{x_1 \ y_1 + x_2 \ y_2}{y_1 + y_2}$$ $$ = \frac{75 \times 15 + 80 \times 18}{15 + 18} = \frac{1125 + 1440}{33} $$ $$ = \frac{2565}{33} = 77.73 $$

Solution: Given values, \(x_1 = 75\), \(x_2 = 80\), \(y_1 = 15\), \(y_2 = 18\), then average age of the students of the class, $$Average \ age = \frac{x_1 \ y_1 + x_2 \ y_2}{y_1 + y_2}$$ $$ = \frac{75 \times 15 + 80 \times 18}{15 + 18} = \frac{1125 + 1440}{33} $$ $$ = \frac{2565}{33} = 77.73 $$

- If the average of six numbers is \(80\), but the average of first four numbers is \(75\) and average of last three numbers is \(85\), then find out the fourth number?
- \(72\)
- \(75\)
- \(70\)
- \(76\)

Answer: (b) \(75\)

Solution: Total value of six numbers \(= 80 \times 6 = 480\)

Total value of first four numbers \(= 75 \times 4 = 300\)

Total value of last three numbers \(= 85 \times 3 = 255\)

total value of first four and last three numbers \(= 300 + 255 = 555\)

Then, forth number $$ = 555 - 480 = 75 $$

Solution: Total value of six numbers \(= 80 \times 6 = 480\)

Total value of first four numbers \(= 75 \times 4 = 300\)

Total value of last three numbers \(= 85 \times 3 = 255\)

total value of first four and last three numbers \(= 300 + 255 = 555\)

Then, forth number $$ = 555 - 480 = 75 $$

- A man covers a certain distance with the speed of \(30 \ km/hr\), \(40 \ km/hr\), and \(50 \ km/hr\), then find the average speed of the man?
- \(50\)
- \(42\)
- \(40\)
- \(35\)

Answer: (c) \(40\)

Solution: Given values, \(k_1 = 30 \ km/hr\), \(k_2 = 40 \ km/hr\), and \(k_3 = 50 \ km/hr\), then average speed of the man, $$Average = \frac{k_1 + k_2 + k_3}{3}$$ $$= \frac{30 + 40 + 50}{3} = \frac{120}{3} = 40 \ km/hr$$

Solution: Given values, \(k_1 = 30 \ km/hr\), \(k_2 = 40 \ km/hr\), and \(k_3 = 50 \ km/hr\), then average speed of the man, $$Average = \frac{k_1 + k_2 + k_3}{3}$$ $$= \frac{30 + 40 + 50}{3} = \frac{120}{3} = 40 \ km/hr$$

- If a student scored average \(20\) marks in \(2\) tests of History, \(45\) marks in \(3\) tests of Geography and \(55\) marks in \(3\) tests of English. Find average marks obtained by the student?
- \(40.2\)
- \(41.2\)
- \(42.5\)
- \(43.5\)

Answer: (c) \(42.5\)

Solution: Given values, \(x_1 = 20\), \(x_2 = 45\), \(x_3 = 55\), \(y_1 = 2\), \(y_2 = 3\), \(y_3 = 3\), then average marks obtained by the student $$ Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3} $$ $$ = \frac{20 \times 2 + 45 \times 3 + 55 \times 3}{2 + 3 + 3}$$ $$ = \frac{40 + 135 + 165}{8}$$ $$= \frac{340}{8} = 42.5 \ marks $$

Solution: Given values, \(x_1 = 20\), \(x_2 = 45\), \(x_3 = 55\), \(y_1 = 2\), \(y_2 = 3\), \(y_3 = 3\), then average marks obtained by the student $$ Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3} $$ $$ = \frac{20 \times 2 + 45 \times 3 + 55 \times 3}{2 + 3 + 3}$$ $$ = \frac{40 + 135 + 165}{8}$$ $$= \frac{340}{8} = 42.5 \ marks $$

- An Indian cricketer scored average \(40\) runs in three matches against England, \(60\) runs in four matches against Sri-Lanka, and \(70\) runs in five matches against Australia. Find average runs scored by the cricketer?
- \(55.78\)
- \(57.56\)
- \(58.17\)
- \(59.17\)

Answer: (d) \(59.17\)

Solution: Given values, \(x_1 = 40\), \(x_2 = 60\), \(x_3 = 70\), \(y_1 = 3\), \(y_2 = 4\), \(y_3 = 5\), then average runs scored by the cricketer $$ Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3} $$ $$ = \frac{40 \times 3 + 60 \times 4 + 70 \times 5}{3 + 4 + 5}$$ $$ = \frac{120 + 240 + 350}{12}$$ $$= \frac{710}{12} = 59.17 \ runs $$

Solution: Given values, \(x_1 = 40\), \(x_2 = 60\), \(x_3 = 70\), \(y_1 = 3\), \(y_2 = 4\), \(y_3 = 5\), then average runs scored by the cricketer $$ Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3} $$ $$ = \frac{40 \times 3 + 60 \times 4 + 70 \times 5}{3 + 4 + 5}$$ $$ = \frac{120 + 240 + 350}{12}$$ $$= \frac{710}{12} = 59.17 \ runs $$

- If the average of eight numbers is \(65\), but the average of first three numbers is \(60\) and average of last four numbers is \(70\), then find out the fourth number?
- \(56\)
- \(55\)
- \(60\)
- \(58\)

Answer: (c) \(60\)

Solution: Total value of eight numbers \(= 65 \times 8 = 520\)

Total value of first three numbers \(= 60 \times 3 = 180\)

Total value of last four numbers \(= 70 \times 4 = 280\)

total value of first three and last four numbers \(= 180 + 280 = 460\)

Then, forth number $$ = 520 - 460 = 60 $$

Solution: Total value of eight numbers \(= 65 \times 8 = 520\)

Total value of first three numbers \(= 60 \times 3 = 180\)

Total value of last four numbers \(= 70 \times 4 = 280\)

total value of first three and last four numbers \(= 180 + 280 = 460\)

Then, forth number $$ = 520 - 460 = 60 $$

- Find the average of first four natural numbers, multiple of five?
- \(13.5\)
- \(12\)
- \(12.5\)
- \(13\)

Answer: (c) \(12.5\)

Solution: $$Average = \frac{5 \ (1 + 2 + 3 + 4)}{4} $$ $$ = \frac{5 + 10 + 15 + 20}{4} = \frac{50}{4} = 12.5 $$

Solution: $$Average = \frac{5 \ (1 + 2 + 3 + 4)}{4} $$ $$ = \frac{5 + 10 + 15 + 20}{4} = \frac{50}{4} = 12.5 $$

- If there is a class of \(30\) students, Average weight of class is \(60 \ kg\) and weight of the monitor of the class also included. If weight of class decreases by \(1 \ kg\), then find out the weight of the monitor?
- \(29 \ kg\)
- \(28.5 \ kg\)
- \(30 \ kg\)
- \(27.25 \ kg\)

Answer: (a) \(29 \ kg\)

Solution: Let the weight of the monitor \(= x \ kg\)

Average weight of the class after \(1 \ kg\) weight decreases \(= 60 - 1 = 59 \ kg\), then weight of the monitor, $$ 59 = \frac{30 \times 60 + x}{31} $$ $$ x = 31 \times 59 - 1800 $$ $$ x = 1829 - 1800 = 29 \ kg $$

Solution: Let the weight of the monitor \(= x \ kg\)

Average weight of the class after \(1 \ kg\) weight decreases \(= 60 - 1 = 59 \ kg\), then weight of the monitor, $$ 59 = \frac{30 \times 60 + x}{31} $$ $$ x = 31 \times 59 - 1800 $$ $$ x = 1829 - 1800 = 29 \ kg $$

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