# Average Aptitude Questions with Solutions:

#### Overview:

 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.

1. 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?

1. $$45.124 \ kg$$
2. $$55.417 \ kg$$
3. $$50.417 \ kg$$
4. $$60.124 \ kg$$

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, $$Average \ weight = \frac{x_1 \ y_1 + x_2 \ y_2}{y_1 + y_2}$$ $$= \frac{50 \times 55 + 60 \times 65}{55 + 65} = \frac{2750 + 3900}{120}$$ $$= \frac{6650}{120} = 55.417 \ kg$$

1. 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?

1. $$61.34$$
2. $$63.34$$
3. $$53.63$$
4. $$60.35$$

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, $$Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3}$$ $$= \frac{50 \times 20 + 60 \times 25 + 70 \times 30}{20 + 25 + 30}$$ $$= \frac{1000 + 1500 + 2100}{75} = \frac{4600}{75} = 61.34$$

1. 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?

1. $$76.25$$
2. $$77.73$$
3. $$78.73$$
4. $$75.05$$

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, $$Average \ age = \frac{x_1 \ y_1 + x_2 \ y_2}{y_1 + y_2}$$ $$= \frac{75 \times 15 + 80 \times 18}{15 + 18} = \frac{1125 + 1440}{33}$$ $$= \frac{2565}{33} = 77.73$$

1. 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?

1. $$72$$
2. $$75$$
3. $$70$$
4. $$76$$

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 $$= 555 - 480 = 75$$

1. 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?

1. $$50$$
2. $$42$$
3. $$40$$
4. $$35$$

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, $$Average = \frac{k_1 + k_2 + k_3}{3}$$ $$= \frac{30 + 40 + 50}{3} = \frac{120}{3} = 40 \ km/hr$$

1. 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?

1. $$40.2$$
2. $$41.2$$
3. $$42.5$$
4. $$43.5$$

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 $$Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3}$$ $$= \frac{20 \times 2 + 45 \times 3 + 55 \times 3}{2 + 3 + 3}$$ $$= \frac{40 + 135 + 165}{8}$$ $$= \frac{340}{8} = 42.5 \ marks$$

1. 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?

1. $$55.78$$
2. $$57.56$$
3. $$58.17$$
4. $$59.17$$

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 $$Average = \frac{x_1 \ y_1 + x_2 \ y_2 + x_3 \ y_3}{y_1 + y_2 + y_3}$$ $$= \frac{40 \times 3 + 60 \times 4 + 70 \times 5}{3 + 4 + 5}$$ $$= \frac{120 + 240 + 350}{12}$$ $$= \frac{710}{12} = 59.17 \ runs$$

1. 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?

1. $$56$$
2. $$55$$
3. $$60$$
4. $$58$$

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 $$= 520 - 460 = 60$$

1. Find the average of first four natural numbers, multiple of five?

1. $$13.5$$
2. $$12$$
3. $$12.5$$
4. $$13$$

Answer: (c) $$12.5$$

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

1. 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?

1. $$29 \ kg$$
2. $$28.5 \ kg$$
3. $$30 \ kg$$
4. $$27.25 \ kg$$

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, $$59 = \frac{30 \times 60 + x}{31}$$ $$x = 31 \times 59 - 1800$$ $$x = 1829 - 1800 = 29 \ kg$$