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 the average of prime numbers between \(10\) and \(20 \ ?\)
- \(18\)
- \(16\)
- \(17\)
- \(15\)

Answer: (d) \(15\)

Solution: Prime numbers between \(10\) and \(20\) = \(11, 13, 17, 19\), then $$ Average = \frac{11 + 13 + 17 + 19}{4} $$ $$ = \frac{60}{4} = 15 $$

Solution: Prime numbers between \(10\) and \(20\) = \(11, 13, 17, 19\), then $$ Average = \frac{11 + 13 + 17 + 19}{4} $$ $$ = \frac{60}{4} = 15 $$

- If a man received salary of \(20,000 \ Rs.\) in first quarter of the year, \(30,000 \ Rs.\) in second quarter of the year, \(50,000 \ Rs.\) in third quarter of the year and \(70,000 \ Rs.\) in last quarter of the year. Find the average salary of the man for whole year?
- \(43,000\)
- \(42,000\)
- \(43,500\)
- \(42,500\)

Answer: (d) \(42,500\)

Solution: $$Average = \frac{20,000 + 30,000 + 50,000 + 70,000}{4}$$ $$= \frac{170,000}{4} = 42,500 \ Rs.$$

Solution: $$Average = \frac{20,000 + 30,000 + 50,000 + 70,000}{4}$$ $$= \frac{170,000}{4} = 42,500 \ Rs.$$

- If the ratio of age of two friends is \(3 : 5\) and after \(2\) years ratio of their age becomes \(4 : 6\) then find the present age of friends and average age of the friends after five years?
- Present age \(6\) and \(10\) years, average age after five years is \(13\) years
- Present age \(8\) and \(12\) years, average age after five years is \(15\) years
- Present age \(6\) and \(13\) years, average age after five years is \(12\) years
- Present age \(6\) and \(11\) years, average age after five years is \(13\) years

Answer: (a) Present age \(6\) and \(10\) years, average age after five years is \(13\) years.

Solution: Let the present age of the friends are \(3x\) and \(5x\).

then ages after \(2\) years $$ \frac{3x + 2}{5x + 2} = \frac{4}{6} $$ $$18x + 12 = 20x + 8 $$ $$ x = 2 $$ The present age of the friends are \(6\) years and \(10\) years.

After \(2\) years the age of the friends will be \(8\) years and \(12\) years.

After \(5\) years the age of the friends will be \(6 + 5 = 11\) years and \(10 + 5 = 15\) years.

The average age of the friends after \(5\) years $$ = \frac{11 + 15}{2} = \frac{26}{2} = 13 \ years$$

Solution: Let the present age of the friends are \(3x\) and \(5x\).

then ages after \(2\) years $$ \frac{3x + 2}{5x + 2} = \frac{4}{6} $$ $$18x + 12 = 20x + 8 $$ $$ x = 2 $$ The present age of the friends are \(6\) years and \(10\) years.

After \(2\) years the age of the friends will be \(8\) years and \(12\) years.

After \(5\) years the age of the friends will be \(6 + 5 = 11\) years and \(10 + 5 = 15\) years.

The average age of the friends after \(5\) years $$ = \frac{11 + 15}{2} = \frac{26}{2} = 13 \ years$$

- Find the average of first \(20\) natural numbers?
- \(13.5\)
- \(11.5\)
- \(10.5\)
- \(12.5\)

Answer: (c) \(10.5\)

Solution: Sum of the first \(n\) natural numbers $$ = \frac{n \ (n + 1)}{2}$$ $$ = \frac{20 \ (20 + 1)}{2} = \frac{20 \times 21}{2} $$ $$ = \frac{420}{2} = 210 $$ $$ Average = \frac{210}{20} = 10.5 $$

Solution: Sum of the first \(n\) natural numbers $$ = \frac{n \ (n + 1)}{2}$$ $$ = \frac{20 \ (20 + 1)}{2} = \frac{20 \times 21}{2} $$ $$ = \frac{420}{2} = 210 $$ $$ Average = \frac{210}{20} = 10.5 $$

- Find the average of all prime numbers between \(20\) and \(30 \ ?\)
- \(25\)
- \(26\)
- \(28\)
- \(24\)

Answer: (b) \(26\)

Solution: Prime numbers between \(20\) to \(30\) = \(23, 29\). Then $$ Average = \frac{23 + 29}{2} = \frac{52}{2} = 26 $$

Solution: Prime numbers between \(20\) to \(30\) = \(23, 29\). Then $$ Average = \frac{23 + 29}{2} = \frac{52}{2} = 26 $$

- Find the average of first \(50\) odd natural numbers?
- \(45\)
- \(48\)
- \(50\)
- \(49\)

Answer: (c) \(50\)

Solution: Sum of first \(n\) natural numbers \(S_n = n^2\), then $$ S_{50} = (50)^2 = 2500 $$ $$ Average = \frac{2500}{50} = 50 $$

Solution: Sum of first \(n\) natural numbers \(S_n = n^2\), then $$ S_{50} = (50)^2 = 2500 $$ $$ Average = \frac{2500}{50} = 50 $$

- Find the average of first \(70\) even natural numbers?
- \(70\)
- \(71\)
- \(72\)
- \(73\)

Answer: (b) \(71\)

Solution: Sum of first \(n\) even natural numbers \(S_n = n \ (n + 1)\), then $$ S_{70} = 70 \ (70 + 1) = 70 \times 71 = 4970 $$ $$ Average = \frac{4970}{70} = 71 $$

Solution: Sum of first \(n\) even natural numbers \(S_n = n \ (n + 1)\), then $$ S_{70} = 70 \ (70 + 1) = 70 \times 71 = 4970 $$ $$ Average = \frac{4970}{70} = 71 $$

- Find the average of squares of first \(20\) natural numbers?
- \(142.3\)
- \(143.5\)
- \(141.5\)
- \(140.3\)

Answer: (b) \(143.5\)

Solution: Sum of squares of first \(n\) natural numbers, $$ S_n = \frac{n \ (n + 1) \ (2n + 1)}{6} $$ $$ S_{20} = \frac{20 \ (20 + 1) \ (40 + 1)}{6} $$ $$ = \frac{20 \times 21 \times 41}{6} $$ $$ = \frac{17220}{6} = 2870 $$ $$ Average = \frac{2870}{20} = 143.5 $$

Solution: Sum of squares of first \(n\) natural numbers, $$ S_n = \frac{n \ (n + 1) \ (2n + 1)}{6} $$ $$ S_{20} = \frac{20 \ (20 + 1) \ (40 + 1)}{6} $$ $$ = \frac{20 \times 21 \times 41}{6} $$ $$ = \frac{17220}{6} = 2870 $$ $$ Average = \frac{2870}{20} = 143.5 $$

- Find the average of cubes of first \(15\) natural numbers?
- \(850\)
- \(900\)
- \(955\)
- \(960\)

Answer: (d) \(960\)

Solution: Sum of cubes of first \(n\) natural numbers, $$ S_n = \left[\frac{n \ (n + 1)}{2}\right]^2 $$ $$ S_{15} = \left[\frac{15 \ (15 + 1)}{2}\right]^2 $$ $$ = \left(\frac{15 \times 16}{2}\right)^2 $$ $$ = (120)^2 = 14,400 $$ $$ Average = \frac{14,400}{15} = 960 $$

Solution: Sum of cubes of first \(n\) natural numbers, $$ S_n = \left[\frac{n \ (n + 1)}{2}\right]^2 $$ $$ S_{15} = \left[\frac{15 \ (15 + 1)}{2}\right]^2 $$ $$ = \left(\frac{15 \times 16}{2}\right)^2 $$ $$ = (120)^2 = 14,400 $$ $$ Average = \frac{14,400}{15} = 960 $$

- A train going from Delhi to Kanpur at the speed of \(60 \ km/hr\) and returns from Kanpur to Delhi at the speed of \(80 \ km/hr\). Find the average speed of the train during the whole journey?
- \(67.57 \ km/hr\)
- \(68.57 \ km/hr\)
- \(69.57 \ km/hr\)
- \(70.52 \ km/hr\)

Answer: (b) \(68.57 \ km/hr\)

Solution: Let the distance between Delhi to Kanpur is \(x \ km\)

Then the average speed of the train, $$ = \frac{2x}{\frac{x}{60} + \frac{x}{80}} $$ $$ = \frac{2x \times 240}{7x} = \frac{480}{7} = 68.57 \ km/hr $$

Solution: Let the distance between Delhi to Kanpur is \(x \ km\)

Then the average speed of the train, $$ = \frac{2x}{\frac{x}{60} + \frac{x}{80}} $$ $$ = \frac{2x \times 240}{7x} = \frac{480}{7} = 68.57 \ km/hr $$

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