Quantitative Aptitude

Averages

1,527 Questions
All Basic average and reverse calculation 405 Adding or removing an observation 342 Weighted and combined averages 320 Replacement and correction of a recorded value 233 Ratio shortcut for two-group mixtures 111 Equal-distance average speed 59 Age, score, weight, and spending applications 32 Consecutive terms and arithmetic progressions 25

Averages Questions

Multiple choice

Soumya and Aarushi wanted to find the average of their marks in five different subjects. Average marks obtained by Aarushi was 5 more than Soumya. Average marks of Soumya was 36 but later on it was discovered that one of the marks was read as 14 instead of 41. What would be the average marks of Aarushi?

  1. 46.4

  2. 41.2

  3. 36.6

  4. 25.4

  5. 22.8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Soumya's correct sum = (36 * 5) - 14 + 41 = 180 - 14 + 41 = 207. Correct average = 207 / 5 = 41.4. Aarushi's average = 41.4 + 5 = 46.4.

Multiple choice

The average weight of ten students of class is 40 kg. If the lightest and heaviest student are not taken into account then average weight of remaining students is 41 kg. If the weight of heaviest student is 50 kg, find the weight of lightest student. (in Kg)

  1. 21

  2. 22

  3. 23

  4. 24

  5. 25
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total weight of 10 students = 400. Weight of 8 students (excluding lightest and heaviest) = 41 * 8 = 328. Sum of lightest and heaviest = 400 - 328 = 72. Since heaviest is 50, lightest = 72 - 50 = 22.

Multiple choice

In each of the following questions, read the given statement and compare the Quantity I and Quantity II on its basis. (only quantity is to be considered) Quantity I : The average marks obtained by 3 departments is 848. If the average marks obtained by the students of physics department of 15 students is 64 and the average marks of the students of chemistry department of 12 students is 72. Find the average marks of the students of Maths department having 9 students. Quantity II : The average marks of four students in Mathematics is 68. Later it was found that, one of the students got 88 marks and due to error in marking he got 76. Find the actual average marks of the four students.

  1. Quantity : I > Quantity : II

  2. Quantity : I ≥ Quantity : II

  3. Quantity : I < Quantity : II

  4. Quantity : II ≥ Quantity : I

  5. Quantity I = Quantity II or relation can't be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice

Robin gets an average aggregate of 72% marks in 5 subjects in the ratio 10:9:8:7:6. If the passing marks are 65% of maximum marks and each subject has the same maximum marks, in how many subjects did he pass?

  1. 3

  2. 1

  3. 2

  4. 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Average 72% for 5 subjects means total marks = 360%. Ratio 10:9:8:7:6 means parts are 40. Each part = 360/40 = 9. Marks: 90, 81, 72, 63, 54. Passing marks = 65. Subjects with >= 65 are 90, 81, 72. Total 3 subjects.

Multiple choice

Alastair and Ben study in the same class. The average score of the class in a Sanskrit exam for which only Ben was absent was 52 while the average score of the class in a Marathi exam for which only Alastair was absent was 68. If the total score of the class in Sanskrit and Marathi exam was 3840, then how many students were there in the class?

  1. 32

  2. 65

  3. 66

  4. 33
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let n be the number of students. Total score = (n-1)*52 + (n-1)*68 = 3840. So (n-1)*120 = 3840. n-1 = 32, so n = 33.

Multiple choice

[CAT 1994] The question is followed by two statements, I and II. Mark the answer as per the given conditions. What is the average weight of the 3 new team members who are recently included into the team? I. The average weight of the team increases by 20 kg. II. The 3 new men substitute earlier members whose weights are 64 kg, 75 kg and 66 kg.

  1. Statement I alone is required to answer the question, but not statement II.

  2. Statement II alone is required to answer the question, but not statement I.

  3. Cannot be answered even with the help of both the statements taken together.

  4. Both statements I and II together are needed to answer the question.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let n be the number of team members. Let S be the sum of weights. New average = (S - 64 - 75 - 66 + W1 + W2 + W3) / n. We have two unknowns (n and the sum of new weights) and only one equation from the average increase. We cannot determine the average weight of the 3 new members.

Multiple choice

A batsman scores 92 runs in the 15th innings and thus increases his average by 4. What is his average after the 15th innings?

  1. 40

  2. 32

  3. 35

  4. 36
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let x be the average after 14 innings. Total runs after 14 innings = 14x. After the 15th innings, the total is 14x + 92, and the new average is x + 4. So, (14x + 92) / 15 = x + 4. 14x + 92 = 15x + 60, which gives x = 32. The average after the 15th innings is x + 4 = 36.

Multiple choice

4 years ago, the average age of a family of 5 memberswas 20 years. A baby having been born, the average age ofthe family is same today. The present age of the baby is?

  1. 0 year

  2. (3/2) years

  3. (7/2) years

  4. 2 years

  5. 4 years
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Four years ago, the sum of the ages of the 5 family members was 100 years. Today, each of those 5 members is 4 years older, making their sum 120 years. Since the average age of the 6-member family today is still 20 years, the total sum of their ages is 120 years, meaning the baby's age is 0 years.

Multiple choice

Find the average of the three original numbers, if three different numbers are chosen such that when each of the numbers is added to the average of the remaining two, it gives 35, 15 and 45 as result.

  1. 15.83

  2. 18.83

  3. 25.83

  4. 35.93

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the numbers be x, y, z and their average be A = (x+y+z)/3. The equations are x + (y+z)/2 = 35, y + (x+z)/2 = 15, and z + (x+y)/2 = 45. Substituting (y+z) = 3A - x into the first gives x + (3A-x)/2 = 35, which simplifies to x + 3A = 70. Summing the three such equations gives 3(x+y+z) + 3A = 190, or 9A + 3A = 190, so 12A = 190, A = 15.833.

Multiple choice

Directions: The question below contains a statement followed by Quantity I and Quantity II. Find both the quantities to determine the relationship among them and mark your answer accordingly. In a cricket match, the average score of a cricketer in first five matches was 60. The average for the second, third, fourth, fifth and six matches was 70. His score in the first and sixth matches was in the ratio 1 : 3. Quantity I: Average of the score in the first and sixth matches Quantity II: 50

  1. Quantity I > Quantity II

  2. Quantity I < Quantity II

  3. Quantity I ≤ Quantity II

  4. Quantity I = Quantity II, or the relationship cannot be established from the information given.

  5. Quantity I ≥ Quantity II
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Sum of first 5 matches = 300. Sum of matches 2 to 6 = 350. Let M1 be first and M6 be sixth. M6 - M1 = 50. Given M1/M6 = 1/3, M6 = 3*M1. 2*M1 = 50, M1 = 25, M6 = 75. Average of M1 and M6 = (25+75)/2 = 50. Quantity I = 50, Quantity II = 50.