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CAT Exam  >  CAT Test  >  Level-wise Tests  >  Test Level 1: Averages - 1 - CAT MCQ

Test Level 1: Averages - 1 CAT MCQs & solutions - Free


MCQ Practice Test & Solutions: Test Level 1: Averages - 1 (10 Questions)

You can prepare effectively for CAT Level-wise Tests for CAT with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test Level 1: Averages - 1". These 10 questions have been designed by the experts with the latest curriculum of CAT 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 20 minutes
  • - Number of Questions: 10

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Test Level 1: Averages - 1 - Question 1
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Find the average of five consecutive even numbers a, b, c, d and e.

Detailed Solution: Question 1

The consecutive even numbers are a, b, c, d, e.
Clearly, b = a + 2, c = a + 4, d = a + 6, e = a + 8

Note: Whenever difference between any two consecutive terms is constant (means the series is an arithmetic series), the average of all the numbers is

(i) the middle number if the number of terms is odd.

(ii) the average of the middle two numbers if the number of terms is even.

Test Level 1: Averages - 1 - Question 2
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At present, the average age of a father and his son is 29 years. The average age of father, mother and son five years from now will be 37 years. Find the mother's present age.  

Detailed Solution: Question 2

Let present age of father = F years
Let present age of son = S years
Let present age of mother = M years
Now, F + S = 29 × 2 = 58 years
Also, F + M + S + 15 = 3 × 37 = 111 years
On solving, we get
M = 38 years
Thus, present age of mother = 38 years

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Test Level 1: Averages - 1 - Question 3
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A student scores an average of 80 marks in six subjects. If the subjects with the highest and the lowest scores are excluded, the average decreases by 1. If his highest score in a subject is 86, what is his lowest score?

Detailed Solution: Question 3

According to the question,
480 - 86 - x = 79  4 = 316
Or, lowest score = x = 480 - 86 - 316 = 78

Test Level 1: Averages - 1 - Question 4
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The mean of 19 observations is 4. If one more observation of 24 is added to the data, the new mean will be

Detailed Solution: Question 4

The mean of 19 observations is 4.

New data when 24 is added:
76 + 24 = 100
New number of observations = 19 + 1 = 20

Test Level 1: Averages - 1 - Question 5
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The average monthly salary of employees, consisting of officers and workers, of an organisation is Rs. 3000. The average salary of an officer is Rs. 10,000, while that of a worker is Rs. 2000 per month. If there are total 400 employees in the organisation, find the number of officers.

Detailed Solution: Question 5

Let the number of officers be x.
Number of workers = 400 - x
10,000x + 2000(400 - x) = 3000(400)
10x + 800 - 2x = 1200
8x = 400
x = 50
Hence, the number of officers is 50.

Test Level 1: Averages - 1 - Question 6
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The average of ten numbers is 27. If the smallest number is deleted from the list, the average of the remaining 9 numbers will be x and if the largest number is deleted, the average will be y. What is the average of the smallest and the largest numbers, if x + y = 10?

Detailed Solution: Question 6

Let the sum of all 10 numbers be S.
S = 10 × 27 = 270
Let the smallest number be s and the largest number be L.
So, S - s = 9x ... (i)
Also, S - L = 9y ... (ii)
Adding (i) and (ii), we get
2S - (s + L) = 9(x + y) = 9(x + y) = 9(10) = 90
Or 2S - 90 = (s + L)

Or 270 - 45 = 225

= 

Thus, answer option (1) is correct.

Test Level 1: Averages - 1 - Question 7
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How many pairs of positive integers, not more than 100, will have an average greater than 50?

Detailed Solution: Question 7

Average of 2 numbers is greater than 50 means that the sum should be greater than 100.
Different pairs are as follows:
(1, 100) … (1 pair)
(2, 99), (2, 100) … (2 pairs)
(3, 98), (3, 99), (3, 100) … (3 pairs)
... so on
(49, 52), (49, 53) ... (49, 100) ... (49 pairs)
(50, 51), (50, 52) … (50, 100) ... (50 pairs)
So, 1 + 2 + 3 + 4 + ... + 50 = 1275
Again
(51, 51), (51, 52), (51, 53), ... (51, 100) ... (50 pairs)
(52, 52), (52, 53), (52, 54), ... (52, 100) ... (49 pairs)
(53, 53), (53, 54), (53, 55), ... (53, 100) ... (48 pairs)
... so on
(99, 99), (99, 100) ... (2 pairs)
(100, 100) ... (1 pair)
So, 1 + 2 + 3 + 4 + ... + 50 = 1275
Hence, total = 1275 + 1275 = 2550 pairs

Test Level 1: Averages - 1 - Question 8
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The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.

Detailed Solution: Question 8

Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 - 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

Test Level 1: Averages - 1 - Question 9
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Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

Detailed Solution: Question 9

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

⇒ y = 500 and x = 5000

X + 100y = 5000 + 50000 = 55000

Average expense = 55000/100 = Rs. 550

Test Level 1: Averages - 1 - Question 10
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The average of the first ten prime numbers is

Detailed Solution: Question 10

Required average = (2 + 3 + 5 + 7 + 11 + 13 + 17
+ 19 + 23 + 29)/10 = 129/10 = 12.9.

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About this CAT Practice Test

This test contains 10 MCQs on Test Level 1: Averages - 1 with detailed solutions for each question. It is part of the Level-wise Tests for CAT course on EduRev, designed to align with the current CAT syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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