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Test Level 1: Averages - 1 - CAT MCQ


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10 Questions MCQ Test Level-wise Tests for CAT - Test Level 1: Averages - 1

Test Level 1: Averages - 1 for CAT 2024 is part of Level-wise Tests for CAT preparation. The Test Level 1: Averages - 1 questions and answers have been prepared according to the CAT exam syllabus.The Test Level 1: Averages - 1 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test Level 1: Averages - 1 below.
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Test Level 1: Averages - 1 - Question 1

Find the average of five consecutive even numbers a, b, c, d and e.

Detailed Solution for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

How many pairs of positive integers, not more than 100, will have an average greater than 50?

Detailed Solution for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

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 for Test Level 1: Averages - 1 - 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

The average of the first ten prime numbers is

Detailed Solution for Test Level 1: Averages - 1 - Question 10

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

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