Quantitative Aptitude – Arithmetic – Averages – In an apartment complex, the number

Quantitative Aptitude – Arithmetic - Averages – In an apartment complex, the number

Slot -1 – Quantitative Aptitude – Arithmetic – Averages – In an apartment complex, the number

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

a) 27

b) 25

c) 26

d) 28

Answer: d) 28

Solution:
Let there are n people whose ages are below 51 years. So

Sum of ages of all people = 38×(30+n)

The average age of the people whose ages are below 51 years will be maximum if average age of people aged 51 years or above will be 51 years.

So Sum of ages of these n people = 38(30+n)-51×30=38n-390

So required average age = (38n-390)/n=38-390/n

This will be maximum when n will be maximum and maximum possible value of n = 39

So maximum possible average age of the people whose ages are below 51 years = 38 –390/39=28

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Quantitative Aptitude – Arithmetic – Averages – A CAT aspirant appears for a certain number

Slot -1 – Quantitative Aptitude – Arithmetic – Averages – A CAT aspirant appears for a certain number

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is?

Answer: 60

Solution: let total number test taken by him = n and average score = x

So average score of last (n-10) test = x+1

Sum of scores of last n-10 test = (n-10)*(x+1) = nx – ( 10x – n + 10)

So total score of first 10 test = 10x – n + 10

20*10 = 10x – n + 10

10x – n = 190  —————1)

average score of first (n-10) test = x-1

Sum of scores of first n-10 test = (n-10)*(x-1) = nx – ( n +10x – 10)

So total score of first 10 test = ( n +10x – 10)

30*10 = ( n +10x – 10)

n + 10x = 310 —————2)

by solving eq 1 ) & eq 2)

n = 60, x = 25

so total number of test taken by him = 60

Other posts related to Quantitative Aptitude – Modern Maths

Permutation and Combination – Fundamental Principle of Counting
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a) 750+ Videos covering entire CAT syllabus
b) 2 Live Classes (online) every week for doubt clarification
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d) 10 Mock Tests in the latest pattern
e) Previous Year Questions solved on video

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