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The following question based on average topic of quantitative aptitude

Questions : The average of n numbers $x_1 , x_2 , .... x_n$ is $\ov{x}$ . Then the value of $∑↙{i=1} ↖n $$(x_i - \ov{x})$ is equal to

(a) n$\ov{x}$

(b) $\ov{x}$

(c) 0

(d) n

The correct answers to the above question in:

Answer: (c)

Using Rule 1,

${x_1+x_2+…x_n}/n$ = $\ov{x}$

∴ $??{i=1}?n $$(x_i - \ov{x})$

= $(x_1 - \ov{x})$+$(x_2 - \ov{x})$+…$(x_n - \ov{x})$

= $({x_1+x_2+…+x_n})$ - n.$\ov{x}$

=n $({x_1+x_2+…+x_n}/n)$ - n.$\ov{x}$

= $n\ov{x} – n\ov{x}$ = 0

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

The average of all the numbers between 6 and 50 which are divisible by 5 is

a) 28.5

b) 22

c) 30

d) 27.5

Answer: (d)

Using Rule 1,

Average of two or more numbers/quantities is called the mean of these numbers, which is given by

$\text"Average(A)" = \text"Sumof observation / quantities"/\text"No of observation / quantities"$

∴ S = A × n

Numbers are : 10, 15, 20, 25, 30, 35, 40, 45

Sum = 220

Average = $220/8$ = 27.5

Question : 2

The mean high temperature of the first four days of a week is 25°C whereas the mean of the last four days is 25.5°C. If the mean temperature of the whole week is 25.2°C then the temperature on the 4th day is

a) 25.2°C

b) 25.6° C

c) 25°C

d) 25.5°C

Answer: (b)

Temperature on 4th day

= (4 × 25 + 4 × 25.5 –25.2 × 7)°C

= (100 + 102 – 176.4)°C

= 25.6°C

Question : 3

The average marks of 50 students in a class is 72. The average marks of boys and girls in that subject are 70 and 75 respectively. The number of boys in the class is

a) 25

b) 30

c) 35

d) 20

Answer: (b)

Number of students in the class = x (let)

∴ Number of girls = 50 – x

According to the question,

x × 70 + (50 – x) × 75

= 50 × 72

⇒ 70x + 3750 – 75x = 3600

⇒ 3750 – 5x = 3600

⇒ 5x = 3750 – 3600 = 150

⇒ x = $150/5$ = 30

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