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In a class of 35 students, the averages age of all the students and cl

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In a class of 35 students, the averages age of all the students and cl  [#permalink]

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New post Updated on: 12 Aug 2018, 22:58
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Solve any Mean and Standard deviation question under a minute- Exercise Question #1

1- In a class of 34 students, the averages age of all the students and class teacher combined is ‘b’ years. Now, if the class teacher, age 43, retires in the same year and is replaced by experienced teacher age 68. Then what will be the new average of the class including the class teacher?

Options

    a) b+\(\frac{5}{9}\)
    b) 7b+5
    c) b-\(\frac{2}{3}\)
    d) b+\(\frac{5}{7}\)
    e) b+\(\frac{7}{5}\)

To solve question 2: Question 2


To read the article: Solve any Mean and Standard deviation question under a minute

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Originally posted by EgmatQuantExpert on 04 Jul 2018, 06:47.
Last edited by EgmatQuantExpert on 12 Aug 2018, 22:58, edited 3 times in total.
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Re: In a class of 35 students, the averages age of all the students and cl  [#permalink]

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New post Updated on: 07 Jul 2018, 07:02
2

Solution



Given:

    • Class has 34 students.
    • Average age of the all the students and teacher= ‘b’ years
    • Age of the retired class teacher= 43 years
    • Age of the new class teacher= 68 years

To find:
    • New average age of the class including the age of new class teacher.

Approach and Working:

Method-1

    • When the age of class teacher is 43 then:
      o \(\frac{Sum\ of\ age\ of\ all\ students+43}{34+1}\)= b
      o Sum of all the students+43= 35b

    • Now, when new class teacher, aged 68 joined then:
      o Average=\(\frac{Sum\ of\ age\ of\ all\ students+68}{34+1}\)
      o =\(\frac{Sum\ of\ age\ of\ all\ students+(43+25)}{35}\)
      o = \(\frac{Sum\ of\ age\ of\ all\ students+43}{35}+ \frac{25}{35}\)
      o = b+\(\frac{5}{7}\)

Hence, the correct answer is option D.

Answer: D
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Originally posted by EgmatQuantExpert on 04 Jul 2018, 06:48.
Last edited by EgmatQuantExpert on 07 Jul 2018, 07:02, edited 1 time in total.
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Re: In a class of 35 students, the averages age of all the students and cl  [#permalink]

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New post 04 Jul 2018, 07:44
2
let the average age of students be x.
(34x + 43)/35 = b
Now ,
35b = total of all age at the moment.
(35b - 43 + 68)/35 is the new average.
=35b/35 + 25/35
= b + 5/7
(option D)
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Re: In a class of 35 students, the averages age of all the students and cl  [#permalink]

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New post 04 Jul 2018, 07:57
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EgmatQuantExpert wrote:
Solve any Mean and Standard deviation question under a minute- Exercise Question #1

1- In a class of 34 students, the averages age of all the students and class teacher combined is ‘b’ years. Now, if the class teacher, age 43, retires in the same year and is replaced by experienced teacher age 68. Then what will be the new average of the class including the class teacher?

Options

    a) b+\(\frac{5}{9}\)
    b) 7b+5
    c) b-\(\frac{2}{3}\)
    d) b+\(\frac{5}{7}\)
    e) b+\(\frac{7}{5}\)


\(Average = \frac{35b - 43 + 68}{35}\)

Or, \(Average = \frac{5(7b + 5)}{5*7}\)

Or, \(Average = b + \frac{5}{7}\), Answer must be (D)
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Re: In a class of 35 students, the averages age of all the students and cl  [#permalink]

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New post 04 Jul 2018, 09:26
EgmatQuantExpert wrote:
Solve any Mean and Standard deviation question under a minute- Exercise Question #1

1- In a class of 34 students, the averages age of all the students and class teacher combined is ‘b’ years. Now, if the class teacher, age 43, retires in the same year and is replaced by experienced teacher age 68. Then what will be the new average of the class including the class teacher?

Options

    a) b+\(\frac{5}{9}\)
    b) 7b+5
    c) b-\(\frac{2}{3}\)
    d) b+\(\frac{5}{7}\)
    e) b+\(\frac{7}{5}\)




LOGIC will get you to the answer immediately..
numbers remain same = \(34+1=35\) ...
only \(68-43=25\) years are added to total
so \(\frac{25}{35}\) will be added to the earlier average b...
ans \(b+ \frac{25}{35} = b+\frac{5}{7}\)

substitution..
take b as 10, so total age is 35*10=350
add 68 and subtract 43 because of changeover so 350+68-43=375
new average =\(\frac{375}{35}= 10\frac{25}{35}= 10+\frac{25}{35} = 10+\frac{5}{7}\)
now substitute b as 10 and D is the answer

Algebraic method
total age = 35*b
add 68 and subtract 43 because of changeover so \(35b+68-43=35b+25\)
new average = \(\frac{35b+25}{35}\frac{[}{fraction] = b+[fraction]}5/7\)
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Re: In a class of 35 students, the averages age of all the students and cl &nbs [#permalink] 04 Jul 2018, 09:26
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