The average age of 80 boys in a class is 15. The average age of group of 15 boys in the class is 16 and the average of another 25 boys in the class is 14. What is the average age of the remaining boys in the class ?
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otal ages of 80 boys = 15 x 80 = 1200 yrs.
Total age of 16 boys = 15 x 16 = 240 yrs
Total age of 25 boys = 14 x 25 = 350 yrs.
Average age of remaining boys = \(1200 - \frac{240+350}{80}-\left ( 25+15 \right )\)= \(\frac{610}{41}\)
= 15.25 yearss
A is three times as old as B. C was twice-as old as A four years ago. In four years' time, A will be 31. What are the present ages of B and C ?
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Let the age of A = \(A\), B = \(B\) and C = \(C\)
From given,
\(A = 3B\)
4 years ago,
\(A + 4 = 31\)
\(\Rightarrow A=27\)
sub in above \(B\) , we get \(B\) = 9
\(C = 2(27 - 4) + 4 = 46 + 4 = 50 \)
Hence B = 9years and C = 50years.
Today is Varun's birthday. One year, from today he will be twice as old as he was 12 years ago. How old is Varun today ?
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Let varun's age = \(x\) years.
From the given conditions,
\(\Rightarrow x+1 = 2(x-12)\)
\(\Rightarrow x=25\) years.
The sum of the ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child ?
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File: /var/www/html/application/controllers/Questions.php
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Let the age of the youngest child is '\(x \)' years.
Then the child elder than the yongest is '\(x + 3\)' yrs
Since each have 3yrs difference upto 5 children
And given that,
\(\Rightarrow x +x +3 + x+6 +x+9 + x+12 = 50\)
\(\Rightarrow 5x+30=50\)
\(\Rightarrow x= 4\) years.
Rajan's present age is three times his daughter's and nine-thirteenth of his mother's present age. The sum of the present ages of all three of them is 125 years. What is the difference between the present ages of Rajan's daughter and Rajan's mother ?
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File: /var/www/html/application/controllers/Questions.php
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File: /var/www/html/index.php
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Let Rajan's present age be '\(x \)' years.
Then his daughter's present age is = \(\frac{x}{3}\) years.
His mother's present age = \(\frac{13x}{9}\) years.
Now, according to the question,
\(\therefore x+\frac{x}{3}+\frac{13x}{9}=125\)
\(\Rightarrow x=\)\(\frac{125*9}{25}\)\(=45\)
Therefore, required difference = \(\frac{13x}{9}-\)\(\frac{x}{3}=13x -\)\(\frac{3x}{9}=\)\(\frac{10x}{9}\)
\(\Rightarrow \frac{10*45}{9}\)= 50 years