Question:
The average weight of ‘b’ boys in a group is 63 kg. A boy whose weight is 147 kg joins the group such that the average weight of the group becomes a prime number in between 65 and 91. Find the value of ‘b+1’.
The average weight of ‘b’ boys in a group is 63 kg. A boy whose weight is 147 kg joins the group such that the average weight of the group becomes a prime number in between 65 and 91. Find the value of ‘b+1’.
Updated On: Aug 31, 2024
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Solution and Explanation
Sum of the weights of 'b' students = 63b kg
After the new boy joins the group
Sum of the weights of (b + 1) boys = (63b + 147) kg
Therefore, new average age of the group = \(\frac{(63b + 147)}{ (b + 1)}\)
= \(\frac{(63b +63 +84)}{ (b + 1)}\)
={ \(\frac{63(b + 1)}{ (b + 1)} \)}+ \(\frac{84}{ (b + 1)}\)
= 63 + \(\frac{84}{ (b + 1)}\)
Since, the new average age is an integer; therefore (b + 1) should be factor of 84
Let b + 1 = 2, then new average age of the group = 63+ (\(\frac{84}{2}\)) = 105 (not possible)
Let b + 1 = 3, then new average age of the group = 63+ (\(\frac{84}{3}\)) = 91 (not possible)
Let b + 1 = 4, then new average age of the group = 63+ (\(\frac{84}{4}\)) = 84 (not a prime number)
Let b + 1 = 6, then new average age of the group = 63+ (\(\frac{84}{6}\)) = 77 (not a prime number)
Let b + 1 = 7, then new average age of the group = 63+ (\(\frac{84}{7}\)) = 75 (not a prime number)
Let b + 1 = 12, then new average age of the group = 63+ (\(\frac{84}{12}\)) = 70 (not a prime number)
Let b + 1 = 14, then new average age of the group = 63+ (\(\frac{84}{14}\)) = 69 (not a prime number)
Let b + 1 = 21, then new average age of the group = 63+ (\(\frac{84}{21}\)) = 67 (possible)
Let b + 1 = 42, then new average age of the group = 63+ (\(\frac{84}{42}\)) = 65 (not possible)
Next all factors will give average less than 70
Therefore, only possible number of students = b + 1 = 21
After the new boy joins the group
Sum of the weights of (b + 1) boys = (63b + 147) kg
Therefore, new average age of the group = \(\frac{(63b + 147)}{ (b + 1)}\)
= \(\frac{(63b +63 +84)}{ (b + 1)}\)
={ \(\frac{63(b + 1)}{ (b + 1)} \)}+ \(\frac{84}{ (b + 1)}\)
= 63 + \(\frac{84}{ (b + 1)}\)
Since, the new average age is an integer; therefore (b + 1) should be factor of 84
Let b + 1 = 2, then new average age of the group = 63+ (\(\frac{84}{2}\)) = 105 (not possible)
Let b + 1 = 3, then new average age of the group = 63+ (\(\frac{84}{3}\)) = 91 (not possible)
Let b + 1 = 4, then new average age of the group = 63+ (\(\frac{84}{4}\)) = 84 (not a prime number)
Let b + 1 = 6, then new average age of the group = 63+ (\(\frac{84}{6}\)) = 77 (not a prime number)
Let b + 1 = 7, then new average age of the group = 63+ (\(\frac{84}{7}\)) = 75 (not a prime number)
Let b + 1 = 12, then new average age of the group = 63+ (\(\frac{84}{12}\)) = 70 (not a prime number)
Let b + 1 = 14, then new average age of the group = 63+ (\(\frac{84}{14}\)) = 69 (not a prime number)
Let b + 1 = 21, then new average age of the group = 63+ (\(\frac{84}{21}\)) = 67 (possible)
Let b + 1 = 42, then new average age of the group = 63+ (\(\frac{84}{42}\)) = 65 (not possible)
Next all factors will give average less than 70
Therefore, only possible number of students = b + 1 = 21
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