If $a:b = 1:3$ and $b$ is a positive, one-digit integer, what is the greatest possible value of $a$?
1 answer:
You might be interested in
Answer :
x = -1.43
Explanation :
As given ,
The first box has a height (x+5) has a square base with side x
So, the Length and Breadth of the box are x and x respectively.
Now,
The volume of first box = Length×Breadth×Height
= (x).(x).(x+5)
= x²(x+5)
⇒The volume of first box = x²(x+5)
Now,
Given that , the second box with height (x+2) has a square base with side (x+5).
So, the Length and Breadth of the box are (x+5) and (x+5) respectively.
Now,
The volume of second box = Length×Breadth×Height
= (x+5). (x+5).(x+2)
= (x+5)²(x+2)
⇒The volume of second box = (x+5)²(x+2)
Now,
Given that, the two boxes have the same volume
⇒x²(x+5) = (x+5)²(x+2)
⇒x² = (x+5)(x+2)
⇒x² = x² + 2x + 5x + 10
⇒x² - x² = 7x + 10
⇒0 = 7x + 10
⇒ 7x = -10
⇒ x = - = -1.43
⇒ x = -1.43
As volume of first box = x²(x+5) = (-1.43)²(-1.43+5) = 7.30 ≈ 7
As volume of second box = (x+5)²(x+2) = (-1.43+5)²(-1.43+2) = 7.26 ≈ 7
At x = -1.43, Volume of both the boxes are same.