chetan2u wrote:
The class teacher divided 100 pencils amongst five students, A, B, C, D and E. The number of pencils received by A is equal to the sum of the pencils received by the three students, C, D and E. It is also known \(a \geq b \geq c\geq d \geq e\), where a, b, c, d and e are pencils received by A, B, C, D and E respectively.
On the basis of the information provided, select for Max A-B, the maximum difference between the number of the pencils received by A and the number of the pencils received by B, and select for Min B the least number of the pencils that B could have got. Make only two selections, one in each column.
We know a = c+d+e, and a+b+c+d+e = 100 or a+b+a = 100, that is 2a+b=100
Max A-B: We are looking into the difference of the two largest numbers, that is a-b. To make it the largest, we have to make a the largest possible and b the smallest possible.
The smallest value of b will be when it is equal to c, and c is also least possible.Now, c+d+e = a
For c to be least, let it be equal to d and e, so c+c+c = a or \(c=\frac{a}{3}\).
Therefore, a+b+(c+d+e) = 100 or a+c+(a) = 100..........\(2a+\frac{a}{3}=100.......7a=300.....a=42.89\)
As a is an integer, the maximum value of a is 42.
Value of b = 42+b+42 = 100.......b=16
Max A-B = 42-16 = 26Max B: We know 2a+b = 100, so to get maximum value of b, we have to minimize a.
Let b be equal to a, then 3a = 100 and a = 33.33
If a = 33, then b becomes 100-2*33 or 34, making b>a.
So a = 34, and b = 100-2*34 or 32.