x,y and z are positive integers such that x<y,x+y=201,zāx=200. What is the maximum value of x+y+z?
x + y = 201
y = 201 - x
y > x that's why x < 201 / 2
x < 100.5
Nearest integer:
x = 100
x + y = 201
y = 201 - x
y = 201 - 100
y = 101
z ā x = 200
z = 200 + x
z = 200 + 100
z = 300
x + y + z = 100 + 101 + 300 = 501
OR
x + y = 201
y = 201 - x
z ā x = 200
z = 200 + x
x + y + z = x + 201 - x + 200 + x
x + y + z = 401 + x
x + y = 201
mean :
x < 201 / 2
x < 100.5
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For example :
If x = 101
y = 201 - x = 201 - 101 = 100
That not satisfie condition :
x < y
If x = 102
y = 201 - x = 201 - 102 = 99
That not satisfie condition :
x < y
If x = 103
y = 201 - x = 201 - 103 = 98
That not satisfie condition :
x < y
etc.
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That's why :
x < 201 / 2
x < 100.5
Largest integer less of 100.5 are 100
so
x = 100
x + y + z = 401 + x = 401 + 100 = 501
To find the maximum value of x+y+z, we need to maximize the values of x, y, and z while still satisfying the given conditions.
We are given that x < y and x + y = 201. Thus, we need to find the maximum values of x and y such that their sum is equal to 201.
Let's start by rearranging the equation x + y = 201 to solve for y:
y = 201 - x
Next, we are given that z - x = 200. By rearranging this equation, we find:
z = x + 200
To find the maximum value of x+y+z, we substitute the values of y and z into the expression:
x + (201 - x) + (x + 200)
Simplifying this expression, we get:
201 + 200
= 401
Therefore, the maximum value of x+y+z is 401.