**Slot -2 – Quantitative Aptitude – Modern Maths – Progressions – The arithmetic mean of x, y**

**The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is?**

**Answer:** 105

**Solution:**

Given, (x+y+z)/3=80

x+y+z=240——1)

And (x+y+z+u+v)/4=75

x+y+z+u+v =375——2)

From eq 1) & eq 2) u+v =135

And from question , u+v =(x+2y+z)/2

Or 270=x+2y+z

y = 30 and x+z = 210

as x ≥ z

so x will be minimum if x =z

minimum value of x =210/2 = 105

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