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25/9 Friday Math Problem
- Avish Dumir
- Sep 25, 2020
- 1 min read
There are three variables:
x y z
You are given that:
x+y=20
y+z=18
x+z=22
Evaluate (x*y*z)/5
Answer on 26/9
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See AllProve that: 0^0 = 0, 0^0 = 1, 0^0 is undefined.
In celebration of no occasion whatsoever, I have taken it upon myself to offer my own puzzle.
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Sure that works, when I was creating the question I had checked it like this, and I was trying to make it easy to follow/step by step.
Bruh. u don't needa do that.
sum all three equations:
2(x+y+z)=60
x+y+z=30
from this we subtract all three original equations, like so:
x+y+z-(x+y)=30-20
therefore z= 10.
do the same with the rest and you get
x=12
y=8
z=10
then, evaluate 12*8*10/5 and you get the answer, which is 192... wouldn't you say that this is better.
As so many people have viewed and replied to this question, here's the answer: So first we take x+y=20 and subtract it by y+z=18
which means that we have:
x+y-y-z=20-18
x-z=2 We now know that x-z=2
We are going to add x+z=22 to x-z=2
x-z+z+x=24
2x=24
x=12
We now know that x is equal to 12, we can substitute its value into the equations to solve for y and z
x+y=20
y+12=20
y=8
x+z=22
z+12=22
z=10
So we now know that:
x=12
y=8
z=10
We have to calculate: (x*y*z)/5
(12*8*10)/5
(96*10)/5
960/5
192
(x*y*z)/5=192
And that is the answer, as even Marcus has not solved it, I realize that the 5 people who looked at this question didn't try to…