Question 18
SSC-CGL 2018 June 4 Shift 1
If $\displaystyle x + y + z = 19$, $\displaystyle x^2 + y^2 + z^2 = 133$ and $\displaystyle xz = y^2$, then the difference between $\displaystyle x$ and $\displaystyle z$ is?
- 3
- 4
- 6
- 5
Solution 1
x, y and z can be easily guessed because they could be small numbers adding to 19, and also satisfying $\displaystyle xz = y^2$. With a slight hit and trial, x = 4, y = 6 and z = 9 giving the difference between x and z as 5 Ans
Solution 2
First solve the equations for one of the variables.
$ \begin{aligned} x + z &= 19 - y\\ \implies (x + z)^2 &= (19 - y)^2\\ (x^2 + z^2) + 2(xz) &= (19 - y)^2\\ \implies (133 - y^2) + 2(y^2) &= (19 - y)^2\\ y^2 \text{cancels to give } y &= 6\\ \end{aligned} $
Now proceed like this
$ \begin{aligned} (x - z)^2 &= x^2 + z^2 -2(xz) \\ &= (133 - y^2) - 2 (y^2)\\ & = (133 - 6^2) - 2 (6^2)\\ & = 25 \implies (x - z ) = 5\:\underline{Ans} \end{aligned} $
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