(solved)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?
(Rev. 03-Aug-2022)

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Parveen,

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?

  1. 3
  2. 4
  3. 6
  4. 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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