(solved)Question 14 SSC-CGL 2020 March 3 Shift 1

If x2a=y2b=z2c\displaystyle x^{2a} = y^{2b} = z^{2c} and x² = yz, then (ab + bc + ca)/bc = ?
(Rev. 01-May-2025)

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Parveen,

Question 14
SSC-CGL 2020 Mar 3 Shift 1

If x2a=y2b=z2c\displaystyle x^{2a} = y^{2b} = z^{2c} and x² = yz, then (ab + bc + ca)/bc = ?

Solution in Brief

Observe that if a =b = c = x = y = z = 1, all the given conditions are satisfied. So with a = b = c = 1, the required expression (ab + bc + ca)/bc becomes 3 ans!

Solution in Detail

Let x2a=y2b=z2c=k\displaystyle x^{2a} = y^{2b} = z^{2c} = k

Take x2a=k\displaystyle x^{2a} = k

    x=k1/2a\displaystyle \implies x = k^{1/2a}

Likewise y=k1/2b\displaystyle y = k^{1/2b} and z=k1/2c\displaystyle z = k^{1/2c}

Given x2=yz\displaystyle x^2 = yz

    (k1/2a)2=k1/2bk1/2c\displaystyle \implies (k^{1/2a})^2 = k^{1/2b} \cdot k^{1/2c}

    k2/2a=k(1/2b+1/2c)\displaystyle \implies k^{2/2a} = k^{(1/2b + 1/2c)}

Bases are same, so powers must also be same

22a=12b+12c\displaystyle \therefore \frac {2}{2a} = \frac{1}{2b} + \frac{1}{2c}

Add 12a\displaystyle \frac{1}{2a} to both sides

32a=12a+12b+12c\displaystyle \frac{3}{2a} = \frac{1}{2a} + \frac{1}{2b} + \frac{1}{2c}

32a=bc+ca+ab2abc\displaystyle \therefore \frac{3}{2a} = \frac{bc + ca + ab}{2abc}

Canceling 2a\displaystyle 2a,

3=bc+ca+abbcAns\displaystyle 3 = \frac{bc + ca + ab}{bc}\:\underline{Ans}

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