JEE Novice - (24) (fixed)

Algebra Level 3

{ 1 x + 1 y = 1 3 1 x + 1 z = 1 5 1 y + 1 z = 1 7 \Large \begin{cases} \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3} \\ \dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{5} \\ \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{7} \end{cases}

If x , y , z x,y,z satisfy the system above , find the value of the ratio z y \dfrac{z}{y} .

This question is a part of JEE Novices .


The answer is 29.

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Nihar Mahajan by Nihar Mahajan , 20, India

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2 solutions

Chew-Seong Cheong
May 28, 2015

{ 1 x + 1 y = 1 3 . . . ( 1 ) 1 x + 1 z = 1 5 . . . ( 2 ) 1 y + 1 z = 1 7 . . . ( 3 ) \begin{cases} \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{3} & ...(1) \\ \dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{5} & ...(2) \\ \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{7} & ...(3) \end{cases}

Eq.1 + + Eq.2 − - Eq.3: 2 x = 1 3 + 1 5 − 1 7 ⇒ 1 x = 41 210 \quad \dfrac{2}{x} = \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{7} \quad \Rightarrow \dfrac{1}{x} = \dfrac{41}{210}

Eq.1: 1 y = 1 3 − 41 210 ⇒ 1 y = 29 210 \quad \dfrac{1}{y} = \dfrac{1}{3} - \dfrac{41}{210} \quad \Rightarrow \dfrac{1}{y} = \dfrac{29}{210}

Eq.2: 1 z = 1 5 − 41 210 ⇒ 1 z = 1 210 \quad \dfrac{1}{z} = \dfrac{1}{5} - \dfrac{41}{210} \quad \Rightarrow \dfrac{1}{z} = \dfrac{1}{210}

⇒ z y = 210 210 29 = 29 \Rightarrow \dfrac {z}{y} = \dfrac{210}{\frac{210}{29}} = \boxed{29}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

There's a slightly easier solution without finding the value of x x . Hint: Add them all up.

1 x + 1 y = 1 3 . . . ( 1 ) \large\frac{1}{x}+\frac{1}{y}=\frac{1}{3} ...(1)

1 x + 1 z = 1 5 . . . ( 2 ) \large\frac{1}{x}+\frac{1}{z}=\frac{1}{5} ...(2)

1 y + 1 z = 1 7 . . . ( 3 ) \large\frac{1}{y}+\frac{1}{z}=\frac{1}{7} ...(3)

Adding all 3 equations gives:

2 x + 2 y + 2 z = 71 105 \large\frac{2}{x}+\frac{2}{y}+\frac{2}{z}=\frac{71}{105}

1 x + 1 y + 1 z = 71 210 \large\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{71}{210}

E q u a t i o n 1 ⇒ 1 z = 1 210 \large Equation 1\quad\Rightarrow\quad\frac{1}{z}=\frac{1}{210}

E q u a t i o n 2 ⇒ 1 y = 29 210 \large Equation 2\quad\Rightarrow\quad\frac{1}{y}=\frac{29}{210}

1 y 1 z = 1 210 29 210 \large\frac{\frac{1}{y}}{\frac{1}{z}}=\frac{\frac{1}{210}}{\frac{29}{210}}

z y = 29 \large\frac{z}{y}=\boxed{29}

Eamon Gupta - 6 years ago

As Challenge Master noted, here's an alternative solution which doesn't require finding the value of any of the unknowns.

z y = y − 1 z − 1 = 2 y − 1 2 z − 1 = [ ( y − 1 + z − 1 ) − ( x − 1 + z − 1 ) + ( x − 1 + y − 1 ) ] [ ( y − 1 + z − 1 ) − ( x − 1 + y − 1 ) + ( x − 1 + z − 1 ) ] = 7 − 1 − 5 − 1 + 3 − 1 7 − 1 − 3 − 1 + 5 − 1 = 105 ( 7 − 1 − 5 − 1 + 3 − 1 ) 105 ( 7 − 1 − 3 − 1 + 5 − 1 ) = 15 − 21 + 35 15 − 35 + 21 = 29 1 = 29 \begin{aligned}\frac{z}{y}=\frac{y^{-1}}{z^{-1}}=\frac{2y^{-1}}{2z^{-1}}&=\frac{\left[(y^{-1}+z^{-1})-(x^{-1}+z^{-1})+(x^{-1}+y^{-1})\right]}{\left[(y^{-1}+z^{-1})-(x^{-1}+y^{-1})+(x^{-1}+z^{-1})\right]}\\&=\frac{7^{-1}-5^{-1}+3^{-1}}{7^{-1}-3^{-1}+5^{-1}}\\&=\frac{105(7^{-1}-5^{-1}+3^{-1})}{105(7^{-1}-3^{-1}+5^{-1})}\\&=\frac{15-21+35}{15-35+21}=\frac{29}{1}=\boxed{29}\end{aligned}

Prasun Biswas - 6 years ago
Ravi Dwivedi
Jul 5, 2015

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Yes, the faster way is to treat 1 x , 1 y , 1 z \frac{1}{x}, \frac{1}{y}, \frac{1}{z} as variables a , b , c a, b, c and solve the system of linear equations. This approach of choosing the correct variables can simplify the working.

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