If 3713 - 2 - 1x-1y-1z- where x- y- z are positive integer- then x - y - z is equal to

The correct option is C 8 3713=2+1x+1y+1z 3713 2=1x+1y+1z 1113=1x+1y+1z ⇒ x+1y+1z=1311 x+1y+1z=1+211 ⇒ x=1 and 1y+1z=211 No ...

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Byju's Answer

Standard XII

Mathematics

Intersection of Sets

If 3713 = 2...

Question

If $\frac{37}{13} = 2 + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}$ , where $x, y, z$ are positive integer, then $x + y + z$ is equal to

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Solution

The correct option is C $8$
$\frac{37}{13} = 2 + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}$

$\frac{37}{13} - 2 = \frac{1}{x + \frac{1}{y + \frac{1}{z}}}$

$\frac{11}{13} = \frac{1}{x + \frac{1}{y + \frac{1}{z}}}$

$\Rightarrow x + \frac{1}{y + \frac{1}{z}} = \frac{13}{11}$

$x + \frac{1}{y + \frac{1}{z}} = 1 + \frac{2}{11}$

$\Rightarrow x = 1 a n d \frac{1}{y + \frac{1}{z}} = \frac{2}{11}$

Now,
$\frac{1}{y + \frac{1}{z}} = \frac{2}{11}$

$\Rightarrow y + \frac{1}{z} = \frac{11}{2} = 5 + \frac{1}{2}$

$\Rightarrow y = 5 a n d \frac{1}{z} = \frac{1}{2}$

$y = 5 a n d z = 2$

So, $x + y + z = 1 + 5 + 2 = 8$

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If 3713=2+1x+1y+1z, where x,y,z are positive integer, then x+y+z is equal to

The correct option is C 83713=2+1x+1y+1z3713−2=1x+1y+1z1113=1x+1y+1z⇒x+1y+1z=1311x+1y+1z=1+211⇒x=1and1y+1z=211Now,1y+1z=211⇒y+1z=112=5+12⇒y=5and1z=12y=5andz=2So, x+y+z=1+5+2=8

If $\frac{37}{13} = 2 + \frac{1}{x + \frac{1}{y + \frac{1}{z}}}$ , where $x, y, z$ are positive integer, then $x + y + z$ is equal to