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# A polyhedron has 32 edges and the number of its faces is 5 less than twice the number of its vertices. What is the number of its faces?

**Given:**

A polyhedron has 32 edges and the number of its faces is 5 less than twice the number of its vertices.

**To do:**

We have to find the number of its faces.

**Solution:**

Let the number of faces be $F$ and the number of vertices be $V$.

Number of faces is 5 less than twice the number of vertices.

This implies,

$F = 2V - 5$

$F + 5 = 2V$

$V = \frac{F + 5}{2}$.............(i)

Number of edges $E = 32$

We know that,

According to Euler's formula $F - E + V = 2$

$F = 32 - \frac{F + 5}{2} + 2$

$F = 34 - \frac{F + 5}{2}$

$2F=68-F-5$

$2F+F = 63$

$F = \frac{63}{3}$

$F=21$

**The number of faces of the polyhedron are 21.**

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