Question 60

If x,y,z are three numbers such that $$x+y=13,y+z=15$$ and $$z+x=16$$, the
value of $$\frac{xy+xz}{xyz}$$ is:

Solution

x+y=13 ---(1)
y+z=15 ---(2)
z+x=16$$ ---(3)

By (1) + (2) + (3),

2(x + y + z) = 13 + 15 + 16

x + y + z = 44/2 = 22

put the value from eq(1),

13 + z = 22

z = 9

From eq(3),

9 + x =16

x = 7

From eq(3),

7 + y = 13

y = 6

Now,

$$\frac{xy+xz}{xyz}$$

= $$\frac{(7)(6)+(7)(9)}{(7)(6)(9)}$$

= $$\frac{6 + 9}{6 \times 9}$$

= $$\frac{5}{18}$$


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