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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$$Â ---(3)
By (1) +Â (2) + (3),
2(x + y + z) = 13 + 15 + 16
x + y + z = 44/2 = 22
put the value from eq(1),
13 + z = 22
z = 9
From eq(3),
9 + x =16
x = 7
From eq(3),
7 + y = 13
y = 6
Now,
$$\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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