The sum of two consecutive multiples

100

Let the two consecutive multiples of 10 be $10 x$ and $10 x+10$. According to the question,

$10 x+10 x+10=210$

$\Rightarrow \quad 20 x+10=210$

$\Rightarrow$ $20 x=210-10$ [transposing 10 to RHS]

$\Rightarrow$ $20 x=200$

$\Rightarrow$ $\frac{20 x}{20}=\frac{200}{20}$ [dividing both sides by 20]

$\therefore$ $x=10$ 

Hence, the smaller multiple is $10 \times 10$, i.e. 100 .