Exercise 33

There is a pair of numbers and one is twice as large as the other. If you subtract 10 from both numbers, their sum is equal to the lowest of the original pair. Thus, the smallest of the original numbers is:

  1. 10

  2. 8

  3. 4

  4. 20

  5. 40

Try to solve it before checking the answer.

  1. 10

Let \( x \) be the lower number, while the other is \( 2x \).

Let’s subtract 10 from each of them:

\[ x – 10, \quad 2x – 10 \]

The sum of these last two numbers is equal to the original minor. Then,

\[ \begin{aligned} &(x – 10) + (2x – 10) = x \\[.5em] &\hspace{5em}\Rightarrow x + 2x – x = 10 + 10 \\[.5em] &\hspace{5em}\Rightarrow 2x = 20 \\[.5em] &\hspace{5em}\Rightarrow x = 10 \end{aligned} \]
\[ \begin{aligned} (x – 10) + (2x – 10) = x &\Rightarrow x + 2x – x = 10 + 10 \\[.5em] &\Rightarrow 2x = 20 \\[.5em] &\Rightarrow x = 10 \end{aligned} \]

Hence, the numbers are 10 and 20, where the smaller of the two is 10.

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