Exercise 63

The sum of two numbers is 113, their quotient is 6, and the remainder is 8. The difference between the larger number and three times the smaller number is:

  1. 59

  2. 38

  3. 53

  4. 7

  5. 56

Try to solve it before checking the answer.

  1. 53

Let \( x \) and \( y \) be the numbers, with \( x \) being the larger. Given that their sum is 113, we have that:

\[ \boldsymbol{(1)} \hspace{2em} x + y = 113 \]

If the quotient of \( x \) divided by \( y \) is 6, with a remainder of 8, then, subtracting the remainder from the dividend, the quotient should be 6. That is:

\[ \frac{x – 8}{y} = 6 \Rightarrow x = 6y + 8 \hspace{2em} \boldsymbol{(2)} \]

De \((1)\):   \( x = 113 – y \)

Replacing this value in (2):

\[ \begin{aligned} 113 – y = 6y + 8 &\Rightarrow 7y = 105 \\[2em] &\Rightarrow y = 15 \\[2em] &\Rightarrow x = 113 – 15 = \boldsymbol{98} \end{aligned} \]
\[ \begin{aligned} 113 – y = 6y + 8 &\Rightarrow 7y = 105 \\[2em] &\Rightarrow y = 15 \\[2em] &\Rightarrow x = 113 – 15 \\[2em] &= \boldsymbol{98} \end{aligned} \]

Hence, the difference between the larger number and three times the smaller is:

\[ x – 3y = 98 – 3(15) = 98-45 = \boldsymbol{53} \]
\[ \begin{aligned} x – 3y &= 98 – 3(15) \\[2em] &= 98-45 = \boldsymbol{53} \end{aligned} \]

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