question_answer 20)

                A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer:

                Let the numbers be \[x\] and \[5x\]. If 21 is added to both the numbers, then first new number \[=x+21\] and, second new number \[=5x+21\] Case I. When first new number is twice the second new number, Then, \[x+21=2(5x+21)\] \[\Rightarrow \]               \[x+21=10x+42\] \[\Rightarrow \]               \[x-10x=10x+42\]                             | Transposing 10x to LHS and 21 to RHS  \[\Rightarrow \]               \[-9x=21\] \[\Rightarrow \]               \[x=-\frac{21}{9}\]                                           | Dividing both sides by ? 3 \[\Rightarrow \]               \[x=\frac{-21\div 3}{9\div 3}\] \[\Rightarrow \]               \[x=-\frac{7}{3}\] \[\Rightarrow \]               \[5x=5\times \left( -\frac{7}{3} \right)\,=-\frac{35}{3}\]                 Hence, the numbers are \[-\frac{7}{3}\] and \[-\frac{35}{3}\]. This case is inadmissible as the required numbers are positive. Case II. When second new number is twice the first new number. Then,   \[5x+21=2(x+21)\] \[\Rightarrow \]               \[5x+21=2x+42\] \[\Rightarrow \]               \[5x-2x=42-21\]                 |Transposing \[2x\] to LHS and 21 to RHS \[\Rightarrow \]               \[3x=21\]             \[\Rightarrow \]               \[x=\frac{21}{3}=7\]                       |Dividing both sides by 3 \[\Rightarrow \]               \[5x=5\times 7=35\] Hence, the numbers and 7 and 35. Check:  \[35=7\times 5\]                 \[7+21=28\] \[35+21=56\]                     \[56=28\times 2\] Hence, the result is verified.