A two digits number is equal to 7 times the sum of its digits and 21 times the difference of its digits. If xy represents the required number then the simultaneous equations are written as: A) 7y – xy = 7x and 21x + xy = 21 B) 7x + xy = 7y and 21 x – xy = 21y C) 7x – xy = 7y and 21x – xy = 21 D) 7x – xy = -7y and 21y + xy = 21x
Question
A two digits number is equal to 7 times the sum of its digits and 21 times the difference of its digits. If xy represents the required number then the simultaneous equations are written as: A) 7y – xy = 7x and 21x + xy = 21 B) 7x + xy = 7y and 21 x – xy = 21y C) 7x – xy = 7y and 21x – xy = 21 D) 7x – xy = -7y and 21y + xy = 21x
Solution
The correct answer is B) 7x + xy = 7y and 21x – xy = 21y.
Here's why:
The problem states that a two-digit number is equal to 7 times the sum of its digits and 21 times the difference of its digits. If we represent the two-digit number as 10x + y (where x is the tens digit and y is the units digit), we can write the two conditions as follows:
- The number is equal to 7 times the sum of its digits: 10x + y = 7(x + y)
- The number is equal to 21 times the difference of its digits: 10x + y = 21(x - y)
Solving these two equations simultaneously will give us the values of x and y, which represent the digits of the number.
So, the correct simultaneous equations are: 7x + xy = 7y and 21x – xy = 21y.
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