
A two-digit number is four times the sum of the digits. It is also equal to three times the product of the digit. Find the number.
Answer
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Hint: Here we will use two different variables. One will be the digit on the unit's place and the other will be the digit on the tens place. Then we will form two equations from the given data. Since we have two variables and two equations. On solving these equations by substituting the method we will get the correct answer as the number.
Complete step by step solution:
Let x and y be the variables used as the digits on the unit’s place and ten’s place.
So the number can be formed as \[10y + x\]
Now according to the given data and our first condition, two-digit number is four times the sum of the digits;
\[10y + x = 4\left( {x + y} \right).....eq1\]
Now we will solve this as, multiply 4 with the bracket,
\[10y + x = 4x + 4y\]
Taking one type of variables on one side,
\[10y - 4y = 4x - x\]
On solving we get,
\[6y = 3x\]
Dividing both sides by 3 we get,
\[2y = x\]
Now from the second condition we get, the number is equal to three times the product of the digit
\[10y + x = 3xy........eq2\]
Now substitute the value of x from the above solution of equation1,
\[10y + 2y = 3 \times 2y \times y\]
On adding and multiplying we get,
\[12y = 6{y^2}\]
Cancelling y from both the sides we get,
\[12 = 6y\]
On dividing both sides by 6 we get,
\[y = 2\]
Thus we have the digit at ten’s place. Now putting this in the value of x from equation 1,
\[x = 2y = 2 \times 2 = 4\]
This is the digit on the unit's place.
Thus the number formed is 24.
So, the correct answer is “ 24”.
Note: Note that students might be confused or in a hassle they may write the number as 42. But we have taken x as units place digit and y as tens place digit. So the number is 24 and not 42.
Also note that, we have taken the original number in the form of \[10y + x\] because the digits we don’t know are the same or different. As well, writing only xy will not be the number because each place has its place value and so does the digit.
Complete step by step solution:
Let x and y be the variables used as the digits on the unit’s place and ten’s place.
So the number can be formed as \[10y + x\]
Now according to the given data and our first condition, two-digit number is four times the sum of the digits;
\[10y + x = 4\left( {x + y} \right).....eq1\]
Now we will solve this as, multiply 4 with the bracket,
\[10y + x = 4x + 4y\]
Taking one type of variables on one side,
\[10y - 4y = 4x - x\]
On solving we get,
\[6y = 3x\]
Dividing both sides by 3 we get,
\[2y = x\]
Now from the second condition we get, the number is equal to three times the product of the digit
\[10y + x = 3xy........eq2\]
Now substitute the value of x from the above solution of equation1,
\[10y + 2y = 3 \times 2y \times y\]
On adding and multiplying we get,
\[12y = 6{y^2}\]
Cancelling y from both the sides we get,
\[12 = 6y\]
On dividing both sides by 6 we get,
\[y = 2\]
Thus we have the digit at ten’s place. Now putting this in the value of x from equation 1,
\[x = 2y = 2 \times 2 = 4\]
This is the digit on the unit's place.
Thus the number formed is 24.
So, the correct answer is “ 24”.
Note: Note that students might be confused or in a hassle they may write the number as 42. But we have taken x as units place digit and y as tens place digit. So the number is 24 and not 42.
Also note that, we have taken the original number in the form of \[10y + x\] because the digits we don’t know are the same or different. As well, writing only xy will not be the number because each place has its place value and so does the digit.
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