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# Seven times a two-digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3. Find the number.  Hint: For a two digit number ‘ab’, the value is equal to (10a + b). When we reverse the digits it becomes ‘ba’, which is equal to (10b + a).

Let the original number be ‘xy, ‘y’ at one’s place and ‘x’ at ten’s place.
$\Rightarrow$ Value of original two digit number = 10x + y
After reversing the digits, we get number ‘yx’, one’s place and ten’s place interchanged.
$\Rightarrow$ Value of reversed number = 10y + x
From the given information, seven times the original number is equal to four times the reversed number.

$\Rightarrow$ 7(10x + y) = 4(10y + x)
On simplification, we get
$\Rightarrow$ 70x + 7y = 40y + 4x
$\Rightarrow$ 66x = 33y
$\Rightarrow$ y = 2x …. (1)
It is clear from the equation (1) that $y > x$.
It is given that the difference of digits is 3. So we can write
y – x = 3
Substituting ‘y’ value from equation (1)
$\Rightarrow$ 2x – x = 3
$\Rightarrow$ x = 3
So y = 2x = 2(3) = 6
$\therefore$ The original two digit number is 36 and reversed number is 63.

Note: We have to clearly understand the given conditions on digits of the number, so that we can form equations with variables to get required values. After solving the equations we will end up with variable values which in turn give us the required number. After getting the numbers we can cross check with the given conditions.
Finally we got the original number as 36 and reversed number is 63.
When we check the first given condition, 7 times the original number is equal to 4 times the reversed number.
7(36) = 252 = 4(63) Verified.
The second condition, the difference between digits 6 – 3 = 3 Verified.
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