Let the digits at units and tens place of the given number be x and y respectively. Thus, the number is ` 10 y + x` The two digits of the number are differing by 2. Thus, we have ` x - y =+- 2` After interchanging the digits, the number becomes `10 x + y.` The sum of the numbers obtained by interchanging the digits and the original number is 66. Thus, we have `(10 x + y ) + ( 10 y + x )= 66` ` ⇒ 10 x + y + 10 y + x = 66` `⇒ 11 x + 11y = 66 ` ` ⇒ 11 ( x + y) = 66/11` ` ⇒ x + y = 66/11` ` ⇒ x + y =6` So, we have two systems of simultaneous equations `x - y = 2` ` x + y = 6` ` x - y = -2` ` x + y = 6` Here x and y are unknowns. We have to solve the above systems of equations for xand y. (i) First, we solve the system ` x - y = 2` ` x + y = 6` Adding the two equations, we have `( x - y ) + ( x + y )= 2 + 6` `⇒ x - y + x + y = 8` ` ⇒ 2x = 8` ` ⇒ x = 8/2` ` ⇒ x = 4` Substituting the value of x in the first equation, we have ` 4 - y = 2` `⇒ y = 4 -2` `⇒ y = 2` Hence, the number is ` 10 xx 2 + 4 = 24`. (ii) Now, we solve the system ` x - y= -2` ` x + y = 6` Adding the two equations, we have ` ( x - y) + ( x + y) = -2 + 6` `⇒ x - y + x + y = 4 ` ` ⇒ 2x = 4` ` ⇒ x = 4/2` ` ⇒ x = 2` Substituting the value of x in the first equation, we have ` 2 - y = -2` ` ⇒ y = 2 + 2` ` ⇒ y = 4` Hence, the number is ` 10 xx 4 + 2 = 42`
There are two such numbers. Let’s assume the digit at unit’s place as x and ten’s place as y. Thus from the question, the number needed to be found is 10y + x. From the question it’s told as, the two digits of the number are differing by 2. Thus, we can write x – y = ±2………….. (i) Now after reversing the order of the digits, the number becomes 10x + y. Again from the question it’s given that, the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, this can be written as; (10x+ y) + (10y+x) = 66 ⇒ 10x + y + 10y + x = 66 ⇒ 11x +11y = 66 ⇒ 11(x + y) = 66 ⇒ x + y = 66/11 ⇒ x + y = 6………….. (ii) Now, we have two sets of systems of simultaneous equations x – y = 2 and x + y = 6 x – y = -2 and x + y = 6 Let’s first solve the first set of system of equations; x – y = 2 …………. (iii) x + y = 6 ………….. (iv) On adding the equations (iii) and (iv), we get; (x – y) + (x + y) = 2+6 ⇒ x – y + x + y = 8 ⇒ 2x =8 ⇒ x = 8/2 ⇒ x = 4 Putting the value of x in equation (iii), we get 4 – y = 2 ⇒ y = 4 – 2 ⇒ y = 2 Hence, the required number is 10 × 2 +4 = 24 Now, let’s solve the second set of system of equations, x – y = -2 …………. (v) x + y = 6 ………….. (vi) On adding the equations (v) and (vi), we get (x – y)+(x + y )= -2 + 6 ⇒ x – y + x + y = 4 ⇒ 2x = 4 ⇒ x = 4/2 ⇒ x = 2 Putting the value of x in equation 5, we get; 2 – y = -2 ⇒ y = 2+2 ⇒ y = 4 Hence, the required number is 10×4+ 2 = 42 Therefore, there are two such possible numbers i.e, 24 and 42.
Text Solution Solution : let tens digite of number is`x` <br> let unit digit be`y` <br> number= `(10x+y)` <br> if no is reversed then <br> unit digit `x` & tens digit `y`<br> reversed number= `(10y+x)` <br> acc to question <br> `(10x+y) + (10y+x)= 66` <br> `11x+11y=66` <br> so, `x+y=6` <br> now `x-y=2` <br> or `y-x=2` <br> case-1 `x-y=2 & x+y=6` <br> by adding both, we get`2x= 8` <br> `x=4` <br> `y=2` <br> so number will be `4*10=2 = 42` <br> case 2 : `y-x=2 & x+y=6` <br> by adding both, we get `2y= 8` <br> `y=4` <br> `x=2` <br> now number will be `24` <br> answer is 24 and 42 |