Question

the sum of the digits of a two-digit number is 15 if the number formed by reversing the digits is less than the original number by 27 find the original number​

Answer 1

Answer:

96

Step-by-step explanation:

Let the digit at unit's place be y,tens place be x

So the original no is 10x+y

No. obtained after reversing the digits is 10y+x

Now sum of digits = 15

x+y=15 ----------------(i)

original no. exceeds no. obtained after reversing the order of the digits by 27

10x+y -( 10y+x) = 27

9x-9y=27

9(x-y)=27

x-y = 3 ----------------------(ii)

on adding equation(i)&(ii)

x+y+x-y= 15+3

2x=18

x = 18/2 = 9

putting in eq(1)

9+y=15

y=15-9

y=6

So the original no. is 10*9+6=96

Answer 2

Let the unit's place = x

The ten's place = 15

$\bull \:  \sf{Original \:  Number  =10(15−x)+x}$

$\:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:   \:  \sf   =150−10x+x$

$\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \sf \: =150−9x$

By reversing the digits, we get

$\sf {New \: number=10x+(15−x)}$

$\:  \: \:  \:  \:  \: \:  \:  \:  \sf=10x+15−x$

$\:  \:  \:  \:  \:  \:  \:  \:  \:   \:  = \boxed{ \sf 9x−15} \blue\bigstar$

According to the question

$\sf \: Original \:  number−New \:  number=27$

$: \implies \sf \: 150−9x−9x+15=27$

$: \implies \sf{−18x+165=27}$

$: \implies \sf{−18x=27−165=(−108)}$

$: \implies \sf{x= \frac{−18}{−108}=6}$

$\sf \: original  \: number=150−9x$

$\:  \: \:  \:  \:  \:  \: \:  \:  \:  \:  \: \:  \:  \sf  = 150−9×6$

$\:  \: \:  \:  \:  \:  \: \:  \:  \:  \:  \: \:  \: \sf  = 150- 54$

$\:  \: \:  \:  \:  \:  \: \:  \:  \:  \:  \: \:  \:  = \underline{\boxed{\frak{96}}} \: \pink{ \bigstar}$

• Hence, the original number 96.

Thank you!

@itzshivani