We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+5
1719
2
avatar+564 

The sum of the digits of a two-digit counting number is 10. If the digits are reversed, the new number is one less than twice the original number. What was the original number?

 Apr 25, 2016

Best Answer 

 #1
avatar+65 
+5

I think it's 91. 

 

9+1=10 

10x2=20 

We reverse 91 to 19

 

19 is one less than 20.

 

 

I am not sure though.

 Apr 25, 2016
edited by Lunar  Apr 25, 2016
 #1
avatar+65 
+5
Best Answer

I think it's 91. 

 

9+1=10 

10x2=20 

We reverse 91 to 19

 

19 is one less than 20.

 

 

I am not sure though.

Lunar Apr 25, 2016
edited by Lunar  Apr 25, 2016
 #2
avatar
+5

 

A + B=10

2[10A + B] - 1=10B + A

So the original number is=37

 

Solve the following system:
{A+B = 10 |     (equation 1)
2 (10 A+B)-1 = A+10 B |     (equation 2)
Express the system in standard form:
{A+B = 10 |     (equation 1)
19 A-8 B = 1 |     (equation 2)
Swap equation 1 with equation 2:
{19 A-8 B = 1 |     (equation 1)
A+B = 10 |     (equation 2)
Subtract 1/19 × (equation 1) from equation 2:
{19 A-8 B = 1 |     (equation 1)
0 A+(27 B)/19 = 189/19 |     (equation 2)
Multiply equation 2 by 19/27:
{19 A-8 B = 1 |     (equation 1)
0 A+B = 7 |     (equation 2)
Add 8 × (equation 2) to equation 1:
{19 A+0 B = 57 |     (equation 1)
0 A+B = 7 |     (equation 2)
Divide equation 1 by 19:
{A+0 B = 3 |     (equation 1)
0 A+B = 7 |     (equation 2)
Collect results:
Answer: |  A = 3             and                  B = 7

 Apr 25, 2016

18 Online Users

avatar
avatar