231k views
1 vote
Eight times the sum of the digits of a certain two-digit number exceeds the number by 19. The tens digit is two less than the units digit. Find the number.

User JRL
by
7.1k points

2 Answers

4 votes

Answer:

13

Explanation:

I checked on RSM

User Mark Dowell
by
6.2k points
2 votes

Answer:

number is 13

Explanation:

Let

the digit at units place is b

the digit at tens place is a

The number is: ab

Now According to condition: The tens digit is two less than the units digit

a = b -2 ---------- eq1

From the condition: Eight times the sum of the digits of a certain two-digit number exceeds the number by 19.

8(a+b)=10a+b+19

By simplifying

Putting a= b-2

8a+8b=10(b-2)+b+19

8(b-2)+8b=10b-20+b+19

8b-16+8b=10b-20+b+19

By adding like terms we get:

5b=15

Dividing both sides by 5

b = 3

Now putting value of b in eq1

a = b - 2

a = 3 -2

a =1

hence the number is : 13

i hope it will help you!

u=3

t=u-2=1

User Jeach
by
5.5k points