Question

The ten's digit of a two digit number is three times the unit's digit. The sum of the number and its unit's digit is 32. Find the number​

Answer 1

Answer:

31

Step-by-step explanation:

`let unit digit=x and 10's  is y

then y=3x

no=10y+x

no+x=32

10y+x+x=32

10*3x+2x=32

32x=32

x=1

uniy digt=1

tens=3*1=3

no=31

Answer 2

Given:

• ten's digit of a two digit number is three times the unit place
• the sum of the number and its unit's digit is 32

To Find:

the original number

Solution:

Let the number at unit place be "x" and the number at ten's place be "y"

Required number:

$\sf{• 10y + x}$

according to the question:

$\sf{y = 3x}$

$\implies\sf{ 3x - y = 0 ........(1)}$

and,

$\sf{10x + x + x = 32}$

$\sf{10x + 2x = 32 ........(2)}$

multiplying equation (1) by 10:

$\sf{30x - 10y = 0 ........(3)}$

now,

adding equation (3) and (2):

⠀$\sf{30x - 10y = 0}$

+ $\sf\underline{ 2x + 10y = 32}$

⠀⠀$\sf{32x = 32}$

$⟹$$\bf{x = 1}$

From (1),we get,

$⟹$ $\sf{ y = 3(1) = 3}$

Required number:

$\implies\sf{ 10(3) + 1 = 31}$

Hence,

the original number is $\sf\underline\red{31}$