Question

The sum of the digits in a two digit number is 14 . The number itself is 2 greater than 11 times the tens digit .Then , what is the number ?
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Answer 1

Gɪᴠᴇɴ :-

The sum of the digits in a two digit number is 14 . The number itself is 2 greater than 11 times the tens digit .

ᴛᴏ ғɪɴᴅ :-

• Number

sᴏʟᴜᴛɪᴏɴ :-

Let the tens place digit be x and ones place digit be y

then,

➦ According to 1st condition :-

• Ones place digit + Tens place digit = 14

• x + y = 14

• x = 14 - y. ---(1)

➦ According to 2nd Condition :-

• Number = 11 × Tens place digit + 2

→ (10x + y) = 11 × x + 2

→ 10x - 11x + y = 2

→ -x + y = 2

→ x = y - 2. ----(2)

From (1) and (2) , we get,

→ 14 - y = y - 2

→ 14 + 2 = y + y

→ 2y = 16

→ y = 16/2

→ y = 8

Put y = 8 in (1) , we get,

→ x = 14 - y

→ x = 14 - 8

→ x = 6

Hence,

• Tens place digit = x = 6
• Ones place digit = y = 8

Therefore,

• Number = (10x + y) = 68

Answer 2

S O L U T I O N :

Let the ten's digit number be r

Let the one's digit number be m

$\boxed{\bf{The\:original\:number=10r+m}}}}}\\\boxed{\bf{The\:reversed\:number=10m+r}}}}}$

A/q

$\longrightarrow\rm{r+m=14}\\\\\longrightarrow\rm{m=14-r....................(1)}$

&

$\longrightarrow\rm{10r+m=2+11r}\\\\\longrightarrow\rm{10r-11r=2-m}\\\\\longrightarrow\rm{-r=2-m}\\\\\longrightarrow\rm{-r=2-(14-r)\:\:\:[from(1)]}\\\\\longrightarrow\rm{-r=2-14+r}\\\\\longrightarrow\rm{-r-r=-12}\\\\\longrightarrow\rm{-2r=-12}\\\\\longrightarrow\rm{r=\cancel{-12/-2}}\\\\\longrightarrow\bf{r=6}$

Putting the value of r in equation (1),we get;

$\longrightarrow\rm{m=14-6}\\\\\longrightarrow\bf{m=8}$

Thus;

$\bigstar\:\underbrace{\sf{The\:number\:(10r+m)=[10(6)+8]=[60+8]=\boxed{\bf{68}}}}}}}$