Question

the digits of two numbers differ by 3. if the digit are interchange, and the resulting number is added to the original number, we get 143. what can be the original number?

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Answer 1

Answer:

$\star\: \textbf{ Original \: number = 58 }$

Step-by-step explanation:

Given:-

• Digits of a number differs by 3
• If digits interchanged, added to original number,sum = 143

To find:-

• Original number = ?

Let the digit in one's place be x and ten's place be y.

$\therefore\sf\blue{Original \: Number = x + 10y }$

A/Q,

$\Rightarrow\sf x - y = 3 \\\\\Rightarrow\sf x = y + 3 - - - - - - - (i)$

2nd condition:-

$\Rightarrow\sf 10x + y + x + 10y = 143 \\\\\Rightarrow\sf 11x + 11y = 143 \\\\\Rightarrow\sf 11(x + y) = 143 \\\\\Rightarrow\sf x + y =\dfrac{ 143}{11} \\\\\sf \: \: Putting \: value \: from \: (i) \: :- \\\\\Rightarrow\sf y + 3 + y = 13 \\\\\Rightarrow\sf 2y + 3 = 13 \\\\\Rightarrow\sf 2y = 13 - 3 \\\\\Rightarrow\sf 2y = 10 \\\\\Rightarrow\sf y =\cancel{\dfrac{10}{2}}$

$\Rightarrow\sf y$ = $\large{\boxed{\bf\green{5 }}}$

$\rule{100}{1}$

$\textbf{Putting \: value \: of \: y \:in \: (i) :-}$

$\Rightarrow\sf x = 5 + 3$

$\Rightarrow\sf x$ = $\large{\boxed{\bf\pink{8}}}$

$\rule{200}{1}$

$\sf Original \: number = x + 10y \\\\\Rightarrow\sf Original \: number = 8 + 10(5) \\\\\Rightarrow\sf Original \: number = 8 + 50$

$\Rightarrow\sf\red{Original \: number}$ = $\large{\boxed{\bf\blue{58 }}}$

${\underline{\therefore{\rm{ Original \: number = 58 }}}}$

Answer 2

❂Question❂

the digits of two numbers differ by 3. if the digit are interchange, and the resulting number is added to the original number, we get 143. what can be the original number?

☢Concept☢

☞ Take, for a example, a two digit, say, 56. it can be writting as 56 = (10 × 5 )+ 6

☞ if the digit in 56 are interchanged, we get 65, which can be written as (10×6)+5.

☢Answer☢

☞ Let us take two digit number such that the digit in the unit place is b. the digit in ten place differ from b by 3.

☞ Let us take it as b+3. so the two digit number is 10(b+3)+b = 10b+30+b=11b+30.

with interchange of digit, the resulting number,

their sum is,

☞ 10b +(b+3)=11b+3

if we add these two digit numbers,their sum is,

☞ (11b +3)+(11b+3)=11b+11b+30+3=22b+33

it is given the sum is 143.

therefore,

☞ 22b + 3 =143

☞ 22b = 143 - 33

☞ 22b = 110

☞ b = 110/22

☞ b = 5

the unit digit is 5 and therefore the ten digit is 5 + 3 which is 8. the number is 85.

on interchange of digit the number we get is 58 .

so, the original number is 58