Question

One of the digit number is 5 times the other, if you interchange the digit of the two digit number. When you add original and new number result is 100

Answer 1

ᴄᴏʀʀᴇᴄᴛ ǫᴜᴇsᴛɪᴏɴ :-

The digit at the tens place of the two - digit number is 4 times the ones place digit , if you interchange the digit of the two digit number. When you add original and new number result is 110. Find the number

ᴛᴏ ғɪɴᴅ :-

• Original number
• Interchanged number

sᴏʟᴜᴛɪᴏɴ :-

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

then ,

According to 1st condition :-

• Tens place digit = 5 × Ones place digit

• x = 5y. --(1)

According to 2nd condition :-

• Original number = (10x + y)
• Interchanged number = (10y + x)
• Original no + Interchanged no = 110

➭ (10x + y) + (10y + x) = 110

➭ 11x + 11y = 110

➭ 11(x + y) = 110

➭ (x + y) = 110/11

➭ (x + y) = 10. ----(2)

Put value of (1) in (2) , we get,

➭ x + y = 10

➭ 4y + y = 10

➭ 5y = 10

➭ y = 10/5

➭ y = 2

Put y = 2 in (1) , we. get

➭ x + y = 10

➭ x + 2 = 10

➭ x = 10 - 2

➭ x = 8

Hence,

• Tens place = x = 8
• Ones place = y = 2

Therefore,

• Original number(10x + y) = 82
• Interchanged number (10y + x) = 28

Answer 2

$\bf{\underline{\underline\green{SOLUTION:-}}}$

$\underline\mathtt \orange {Correct \: Question:}$

• One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two digit number and add the resulting number to the original number, you get 88. What is the original number?

$\underline\mathtt \orange {AnswEr:}$

• The original number = 62.

$\underline\mathtt \orange {Given:}$

• One of the two digits of a two digit number is three times the other digit.

• If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88.

$\underline\mathtt \orange {Need \: To \: Find:}$

• The original Number = ?

$\bf{\underline{\underline\green{ExPlanation:-}}}$

$\textbf {Let the digit in units place be x. }$

$\textbf {Then, digit in ten's place = 3x}$

$\textbf {So, original number = 10(3x) + x = 31x}$

$\textbf {On interchanging the digits,}$

$\textbf {New number = 10x + 3x = 13x}$

$\underline\mathtt \orange {According \: to \: the \: given \: condition,}$

$\sf \: 31x + 13x = 88 \\ \\ \\ \longrightarrow\sf \: 44x = 88 \\ \\ \\ \ \longrightarrow\sf \: x = 2$

$\underline\mathtt \orange {So,}$

$\mathsf {The \: original \: number \: is \: 31x \: or \: 31 \times 2 = 62.}$

On the other hand, if we consider the digit in ten's place as x, then the digit in unit's place will be 3x.

$\underline\mathtt \orange {So,}$

$\textbf {The resulting number we get is 26.}$

$\underline\mathtt \orange {Here,}$

$\textbf {Both answers are correct, as 26 + 62 = 88.}$

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