Question

A number consists of two digits.The digit in the units place is one-third of the digit in the tens place.If the digits are reversed,the new number falls short of the original number by 36. find the number.

Answer 1

Answer:

Let the Ten Digit be a and Unit Digit be b, so Number Formed will be (10a + b).

☯ By First Statement :

↠ Unit Digit = One Third of Tens Digit

↠ Unit Digit is Thrice = Tens Digit

↠ 3b = a — eq. ( I )

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☯ According to the Question :

⇢ Reverse No. = Original No. – 36

⇢ (10b + a) = (10a + b) – 36

⇢ 36 = 10a + b – 10b – a

⇢ 36 = 9a – 9b

⇢ 36 = 9(a – b)

• Dividing both term by 9

⇢ 4 = a – b

• putting value of a from eq. ( I )

⇢ 4 = 3b – b

⇢ 4 = 2b

• Dividing both term by 2

⇢ b = 2

⋆ Using value of b in eq. ( I ) :

⇢ 3b = a

⇢ 3(2) = a

⇢ a = 6

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☯ Number Formed :

⇴ Number = (10a + b)

⇴ Number = 10(6) + 2

⇴ Number = 60 + 2

⇴ Number = 62

∴ Original Number Formed will be 62.

Answer 2

Given :-

• Digit in unit's place = ⅓ of digit in ten's place.

• Digits interchanged, new number = 36 less than original number.

To find :-

• Original number = ?

Solution :-

Let the digit at ten's place be y so, digit at unit's place is = y × 1/3 = y/3

→ Original number = 10y + y/3

★ Interchanged number:-

→ New number = 10(y/3) + y

A/q,

→ (10y + y/3) - [ 10(y/3) + y] = 36

→ (30y + y)/3 - [(10y/3) + y]= 36

→ (31y/3) - [(10y + 3y)/3] = 36

→ 31y/3 - 13y/3 = 36

→ (31y - 13y)/3 = 36

→ 18y/3 = 36

→ 18y = 3 × 36

→ 18y = 108

→ y = 108/18

→ y = 6

So, digit at ten's place = 6

Now finding digit at unit's place:-

→ Digit at unit's place = y/3

→ Digit at unit's place = 6/3

→ Digit at unit's place = 2

Now finding original number:-

→ Original number = 10y + y/3

→ Original number = 10(6) + 2

→ Original number = 60 + 2

→ Original number = 62

Therefore,

$\therefore\underline{\boxed{\textsf{Original number = {\textbf{62 }}}}}$