Question

The digit at the ten's place of a two-digit number is twice the digit at the unit's place. If the sum of the number formed by reversing the digits is 66, find the original number​

Answer 1

Given :-

The digit at the ten's place of a two-digit number is twice the digit at the unit's place. If the sum of the number formed by reversing the digits is 66.

To Find :-

• Original number

Solution :-

Let the tens digit be x and ones digit be y

• Original number = 10x + y

According to question :-

★ The digit at the ten's place of a two-digit number is twice the digit at the unit's place.

• x = 2y ------(i)

★ If the sum of the number formed by reversing the digits is 66

• Reversed number = 10y + x

→ (10x + y) + (10y + x) = 66

→ 11x + 11y = 66

→ 11(x + y) = 66

→ x + y = 66/11

→ x + y = 6 ------(ii)

Now, substitute the value of x in equation (ii)

→ x + y = 6

→ 2y + y = 6

→ 3y = 6

→ y = 6/3 = 2

Put the value of y in equation (i)

→ x = 2y

→ x = 2 × 2

→ x = 4

Hence,

• Original number = 10x + y = 42
• Reversed number = 10y + x = 24

Answer 2

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⏹ Required Answer:

✏ GiveN:

• The digit at ten's place is twice the digit at one's place.
• Sum of these numbers is 66.

✏ To FinD:

• The original number.

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⏹ How to solve?

We have to take two variables for the ten's place digit and unit's place digit. Then for any two digit number, we write it as: 10 × (ten's place digit) + unit's digit i.e.

❇ For Example:

• 28 = 2 × 10 + 8
• 67 = 6 × 10 + 7

Similarly, we have to multiply the variable for tens digit with 10 + unit's digit for getting the two digit number. In this way, we will solve this question.

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⏹ Solution:

Let the tens digit be x and unit's digit be y.

Then the two digit number is 10x + y

✒ After reversing:

The tens digit is now y and unit's digit is x

The two digit number will be 10y + x

✒ According to question:

Original Number + Reversed Number = 66

$\large{ \rm{ \longrightarrow \: 10x + y + 10y + x = 66}}$

✒ Solving the equation,

$\large{ \rm{ \longrightarrow \: 11x + 11y = 66}} \\ \\ \large{ \rm{ \longrightarrow \: 11(x + y) = 66}} \\ \\ \large{ \rm{ \longrightarrow \: x + y = 6..............(1)}}$

☣ Another thing given here, is Tens digit is twice the unit digit of original number. So,

$\large{ \rm{ \longrightarrow \: x = 2y}} \\ \\ \large{ \rm{ \longrightarrow \: x - 2y = 0.............(2)}}$

✒Subtracting eq.(2) from eq.(1),

$\large{ \rm{ \longrightarrow \: x + y - (x - 2y) = 6 - 0}} \\ \\ \large{ \rm{ \longrightarrow \: \cancel{ x} + y - \cancel{ x} + 2y = 6}} \\ \\ \large{ \rm{ \longrightarrow \: 3y = 6}} \\ \\ \large{ \rm{ \longrightarrow \: y = 2}}$

✒ Putting value of y in eq.(1),

$\large{ \rm{ \longrightarrow \: x + 2 = 6}} \\ \\ \large{ \rm{ \longrightarrow \: x = 4}}$

☣ Our original fraction was 10x + y, So putting values of x and y, to get original no.

$\large{ \rm{ \longrightarrow \: Original \: number = 10(4) + 2}} \\ \\ \large{ \rm{ \longrightarrow \: Original \: number = \boxed{ \red{ \rm{42}}}}} \\ \\ \large{ \therefore{ \underline{ \underline{ \rm{ \purple{Hence \: solved \: \dag}}}}}}$

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