Question

A number consists of two digit of which tens digit exceeds the unit digit by 3. The number is ten time sum of it's digit.find the number.​

Answer 1

Take the tens digit as x and ones digit as y

x=y+3

x-y=3. this is the first equation

second equation is the Number is 10x+y and it is equal to 7(x+y)

10x+y=7x+7y

3x=6y

3x-6y=0

Solve for the both equation

y=3

x=6

no. is 10x+y

63..

Answer 2

$\huge\underline\mathrm{SOLUTION:-}$

AnswEr:

• Required number will be 30.

Given:

• A number consists of two digit of which tens digit exceeds the unit digit by 3. The number is ten time sum of it's digit.

Need To Find:

• Required number will be = ?

ExPlanation:

Let the ten's digit be x.

And the unit's digit be y.

Therefore:

➠ Number = 10x + y

According to the first statement:

➠ Number = 10 × Sum of digits

➠ 10x + y = 10(x + y)

➠ 10x + y = 10x + 10y

➠ 10x - 10x = 10y - y

➠ 0 = 9y

Now, divide both terms by 9.

➠ y = 0

Hence:

• y = 0

______________________________________

According to the second statement:

➠ Ten's digit = 3 + Unit's digit

➠ x = 3 + y

➠ x = 3 + 0

➠ x = 3

Hence:

• x = 3

ThereFore:

• Required number will be 30.

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