Question

a two digit number is 2 more than 5 times the sum of the digits. if the digit in the tens place is 1 less than the digit in the unit place. find the number.

Answer 1

Answer:

10y+x-5(x+y)=2

10y+x-5x-5y=2

-4x+5y=2. 1st equation

2nd equation= x-y=1

1st equation + 2nd equation *4 =

x-6=1

x=7

7-y=1

y=7-1=6

Number= 10y + x

10*6+7=67

Step-by-step explanation:

Answer 2

The number is 67.

Given:-

Two digit number + 2 = 5 × sum of digits

Digit in tens place = 1 - digit in unit place

To Find:-

The number.

Solution:-

We can easily find out the number by using these simple steps.

As

Two digit number + 2 = 5 × sum of digits

Digit in tens place = 1 - digit in unit place

Here, let the unit digit of number be x and tens digit be y.

Then,

number = (10y+x)

Also, sum of digits = x + y

Difference of digits = x - y

Now, According to the question we will make the equation as.

Case 1,

(10y+x) - 2 = 5(x+y)

10y + x - 2 = 5x + 5y

10y-5y + x-5x - 2 = 0

5y - 4x = 2

Case 2,

y = 1 - x

Putting the value of y in above equation, we get

5y - 4x = 2

5(1-x) -4x = 2

-5 + 5x - 4x = 2

x - 7 = 0

x = 7

Also, y = 1 - x

y = 6

So, the number is 10y + x = 10×6 + 7 = 60+7 = 67

Hence, The number is 67.

#SPJ2