Question

the digits of a two-digit number differ by 3 if the digits are interchanged and the two numbers are added the sum is 77 find the numbers ​

Answer 1

Let the digit at units place be x

Let the digit at tens place be x + 3

10(x + 3) + x

When digits interchanged, the number becomes 10x + x + 3

According to the question,

10(x + 3) + x + 10x + x + 3 = 77

10x + 30 + 12x + 3 = 77

22x + 33 = 77

22x = 77 - 33

22x = 44

x = \frac{44}{22}

x = 2

∴ The number = 10(x + 3) + x = 10 * 5 + 3 + 5 = 50 + 3 + 5 = 53 + 5 = 58

∴ The number is 58

Answer 2

Answer:

Let the two digits of the number be:

x and y

x - y = 3— (i)

The number formed:

xy

xy= 10x+y (because x is in ten’s place)

Reversed number:

yx

yx= 10y+x (because y is in ten’s place)

ATQ:

When the number s reversed and it is added to the original number, the sum is 77.

Original number= 10x+y

Reversed number= 10y+x

⇒ 10x+y+10y+x= 77

11x+11y= 77

11(x+y)= 77

x+y= 7— (ii) (77÷11)

Now we will compare equations (i) and (ii) using elimination method.

x-y= 3

x+y= 7

2x= 10

x= 5

Now to find value of y.

x-y= 3

5-y= 3

-y= 3-5

-y= -2

y= 2

The numbers formed:

Original number= xy

= 52

Reversed number= yx

= 25

VERIFY:

xy+yx= 77

52+25= 77

77=77

LHS=RHS

Hence, verified

Hope it helps

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