Maths problem?

Question

Sum of digits of 2two digit no is 7. If the digits are reversed, the no. formed is 9 less than the original no. Find the original no. Please give step wise details

Answer 1

10d₁ + d₂ = a two-digit number
"Sum of digits is 7"
d₁+d₂ = 7

"Number formed by reversing digits" = 10d₂ + d₁
"Number formed is 9 less than the original"
10d₂ + d₁ = 10d₁ + d₂ - 9
9d₂ = 9d₁ - 9
d₂ = d₁ - 1

d₁+d₂ = 7
d₁+(d₁ - 1) = 7
2d₁ - 1 = 7
d₁ = 4
d₂ = 3

Original number is 43.

Answer 2

We have the first condition given in the question

ie "Sum of digits of the two digits is 7.

ie y + x = 7 ........................................................ (eqn 1)

Now assume that -

Digit at unit's place = x and
digit at Ten's place = y

Immediately find out the original number by adding the place values of x and y. Here in your case

Original number = 10y + x (note this step carefully)

When digits are inter changed,

Digit at unit's place = y and
digit at Ten's place = x

Again find out new place values of x and y, and add them to find out the new number after reversal.

The new number = 10x + y ....................

Now it is given that the new number is 9 less than the original number. It implies that (Original Number) - (new Number) = 9

Substitute the values to get -

(10y + x) - (10x+y) = 9
ie 9y - 9x = 9

ie y - x = 1 .................... (2)
We now have y + x = 7 .................... (1)

Add eqn (1) and (2) you get

2y = 8. ie y = 4.

from (eqn 1) we get -
4 + x = 7 ie x = 3

Hence the original number = 43 .............. Answer

Answer 3

Let no be 10 a + b

Now :

a + b = 7 ==> Eq 1

10b + a - (10a+b) = -9
10 a + b - 10 b - a = 9
9a - 9 b =9
a - b = 1
a = b + 1 == >> Eq 2

Pu t a = b + 1 in Eq 1

b + 1 + b = 7

2b = 6

b = 3

Put b = 3 in Eq 2

a = 3 + 1 = 4

Required number is 10a + b = 10 X4+3= 43

Testing:
43 - 34 = 9

Answer 4

Assume, Digit at 10s place = x
and Digit at unit's place = y

x + y = 7
x = 7 - y

So, original number is 10x + y
= 10(7-y) + y
= 70 - 10y +y
= 70 - 9y

When digits are reversed,
10y + x = 70 - 9y - 9
10y + 7 - y = 70 - 9y - 9
9y + 9y = 61 -7
18y = 54
y = 3

x = 7 - y = 7 - 3 = 4

So digit at tens place is 4
digit at units place is 3

and The number is 43

Answer 5

let the number be AB where A is 10's digit and B is unit digit
Then the number is 10A+B
When the digits are reversed, the new number is BA or 10B + A
ATQ 10A + B = 10B + A + 9
hence 9A - 9B = 9
or A - B = 1 .................... (1)
also sum of digits is 7
so A + B = 7 .................... (2)
solving (1) & (2)
A = 4 and B = 3 so the number is 43

Note: You can use this approach for any similar question.

Answer 6

Solution depends on your standard.
Solution for lower std.
Digits --------No.-------Reversed ----- Diff
1 + 6 ---------16 ----------61-------------61 – 16
2 + 5 ---------25 ---------52 ------------ 52 – 25
3 + 4 ---------34 ---------43 ------------ 43 – 34 = 9
Now the reversed number is smaller so digits are
4 + 3 ---------43 --------- 34
So number is 43
For higher std.
Let digits be a and b so number is ab
No. = 10a + b
Reversed No. = ba
No. = 10b + a
(10a + b) – (10 b + a) = 9
10 a + b – 10 b – a = 9
9 a – 9 b = 9
a – b = 1
a + b = 7
2 a = 8 --------- adding two eq.
a = 4
and b = 3
No. is 43

Answer 7

43

Answer 8

43

Answer 9

43

Answer 10

assume first digit is 'x' and second digit is 'y'
so X+Y=7
The actual number is 10X+Y (We use 10 coz it is a 2-digit number)
So with the second condition
10Y+X=10X+Y-9 --> X-Y=1
Solve X+Y=7 and X-Y=1 to get answer '43'