Question

A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number its digits are reversed. Find the number.

Answer 1

Let the digits be x and y.

Thus, number=10x+y

Now,.

(10x+y) - 4(x+y)= 3.....(i)

and

(10x+y)+ 18=10y+x.....(ii)

Solving (i)

10x+y-4x-4y=3

or,6x-3y=3

or,2x-y=1....(1)...

Solving (ii)

10x+y-10y-x=-18

or, 9y-9x=18

or,y-x=2...(2)

(1)+(2)

2x-y+y-x=1+2

or,x=3

Thus, y=x+2 (from (2))

y=5

Therefore the number=

10x+y=10x3+5=35

Let the tens place be x and unit place be y

Now two digit no. is 3 more than 4 times the sum if its digits

10x+y=4(x+y)+3

10x+y=4x+4y+3

10x-6x+y-4y=3

4x-3y=3

2x-y=1 .........i)

If 18 added to the no. its digits are reversed

10x+y+18=10y+x

10x-x+y-10y=-18

9x-9y=-18

x-y=-2 .........ii)

On subtracting ii)from i)

x-y-2x-y=-2-1

-x-y-2x-y=-2-1

-x=-3

x=3

Now

2x-y=1

2×3-y=1

6-y=1

-y=-5

y=5

------------

Then the number=10x+y =10×3+5

=35

Answer 2

Answer:

35

Step-by-step explanation:

Let the tens and the units digits of the required number be x and y, respectively.

Required number = (10x + y) 10x + y = 4(x + y) + 3

⇒10x + y = 4x + 4y + 3

⇒ 6x – 3y = 3 ⇒ 2x –y = 1 .(i) Again, we have:

10x + y + 18 = 10y + x

⇒9x – 9y = -18

⇒x – y = -2 ……..(ii)

On subtracting (ii) from (i),

we get: x = 3 On substituting x = 3 in (i)

we get 2 × 3 –y = 1

⇒ y = 6 – 1 = 5

Required number = (10x + y) = 10 × 3 + 5 = 30 + 5 = 35 Hence, the required number is 35