Question

In a two-digit number, the square of the units digit is greater than the tens digit by 2. The sum of the number and the number formed by reversing its digits is 110. What is the two-digit number? Solve using simultaneous equation.

Answer 1

Let the first digit be `a` and the second digit be `b` in the two digit number.

Number`=10a+b`
Reversed number`=10b+a`

`10a+b+10b+a=110`
`11a+11b=110`
`a+b=10`

The first digit ( `a` ) is greater than the square of the second digit ( `b` ) by `2`.

`a=b^2-2`

Substitute `a=b^2-2` into `a+b=10`,

`b^2-2+b=10`
`b^2+b-12=0`
`b=3 or -4` ( reject )

Substitute in `b=3` into `a+b=10`,

`a+3=10`
`a=7`

So, the number is `73`.

Answer 2

`73`.

Explanation:

Suppose that the digits in places of 10 and 1 are, resp.,

`x and y`.

Clearly, the 2-digit number (no.) is `10x+y,` so that, the no.

formed by reversing digits is `10y+x`.

From the first condition, we get, `y^2=x+2..........<<1>>`, and,

from the second one, `(10x+y)+(10y+x)=110, i.e.,`

`11x+11y=110 rArr x+y=10.........................<<2>>`.

`<<1>> rArr x=y^2-2....................................<<1'>>`.

`<<1', &, 2>> rArr (y^2-2)+y=10, i.e., y^2+y-12=0`.

`:.(y+4)(y-3)=0 rArr y=-4," impossible; ":. y=3`.

`<<1'>> rArr x=7`.

So, the no. is `73`.

Answer 3

The two digit number is `73`

Explanation:

Let the units digit is `B` and tens digit is `A`

The number is `10A+B` , when the digits are reversed

the number becomes `10B+A`.

By give conditions `B^2= A+2 ; (1)` and

`10A+B +10B+A=110 or 11 A +11B =110` or

`11(A+B)=110 or A+B=10;(2) :. A=10-B ;`

Putting `A=10-B ;` in equation (1) we get

`B^2= 10-B+2 or B^2 +B -12=0` or

`B^2+4B-3B-12=0 or B(B+4)-3(B+4)=0` or

`(B+4)(B-3)=0 :. B=-4 or B=3` Units digit cannot

be negative , so `B=3 : A =10-B=10-3=7`

Hence the two digit number is `73` [Ans]