Question

5. A number consists of three digits, the right hand being zero. If the left hand and the middle digits
are interchanged the number is diminished by 180. If the left hand digit is halved and the middle and
right hand digits are interchanged the number is diminished by 454. Find the number.
[Hint. Let the original number be 100a+10b+0. ie. 100a+10b, then, by the first condition
(100a+10b) - (100b + 10a) = 180a-b-2
By the second condition, new number = 100* +0+ b = 50a + b
2
and so (100a+10b) - (50a + b) = 45450a + 9 =454)​

Answer 1

Answer:

Since the number contains three digits, it has a hundreds place, a tens place and a ones place.

It is given that the right most digit is 0. Let the digits in the hundreds place and tens place be x and y respectively. Thus, the number is

100x+10y+0=100x+10y.

It is also given that if the digits in the hundreds place and tens place are interchanged, the difference between the original number and the new number after interchange is 180.

=>100x+10y−(100y+10x)=180

=>100x+10y−100y−10x=180

=>90x−90y=180

=>90(x−y)=180

=>x−y=2 ......(i)

It is also given that if the digit in the hundreds place is halved and digits in tens place and ones place are interchanged, the difference between the original number and the new number after interchange is 180.

=>100x+10y−(100x/2+0+y)=454

=>100x+10y−50x−y=454

=>50x+9y=464 .....(ii)

Multiplying both sides of Eqn (i) by 50, we get

50x−50y=100

Eqn(i)−Eqn(ii)=>(50x−50y)−(50x+9y)=100−454

=>50x−50y−50x−9y=−354

=>−59y=−354

=>y=6

Now, substitute the value of y in the equation x−y=2.

=>x−6=2

=>x=8

Therefore, the three digit number xy0=860

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