A no. consists of 3 digits , the right hand being zero. If the left hand and middle digits be interchanged the no. is diminished by 180.If the left hand digit be halved and middle and right hand digits be interchanged the no. is diminished by 454. Find the no. .
Answers
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.
ANSWER :let the original number
Be 100a+10b+0
(100a+10b)-(100b+10a)
180=a-b=2
(100+10b)-(50+b)=454
50a+9b=454
................................
=50x-50y=100
0+59y=354
Soy=6
Putt the value of eqution
X=8
100x+y
=800+60
=860