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
ANSWER:
860
EXPLANATION:
Given:
A number consists of 3 digits
Any number having three digits is of the form hundreds, tens and units starting from the LHS.
Also stated that, the right hand being zero.
Let's get it clear with the help of variables at hundreds and tens place and considering a zero at units place.
Suppose, x is any digit of a number at hundreds place and y is another digit of that number at tens place. Now, according to the the right hand being zero, so it forms as x+y+0.
The value of x as being in the hundreds place is undoubtedly 100x and the value of y as being at tens place is 10y.
So, the equation formed is,
100x+10y+0= 100x+10y
Now, it is stated about interchanging the digits which results in number being diminishing. Given in the question as; if the left hand and middle digits be interchanged the no. is diminished by 180.
Left hand side i.e hundreds place
and Middle digit i.e tens place is interchanged the difference between the real number and new number obtained by interchanging will be 180.
Let's solve it out,
100x+10y -(100y+10x)=180
(the numbers are interchanged)
100x+10y-100y-10x=180
100x - 10x + 100y-10y=180
90x-90y=180
90(x-y) =180
(90 being a common number)
x - y = 180/90
x- y = 60/30(dividing by 3)
x- y = 2 -------> (1)
Now, moving to the next part of the given question, which states that;
if the left hand digit be halved and middle and right hand digits be interchanged the no. is diminished by 454.
Left hand i.e hundreds place digit is halved and middle i.e tens place and right hand digit i.e 0 is interchanged the difference between the real number and new number obtained by interchanging will be 454
Let's solve it out;
100x +10y - (100x÷2+0+y)=454
100x + 10y - 50x - y = 454
100x - 50x + 10y - y=454
50x + 9y= 454 ------->2
We got tow equations, now multiply equation 1 by 50.
The result obtained is;
50x-50y=100
Now, subtract equation 1 from equation 2.
(50x-50y) -(50x+9y)= 100-454
50x-50y -50x-9y= -354
50x-50x -50y-9y= -354
-59y= -354
y= -354/ -59
y= 354/ 59
y= 6
Substitute y=6 in equation 1 i.e
x - y = 2
x - 6 = 2
x = 2 + 6
x=8
We now, have obtained the digit at hundreds place i.e x as 8 and the digit at tens place i.e y as 6 and as stated in the question the right hand being 0, so the number that can be formed after solving is 860.