Question

A number of three digits has five in the unit’s place and the middle digit is half
the sum of other two. If 108 be added to the number hundredth figure will take
the unit’s place and the unit’s the ten’s and the ten’s the hundred’s. Find the
number.

Answer 1

Answer:

345

Step-by-step explanation:

let the 3 digits of the given number be a,b, and c respectively.

so, the number is 100a + 10b + c

given that c = 5

also, given b = (a+c)/2 = (a+5)/2

ie., 2b = a+5

=> a = 2b-5 ------- (1)

then, given that,

(100a + 10b + c) + 108 = 100b + 10c + a

ie., (100a + 10b + 5) + 108 = 100b + 10*5 + a

=> 100a-a +10b-100b = 50-113

=> 99a-90b = -63

by (1)

=>  99(2b-5) +90b = -63

=> 198b - 495 + 90b = -63

=> 108b = 495-63 = 432

=> b = 432/108 = 4

b=4

so, (1) => a = 2*4 - 5

                 = 8-5

                 = 3

ie., a = 3

so, the number is ,

(100*3) + (10*4) + 5

=> 300+40+5

=> 345

Answer 2

Given : Digit at unit's place = 5

Solution : Let the digit at tens and hundreds place be x and y respectively.

According to the first condition,

x=1 / 2( 5 + y )

or

2x - y = 5 =====> 1 )

• Now the given number is = 100y + 10x + 5

When 108 is added to the original number,then the new number is 100x + 50 + y .

According to the second condition,

100y + 10x + 5 + 108 = 100x + 50 + y

or

10x - 11y = 7 =======> 2)

•Solving 1) and 2) ,we get x = 4 ; y = 3

*Substituting the value of x and y in the given number , we get :-

#Given number : 100 × 3 + 10 × 4 + 5

= 345 .

•Therefore, the number is 345.

-@BrainlyPoison...ᥫ᭡~