a number consists of two digits whose sum is 8. If 18 is added to the number, its digits are reveresed. Find the number.

 

can anyone has the solution to it??? i could not get it...please help me

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 Let the ten's digit of the no.be x and the unit's digit be y.

Therefore,x+y=8.

The number=10x+y.

New number=10x+y+18.

10x+y+18=10y+x(as the digits het reversed)

9x-9y=-18=>x-y=-2(dividing all the terms by 9)

Now we have a pair of simultaneous linear equations.

x+y=8 and x-y=-2.

Solve them.You'll get x=3 and y=5.

Thus the number =10x+y=35.

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the first sentence say the sum of the 2 digits is 8, so x+y = 8 
if we add 18 to the number, ie 10x+y+18
 
the result is then yx (ie 10y+x)
 10x+ y +18 = 10y+ x
 9x + y + 18 = 10y                  
 9x - 9y = -18
 x - y = -2 
So we have 2 equations:
x+ y=8
x- y=-2 
solve these to give x=3 and y=5 ie we had 35 and 53. 
  • 0

the two digit number is 35

because 3+5=8

if 18 is added to 35 it digits get interchanged and the answer is 53

  • 1

first find all the two digit numbers that have the sum of their digits as 8 less than 50. they are-

17, 26, 35 and 44.

then add 18 to each of them and see in which case the digits are reversed.

the answer is 35 as 35+18 = 53

hope it helps!

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