Let unit's digit be x and Ten's digit be y and Number 10y+x
Now according to the question
y=7+x (i)
x+y=1/9(10y+x) (ii)
Solving (i) and (ii)
we get y=8 and x=1
Number 10y+x = 10*8+1
81
option 5
Find two numbers such that one third of greater exceeds one half on the lesser by 1, and one fifth of the greater added to one sixth of the lesser equals one half of the lesser.
Now, according to question
(1/3)x=(1/2)y+1 (let larger no x, smallest no y)
and (1/5)x+(1/6)y=(1/2)y
On solving (i) and (ii), we get x=30 and y =18
option 3
Sum of 50 numbers is 490. If greatest number is 20 more than the least. Least number is 0 and repeated twice then the average of remaining 47 numbers will be
greatest number = 20
least number = 0 and repeated twice so sum of these three numbers = 20+0+0 = 20
so sum of 47 numbers = 490-20 = 470
so avg = 470/47 = 10
5x^2-3y^2=(11/2)xy
10x^2-6y^2=11xy
(10x^2/y^2)-6=11x/y (Dividing each term by y^2)
let x/y=k
10k^2-11k-6=0
solve
k=3/2 or -2/5
Taking the positive value for k we get,
x/y=3/2
option 7