Question

Find the 2 digit no. Such that sum of twice the first and thrice the second is 92 and four times the first exceeds seven times the second by two

Answer 1

Let the numbers be x and y..

Therefore,

2x +3y =92 and 4x=7y+2 => x=(7y+2)÷4

Now , 2x +3y=92

Or 2×(7y+2)÷4 +3y =92

Or (7y+2)÷2 +3y =92

Or (7y +2 +6y)÷2 =92

Or 13 y=184-2

Or y=182÷13

Or y =14….

So now again, x=(7y+2)÷4

Or x= (7×14 +2)÷4

Or x= 100÷4

Or x=25…