JAMB - Mathematics (1981 - No. 4)
Find a two-digit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits
24
42
74
47
72
Explication
Let the tens digits of the number be x and the unit digit be y
3x = 2y - 2
3x - 2y = -2.......(i)
If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20
(20x + 2y) - (10y + x) = 20
19x - 8y = 20.....(ii)
Multiply eqn.(i) by 8 and eqn.(ii) by 2
24x - 16y = -16......(iii)
38x - 16y = 40........(iv)
eqn(iv) - eqn(iii)
14x = 56
x = 4
Sub. for x = 4 in eqn(i)
3(4) - 2y = -2
14 = 2y
y = 7
So the original number is 10(4) + 7
i.e. 10x + y
= 47
3x = 2y - 2
3x - 2y = -2.......(i)
If the digits are interchanged, the tens digit becomes y, the unit digit becomes x. Hence 2(10x + y) = 10y + x + 20
(20x + 2y) - (10y + x) = 20
19x - 8y = 20.....(ii)
Multiply eqn.(i) by 8 and eqn.(ii) by 2
24x - 16y = -16......(iii)
38x - 16y = 40........(iv)
eqn(iv) - eqn(iii)
14x = 56
x = 4
Sub. for x = 4 in eqn(i)
3(4) - 2y = -2
14 = 2y
y = 7
So the original number is 10(4) + 7
i.e. 10x + y
= 47