Past papers › NECO · 2023 · SSCE · General Maths · Paper 1 › Question 18 Question NECO General Maths 2023 Objective Linear & simultaneous equations Linear & simultaneous equations
NECO 2023 · Paper 1 · Q18 If 1 is added to the denominator of a fraction, the fraction becomes 1 2 \frac12 2 1 3 4 \frac34 4 3
A 2 5 \frac25 5 2 B 1 2 \frac12 2 1 C 3 5 \frac35 5 3 D 3 4 \frac34 4 3 E 4 5 \frac45 5 4
Worked solution (try it first) Let the fraction be
x y \frac xy y x .
Adding 1 to the denominator gives
x y + 1 = 1 2 \frac{x}{y + 1} = \frac12 y + 1 x = 2 1 , so
y + 1 = 2 x y + 1 = 2x y + 1 = 2 x and
y = 2 x − 1 y = 2x - 1 y = 2 x − 1 .
Adding 3 to both gives
x + 3 y + 3 = 3 4 \frac{x + 3}{y + 3} = \frac34 y + 3 x + 3 = 4 3 .
Cross-multiply:
4 x + 12 = 3 y + 9 4x + 12 = 3y + 9 4 x + 12 = 3 y + 9 .
Substitute
y = 2 x − 1 y = 2x - 1 y = 2 x − 1 :
4 x + 12 = 6 x + 6 4x + 12 = 6x + 6 4 x + 12 = 6 x + 6 , so
2 x = 6 2x = 6 2 x = 6 and
x = 3 x = 3 x = 3 .
Then
y = 5 y = 5 y = 5 , so the fraction is
3 5 \frac35 5 3 , option C.
Watch out
1 2 \frac12 2 1 3 4 \frac34 4 3 3 5 \frac35 5 3 3 6 = 1 2 \frac{3}{6} = \frac12 6 3 = 2 1 6 8 = 3 4 \frac{6}{8} = \frac34 8 6 = 4 3 Report a problem with this question