PSLE
Mr Choy used ²₇ of his money to buy 4 lemons and 3 pomegranates. The cost of 2 lemons was the same as that of 4 pomegranates. What was the most number of pomegranates that Mr Choy could buy with the money he had left?
Fraction of his money that he had left
= 1 - ²₇ 
= ⁵₇

Let the cost of 1 lemon be 1 L.
Let the cost of 1 pomegranate be 1 P.

2 L = 4 P _(x2)
4 L = 8 P

Since Mr Choy wants to buy pomegranates with the money he had left, 4 lemons will be substituted as 8 pomegranates.

²₇ of his money
= 4 L + 3 P
= 8 P + 3 P
= 11 P

²₇ of his money → 11 P
¹₇ of his money → ¹¹₂ P
⁵₇ of his money → 5 x ¹¹₂ P = 27¹₂ P

Maximum number of pomegranates he could buy with the money that he had left = 27

Answer(s): 27