PSLE
Mr Ng used ³₁₀ of his money to buy 4 lemons and 11 pomegranates. The cost of 2 lemons was the same as that of 3 pomegranates. What was the most number of pomegranates that Mr Ng could buy with the money he had left?
Fraction of his money that he had left
= 1 - ³₁₀ 
= ⁷₁₀

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

2 L = 3 P _(x2)
4 L = 6 P

Since Mr Ng wants to buy pomegranates with the money he had left, 4 lemons will be substituted as 6 pomegranates.

³₁₀ of his money
= 4 L + 11 P
= 6 P + 11 P
= 17 P

³₁₀ of his money → 17 P
¹₁₀ of his money → ¹⁷₃ P
⁷₁₀ of his money → 7 x ¹⁷₃ P = 39²₃ P

Maximum number of pomegranates he could buy with the money that he had left = 39

Answer(s): 39