PSLE
Mr Soh used ²₇ of his money to buy 10 oranges and 9 pomegranates. The cost of 5 oranges was the same as that of 2 pomegranates. What was the most number of pomegranates that Mr Soh could buy with the money he had left?
Fraction of his money that he had left
= 1 - ²₇ 
= ⁵₇

Let the cost of 1 orange be 1 O.
Let the cost of 1 pomegranate be 1 P.

5 O = 2 P _(x2)
10 O = 4 P

Since Mr Soh wants to buy pomegranates with the money he had left, 10 oranges will be substituted as 4 pomegranates.

²₇ of his money
= 10 O + 9 P
= 4 P + 9 P
= 13 P

²₇ of his money → 13 P
¹₇ of his money → ¹³₂ P
⁵₇ of his money → 5 x ¹³₂ P = 32¹₂ P

Maximum number of pomegranates he could buy with the money that he had left = 32

Answer(s): 32