The table below shows the price of economical rice at four different stalls at a hawker centre.

1. Identify the lowest possible cost for buying from two stalls

To buy from two stalls with the least amount of money, we must look at the lowest prices of the stalls.
Lowest prices:
A: $\char36{}4.20$
B: $\char36{}6.00$
C: $\char36{}3.70$
D: $\char36{}4.60$

2. Find the combination of two stalls whose lowest prices sum to $8 or less

Sum of lowest prices for A and C: $\char36{}4.20 + \char36{}3.70 = \char36{}7.90$
Since $\char36{}7.90 \le \char36{}8.00$, Raymond can buy from stalls A and C.
Checking other combinations with C (the cheapest):
C and D: $\char36{}3.70 + \char36{}4.60 = \char36{}8.30$ (more than $\char36{}8$)
Therefore, only A and C is possible.

1. Calculate the amount Xue Ling spent on food

Amount spent = $\char36{}20 - \char36{}9.50 = \char36{}10.50$

2. Identify the stall with a price range that includes the amount spent

Comparing $\char36{}10.50$ with the price ranges:

• Stall A: $\char36{}4.20$ to $\char36{}8.80$
• Stall B: $\char36{}6.00$ to $\char36{}10.30$
• Stall C: $\char36{}3.70$ to $\char36{}6.20$
• Stall D: $\char36{}4.60$ to $\char36{}12.50$

Only Stall D's price range includes $\char36{}10.50$.