Question

Mr Tan had some money. He bought some regular-sized burgers and some extra-large burgers. He bought 5 extra-large ones and 4 regular-sized ones with \$63.35. He could not buy another extra-large burger with the remaining amount as he was short of \$0.30. Instead, he bought 2 more regular-sized burgers and had \$0.75 left.

a) How much more did each extra-large burger cost than 2 regular-sized burgers?

b) How much did Mr Tan have at first?

This is a simultaneous equation question.

 R1 R2 R2 × 2 CR  + Extra 5 1 2 7 Regular 4 -2 -4 0 Amount 63.35 1.05 2.10 65.45

a)

0.30 + 0.75 = 1.05

b)

7u = 65.45

1u = 65.45 ÷ 7 = 9.35

9.35 – 0.3 = 9.05

9.05 +63.35 = \$72.40

0 Replies 1 Like ✔Accepted Answer

(a)
1 extra-large burgers – 0.30 ——–> 2 regular-sized burgers + 0.75
1 extra-large burgers – 2 regular-sized burgers = 1.05

(b)
5 extra-large ones + 4 regular-sized ones ——-> 63.35
(10 regular-sized burgers + 5.25) + 4 regular-sized ones ——-> 63.35
1 regular-sized burger ——-> (63.35 – 5.25)/14 = 4.90
63.35 + 2 x 4.90 + 0.75 = 73.90

Ans: (a) \$1.05 more; (b) \$73.90.

0 Replies 0 Likes