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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?

Answer

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.

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