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Question

Mrs Tay sold basketballs and netballs. Each basketball cost $21 each and each netball cost 3/7  as much as the basketball. Mrs Tay sold 1/3 of the balls and collected $2070. 2/5 of the balls sold were basketballs.

a) How many netballs were sold?

b) How much more money did Mrs Tay receive from the sales of the basketball?

Answer

Mrs Tay sold basketballs and netballs. Each basketball cost $21 each and each netball cost 3/7  as much as the basketball. Mrs Tay sold 1/3 of the balls and collected $2070. 2/5 of the balls sold were basketballs.

a) How many netballs were sold?

b) How much more money did Mrs Tay receive from the sales of the basketball?

Total balls sold = X

0.4X * 21 + 0.6X * $9 = $2070

8.4X + 5.4X = 2070

2070 = 13.8X

X=150 (150 balls sold)

(8.4 x 150) – 5.4 x 150)= $450

 

 

 

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Hope this answers your question. 

Owlinker
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Hope this answers your question. 

Owlinker
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FB Owlinker

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(a)
21 x (3/7) = 9 (cost of each netball)
basketballs sold : 2 units
2 x 21 = 42
netballs sold : 3 units
3 x 9 = 27
27 + 42 = 69
2070/69 = 30
30 x 3 = 90

(b)
90 x 9 = 810 (total money received form sales of 3 units of netballs)
2070 – 810 = 1260 (total money recfrom sales of 2 units of basketball)
1260 – 810 = 450

Ans : (a) 90 netballs; (b) $450 more.

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Hi,

You also not using the 1/3?

So it must be question error.

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Hi

Is there a need to do so?

If there is not, it, 1/3 that is, must be redundant in solving the question.

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This is a Quantity Each Product question. 

  Quantity  Each Product 
Basketball  2u 21 42u
Net ball 3u 9 27u
Total     69u

21 ×  3/7 = 9

69u = 2070

1u = 2070 ÷ 69 = 30

a)

3u = 30 × 3 = 90 

b)

42u – 27u = 15u

15u = 30 × 15 = $450 more

 

Note

1/3 is not used.

Can anyone using the school method show this is used?

 

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