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 P6 Maths problems
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By: Dahlia98 (offline)  Monday, September 27 2010 @ 07:47 PM CDT (Read 2660 times)  
Dahlia98

Need help on these questions.

During a sale, stall A and stall B were selling durians, each at $5 and $3 respectively. Before the sale, the price of the durians was the same at both stalls. A sum of $96 could be saved by buying 8 durians from each stall during the sale. How much could I save if I were to buy 4 durians from stall B?

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Cinema admission tickets for children were priced at $8 each. There were 20 more boys than girls in the cinema from the first movie. For the second movie, the number of boys decreased by 10% but the number of girls increased by 5%. If there were 408 children at the second movie, how much money was collected from the sales of admission tickets for the first movie?

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Thank you for helping!

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By: Laks (offline)  Tuesday, September 28 2010 @ 12:54 AM CDT  
Laks

1. Cinema admission tickets for children were priced at $8 each. There were 20 more boys than girls in the cinema from the first movie. For the second movie, the number of boys decreased by 10% but the number of girls increased by 5%. If there were 408 children at the second movie, how much money was collected from the sales of admission tickets for the first movie?

Girls : Boys
1st movie : 100% : 100% + 20
2nd movie : 105% : 90 % + 18 ( decrease of 10%) 90 /100 x 20 = 18


105% + 90% + 18 = 408
195% = 408 - 18
1% = 2
(2 x 200) + 20 = 420
420 x $ 8 = $3360

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By: blessed (offline)  Saturday, October 30 2010 @ 12:43 AM CDT  
blessed

save $28 for 4 durians for first question? Anyone with same answer?

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By: mary16 (offline)  Wednesday, November 03 2010 @ 12:24 AM CDT  
mary16

During a sale, stall A and stall B were selling durians, each at $5 and $3 respectively. Before the sale, the price of the durians was the same at both stalls. A sum of $96 could be saved by buying 8 durians from each stall during the sale. How much could I save if I were to buy 4 durians from stall B?


Usual price of each durian
Stall A = $X + $5
Stall B = $Y + $3

$X + $5 = $Y + $3
$X = $Y - $2 --- (1)

8 * (X+Y) = $96
X + Y = $12 --- (2)

Substitute (1) into (2)
($Y - $2) + $Y = $12
2$Y = $14
$Y = $7 --- (3)

4 X $7 = $28

I would save $28 if I were to buy 4 durians from stall B.



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By: kennytan87 (offline)  Wednesday, November 03 2010 @ 02:19 AM CDT  
kennytan87

Mary 16. I don't think Primary allows to solve alegbra and solve using simultaneous equation:

Suggested Solution:

Cost of durians during Sale:
A $5
B $3

Total cost from buying 8 from each stall (during sale):
$5 x 8 + $3 x 8 = $40 + $24
= $64

Total cost from buying 8 from each stall (without sale):
$64 + $96 = $160

Since 16 durians are bought & all are same price:
$160 / 16 = $10

Buying 4 how much can he save:

$10x4 - $3x4 = $40 - $12
= $28

He can save $28.

Hope it helps you in understanding. Smile


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