Welcome to Test-paper.info
Thursday, December 14 2017 @ 12:14 AM CST
 Select a Forum » General Chat » Primary 1 Matters » Primary 2 Matters » Primary 3 Matters » Primary 4 Matters » Primary 5 Matters » Primary 6 Matters » Question, Feedback and Comments » Education Classified
 Forum Index >  Test Paper Related >  Primary 6 Matters P6 Maths problems
 | Printable Version
 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?

-----------------------

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?

------------------------------

Thank you for helping!

Newbie

Registered: 11/03/09
Posts: 2

 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

Newbie

Registered: 12/31/06
Posts: 1

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

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

Newbie

Registered: 12/31/06
Posts: 1

 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.

Newbie

Registered: 10/24/10
Posts: 1

 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.

Newbie

Registered: 05/31/10
Posts: 1

 All times are CST. The time is now 12:14 am.
 Normal Topic Locked Topic Sticky Topic
 New Post Sticky Topic w/ New Post Locked Topic w/ New Post
 View Anonymous Posts Able to Post HTML Allowed Censored Content