Q&A - PSLE Math

[isetan]

kiasuaunt:

isetan:

Hi need help on another question:--


Miss Li sold watches and bracelets. Each watch was sold at $42. Each bracelet was sold at 2/3 of the price of each watch. Miss Li sold 1/3 of the items and collected $3360. 2/5 of the items sold were watches.

A) how many bracelets were sold? (ans- 60)
B) what was the total number of items left unsold? (ans:- 200)


TIA.

Hi - this is how I would do it :

A) Watch - $42
Bracelet - 2/3 x $42 = $28

2/5 of the items sold were watches - 2u x $42 = 84u
Therefore, 3/5 of the items sold we




re bracelets - 3u x $28 = 84u

84u + 84u = 168u

168u = 3360
1u = 3360/168 = 20

Bracelets sold : 3u = 3 x 20 = 60 (ans)

B) 1/3 of items sold, therefore 2/3 of items left unsold.

Watches sold : 2u = 2 x 20 = 40
Total number of items sold : 1/3 = 60 + 40 = 100
Total number of items left unsold : 2/3 = 2 x 100 = 200 (ans)

Hope this helps. God Bless.

Thanks kiasuaunt!

[small]

whitety:

Hi,

I need help on 3 speed qns:

1) A scooter and a lorry were travelling along the same road in the same direction.The scooter travelled at an average speed of 70km/h and the lorry travelled at an average speed of 49km/h. If the scooter overtook the van at 11.57am, at what time will the scooter be 14km ahead of the lorry?

Thanks!

Hi,

HTH..

At the point when the scooter overtook the van and continuing to move for 1 hour, they will be 21km apart.

21km ----- 1 hour
14km ----- 14/21 x 60minutes = 40minutes

40 minutes after 11.57am is 12.37pm

[Tiramisu Coco]

Hi, need help for this question.


Tom only has watermelons. Sam only has oranges. They gave each other half of their fruits. Tom sold 9 watermelons and Sam sold 85 oranges. In the end the ratio of  watermelons to oranges for Tom and Sam is 1:10 and 1:5 respectively. How many watermelons did Tom have at first?

Thanks.

[small]

whitety:

Hi,

I need help on 3 speed qns:

2) Jay took 10min to walk from the train station to the shopping mall at a uniform speed. Kay took 15min to walk from the shopping mall to the train station at a uniform speed. Both of them started at 4.23pm. At what time did Jay and Kay pass each other along the way?

Thanks!

Hi,

HTH..

Every 1 minute:
Jay had covered 1/10 of the distance
Kay covered 1/15 of the distance from opposite side

1/10 + 1/15
= 3/30 + 2/30
= 1/6 of the total distance

So when Jay met Kay, they had covered the total distance.

1/6 of the total distance ----- 1 minute
6/6 of the total distance ----- 6 minutes

6 minutes after 4.23pm is 4.29pm

[MathIzzzFun]

whitety:

Hi,

I need help on 3 speed qns:


2) Jay took 10min to walk from the train station to the shopping mall at a uniform speed. Kay took 15min to walk from the shopping mall to the train station at a uniform speed. Both of them started at 4.23pm. At what time did Jay and Kay pass each other along the way?



Thanks!

Hi

alternate approach for Q2 using time & speed/distance ratio.
Q2.
Time ratio of Jay : Kay --> 10 : 15 = 2 : 3
Speed (or distance ratio) of Jay : Kay --> 3u : 2u, when they meet, Jay completes 3u of distance, Kay completes 2u of distance. Total distance = 5u
Time taken by Jay to complete 3u --> 3/5 x 10 min = 6min
meeting time --> 4.23pm + 6 min = 4.29 pm

cheers.

[MathIzzzFun]

whitety:

Hi,

I need help on 3 speed qns:

3) Ace and Nash started jogging at the same speed along a track. When Ace completed 5/6 of the distance, Nash had only completed 3/8 of the distance. Ace's average speed was 88m/min faster than Nash's. What was Nash's average speed in m/min?

Thanks!

Hi

Q3.
Total distance --> 24 u
distance completed by Ace --> 20 u
distance completed by Nash --> 9 u
speed of Ace : Nash --> 20u : 9u
11u --> 88m/min
Nash's speed = 9/11 x 88 m/min = 72 m/min

cheers.

[MathIzzzFun]

Tiramisu Coco:

Hi, need help for this question.


Tom only has watermelons. Sam only has oranges. They gave each other half of their fruits. Tom sold 9 watermelons and Sam sold 85 oranges. In the end the ratio of  watermelons to oranges for Tom and Sam is 1:10 and 1:5 respectively. How many watermelons did Tom have at first?

Thanks.

Hi

here's a similar question ... http://www.flickr.com/photos/62167097@N02/6943541872/in/photostream

you can use the same approach to solve.

cheers.

[whitety]

Hi small & MathIzzzFun,


Thank you very much!

[Neat]

Hi.


Please assist with the following question:

A gallery owner had three boxes, A, B and C, containing a total of 1500 vintage stamps. The number of vintage stamps in Box A to the total number of vintage stamps was 3:10. He sold 330 vintage stamps from Box B and sold 1/3 of the vintage stamps in Box C. The number of vintage stamps left in Box B to the number of vintage stamps left in Box C was 3:1. How many vintage stamps were there in Box B at first?

Thanks

[small]

Neat:

Hi.


Please assist with the following question:

A gallery owner had three boxes, A, B and C, containing a total of 1500 vintage stamps. The number of vintage stamps in Box A to the total number of vintage stamps was 3:10. He sold 330 vintage stamps from Box B and sold 1/3 of the vintage stamps in Box C. The number of vintage stamps left in Box B to the number of vintage stamps left in Box C was 3:1. How many vintage stamps were there in Box B at first?

Thanks

Hi,

HTH.

10units = 1500
7units = 1050

Number of stamps in Box B and Box C at first is 1050

Given that 2/3 of the vintage stamps left in Box C

Now
B : C ----- 3 : 1
B : C ----- 6p : 2p

At first
B : C ----- 6p+330 : 3p

9p + 330 = 1050
9p = 720
1p = 80

Number of stamps in Box B at first ----- 6p + 330 = 810