Q&A - PSLE Math

[tianzhu]

Essential:

Hi Tian Zhu thank you for your help for earlier question. One more question from Nanyang Prelim 2012. Q13. both shops. ABC and XYZ , offered a 15% discount for the same type of mattress. If Mr Muthu were to buy from shop XYZ ,he would have paid $38.25 less than the discounted price in shop ABC. a) What was the discounted price in the shop ABC ?

b) the original price of the mattress in Shop ABC was $1510. what was the minimum percentage discount it must offer so that the discounted price in shop ABC would be lower than the discounted price in Shop XYZ? Give your answer connect to the nearest percent.Thankyou in adv

Hi

15% ------- 38.25

100% ------- 255 (original price in XYZ)

For ABC ------ 255*100/85 ------- 300

a) Difference -------300 – 255 ------- 45

For (b)

Discounted price in ABC -------- 1510*0.85 ------ 1283.50

Discounted price in XYZ -------- 1283.50 – 38.25 ------- 1245.25

1510 – 1425.25 ------- 264.75

264.75/1510*100% ------- 18%

Best wishes

[zanen]

tianzhu:

zanen:

help ls Two cars race around a circular track at constant speeds starting at the same point. If they travel in opposite directions, then they meet every 30 seconds. If they travel in the same direction, then they meet every two minutes. If the track is 1800m long,what is the speed of each car?

Hi

Could you share with us where you take this question from?

Is this a PSLE type of question?

The answers to your question are 22.5m/s and 37.5m/s.

please help me. it was given by the relief teacher from my math teacher. the relief teacher dont know how to teach math ask me to find out by myself//
Best wishes

[zanen]

Tom and Peter stay at Towns A and B respectively which are 500km apart. They arranged to meet up at Town C which is in between Towns A and B, and is x km away from Town A. Tom travelled at an average speed of 90km/h while Peter was 30km/h faster. They set off at the same time and Tom arrived at Town C 40 min earlier than Peter. Form an equation in x and find the time Peter took to reach Town C.

my hw.

[MathIzzzFun]

zanen:

Tom and Peter stay at Towns A and B respectively which are 500km apart. They arranged to meet up at Town C which is in between Towns A and B, and is x km away from Town A. Tom travelled at an average speed of 90km/h while Peter was 30km/h faster. They set off at the same time and Tom arrived at Town C 40 min earlier than Peter. Form an equation in x and find the time Peter took to reach Town C.

my hw.

This is a typical question for Sec 1 students when they learn algebra.

I believe this is your tuition home work. While students can use algebra to solve problems, solving of algebraic equations are not covered in P6 syllabus and not tested in PSLE. For P6, students are required to form equation in x (or any variable the question dictates) but value of x will be provided for the student to work out the answer.
For eg. A pencil cost $2 and a pen cost $2k more. A pencil case cost half as much as the total cost of a pencil and a pen.
a) express the cost of the pencil case in terms of k.
b) if k =0.5, find the cost of a pencil case.

For P6, the question can be solved using speed ratio/distance ratio/time ratio.

Here's a solution using distance ratio:
Tom's speed : Peter's speed = 90:120=3:4
Over the same time, distance travelled by Tom : Peter --> 3u:4u
When Tom reached Town C, Peter is still 40 min away,
total distance travelled = 3u+4u=7u
40 min x 120 km/h = 80 km --> Peter is still 80 km away from C
Total distance between A and B = 500km --> 7u + 80 km
1u --> 60km
Distance of Town C from Town B --> 4u + 80km = 4 x 60 km + 80 km = 320 km
Time taken by Peter to reach Town C = 320/120 h = 2h 40 min

cheers.

[kay22]

http://i47.tinypic.com/347xpn8.jpg\">

[kay22]

Please help...thanks a lot.


http://i50.tinypic.com/2rqkluc.jpg\">

[MathIzzzFun]

kay22:

Please help...thanks a lot.


http://i50.tinypic.com/2rqkluc.jpg\">

unshaded area = 1/2 area of rectangle
shaded area = area of quadrant - 1/2 area of rectangle

perimeter = circumference of quadrant + 2 x radius - 1/2 perimeter of rectangle + radius (diagonal)

cheers.

[kay22]

MathIzzzFun:

kay22:

Please help...thanks a lot.


http://i50.tinypic.com/2rqkluc.jpg\">

unshaded area = 1/2 area of rectangle
shaded area = area of quadrant - 1/2 area of rectangle

perimeter = circumference of quadrant + 2 x radius - 1/2 perimeter of rectangle + radius (diagonal)

cheers.

Thanks for the clarification...We are stuck at \" Radius (diagonal)\". It don't seem to be touching from one end to another.

[MathIzzzFun]

kay22:

MathIzzzFun:

[quote=\"kay22\"]Please help...thanks a lot.


http://i50.tinypic.com/2rqkluc.jpg\">

unshaded area = 1/2 area of rectangle
shaded area = area of quadrant - 1/2 area of rectangle

perimeter = circumference of quadrant + 2 x radius - 1/2 perimeter of rectangle + radius (diagonal)

cheers.

Thanks for the clarification...We are stuck at \" Radius (diagonal)\". It don't seem to be touching from one end to another.[/quote]draw the other diagonal n you see that the diagonal is equal to radius.

cheers.

[kay22]

[cheers.[/quote]


Thanks for the clarification...We are stuck at \" Radius (diagonal)\". It don't seem to be touching from one end to another.[/quote]

draw the other diagonal n you see that the diagonal is equal to radius.

cheers.[/quote]

Ok, great...Fully understand now! :thankyou: