Q&A - PSLE Math

[small]

enoawng:

The ratio of the number of stamps to number of marbles in Ali's collection was 3:4. He gave aways 99 stamps and bought another 18 marbles. The ratio of the stamps to marbles became 3:7. How many stamps were left in Ali's collection.


Appreciate any help. Thanks.

Hope this helps.

Stamps : marbles
3 u : 4u
-99 : +18
-------------------
3p : 7p


7 (3u-99) = 3 ( 4u+18 )
21u - 693 = 12u + 54
9u = 747
1u = 83

Stamps that left -> 3p = 3u -99
= (3x 83)-99
= 249 -99
= 150 stamps

There were 150 stamps left in Ali's collection.

[James Ang]

James Ang:

enoawng:

The ratio of the number of stamps to number of marbles in Ali's collection was 3:4. He gave aways 99 stamps and bought another 18 marbles. The ratio of the stamps to marbles became 3:7. How many stamps were left in Ali's collection.


Appreciate any help. Thanks.

7u = 4u +132 +18
3u = 150
1u = 50
3u = 150 stamps

I didn't read the question carefully, the stamps is indeed 3 unit as per small's workings. Small's answer of 150 stamps is correct.

[atutor2001]

youngmum:

My boy got this problem. Can someone help me to solve this?


Monkey King had 2000 pears. He gave all the pears to some monkeys. On the first day, the monkeys ate half of the total number of pears. On the second day, they ate 1/3 of the remaining pears. On the third day, they ate 1/4 of the rest of the pears. They went on to eat the pears as 1/5,1/6,1/7,.....,1/2000 of the remainder of every previous day's pears. Find the number of pears left at the end.

Thanks. By the way, please explain clearly because i m a not so smart mum.

The answer is 4

1st Remainder\t4/5 x 2000
2nd Remainder\t4/5 x 5/6 x 2000
3rd Remainder\t4/5 x 5/6 x 6/7 x 2000


2nd last Remainder\t 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 2000

Last Remainder 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 1999/2000 x 2000 = 4

[Full.010894Cream]

atutor2001:

youngmum:

My boy got this problem. Can someone help me to solve this?


Monkey King had 2000 pears. He gave all the pears to some monkeys. On the first day, the monkeys ate half of the total number of pears. On the second day, they ate 1/3 of the remaining pears. On the third day, they ate 1/4 of the rest of the pears. They went on to eat the pears as 1/5,1/6,1/7,.....,1/2000 of the remainder of every previous day's pears. Find the number of pears left at the end.

Thanks. By the way, please explain clearly because i m a not so smart mum.

The answer is 4

1st Remainder\t4/5 x 2000
2nd Remainder\t4/5 x 5/6 x 2000
3rd Remainder\t4/5 x 5/6 x 6/7 x 2000


2nd last Remainder\t 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 2000

Last Remainder 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 1999/2000 x 2000 = 4

think you missed the 1st 3 days of the pear feast!

[Dharma]

Full.Cream:

atutor2001:

[quote=\"youngmum\"]My boy got this problem. Can someone help me to solve this?


Monkey King had 2000 pears. He gave all the pears to some monkeys. On the first day, the monkeys ate half of the total number of pears. On the second day, they ate 1/3 of the remaining pears. On the third day, they ate 1/4 of the rest of the pears. They went on to eat the pears as 1/5,1/6,1/7,.....,1/2000 of the remainder of every previous day's pears. Find the number of pears left at the end.

Thanks. By the way, please explain clearly because i m a not so smart mum.

The answer is 4


1st Remainder\t4/5 x 2000
2nd Remainder\t4/5 x 5/6 x 2000
3rd Remainder\t4/5 x 5/6 x 6/7 x 2000


2nd last Remainder\t 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 2000

Last Remainder 4/5 x 5/6 x 6/7 x .... x 1998/1999 x 1999/2000 x 2000 = 4

think you missed the 1st 3 days of the pear feast! :D[/quote]http://www.postimage.org/

Poor monkeys feasting on rotten pears. Most may have suffered from food poisoning.

[atutor2001]

Full.Cream:

think you missed the 1st 3 days of the pear feast!

Thanks the answer is 1, was dreaming.

[youngmum]

Thanks everyone!

[N3SKiasu]

Hi all,

Need your help on this Rosyth Prelim 2008 Q48 (I have searched thru past pages and could not find any discussion on this question, hope my eyes are working fine by browsing thru the 50+ pages and it is not repeated):

Ning and Zongwei jogged to and fro repeatedly along a straight path in a park between 2 points A and B.
Ning jogged at a uniform speed of 4m/s.
Zongwei jogged at uniform speed of 6m/s.
They started jogging from opposite directions at the same time as shown below.
Ning ------> <----------Zongwei
A---------------------X---------------------Y-------------------B

They first met one another at point X. The second time they met was at point Y.
a) Given that the distance between X and Y is 160m. Find the distance between A and B. (Answer given : 400m)
b) If they started jogging at 8am, how long did they take to meet again for the third time? (Express your answer in minutes and seconds). (Answer given : 3m20s)

Do you consider this is as a challenging question?
A bit worried for DD now as she could not solve this sort of question at this point...and so do I, helpless mom...counting down...
TIA.

[N3SKiasu]

Would like to check on the answer for SCGS Prelim 2008 Q23:


A 200m long MRT train took 20 min to pass through a tunnel completely.
If the MRT train travelled at an average speed of 120 m/min, find the length of the tunnel.

Solution:
120m/min x 20 min = 2400 m
2400m - 200 m = 2200m

But the answer given is 2300 m. Just wondering if the answer given is correct?

[tianzhu]

N3SKiasu:

Hi all,

Need your help on this Rosyth Prelim 2008 Q48 (I have searched thru past pages and could not find any discussion on this question, hope my eyes are working fine by browsing thru the 50+ pages and it is not repeated):

Hi
Dharma had answered this Q.Please go to pg 54.
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=280&postdays=0&postorder=asc&start=795