Q&A - PSLE Math

[Chalupa]

(I need help for this question !) Mr. Lee left town at 6.45am and drove at a uniform speed of 60km/h to Town Q.Mrs Chen left town P at 7:15am and drove towards Town Q along the same expressway as Mr Lee.She passed him at 10:15am and continued driving at the same speed until she reached Town Q.The distance between Town P to Town Q was 560km.How far was Mr Lee from Town Q when Mrs Chen reached Town Q ?

Thanks in advance.

[PapayaDad]

Mr.Lee left town P at 6.45am? did u miss the P?


Also the correct thread for these is here :

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=8950

[PapayaDad]

Mr Lee start 6.45 ; speed 60kmh

At 10.15 (3.5h later) ; Mr Lee travel 60+60+60+30=210km already

Mrs Chen took 3h (7.15 to 10.15) to travel 210km (catch up Mr Lee)
Mrs Chen Speen is 70kmh.
Mrs Chen need to travel another 350km to reach Q ie another 5h

Another 5h, Lee travel 300km only…still 50km away…

[Musicstar]

Hi,


Can someone enlighten me the below question with two types of method to solve the question but why different answer? Please advise the mistake. I suppose no matter which methods should also get the same answer. Thank you.

A box contained some twenty-coins and some fifty-cent coins in the ratio 3:4. When 26 fifty-cent coins were taken out and replaced by the same number of twenty-cent coins, the ratio became 4:1. Find the sum of money in the box in the box at first.


Working No 1
Since the number of coins did not change, we make the total the same

Before
T : F
3 : 4 (Total = 7u)
15 : 20 (Total = 35u)

After
T : F
4 : 1 (Total = 5u)
28 : 7 (Total = 35u)

13u = 26
1u = 2

15u of 20 cent = 15 x 2 x 0.2 = $6
20u of 50 cent = 20 x 2 x 0.5 = $20

Total = $26

Working No 2
Using units and parts
20cents : 50cents
3 units : 4 units
+65 -26 x 4
4 parts 1 part

3u +65 = 4p
16u -104 = 4p
3u + 65 = 16u - 104
16u - 3u = 65 + 104
13u = 169
u = 13
3u = 13 x 3 = 39
39 x $0.20 = $7.80
4u = 13 x 4 = 52
52 x $0.50 = $26
$7.80 + $26. = $33.80

[Oldschool]

Musicstar:

Hi,


Can someone enlighten me the below question with two types of method to solve the question but why different answer? Please advise the mistake. I suppose no matter which methods should also get the same answer. Thank you.

A box contained some twenty-coins and some fifty-cent coins in the ratio 3:4. When 26 fifty-cent coins were taken out and replaced by the same number of twenty-cent coins, the ratio became 4:1. Find the sum of money in the box in the box at first.


Working No 1
Since the number of coins did not change, we make the total the same

Before
T : F
3 : 4 (Total = 7u)
15 : 20 (Total = 35u)

After
T : F
4 : 1 (Total = 5u)
28 : 7 (Total = 35u)

13u = 26
1u = 2

15u of 20 cent = 15 x 2 x 0.2 = $6
20u of 50 cent = 20 x 2 x 0.5 = $20

Total = $26

Working No 2
Using units and parts
20cents : 50cents
3 units : 4 units
+65 -26 x 4
4 parts 1 part

3u +65 = 4p
16u -104 = 4p
3u + 65 = 16u - 104
16u - 3u = 65 + 104
13u = 169
u = 13
3u = 13 x 3 = 39
39 x $0.20 = $7.80
4u = 13 x 4 = 52
52 x $0.50 = $26
$7.80 + $26. = $33.80

Hi,

This is because in (b), instead of replaced with same number of coins, it is substituted with same amount of coins.

i.e. 65x20cents= 26x50cents in amount


But the question said \"same number\" though.

Regards

[Musicstar]

Oldschool,


Thank you for pointing out to me. I am really careless and did not read the question properly. The two words are really important. If don't have these two words \"the same\" which mean I need to change to the number of coin to the amount.

Oldschool:

Musicstar:

Hi,

Can someone enlighten me the below question with two types of method to solve the question but why different answer? Please advise the mistake. I suppose no matter which methods should also get the same answer. Thank you.

A box contained some twenty-coins and some fifty-cent coins in the ratio 3:4. When 26 fifty-cent coins were taken out and replaced by the same number of twenty-cent coins, the ratio became 4:1. Find the sum of money in the box in the box at first.


Working No 1
Since the number of coins did not change, we make the total the same

Before
T : F
3 : 4 (Total = 7u)
15 : 20 (Total = 35u)

After
T : F
4 : 1 (Total = 5u)
28 : 7 (Total = 35u)

13u = 26
1u = 2

15u of 20 cent = 15 x 2 x 0.2 = $6
20u of 50 cent = 20 x 2 x 0.5 = $20

Total = $26

Working No 2
Using units and parts
20cents : 50cents
3 units : 4 units
+65 -26 x 4
4 parts 1 part

3u +65 = 4p
16u -104 = 4p
3u + 65 = 16u - 104
16u - 3u = 65 + 104
13u = 169
u = 13
3u = 13 x 3 = 39
39 x $0.20 = $7.80
4u = 13 x 4 = 52
52 x $0.50 = $26
$7.80 + $26. = $33.80

Hi,

This is because in (b), instead of replaced with same number of coins, it is substituted with same amount of coins.

i.e. 65x20cents= 26x50cents in amount


But the question said \"same number\" though.

Regards

[rocklee]

Anyone can help?


There were a total of 20800 toys in Factory A and Factory B. After 3/4 of toys in Factory A and 3/5 in Factory B were sold, there were 1040 more toys in B than A. How many toys were there in each factory at first?

[CloudeeDaz]

rocklee:

Anyone can help?


There were a total of 20800 toys in Factory A and Factory B. After 3/4 of toys in Factory A and 3/5 in Factory B were sold, there were 1040 more toys in B than A. How many toys were there in each factory at first?

hihi
there is a solution http://www.kiasuparents.com/kiasu/forum/viewtopic.php?p=775568#p775568/

[CloudeeDaz]

and http://www.kiasuparents.com/kiasu/forum/viewtopic.php?p=23793#p23793%20/

[Musicstar]

Please help.Thank you


Q11) Sally , Yiling and Ali shared a bag of beads. Sally received 3/4 of the bag of beads and 3/4 of a bead. Yiling received 3/4 of the remaining bag of beads and 3/4 of a bead. Ali received the last bead in the bag. How many beads did Sally receive? (No bead was to be divided in any way)

Q17 ) Jane and Iris had 255 sweets altogether. Jane had 15 more sweets than Iris. Jane gave away 25% as many sweets as Iris. She was left with twice as many sweets as Iris.
(a) How many sweets did Iris give away?
(b) How many sweets did Jane have in the end?